Calculus II

by

Neal Koblitz (Chapters 1 - 18) and J.H. Wells (Chapters 19 - 21)

A Text for Math 114

(1997–1998)

[∑_{n=0}^{∞} \frac{(−1)^{n}}{2 n + 1}, ∑_{n=0}^{∞} \frac{(−1)^{n}}{2 n + 2}] = [1/4 π, 1/2 ln(2)]

Edited for the University of Kentucky, Fall 1997 Copyright 1997 by Neal Koblitz (chapters 1 through 18) Copyright 1997 by J.H. Wells (chapters 19 through 21)

Permission is granted to duplication services (e.g. Kinkos) or

individuals to make copies for student use.

Send comments to: koblitz@math.washington.edu, wells@ms.uky.edu

Printed on January 8, 1998

ii

Preface

Introduction to the Kentucky Text:

This text is a minor adaptation of Professor Neal Koblitz’s notes from his University of Washington Math 125 course and material from two chapters of a book by Professor J. H. Wells. Professors Koblitz and Wells have generously permitted their use in this extension to Kentucky of Koblitz’s idea of a complete calculus text for which students pay only the reproduction costs.

Errors which have been introduced into the text during the adaptation process are the responsibility of the Affordable Text Project of the University of Kentucky. This project has been supported by the National Science Foundation ( NSF) Calculus and Advanced Technical Education ( ATE) programs.

Advice for Students

The overwhelming emphasis in this course is on applications — story problems. To master story problems one needs a tremendous amount of practice — doing homework, worksheet, and practice exam problems. For this reason, this course will require much more of your time than would a high school or community college calculus course that does not have such a focus on word problems.

Here are some pointers for doing story problems:

1. Carefully read each problem twice before writing anything.

2. Assign letters to quantities that are described only in words; draw a diagram if appropriate.

3. Decide which letters are constants and which are variables. A letter stands for a constant if its value remains the same throughout the problem.

4. Using math notation, write down what you know and then write down what you want to find.

5. Decide what category of problem it is (this might be obvious if the problem comes at the end of a particular chapter, but will not necessarily be so obvious if it comes on an exam covering several chapters).

6. Double check each step as you go along; don’t wait until the end to check your work.

7. Use common sense; if an answer is out of the range of practical possibilities, then check your work to see where you went wrong.

Suggestions for Using These Notes

1. Read the example problems carefully, filling in any steps that are left out (ask someone if you can’t follow the solution to a worked example).

2. Later use the worked examples to study by covering the solutions, and seeing if you can solve the problems on your own.

3. Give yourself the practice midterms and finals in an exam situation (without access to material other than a single sheet of notes, and with the same time limit as on the exam).

4. The answers to homework and practice exam problems that are in the back of the notes should be used only as a final check on your work, not as a crutch. Keep in mind that sometimes an answer could be expressed in various ways that are algebraically equivalent, so don’t assume that your answer is wrong just because it doesn’t have exactly the same form as the answer in the back.

Formulas Used in This Course

Algebra Completing the square: X^{2} + bX + c = (X + \frac{b}{2})^{2} − \frac{b^{2}}{4} + c . Quadratic formula: roots of aX^{2} + bX + c are \frac{−b ± \sqrt{b^{2} − 4ac}}{2a} .

Exponents: a^{b} · a^{c} = a^{b+c}; \frac{a^{b}}{a^{c}} = a^{b−c}; (a^{b})^{c} = a^{bc}; a^{1/b} = \sqrt[b]{a} .

Geometry Circle: circumference = 2πr; area = πr^{2} . Sphere: vol = \frac{4}{3}πr^{3}; surface area = 4πr^{2} . Cylinder: vol = πr^{2}h; lateral area = 2πrh; total surface area = 2πrh + 2πr^{2}. Cone: vol = \frac{1}{3}πr^{2}h; lateral area = πr\sqrt{r^{2} + h^{2}}; total surface area = πr\sqrt{r^{2} + h^{2}} + πr^{2}. Analytic geometry Point-slope formula for straight line: y = y_{0} + m(x − x_{0}). Circle centered at (h, k): (x − h)^{2} + (y − k)^{2} = r^{2}. Ellipse with semimajor axis along x-axis and semiminor axis along y-axis: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1. Trigonometry sin = \frac{opposite}{hypotenuse}; cos = \frac{adjacent}{hypotenuse}; tan = \frac{opposite}{adjacent}; sec = \frac{1}{cos}; csc = \frac{1}{sin}; cot = \frac{1}{tan}; tan = \frac{sin}{cos}; cot = \frac{cos}{sin}; sin x = cos (\frac{π}{2} − x); cos x = sin (\frac{π}{2} − x); sin(x + π) = − sin x; cos(x + π) = − cos x. Law of cosines: a^{2} = b^{2} + c^{2} − 2bc cos A. Law of sines: \frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} . Sine of sum of angles: sin(x + y) = sin x cos y + cos x sin y. sin^{2} and cos^{2} formulas: sin^{2} + cos^{2} = 1; sin^{2} x = \frac{1 − cos(2x)}{2}; cos^{2} x = \frac{1 + cos(2x)}{2}.

vi PREFACE

Contents

Preface iii Introduction to the Kentucky Text: iii Advice for Students iii

Chapter 1. Area Between Curves; Distance/Velocity/Acceleration 1 1. Area Between Curves 1 2. Distance/Velocity/Acceleration 3 3. Total vs. Net Distance 4

Chapter 2. More on Falling Bodies; Volumes of Revolution 7 1. Volume of a Solid of Revolution (Slice Method) 7 2. Some Variants 11

Chapter 3. Related Rate Volume Problems 13

Chapter 4. Arclength, Parameterized Curves, Ellipses 17 1. Review of Parameterized Curves 17 2. Ellipses 18 3. Arclength 20

Chapter 5. Numerical Integration: The Trapezoid Rule 25 1. Riemann sums 25 2. The Trapezoid Rule 26 3. Functions Obtained by Interpolating Experimental Data 28

Chapter 6. Center of Mass 33

Chapter 7. Inverse Trig Functions 41 1. The Inverse Tangent Function 42 2. The Derivative of y = Arctan x 43 3. The Inverse Sine Function 45 4. The Derivative of y = Arcsin x 45 5. The Inverse Cosine Function 46

Chapter 8. The Natural Logarithm 49 1. Appearance of the Curve y = ln x. 52 2. Relation with Logarithm-to-the-Base-a 53 3. Logarithmic Differentiation 53

Chapter 9. Improper Integrals 57

Chapter 10. The Number e and the Exponential Function 63 1. Some Properties of the Exponential Function 65 Chapter 11. Exponential Growth and Decay 69

1. Continuously Compounded Interest 69 2. Other Examples of Exponential Growth. 71 3. Exponential Decay 71 4. Examples of Exponential Decay. 72

Chapter 12. Techniques of Integration I: Substitution, Powers of Trig Functions 75 1. Substitution 75 2. Odd Powers of Sine and Cosine 75 3. Even Powers of Sine and Cosine 76

Chapter 13. Techniques of Integration II: Integration by Parts 79 1. How to Choose u and v 82

Chapter 14. Techniques of Integration III: Rational Functions 85 1. Quadratic Denominator That Factors 86 2. Quadratic Denominator That Does Not Factor 87

Chapter 15. Techniques of Integration IV: Inverse Trig Substitution 91 1. A Final Word 95

Chapter 16. Differential Equations I: Separation of Variables 97 1. Installment Buying 97 2. Newton’s Law of Cooling and Heating 101 3. Falling Bodies with Air Resistance 102 4. Mixing Problems 103 5. The Constant Solution 104 6. A Final Remark 105

Chapter 17. Differential Equations II: More Separation of Variables 111 1. Spread of an Epidemic 111 2. More Mixing Problems 113 3. Harmonic Oscillation 114

Chapter 18. More Differential Equations 121

Chapter 19. Taylor’s Theorem 125 1. Taylor Polynomials 125 2. Taylor’s Theorem 127

3. L’Hospital’s Rule: The Indeterminate Form \frac{0}{0} 134 4. Other Indeterminate Forms 137

Chapter 20. Infinite Series 145 1. Infinite Sequences 145 2. Monotonic and Bounded Sequences 152 3. Convergent and Divergent Series, Partial Sums 154 4. Series of Positive Terms 163 5. Alternating Series 169 6. Absolute Convergence 180

Chapter 21. Power Series 185 1. Power Series 185

CONTENTS ix

2. Power Series Calculus 188 3. Taylor Series 192

Lab Projects for Ma114 201 Review Problem Sets: First Midterm 209 Review Problem Sets : Second Midterm 215 Review Problem Sets: Final Exam 219 Answers 225

4. Answers to Selected Homework Problems 225 5. Answers to Review Problem Sets 228

Index 235

x CONTENTS

CHAPTER 1 Area Between Curves; Distance/Velocity/Acceleration

1. Area Between Curves

Suppose that we are interested in the area enclosed by two curves y = f_{1}(x) and y = f_{2}(x) between the two vertical lines x = a and x = b, then we can proceed exactly as before. Recall that we first created an area function:

A(u) = area caught between y = f_{1}(x) and y = f_{2}(x) between x = a and x = u.

As before we can demonstrate the derivative of the function to be |f_{1}(x) − f_{2}(x)| and as long as this function is continuous, the Fundamental Theorem of Calculus will guarantee that the desired area is given by

A(b) = ∫_{a}^{b} |f_{1}(x) − f_{2}(x)|dx.

Note that in our old formulation of area under a curve y = f(x), we did not have the absolute value sign, but for proper calculations, we had to make sure that the curve y = f(x) was never below the x-axis or the curve y = 0. In other words, we needed f(x) ≥ 0 in the desired interval [a, b]. So f(x) = |f(x)| in the interval [a, b] and we did not need to write the absolute value sign. Similarly, we can omit it here, if we can ensure that f_{1}(x) ≥ f_{2}(x) during the interval [a, b].

On the other hand, we did interpret ∫_{a}^{b} f(x)dx as an algebraic area between the curve y = f(x) and the x-axis and it is possible to make a similar algebraic interpretation of ∫_{a}^{b}(f_{1}(x) − f_{2}(x))dx as well.

To work out concrete problems, it is better to remember the area formula in words as follows:

The area enclosed between y = f_{1}(x) and y = f_{2}(x) above the interval x = a to x = b is computed by integrating from a to b the expression l(x)dx. The expression l(x) is the length of a typical vertical section of the area at x.

Typically, l(x) = f_{1}(x)−f_{2}(x) provided you have f(1) ≥ f_{2}(x) during the interval [a, b]. If only the two functions f_{1}(x) and f_{2}(x) are given, then the values of a, b will be the

x-coordinates of common points of the two curves y = f_{1}(x) and y = f_{2}(x).

As before some shapes can best be integrated by considering exchanging the roles of x, y. In other words by letting x = g_{1}(y) be the curve on the right and x = g_{2}(y) the curve on the left, and computing Area= ∫_{c}^{d} g_{1}(y)dy − ∫_{c}^{d} g_{2}(y)dy, where c and d are the y-coordinates of the intersection points.

y = f_{1}(x) y = f_{2}(x) a  b Area= ∫_{a}^{b}(f_{1}(x) − f_{2}(x))dx  x = g_{1}(y) x = g_{2}(y) cd Area= ∫_{c}^{d}(g_{1}(y) − g_{2}(y))dy

Example 1.1. Find the area in the first quadrant between the unit circle and the parabola y = 1 − x^{2}.

By sketching a graph of the two curves, we can see that they intersect at (0,1) and (1,0). Alternatively, we can find the points of intersection algebraically by setting the two functions equal: 1 − x^{2} = \sqrt{1 − x^{2}}. Squaring both sides and moving everything to the right of the equality gives 0 = x^{2} − x^{4} = x^{2}(1 − x^{2}), so that x = 0, ±1 (we ignore x = −1, since that is

not in the first quadrant). This gives us the two points. Which is the top curve? The sim- plest way to answer this is to take an intermediate point, say x = 0.5: the y-coordinate of the corre- sponding point on the circle is 0.866, while the y- coordinate of the point of the parabola is 0.75; thus, the circle is the top curve. Finally, the area we want is:

area under circle − area under parabola
= π/4 − ∫_{0}^{1}(1 − x^{2}) dx
= (π/4) − (2/3) = 0.11873.

Another way to make sure that you have the right formula is as follows. We have determined the limits of integration to be 0, 1 already. For any typical x-value in between, the circle has the point (x, \sqrt{1 − x^{2}}) and the parabola has the point (x, x^{2}). The desired length function is the |\sqrt{1 − x^{2}} − x^{2}|. How do we know that the expression inside the | | is already nonnegative? As above, we can check it at some value like x = 0.5 and see that it is positive. To finish the argument, we need to know that it stays nonnegative in [0, 1]. But if it did become negative somewhere during this interval, then it will become 0 during the interval also. What Theorem are we using? But such a value of x must be the x-coordinate of a common point of the circle and the parabola Why?. Hence we are done!

What is the purpose of such abstract and complicated arguments, when we can just “see” which curve is above and which is below? The idea is to double check the sketching for messy curves and to possibly bypass the sketching altogether when it might be too complicated.

Note that we did not use the full formula:

∫_{0}^{1} |\sqrt{1 − x^{2}} − x^{2}|dx = ∫_{0}^{1}(\sqrt{1 − x^{2}} − x^{2})dx = ∫_{0}^{1} \sqrt{1 − x^{2}}dx − ∫_{0}^{1} x^{2}dx

in our calculations. We short-circuited the first difficult integral by interpreting it as the area of a quarter circle! You should use such shortcuts whenever possible!

Example 1.2. Find the area in the first quadrant bounded by 2x + y = 2 and y = sin(πx).

Here one cannot find the intersection points algebraically by solving the equation 2 − 2x = sin(πx). However, if we graph both y = 2−2x and y = sin(πx) carefully, we see that they both cross the x-axis at (1,0) and they have another point of intersection at (0.5,1). Since the sine curve is above the line joining these two points, it follows that the area we want is

∫_{0.5}^{1} sin(πx)dx − ∫_{0.5}^{1}(2 − 2x) dx =
− (1/π) cos(πx)|_{0.5}^{1} − (2x − x^{2})|_{0.5}^{1}
= (1/π) − 0.25 = 0.0683.

As above, let us give indirect arguments to verify our work. We have somehow guessed the limits to be 0.5, 1 as above and we need to show that l(x) = sin(πx) − (2 − 2x) ≥ 0 on the interval [0.5, 1]. Note that h(0.5) = h(1) = 0. We claim that h(x) does not have another root in [0.5, 1]. This will give us what we want. (Why?).

Suppose, if possible, h(c) = 0 for some 0.5 < c < 1. Then h′(x) must have at least two roots in [0.5, 1]; at least one in (0.5, c) and at least one in (c, 1) (Why?). We show that this is not possible! This will finish our claim.

The derivative h′(x) = π cos(πx) + 2 and its derivative h′′(x) = −π^{2} sin(πx). Clearly h′′(x) < 0 on the interval [0.5, 1] (Why?) and so h′(x) steadily decreases from h′(0.5) = 2 to h′(1) = 2 − π < 0. Hence, h′(x) can have exactly one root in [0.5, 1].

Notice how knowledge of higher derivatives of a function yields useful information about the function itself. This technique is often used.

2. Distance/Velocity/Acceleration

We next recall a general principle that will later be applied to distance-velocity-acceleration problems, among other things. If F (u) is an anti-derivative of f(u), then ∫_{a}^{b} f(u) du = F (b) − F (a). Suppose that we want to let the upper limit of integration vary, i.e., we replace b by some variable x. We think of a as a fixed starting value x_{0}. In this new notation the last equation (after adding F (a) to both sides) becomes:

F (x) = F (x_{0}) + ∫_{x_{0}}^{x} f(u) du.

(Here u is the variable of integration, called a “dummy variable,” since it is not the variable in the function F (x). In general, it is not a good idea to use the same letter as a variable of integration and as a limit of integration. That is, ∫_{x_{0}}^{x} f(x)dx is bad notation , and can lead to errors and confusion.)

An important application of this principle occurs when we are interested in the distance an object travels (say, along the x-axis) between time t_{0} and time t. Let s(t) denote the position of the object at time t (its distance from a reference point, such as the origin on the x-axis). Then the displacement of the object during the interval is s(t) − s(t_{0}). If the object moves in a single direction along a line, then the displacement is equal to the distance traveled.

Since s(t) is an anti-derivative of the velocity function v(t), we can write

s(t) − s(t_{0}) = ∫_{t_{0}}^{t} v(u)du.

Similarly, since the velocity is an anti-derivative of the acceleration function a(t), we have

v(t) = v(t_{0}) + ∫_{t_{0}}^{t} a(u)du.

As an example, let us take the case when an object is acted upon by a constant force F. By Newton’s laws of motion, the acceleration is the constant F/m, where m is the mass. Then we first have

v(t) = v(t_{0}) + ∫_{t_{0}}^{t} F/m du = v_{0} + (F/m)u|_{t_{0}}^{t} = v_{0} + \frac{F}{m}(t − t_{0});

and then we have

s(t) = s(t_{0}) + ∫_{t_{0}}^{t} (v_{0} + \frac{F}{m}(u − t_{0})) du = s_{0} + (v_{0}u + \frac{F}{2m}(u − t_{0})^{2})|_{t_{0}}^{t}

= s_{0} + v_{0}(t − t_{0}) + \frac{F}{2m}(t − t_{0})^{2}.

For instance, when F/m = −g is the constant of gravitational acceleration, then this is the falling body formula (where we neglect air resistance).

3. Total vs. Net Distance

The integral of the velocity function gives the displacement or the net distance traveled. This is because the integral of negative velocity (area underneath the t-axis) counts negatively. If you want to know the total distance traveled, you must integrate the |v(t)| or the speed function. To do this, the best method is to find out where the velocity function crosses the t-axis, integrate separately over the time intervals when v(t) is positive and when v(t) is negative, and add up the absolute values of the different integrals. For example, if an object is thrown straight upward at 19.6 m/sec, its velocity function is v(t) = −9.8t + 19.6. This is a straight line which is positive for t < 2 and negative for t > 2. The net distance traveled in the first 4 seconds is ∫_{0}^{4}(−9.8t + 19.6)dt = 0, while the total distance traveled

in the first 4 seconds is ∫_{0}^{2}(−9.8t + 19.6)dt + |∫_{2}^{4}(−9.8t + 19.6)dt| = 39.2 meters (i.e., 19.6 meters up and 19.6 meters down).

Example 1.3. The acceleration of an object is given by a(t) = cos(πt), and its velocity at time t = 0 is 1/2π. Find the displacement (net distance) as well as the total distance traveled in the first 1.5 sec.

We compute

v(t) = v(0) + ∫_{0}^{t} cos(πu)du = \frac{1}{2π} + \frac{1}{π} sin(πu)|_{0}^{t} = \frac{1}{π} (0.5 + sin(πt)).

The net distance traveled is then

s(1.5) − s(0) = ∫_{0}^{1.5} (\frac{1}{π} (0.5 + sin(πt)) dt
=\frac{1}{π} (0.5t − \frac{1}{π} cos(πt)) |_{0}^{1.5}= \frac{0.75}{π} + \frac{1}{π^{2}} = 0.340.

To find the total distance traveled , we ask when (0.5 + sin(πt)) is zero, and when it is positive and negative. This function is 0 when sin(πt) is −0.5, i.e., when πt = 7π/6, 11π/6, etc. The value πt = 7π/6, i.e., t = 7/6, is the only value in the range 0 ≤ t ≤ 1.5. We see that v(t) > 0 for t < 7/6 and v(t) < 0 for t > 7/6. Hence, the total distance traveled is

∫_{0}^{7/6} (\frac{1}{π} (0.5 + sin(πt)) dt + |∫_{7/6}^{1.5} (\frac{1}{π} (0.5 + sin(πt)) dt|
=\frac{1}{π} (\frac{3.5}{6} − \frac{1}{π} cos(7π/6) + \frac{1}{π} + |0.75 − \frac{3.5}{6} + \frac{1}{π} cos(7π/6)|) = 0.409.

Homework 1.

1. Find the area between each of the following pairs of curves:

(a) y = x^{4} − x^{2} and y = x^{2} (the part on the right of the y-axis);

(b) x = y^{3} and x = y^{2};

(c) y = x^{n} and x = y^{n} (in the first quadrant), where n > 1;

(d) x = 1 − y^{2} and y = −x − 1;

(e) x = 3y − y^{2} and x + y = 3;

(f) y = cos(πx/2) and y = 1 − x^{2} (in the first quadrant);

(g) y = sin(πx/3) and y = x (in the first quadrant).

2. Find the “triangular-looking” area bounded by the three curves:

(a) y = sin x, y = cos x, and x = 0 (in the first quadrant);

(b) y = \sqrt{x}, y = 20 − x, and y = 2.

3. For each velocity function find both the net distance and the total distance traveled during the indicated time interval (graph v(t) to determine when it’s positive and when it’s negative):

(a) v = cos(πt), 0 ≤ t ≤ 2.5;  (b) v = −9.8t + 49, 0 ≤ t ≤ 10;
(c) v = 3(t − 3)(t − 1), 0 ≤ t ≤ 5; (d) v = sin(πt/3) − t, 0 ≤ t ≤ 1.

4. Two parallel charged plates create a force of constant magnitude F on a charged particle of mass m. Suppose that the plates are rotating at the constant angular velocity of ω radians/sec, so that the direction of the force is not constant (see the figure at the right). Assume that the particle is at rest at the origin of the xy-plane at time t = 0, and that the force pulls in the positive x-direction at time t = 0 and rotates counterclockwise. (Also assume that ω is small enough so that the effects of the magnetic field created by the varying electric field can be neglected.) To describe the motion of the particle, consider the x- and y-directions separately.

(a) Express the x-component of the force vector in terms of the angle θ through which the plates have rotated, and express the latter angle in terms of t. Then find the x-acceleration function and the y-acceleration function (by dividing the force by the mass m). Here F, m, and ω are constants.

θ

(b) Find the horizontal component of velocity and the vertical component of velocity as functions of time.

(c) Find formulas for the x-coordinate of the particle at time t and for the y-coordinate of the particle at time t.

(d) Set the constants all equal to 1, and sketch the parameterized curve in part (c).

(e) How is the trajectory affected if the particle has an initial velocity v_{0} in the ydirection? For what value of v_{0} (expressed in terms of F, m, and ω) will the particle travel in a circle?

CHAPTER 2 More on Falling Bodies; Volumes of Revolution

If we know the initial velocity v_{0} and the initial height s_{0} of a falling body at time t = 0, then the formula

s(t) = −\frac{1}{2}gt^{2} + v_{0}t + s_{0}

tells us the height above the ground of the falling body at any time t. The formula for the velocity is given by its derivative:

v(t) = −gt + v_{0}.

Note that we are neglecting air resistance and taking upward as the positive direction. We usually use the traditional value g = 9.8 m/sec^{2} for the gravitational acceleration.

But in some situations we do not know v_{0} and/or s_{0}, and must determine these values from other given information. Generally speaking, any two bits of information can be used to determine v_{0} and s_{0}.

Example 2.1. You see an object thrown downward from a building. It hits the ground 5 sec after it was thrown with an impact velocity of 60 m/sec. What was the height from which it was thrown, and with what initial velocity was it thrown?

We translate the information given into two equations. Since “hits the ground” means that s = 0, we have s(5) = 0; we also know that v(5) = −60 (i.e., 60 m/sec downward). We plug 5 into the s(t) and v(t) formulas to obtain two equations with the two unknowns s_{0} and v_{0}:

0 = s(5) = −4.9 · 5^{2} + v_{0} · 5 + s_{0} = −122.5 + 5v_{0} + s_{0};
−60 = v(5) = −9.8 · 5 + v_{0} = −49 + v_{0}.

The second equation immediately gives v_{0} = −60 + 49 = −11 m/sec. Substituting this value for v_{0} in the first equation and solving for s_{0}, we obtain:

s_{0} = 122.5 − 5v_{0} = 122.5 − 5 · (−11) = 177.5.

Thus, the object was thrown downward at 11 m/sec from a height of 177.5 meters.

1. Volume of a Solid of Revolution (Slice Method)

A solid of revolution is a solid which is symmetrical with respect to a line called its axis. Such a solid can be obtained by taking some two-dimensional shape and revolving it around that axis. Familiar examples of solids of revolution include spheres, hemispheres, cylinders, and cones.

The basic situation to be dealt with first is the case when the axis of symmetry of the solid is the x-axis and the shape being revolved around the x-axis is the area under a curve y = f(x) between x = a and x = b. The familiar examples listed above can be obtained in this way.

Example 2.2. (Spheres and hemispheres)

If we place the sphere of radius r so that its center is at the origin of the xy-plane and take f(x) = \sqrt{r^{2} − x^{2}} (the upper semicircle of radius r), then we can get the sphere by rotating f(x) between x = −r and x = r around the x-axis. To get a hemisphere of radius r, we situate the hemisphere so that the x-axis is its axis of symmetry, the y-axis between r and −r is a diameter of its circular base, and the “north pole” of the hemisphere is at the point (r, 0). Then this hemisphere is the solid of revolution obtained by revolving f(x) = \sqrt{r^{2} − x^{2}} around the x-axis between x = 0 and x = r.

−r  y  rx  x y Rotate around x-axis to get sphere of radius r.

Example 2.3. (Right circular cones)

By “ right circular cone ” we mean a cone which is symmetric about an axis of symmetry, and for which any cross-section perpendicular to this axis is a circle. Such a cone is determined if we know its height h and its base radius r. (In what follows, we shall always use the term “cone” to mean a right circular cone.) To describe this cone as a solid of revolution, we first choose an efficient way to situate it with respect to the xy-plane. We place it so that its vertex is at the origin and its axis of symmetry is the part of the x-axis from 0 to h. Then our cone is obtained by taking the line joining the origin to the point (h, r) and rotating it around the x-axis. The advantage in situating the cone in this way is that the function we are revolving around the x-axis — the line from (0, 0) to (h, r) — has a very simple equation: f(x) = mx, where m = r/h is the slope of the line.

y  (h, r) x  x y Rotate around x-axis to get cone of height h, base radius r.

We now return to the general situation of a solid of revolution obtained by revolving y = f(x) around the x-axis between x = a and x = b. We want to find its volume. To do this, we first take the region under the curve, and divide it up into Riemann sum rectangles as we did earlier. Of course, the rectangles do not exactly fill out the region: when the curve increases, the rectangles fall a little short, and when the curve decreases, the rectangles give you a little too much (see the second diagram below). But as the theory works out, the limit of the Riemann sums gives the correct answer.

Recall how the rectangles were constructed for the n-th left endpoint Riemann sum. First, the interval [a, b] is divided into n equal parts, each of length Δx = (b − a)/n. Next, above each subinterval we draw the rectangle whose height is the distance up to the curve at the left endpoint. The heights of the rectangles are successively f(a), f(a + Δx), f(a + 2Δx), …, f(a + (n − 1)Δx).

y = f(x) a  b Graph of y = f(x)  ^{y = f(x) a  b Riemann sum rectangles}

When we rotate our function f(x) around the x-axis, what happens to the rectangles? As each narrow rectangle rotates through 360^{∘}, it forms a sideways cylinder whose “height” (actually, its distance across from left to right) is Δx and whose base radius is the height of the rectangle. Since the volume of a cylinder is equal to its base area (which is π · radius^{2}) times its height, it follows that the volume of the first cylinder is πf(a)^{2}Δx; the volume of the second is πf(a+Δx)^{2}Δx; the volume of the third is πf(a+2Δx)^{2}Δx, and so on. Thus, the total volume of all of the cylindrical slices is

(πf(a)^{2} + πf(a + Δx)^{2} + πf(a + 2Δx)^{2} + ⋯ + πf(a + (n − 1)Δx)^{2})Δx,

which can be written more compactly using the summation notation, as follows:

∑_{i=0}^{n−1} πf(a + iΔx)^{2}Δx.

If Δx is small (i.e., if n is large), this volume is a very close approximation to the true volume of the solid of revolution. As Δx ⟶ 0 (i.e., as n ⟶ ∞), this sum gets closer and closer to the volume of the solid we want. As discussed in the Chapter on Riemann Sums, the limit of this sum will become the definite integral of πf(x)^{2} (rather than the definite integral of f(x)). We conclude that

Volume of solid of revolution = π ∫_{a}^{b} f(x)^{2}dx.

(The constant π that is next to f(x)^{2} can be pulled out in front of the integral sign.)

One way to think of this formula is to look at the contribution to the volume that comes from the part of the region under y = f(x) that lies between x and x + Δx, where we suppose that Δx is very small. This is a thin slice whose thickness is Δx and whose circular base (at the left) has area πf(x)^{2}.

Since Δx is small, the slice is very close to being a cylinder, and so its volume is very close to πf(x)^{2}Δx. Now to find the volume of the entire solid of revolution we have to sum up the contributions as x goes from a to b. In the limit as Δx goes to zero this sum becomes an integral, and we write dx instead of Δx. That is, we get ∫_{a}^{b}(πf(x)^{2})dx, as before.

More generally, we can think of this formula as follows. Suppose that we have a solid caught between the planes x = a and x = b and the cross sectional area corresponding to any value of x is given by a function F (x), then the total volume can be computed by the integral

∫_{a}^{b} F (x) dx

In our current situation the cross sectional area happend to be a circle. This formula can also be understood by creating a variable “volume” func-

tion

V (u) = volume caught between x = a and x = u

and deducing as before that V′(x) = g(x). Then we use the Fundamental Theorem of Calculus to deduce:

V (b) = ∫_{a}^{b} g(x) dx.

Volume of revolution.  Thin slice of solid.

Example 2.4. Find the formula for the volume of a sphere of radius r.

We saw that the sphere is obtained by rotating the function f(x) = \sqrt{r^{2} − x^{2}} around the x-axis between x = −r and x = r. Actually, it turns out to be a little simpler just to rotate the part between x = 0 and x = r — thereby obtaining a hemisphere — and then doubling this volume. That is, using the formula we just derived, we see that the volume of the sphere is

2π ∫_{0}^{r} (\sqrt{r^{2} − x^{2}})^{2} dx = 2π ∫_{0}^{r}(r^{2} − x^{2}) dx = 2π (r^{2}x − \frac{x^{3}}{3} ) |_{0}^{r}

= 2π(r^{3} − r^{3}/3) = \frac{4}{3}πr^{3}.

Example 2.5. Find the formula for the volume of a cone of height h and base radius r.

According to our earlier discussion of how to get this cone as a solid of revolution, we have

Vol = π ∫_{0}^{h}((\frac{r}{h}) x)^{2}dx = πr^{2}/h^{2} ∫_{0}^{h} x^{2}dx = π(r^{2}/h^{2})(h^{3}/3) = \frac{1}{3}πr^{2}h.

Expressed in words, this formula says that the volume of a cone is one third the volume of the cylinder of the same height and base radius. By examining the computations in the last example, we see how the 1/3 arose: when we integrated x^{2} to get (x^{3}/3)|_{0}^{h}.

2. Some Variants

Suppose we want our solid to be “right side up” with the y-axis as its axis of symmetry. Then the curve that is rotated around the y-axis should be viewed from the y-perspective as a function of the form x = g(y) (rather than y = f(x)). Everything works the same with the roles of x and y reversed. Suppose that the solid is obtained by rotating x = g(y) around the y-axis between y = c and y = d. Then its volume is:

π ∫_{c}^{d} g(y)^{2}dy.

As another variant, suppose we want to rotate around the line y = c which is parallel to the x-axis and a distance c above it (if c is negative, a distance |c| below it). The only change is to replace f(x) by f(x) − c, which is the distance from the axis of symmetry to the curve at the position x on the x-axis. Thus, the volume is π ∫_{a}^{b}(f(x) − c)^{2}dx. A similar remark applies to the case when we take a function x = g(y) and revolve it around the line x = c rather than around the y-axis.

Example 2.6. Let S be the shape in the xy-plane that is bounded by the curve y = x^{2}, the line y = 4 and the line x = 1, and which is to the right of the line x = 1. Find the volume of the shape obtained by revolving S around (a) the x-axis, (b) the line y = 4, (c) the y-axis, (d) the line x = 1.

(a) Here we obtain a “hollowed out” shape. That is, to get our solid we first take the cylinder that is obtained by rotating the line y = 4 around the x-axis between x = 1 and x = 2, and from this cylinder we then carve out the solid obtained by rotating y = x^{2} between x = 1 and x = 2. Thus,

Vol = π · 4^{2} · 1 − π ∫_{1}^{2}(x^{2})^{2}dx = 16π − π ∫_{1}^{2} x^{4}dx = (16 − (32 − 1)/5)π = 9.8π.

(b) Here we are taking the full solid of revolution obtained by rotating y = x^{2} around the line y = 4. The result is

Vol = π ∫_{1}^{2}(x^{2} − 4)^{2}dx = π ∫_{1}^{2}(x^{4} − 8x^{2} + 16)dx = 53π/15.

(c) Here we start with the solid of revolution obtained by rotating x = \sqrt{y} around the y-axis between y = 1 and y = 4, and then hollow out a cylinder of height 3 and base radius 1 (obtained by rotating x = 1 around the y-axis between y = 1 and y = 4). The result is:

Vol = π ∫_{1}^{4}(\sqrt{y})^{2}dy − π · 1^{2} · 3 = π ∫_{1}^{4} y dy − 3π = 4.5π.

(d) Here we have the full solid of revolution where we rotate x = \sqrt{y} around the line x = 1. Thus,

Vol = π ∫_{1}^{4}(\sqrt{y} − 1)^{2}dy = π ∫_{1}^{4}(y − 2\sqrt{y} + 1)dy = 7/6π.

Example 2.7. Suppose that a solid has square cross-sections. The centers of the squares are on the x-axis, and the tops of the squares run through the line y = f(x) = 4 − x^{2} between x = −2 and x = 2. (See the diagram below.) Find the volume of this solid.

x y  y = f(x) y = −f(x) Side view  Perspective view

Here the cross-section at x is a square of side 2f(x). So the cross-sectional area is (2f(x))^{2} = 4(4 − x^{2})^{2}. We integrate this from −2 to 2. (Because of the symmetry — i.e., the fact that f(x) = f(−x) — we can take the integral from 0 to 2 and then double the result.) Thus, the volume is

2 ∫_{0}^{2} 4(4 − x^{2})^{2}dx = 8 ∫_{0}^{2}(16 − 8x^{2} + x^{4})dx = 136\frac{8}{15}.

CHAPTER 3 Related Rate Volume Problems

Suppose we have a container that is described by rotating some function x = g(y) around the y-axis between y = c and y = d, and suppose that the container is filled with water to a height h. Then the volume of water is given by the definite integral

V (h) = ∫_{c}^{c+h} πg(y)^{2}dy.

We write V (h) to indicate that the volume of the water is a function of its depth h.

In this situation one type of question we might want to answer is: given the rate at which water is being pumped into the container, what is the rate at which the water level is rising? In symbols this says: given dV/dt, find dh/dt. Alternately, we might know dh/dt (by observing how fast the water level is rising) and want to find dV/dt. These are called “related rate” problems. According to the chain rule, the two rates dV/dt and dh/dt are related as follows:

\frac{dV}{dt} = \frac{dV}{dh} · \frac{dh}{dt}.

Since V (h) is given as an integral with h in the upper limit of integration, we can use the “fundamental theorem of calculus” to find dV/dh, without ever having to actually compute an integral. To see how to do this, let’s review what the fundamental theorem says, using the present notation. Suppose we have a function defined by an integral

V = ∫_{c}^{u(h)} f(y)dy.

Then

\frac{dV}{du} = f(u(h))

(this says that the derivative of an integral with respect to the upper limit of integration is obtained by removing the integral sign and substituting the upper limit of integration in place of the dummy variable in the integrand). Thus

\frac{dV}{dh} = \frac{dV}{du} · \frac{du}{dh} = f(u(h))\frac{du}{dh} .

In our situation with the volume of revolution we have u(h) = h + c, so that du/dh = 1, and the integrand f(y) is πg(y)^{2}. Thus, the fundamental theorem tells us that

\frac{dV}{dh} = πg(c + h)^{2},  and so  \frac{dV}{dt} = πg(c + h)^{2} \frac{dh}{dt}.

Another way to say this is as follows: the ratio of volume rate to height rate is the cross-sectional area, i.e., the area of the top surface of the liquid at the instant in question. This principle also applies when the liquid is filling a container whose cross-sections are not necessarily circles.

Example 3.1. A container is formed by rotating the function x = 4 sin(πy/12) around the y-axis between y = 0 and y = 6 (the units are feet). At a certain instant the container is filled to a depth of 4 ft, and water is being poured in at the rate of 5 ft^{3}/sec. How fast is the water level rising?

It’s generally a good idea (even if not absolutely essential) to draw a picture of what we’re dealing with. As explained earlier in the section on sinusoidal functions, the graph of x = 4 sin(πy/12) is obtained from the graph of x = sin y by stretching in the x-direction by a factor of 4 and in the y-direction by a factor of 12/π (so that its period expands from 2π to 24):

π 1x = sin y  12  4 x = 4 sin (\frac{π}{12}y)

Taking this function from y = 0 to y = 6 means taking the part until x reaches its first maximum at the point (4,6). (See the picture at the right). According to the above formula (here c = 0), we have

\frac{dV}{dt} = π(4 sin(πh/12))^{2} \frac{dh}{dt} .

At the instant in question h = 4 and dV/dt = 5, so we have

5 = 16π sin^{2}(π/3)\frac{dh}{dt} = 16π(0.5\sqrt{3})^{2} \frac{dh}{dt} = 12π \frac{dh}{dt} ,

so that

\frac{dh}{dt} = \frac{5}{12π} = 0.1326\frac{ft}{second} = 1.59\frac{inch}{second}.

In this problem, notice that we never had to evaluate an integral. If we had instead been asked to find the volume of water in the container when h = 4, then we would have had to compute π ∫_{0}^{4} 16 sin^{2}(πy/12) dy. At this point we do not have a technique for finding the anti-derivative of sin^{2} (later we will get a method for doing this type of integral). The point is: there is a big difference between asking about the rates at which the height and volume are increasing and asking about the actual values of the height and volume. In the latter case you need to evaluate an integral, whereas in the former case you do not.

Homework 2.

1. The graph of y = \sqrt{x} is rotated around the y-axis to form a container into which liquid is being poured at the rate of 0.1 cubic units per minute.

(a) How many minutes before the container is filled to a height of 1?

(b) At that point (h = 1), how fast is the water level rising?

2. Same as Problem 1, but with the graph of y = x^{4} rotated around the y-axis.

3. Sketch the graph of x = tan(πy/20). This graph is rotated around the y-axis to form a vat (the units are feet). Liquid is poured into the vat in such a way that the height of liquid is increasing steadily at 0.2 ft/sec. At what rate (in ft^{3}/sec) is liquid being poured when the height is 5 ft?

4. A hemispherical vat of radius 2 meters is being filled with a liquid, starting at time t = 0 sec, in such a way that the depth of the liquid increases at the steady rate of 4 cm/sec. What is the rate of flow of the liquid (in m^{3}/sec) after t seconds (expressed in terms of t)?

5. Same as Problem 4, except that the vat is obtained by rotating around the y-axis the bottom half of the ellipse x^{2}/9 + y^{2}/4 = 1.

6. With the vat in Problem 5, suppose now that it is being filled at the steady rate of 0.15 m^{3}/sec. Find the rate at which the depth of the liquid in the container is increasing at the moment when its depth is 0.5 m.

7. Suppose that the tetrahedron in Problem 9 of the previous chapter is situated upside down with its vertex at the origin and its base perpendicular to the y-axis at y = H. If the tetrahedron is being filled with water, find the ratio of dV/dt to dh/dt at the instant when the water has depth h at the deepest point. Express your answer in terms of h, H, and s.

16 3. RELATED RATE VOLUME PROBLEMS

CHAPTER 4 Arclength, Parameterized Curves, Ellipses

1. Review of Parameterized Curves

We first briefly review parameterized curves. For a more thorough review, reread the two chapters on this topic in the Math 113 notes.

A curve parameterized by time can be thought of as the motion of a particle in the xy-plane whose x- and y-coordinates are each given by a function of t: x = x(t), y = y(t). Then ẋ(t) = dx/dt is the horizontal velocity, ẏ(t) = dy/dt is the vertical velocity, and dy/dx = ẏ(t)/ẋ(t) is the slope of the particle’s path at the particular time t. In some situations we have other parameters besides t, for example, an angle parameter θ. In that case we can still compute the slope of the curve dy/dx by taking the ratio of dy/dθ to dx/dθ.

Some basic examples to understand are:

1. Straight line path from P = (x_{0}, y_{0}) to Q = (x_{1}, y_{1}). If you travel from P to Q at constant speed in time T, then your x- and y-coordinates at time t are:

x(t) = x_{0} + ((x_{1} − x_{0})/T )t, and y(t) = y_{0} + ((y_{1} − y_{0})/T )t .

2. Falling bodies (no air resistance). If the object starts from (x_{0}, y_{0}) with initial velocity of v_{0} directed at an angle α above the horizontal (α is a negative angle if it’s being thrown down), then

x(t) = x_{0} + (v_{0} cos α)t and y(t) = −\frac{1}{2}gt^{2} + (v_{0} sin α)t + y_{0} .

3. Circular motion. If the particle is moving around the circle of radius r centered at the point (h, k), and if we let θ denote the angular position of the particle in radians measured counterclockwise from the positive x-direction, then x(θ) = h + r cos θ, y(θ) = k + r sin θ. If θ is a known function of t, then we can substitute that function here to get x(t), y(t). If the particle’s angular position at time t = 0 is θ_{0} and if it has constant angular velocity (in radians/sec) equal to ±ω — where we take ω when it is moving counterclockwise and −ω when it is moving clockwise — then

x(t) = h + r cos(θ_{0} ± ωt), and y(t) = k + r sin(θ_{0} ± ωt) .

Some important special cases are:

1. h = k = θ_{0} = 0, so that x(t) = r cos(ωt), y(t) = r sin(ωt);

2. θ_{0} = π/2 (i.e., it starts from the positive y-direction) and the rotation is clockwise, so that x(t) = h + r cos(\frac{π}{2} − ωt) = h + r sin(ωt), y(t) = k + r sin(\frac{π}{2} − ωt) = k + r cos(ωt);

3. clockwise rotation starting from the negative y-direction, in which case x(t) = h − r sin(ωt), y(t) = k − r cos(ωt).

4. Combined circular motion with some other motion. Here the particle is whirling around a center which itself is moving. In that case you use the circular motion equations in the preceding paragraph, but with h and k (the coordinates of the center) replaced by the functions x_{center}(t) and y_{center}(t) which describe the motion of the center. For example, the trochoid is the motion of a point P which is stuck at a distance a (where a is between 0 and r) from the center of a wheel (where we assume that at time t = 0 the wheel passes over the origin with P located directly below the center at that instant). In that case the circular motion of P around the center has radius a, and from special case (3) in the preceding paragraph we obtain the equations

x(t) = rωt − a sin(ωt), and y(t) = r − a cos(ωt) .

The cycloid is the special case of this when a = r. If the wheel is traveling along at a known linear velocity (say, in m/sec), then the angular velocity ω (in rad/sec) is obtained by dividing the linear velocity by the radius of the wheel. If you are given ω in units of revolutions/sec, then multiply by 2π to get rad/sec.

Example 4.1. A wheel of radius 0.2 meters rolls without slipping along the flat roof of a building at speed 6 m/sec. At time t = 0 it flies off the roof. The building is 122.5 m high. Let P be the point on the rim of the wheel which is at the top at the instant t = 0 when the wheel leaves the roof. Find parametric equations for the motion of P. Take the side of the building as the y-axis and the ground as the x-axis.

When the wheel rolls along the roof, its angular velocity is: (linear velocity)/radius= 6/0.2 = 30 rad/sec. After leaving the roof it continues spinning at the same rate. So the point P revolves clockwise around the center of the wheel at ω = 30 rad/sec, starting above the center when t = 0. This is special case (2) in the paragraph above on circular motion. Thus, the motion of P relative to the center has equations 0.2 sin(30t) and 0.2 cos(30t). Meanwhile, the center of the wheel starts from

[figure: Path of center; P]

(0, 122.7) with initial horizontal component of velocity 6 m/sec (and zero initial vertical component of velocity). So x_{center}(t) = 6t, y_{center}(t) = −4.9t^{2} + 122.7. Finally, we get the equations for P by adding its equations relative to the center to the equations for the coordinates of the center:

x(t) = 6t + 0.2 sin(30t), y(t) = −4.9t^{2} + 122.7 + 0.2 cos(30t).

2. Ellipses

An ellipse is obtained by stretching the circle of radius 1 by a factor of a in one direction and by a factor of b in the perpendicular direction. An ellipse has two axes of symmetry, the major axis, having length 2a, which joins the two farthest opposite points on the circumference, and the minor axis, having length 2b, which joins the two closest opposite points on the circumference. The half-length a is called the semi-major axis, and b is called the semi-minor axis. If we choose to locate the ellipse in the xy-plane so that its center is at the origin, its major axis is in the x-direction, and its minor axis is in the y-direction, then it has the implicit equation x^{2}/a^{2} + y^{2}/b^{2} = 1. (The special case a = b = r is simply the circle of radius r, having equation x^{2} + y^{2} = r^{2}.) In parametric form, the ellipse can be described by the equations x(θ) = a cos θ, y(θ) = b sin θ, where we go once around the ellipse as θ goes from 0 to 2π. (One difference with a circle, however: the point P = (x(θ), y(θ)) is not located at an angle θ from the positive x-axis. In fact, if α is the angular position of P, then tan α = y(θ)/x(θ) = (b/a) tan θ, which is not the same as tan θ when a ≠ b; hence α is not the same as θ.)

Another description of an ellipse is the locus of all points P for which the sum of the distance from P to two fixed points is a constant. Those two points are called the foci (the plural of focus). If the ellipse is situated in the standard way, with the major axis along the x-axis, then both foci are on the x-axis, equidistant from the origin. We let (c, 0) and (−c, 0) denote the coordinates of the foci. See the drawing at the right.

−a −c a c −b b

We can get an important relationship between a, b and c if we notice that (1) the point (a, 0) on the ellipse is at a distance a + c from the farther focus and a − c from the closer focus, so the sum of the distances is 2a; and (2) the point (0, b) on the ellipse is at a distance \sqrt{b^{2} + c^{2}} from each of the two foci, so the sum of the distances is 2\sqrt{b^{2} + c^{2}}. Since these sums of distances must be the same, it follows that 2a = 2\sqrt{b^{2} + c^{2}}. Dividing by 2 and squaring both sides, we conclude that a^{2} = b^{2} + c^{2}. This relation enables us to find one of the three constants a, b, c if we know the other two.

The eccentricity of an ellipse is the ratio e = c/a, i.e., the fraction of the way from the center to the circumference that a focus is located on the x-axis. The eccentricity is always between 0 and 1: 0 ≤ e < 1. If e = 0, then a = b and we have a circle. The greater e, the farther our ellipse is from looking like a circle. Ellipses with e near 1 are long and thin.

Example 4.2. Find the equation of the ellipse of eccentricity 0.8 whose foci are at (0, ±60).

Here the foci are on the y-axis rather than the x-axis. This means that the major axis is in the y-direction. The distance c from the center to a focus is 60, and e = 0.8 = c/a. Hence, a = c/e = 60/0.8 = 75. Then b = \sqrt{a^{2} − c^{2}} = 45. When the major axis is in the y-direction,

the equation of the ellipse is x^{2}/b^{2} + y^{2}/a^{2} = 1, i.e., our answer is \frac{x^{2}}{2025} + \frac{y^{2}}{5625} = 1.

We now give a second way to describe an ellipse parametrically. We first move the ellipse horizontally a distance c to the right so that it has a focus at the origin. Then the other focus is at the point (2c, 0). (Sometimes it is more natural to put the origin at a focus of the ellipse rather that at its center. For example, the orbit of a planet as it revolves around the sun is an ellipse with the sun at one focus.) See the diagram on the next page.

If (x, y) are the coordinates of a point on the ellipse, then for our parameter we use the angle θ that the line from the origin to (x, y) makes with the xaxis. (As mentioned above, the θ-parameter in our earlier parametric equations does not have this simple geometrical meaning.) Let r = \sqrt{x^{2} + y^{2}} denote the distance of the point from the origin, and let d be its distance from the other focus. Since the sum of the distances of (x, y) from the two foci is equal to 2a, it follows that d = 2a − r. On the other hand, by the law of cosines, d^{2} = r^{2} + (2c)^{2} − 2r(2c) cos θ.

Thus, (2c, 0) θ (x, y) r

(2a − r)^{2} = r^{2} + 4c^{2} − 4rc cos θ.

If we expand the square on the left, then cancel the r^{2} and divide through by 4, we get a^{2} − ar = c^{2} − rc cos θ, and so

r = \frac{a^{2} − c^{2}}{a − c cos θ} .

Since x = r cos θ and y = r sin θ, we immediately obtain formulas for x and y in terms of the parameter θ:

x = \frac{(a^{2} − c^{2}) cos θ}{a − c cos θ} ,  y = \frac{(a^{2} − c^{2}) sin θ}{a − c cos θ} ,  0 ≤ θ ≤ 2π.

3. Arclength

Suppose we have a curve described parametrically by equations x = x(t), y = y(t). By the arclength of the curve between t = a and t = b we mean the actual length of the path (distance traveled). That is, if you were to imagine the path made of string and then straighten out the string and put it along the x-axis, this would be the length along the x-axis. In what follows we shall use the notation s(t) to mean the distance along the path from some reference point (perhaps the beginning of the path) to the point (x(t), y(t)). Then the arclength of the path between t = a and t = b is simply s(b) − s(a).

\begin{aligned}& t = b \\ t = a \end{aligned}

The derivative ds/dt — the rate at which the particle is traveling along its path — is what we normally call its speed. This is the length of the velocity vector, i.e., the hypotenuse of the triangle made up of the horizontal and vertical components of velocity. Thus,

\frac{ds}{dt} = \sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt} )^{2}}.

This means that the distance function s(t) is an antiderivative of the square root on the right, and so we have

(arclength from t = a to t = b) = s(b) − s(a) = ∫_{a}^{b} \sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt} )^{2}} dt.

In words, this says that distance traveled is equal to the area under the graph of speed.

Example 4.3. Find a definite integral for the distance traveled by the point P in Example 4.1 from when the wheel leaves the roof to when it hits the ground.

The wheel leaves the roof at time t = 0. To find out when it hits the ground, we observe that at that instant the center is at height 0.2 (the radius). Setting 0.2 = y_{center}(t) = −4.9t^{2} + 122.7, we find that t = \sqrt{122.5/4.9} = 5 sec. To find the speed, we compute the velocity components dx/dt = 6 + 6 cos(30t), dy/dt = −9.8t − 6 sin(30t), and so the distance traveled is

∫_{0}^{5} \sqrt{(6 + 6 cos(30t))^{2} + (−9.8t − 6 sin(30t))^{2}} dt.

This integral is too complicated to evaluate by the antiderivative method. It must be computed by “numerical integration,” i.e., an approximation method. In the next section we shall present one such method, the trapezoid rule.

If your curve is given in terms of a parameter other than time, you can use the same formula to determine the arclength — there is nothing special about the letter t in the above formula. For example, if θ is your parameter, then the arclength of the curve between θ = a and θ = b is

∫_{a}^{b} \sqrt{(\frac{dx}{dθ})^{2} + (\frac{dy}{dθ} )^{2}} dθ.

Example 4.4. Find the arclength of one arch of the cycloid, i.e., x(θ) = rθ − r sin θ, y(θ) = r − r cos θ, between θ = 0 and θ = 2π. Express your answer in terms of r. This is the actual distance traveled by a pebble in the tread of the wheel each time the wheel makes one revolution. If the bicycle travels 1 mile, how far does the pebble travel?

Here dx/dθ = r(1 − cos θ) and dy/dθ = r sin θ. Before taking the integral, it’s helpful to simplify the sum of the squares of the derivatives. We do this by factoring out r^{2} and then using the trig identity sin^{2} + cos^{2} = 1:

(\frac{dx}{dθ})^{2} + (\frac{dy}{dθ} )^{2} = r^{2}((1 − cos θ)^{2} + sin^{2} θ)
= r^{2}(1 − 2 cos θ + cos^{2} θ + sin^{2} θ) = r^{2}(2 − 2 cos θ).

We must integrate the square root of this: ∫_{0}^{2π} r\sqrt{2 − 2 cos θ} dθ. It is not obvious how to do this integral by the antiderivative method. For example, a u-substitution of the form u = 2 − 2 cos θ would not work. (Such a substitution would work if we had a sin θ term sitting beside the square root — but unfortunately we don’t have that.) Perhaps we should wait until the trapezoid rule, and be satisfied with an approximate value. However, there is a trick that will make the integrand take a simpler form. It uses the double angle formula cos(2A) = 1 − 2 sin^{2}(A). (We will later find that this formula comes up in integration problems in other situations — see the section on integrating powers of sine and cosine.) If we replace A by \frac{1}{2}θ in this double angle formula, and then move the cosine to the right and the sin^{2} to the left and multiply through by 2, we obtain: 4 sin^{2}(θ/2) = 2 − 2 cos(θ). Taking the square root of both sides, we get: \sqrt{2 − 2 cos θ} = 2 sin(θ/2). (Note: In general when we take square roots we have to put a ± in front of the radical; however, in this case, since we have assumed that θ is in the range between 0 and 2π, and so θ/2 is between 0 and π, we know that sin(θ/2) is positive.) Thus, the arclength of the cycloid is

∫_{0}^{2π} 2r sin(θ/2) dθ = −2r · \frac{1}{1/2} cos(θ/2) |_{0}^{2π}= −4r(cos π − cos 0) = 8r.

Thus, the pebble travels a distance 8r during the time the wheel makes one revolution, i.e., during the time the bicycle travels 2πr. Thus, the pebble travels 8/2π times as far as the bicycle, and so when the bicycle travels 1 mile the wheel travels 8/2π = 4/π = 1.27 miles.

Finally, we consider the case when a function is given in the explicit form y = f(x) between x = a and x = b. We can still use the same arclength formula, because the explicit form of a curve is a special case of parameterized equations: simply choose x as your parameter, and take the two equations x = x, y = f(x). Then dx/dx = 1 and dy/dx = f′(x). Using x in place of θ in the above arclength formula, we obtain

(arclength of the curve y = f(x) from x = a to x = b) = ∫_{a}^{b} \sqrt{1 + (f′(x))^{2}} dx.

In most cases — even when dealing with simple basic functions — these arclength formulas lead to integrals that cannot be computed by the antiderivative method, and we must use numerical integration.

Example 4.5. Find the length of one arch of the sine curve.

Here we want the arclength of y = f(x) = sin x from x = 0 to x = π. By the previous formula, since dy/dx = cos x we have

arclength = ∫_{0}^{π} \sqrt{1 + cos^{2} x} dx.

This integral cannot be computed by the antiderivative method. In the next section we will get an approximate value by the trapezoid rule.

Sometimes instead of the arclength itself, one is interested in the rate at which arclength is increasing — i.e., the speed of the particle. For example, given the horizontal component of velocity, we might ask for ds/dt; or given ds/dt, we might want to determine dx/dt or dy/dt. Here is a slightly more complicated example of a related rate problem.

Example 4.6. A bug is crawling up the y = x^{2} parabola. Meanwhile, the sun is shining from the left from an angle of 30^{∘} above the horizontal, casting a shadow on the x-axis (see the figure). When the bug reaches the point (1.5, 2.25), its shadow is moving at the rate of 0.3 units/min. Find the bug’s horizontal and vertical components of velocity and its speed.

Let u(t) denote the location of the bug’s shadow on the x-axis at time t. We know that du/dt = 0.3, and we want to find dx/dt, dy/dt, and ds/dt = \sqrt{(dx/dt)^{2} + (dy/dt)^{2}}. We have a simple relationship between y and x, namely y = x^{2}, but we need to find a relationship between u and x. To do that we observe the geometry of the shadow. Let O be the origin, P be the point on the x-axis directly under the bug, and Q be the location of the bug’s shadow on the x-axis. Then

[figure: bug; shadow; O; P; Q]

u = \overline{OP} + \overline{P Q} = x + y cot(30^{∘}) = x + y\sqrt{3} = x + x^{2}\sqrt{3}. This gives us a relation between u and x. Taking d/dt of this relation, we obtain:

0.3 = \frac{du}{dt} = \frac{d}{dt}(x + x^{2}\sqrt{3}) = \frac{dx}{dt} + 2x\sqrt{3}\frac{dx}{dt} = (1 + 2 · 1.5\sqrt{3})\frac{dx}{dt}.

Solving for dx/dt gives us dx/dt = 0.3/(1 + 3\sqrt{3}) = 0.0484 units/min. Next, dy/dt = (dy/dx)(dx/dt) = 2x(dx/dt) = 2 · 1.5 · 0.0484 = 0.1452 units/min, and the speed is \sqrt{0.0484^{2} + 0.1452^{2}} = 0.1531 units/min.

Homework 3.

1. Find the equation of an ellipse centered at the origin if (a) it passes through (2,0) and also through (0,1/2);

(b) it passes through (10,0) and has foci at (±6,0);

(c) it has foci at (±7,0) and eccentricity 0.28;

(d) it passes through (1,0) and also through (0,3);

(e) it has foci at (0,±3) and passes through (4,0).

2. Find the foci and eccentricity of the ellipses (a) x^{2}/25 + y^{2}/16 = 1; (b) x^{2}/49 + y^{2}/625 = 1; (c) 4x^{2} + 9y^{2} = 1.

3. A particle travels at constant velocity in a straight line from the point (1,3) to the point (7,11) in 2 seconds.

(a) Find parametric equations x(t), y(t) for the motion of the particle, and find the distance the particle travels using the formula for the arclength of a parameterized curve.

(b) Find the equation of the line the particle travels along in the form y = mx + b, and then find the distance the particle travels using the formula for the arclength of an explicit function y = f(x).

(c) Find the distance the particle travels by simply using the distance formula between its starting point and stopping point. Your answers to parts (a), (b) and (c) should be the same.

4. Let n be greater than 1.

(a) Write a definite integral for the arclength of y = x^{n} from x = 0 to x = b.

(b) There is one (and only one) case when this integral can be easily evaluated by the anti-derivative method, namely, when n = 3/2. In that case use a suitable substitution to evaluate the integral and find a formula in terms of b for the arclength in part (a).

5. Write a definite integral for (a) the total distance traveled by a ball that is thrown from a 38.8-meter high window

at 14 m/sec at an angle of 45^{∘} above the horizontal;

(b) the distance one-quarter the way around an ellipse having semi-major axis a and semi-minor axis b, i.e. the part of the circumference that is in the first quadrant (use parametric equations for the ellipse);

(c) the distance traveled by the pebble in Example 4.4 if the pebble is caught in a spoke half-way from the center to the rim.

6. A bug is crawling up the arch of the sine curve y = sin x at speed 0.5. When the bug reaches the point (π/6, 0.5), find

(a) the rate at which its x-coordinate is increasing;

(b) the velocity of its shadow on the x-axis, if the sun is shining from the left from an angle of 21.8^{∘} above the horizontal;

(c) the same as part (b), but no longer assume that the angle from which the sun is shining remains constant. Suppose that the units of time are hours, and the sun is setting at the rate of 15^{∘} per hour.

CHAPTER 5 Numerical Integration: The Trapezoid Rule

Suppose we want to find an integral ∫_{a}^{b} f(x)dx, i.e., the area under the curve y = f(x) between x = a and x = b, and suppose that we are unable to evaluate this integral by the antiderivative method. The purpose of this section is to describe an approximation method which, while not giving us the exact value of the integral, can be used to obtain a value which is accurate enough for practical purposes.

1. Riemann sums

The most basic way to approximate the area under a curve is by Riemann sums. A Riemann sum is the total area of a number of thin rectangles that we draw so as to fill out roughly the area under the curve. More precisely, we first describe the n-th left endpoint Riemann sum. We choose a number n, and divide the interval [a, b] into n equal subintervals, each of length (b − a)/n. We let Δx = (b − a)/n denote the length of a subinterval. If n is large, then we have very small subintervals, and in that case it turns out that the Riemann sum we later get will be close to the true value for the area under the curve.

Next, we draw a vertical line up to the curve at each endpoint of the subintervals. The first line has length f(a), the second line has length f(a + Δx), the third line has length f(a + 2Δx), and so on, until we get to the last line, which has length f(a + nΔx) = f(b). Now form a rectangle by drawing a little horizontal line of length Δx from left to right at the top of each of the vertical lines (except for the last one at b).

Each thin rectangle has height equal to the value of f(x) at the left endpoint of the little subinterval that forms its base; and each rectangle has width equal to Δx. Thus, the area of the first thin rectangle is Δx · f(a), the area of the second rectangle is Δx · f(a + Δx), the area of the third is Δx · f(a + 2Δx), and so on; finally, the area of the n-th thin rectangle is

Δx · f(a + (n − 1)Δx) .

y = f(x) a b Δx Δx Δx Δx Left endpoint Riemann sum

Hence, the sum of the areas of the n rectangles is

Δx(f(a) + f(a + Δx) + f(a + 2Δx) + ⋯ + f(a + (n − 1)Δx)).

This “left endpoint Riemann sum” is approximately equal to the area under the curve. Using the summation notation, we can write:

∫_{a}^{b} f(x)dx ≈ Δx ∑_{i=0}^{n−1} f(a + iΔx), where Δx = \frac{b − a}{n}.

(To interpret the summation notation, let i start at the value written below the ∑, namely 0, and run through the numbers until you get to the top limit of the ∑, namely n − 1; plug each value of i into f(a + iΔx) and add the resulting terms.)

Alternately, we could use the right endpoint Riemann sum, where we take rectangles that go to the opposite side of the curve from the rectangles in the left endpoint Riemann sum. (In

other words, if a rectangle in the left endpoint sum is a little below the curve — which happens when y = f(x) is an increasing func- tion — then the corresponding right endpoint rectangle will be a little above the curve; and if a left endpoint rectangle is a little above the curve — which happens when you have a de- creasing function — then the corre- sponding right endpoint rectangle will be a little below the curve.) In the for- mula for the right endpoint Riemann sum, the f-values start at f(a + Δx) and go through f(b):

y = f(x) a  b Δx Δx Δx Δx Right endpoint Riemann sum

Δx(f(a + Δx) + f(a + 2Δx) + ⋯ + f(a + (n − 1)Δx) + f(b)) .

(Compare this with the formula for the left endpoint sum.)

The accuracy of an approximation is greater if we choose n large. Generally, neither the left endpoint nor the right endpoint Riemann sum method is very accurate. That is, to get a value close to the true area under the curve, you need to use very large n.

2. The Trapezoid Rule

A better method can be obtained by averaging the left and right endpoint Riemann sums. This is equivalent to drawing a slanted line across the top (see the figure below at the right).

The resulting sum is called the trapezoid rule. The trapezoid rule is more accurate than the left or right endpoint Riemann sum. It is easy to find a formula for the trapezoid rule by adding the left endpoint formula and the right endpoint formula and then dividing by 2. That is, we take the sum

y = f(x) a b Δx Δx Δx Δx The Trapezoidal Rule

Δx(f(a)+f(a + Δx) + f(a + 2Δx) + ⋯ + f(a + (n − 1)Δx)) +
+Δx(f(a + Δx) + f(a + 2Δx) + ⋯ + f(a + (n − 1)Δx) + f(b))

and divide by 2. The result is the trapezoid rule:

∫_{a}^{b} f(x)dx ≈ Δx(\frac{1}{2}f(a) + f(a + Δx) + f(a + 2Δx) + f(a + 3Δx) +

+ ⋯ + f(a + (n − 1)Δx) + \frac{1}{2}f(b)).

Notice the difference with the Riemann sum formula: both f(a) and f(b) appear at the two extremes of the sum, each with coefficient \frac{1}{2}.

Example 5.1. Compare the approximations to ∫_{1}^{2} x^{−2}dx using the left endpoint Riemann sum, the right endpoint Riemann sum, and the trapezoid rule — all with n = 4 subdivisions — with the true value of the integral, which is −x^{−1} |_{1}^{2}= 1/2.

Here f(x) = 1/x^{2}, Δx = 1/4, and the 4 subintervals have endpoints 1, 5/4, 3/2, 7/4 and 2. We have

left Riemann sum = \frac{1}{4} (1 + \frac{1}{(5/4)^{2}} + \frac{1}{(3/2)^{2}} + \frac{1}{(7/4)^{2}} ) = 0.603;
right Riemann sum = \frac{1}{4} (\frac{1}{(5/4)^{2}} + \frac{1}{(3/2)^{2}} + \frac{1}{(7/4)^{2}} + \frac{1}{2^{2}} ) = 0.415;

trapezoid rule =  \frac{0.603 + 0.415}{2}  = 0.509.

Note that the left (or right) Riemann sum is pretty inaccurate, while the trapezoid rule is a lot closer, even though the number n = 4 of subdivisions is rather small. We would get an even closer approximation to the true value of the integral if we used a larger value of n.

Example 5.2. Use the trapezoid rule with n = 6 subdivisions to find the length of one arch of the sine curve.

In Example 5 of the last section, we saw that an arch of the sine curve y = f(x) = sin x has length equal to ∫_{0}^{π} \sqrt{1 + (f′(x))^{2}}dx = ∫_{0}^{π} \sqrt{1 + cos^{2} x} dx. We now apply the trapezoid rule to this integral. When using the trapezoid rule, be careful about what function you are substituting for the f(x). In this problem, because we are taking the arclength rather than the area under the sine curve, the function to substitute in the trapezoid rule is \underline{not} our original function sin x. The f(x) in the trapezoid rule is \sqrt{1 + cos^{2} x}. That is, it is a different f(x) from the f(x) we substituted into the arclength formula ∫_{a}^{b} \sqrt{1 + (f′(x))^{2}}dx.

Be careful not to get confused: the f(x) in the arclength formula is not the f(x) in the trapezoid rule.

Taking n = 6, we have Δx = (π − 0)/6 = π/6. Thus,

∫_{0}^{π}\sqrt{1 + cos^{2} x} dx ≈ \frac{π}{6} (\frac{1}{2}\sqrt{1 + cos^{2} 0} + \sqrt{1 + cos^{2}(π/6)} + \sqrt{1 + cos^{2}(π/3)}

+ \sqrt{1 + cos^{2}(π/2)} + \sqrt{1 + cos^{2}(2π/3)} + \sqrt{1 + cos^{2}(5π/6)} + \frac{1}{2}\sqrt{1 + cos^{2}(π)})

= \frac{π}{6} (\frac{1}{2}\sqrt{2} + \sqrt{7/4} + \sqrt{5/4} + 1 + \sqrt{5/4} + \sqrt{7/4} + \frac{1}{2}\sqrt{2}) = 3.8202.

Notice that the straight-line distance from the beginning to the end of the arch is π = 3.1416; so our answer of 3.8202 for the arclength is reasonable. In doing arclength problems, it is a good idea to compare your answer with a rough estimate for the length in question: if the answer makes no sense, then go over your work again looking for careless errors.

3. Functions Obtained by Interpolating Experimental Data

There are two basic uses of numerical integration: (1) finding the definite integral of a function given by a formula for which you cannot find the antiderivative; and (2) finding the definite integral when the function is not given by a formula at all. The latter situation occurs frequently in practice. Suppose, for example, that you have a function f(t) whose values come from instrument readings at regular time intervals from t = a to t = b. You can graph f(t) by tabulating and plotting the points corresponding to the instrument readings, and then drawing a smooth curve connecting these points. In this situation you can use the table of values along with the trapezoid rule to compute ∫_{a}^{b} f(t)dt, or perhaps the integral of some other function made up from f(t). It is generally best to choose the number n of subdivisions to equal the number of time intervals over which you took instrument readings.

Example 5.3. The table below shows the speedometer reading in miles/hour when a car tried to brake before an accident. Use the trapezoid rule to determine how far the car traveled in this time.

t (in seconds) 0 1 2 3 4 5 6 7 8
speed (mph) 70 65 59 52 44 36 27 18 10

First, we note that to convert miles/hour to feet/sec we must multiply by 5280 and divide by 3600. Thus, the distance in feet (using the trapezoid rule with n = 8 and Δt = 1) is

\frac{5280}{3600} · (\frac{1}{2} · 70 + 65 + 59 + 52 + 44 + 36 + 27 + 18 + \frac{1}{2} · 10) = 500.13 ft.

Since your table of speeds is given with only two significant figures of accuracy, in practical terms it is misleading to give the distance traveled to five significant figures. We should simply say that the car traveled 500 ft.

Example 5.4. A wine cork exactly 3 cm high has circular bottom of radius 0.542 cm and circular top of radius 0.859 cm. The following table shows the radius in centimeters of the cork’s circular cross-section at intervals of 0.5 cm from bottom to top:

0.542 0.608 0.668 0.723 0.774 0.820 0.859

Find the volume of the cork.

The table shows the distance g(y) from the axis of symmetry (which we take as the y-axis) to the curve that is rotated around the y-axis to get the cork, taken at the endpoints of the 6 intervals of length Δy = 0.5 from y = 0 to y = 3. Thus, it makes sense to apply the trapezoid rule with n = 6 subdivisions. The function being integrated is g(y)^{2}:

Vol = π ∫_{0}^{3} g(y)^{2} dy

≈ πΔy(\frac{1}{2}g(0)^{2} + g(0.5)^{2} + g(1)^{2} + g(1.5)^{2} + g(2)^{2} + g(2.5)^{2} + \frac{1}{2}g(3)^{2})

= \frac{π}{2} (\frac{1}{2} · 0.542^{2} + 0.608^{2} + 0.668^{2} + 0.723^{2} + 0.774^{2} + 0.820^{2} + \frac{1}{2} · 0.859^{2})

= 4.91 cm^{3}.

Example 5.5. (Measuring cardiac output.^{1}) The rate R at which a person’s heart pumps blood is called the person’s cardiac output. The normal rate is about 5 liters/min, but the cardiac output can be quite different if the person has certain illnesses or is engaged in strenuous activity.

One way to measure R is to inject a known amount D of dye in a vein near the heart. The dyed blood then circulates once through the system in a little less than half a minute. Meanwhile, the concentration of dye in the blood, measured in milligrams per liter, can be monitored in the aorta, where the first of the dye will appear a few seconds after injection. The table below at the left shows concentration readings in mg/liter taken at 1-second intervals until all of the dye has passed the monitoring instrument. Here t = 0 is the instant when D = 5.0 mg of dye was injected into the blood. The graph below on the right is obtained by plotting the points in the table and interpolating a smooth curve between those points.

t C(t) t C(t)
0 0.0 13 4.0
1 0.0 14 3.8
2 0.0 15 3.7
3 0.1 16 2.9
4 0.6 17 2.2
5 0.9 18 1.5
6 1.4 19 1.1
7 1.9 20 0.9
8 2.7 21 0.8
9 3.0 22 0.5
10 3.7 23 0.3
11 4.0 24 0.0
12 4.1

We now derive a relationship between the cardiac output R, the amount of dye D, and the integral under the C(t) curve. To do this, we define a function D(t) to be the amount of dye that has passed by the monitoring point within the first t seconds. Then dD/dt is the rate at which dye is passing. We have

1Taken from Measuring Cardiac Output, by B. Horelick and S. Koont, Project UMAP, Education Development Center, 1978.

\frac{dD}{dt} = R · C(t),

i.e., the number of mg per sec of dye is equal to the number of liters per second of blood times the number of mg of dye per liter of blood. This is an important practical principle, which we shall encounter again when we treat mixing problems: the rate (in mass per unit time) at which an impurity (such as dye) is flowing past a point is equal to the product of the rate (in liters per unit time) that the liquid is flowing past and the concentration (mass per liter). For example, if 2 liters per second are going by, and each liter contains 3 mg of impurity, then 6 mg per second of impurity are going by.

Returning to the dye, we now have an expression for the derivative of our function D(t). Then the total dye that passes — which is the amount D that we originally injected — is equal to the definite integral of that expression, i.e.,

\frac{dD}{dt} = R C(t) implies that D = D(24) − D(0) = ∫_{0}^{24} R C(t) dt = R ∫_{0}^{24} C(t) dt,

where we took the constant R outside of the integral. Suppose we find the integral of C(t). If we knew R, then this formula would tell us what D is. As it is, we do not know R, but we do know D. So we want to use this formula to solve for R in terms of D:

R = \frac{D}{∫_{0}^{24} C(t) dt} = \frac{5.0}{∫_{0}^{24} C(t) dt}.

It remains to find the integral of C(t), which we do using the trapezoid rule with n = 24 subdivisions (so that Δt = 1 sec):

∫_{0}^{24} C(t) dt ≈ \frac{1}{2}f(0) + f(1) + f(2) + f(3) + ⋯ + f(23) + \frac{1}{2}f(24) = 44.1 .

Finally, R = 5.0/44.1 = 0.113 liters/sec = 6.8 liters/min. Notice that if f(a) = 0 and f(b) = 0, as is the case in this example, then there is no difference between the left (or right) endpoint Riemann sum and the trapezoid rule.

Homework 4.

1. Suppose that at time t = 10 sec an object is traveling at 30.0 m/sec. Its acceleration is measured at two-second intervals until time t = 20, with the following results (the units of acceleration are m/sec^{2}):

\begin{matrix}t & 10 & 12 & 14 & 16 & 18 & 20 \\ a(t) & 2.32 & 2.43 & 2.52 & 2.59 & 2.65 & 2.70 \end{matrix}

Use the trapezoid rule to find the velocity of the object at time t = 20.

2. Using n = 4 subdivisions, see how close to the true value of ∫_{0}^{π/2} sin x dx you get using (a) the left endpoint Riemann sum, (b) the right endpoint Riemann sum, (c) the trapezoid rule, (d) Find how much closer you get to the true value using the trapezoid rule with 10 rather than 4 subdivisions.

3. Find the arclength of the curve y = x^{3} from x = 0 to x = 2 using the trapezoid rule with n = 4 subdivisions.

4. Find the arclength of the ellipse 4x^{2} + y^{2} = 4 between the points (0, 2) and (0.5, \sqrt{3}) by (a) solving for y as a function of x and applying the trapezoid rule with n = 5 subdivisions to ∫_{a}^{b} \sqrt{1 + (f′(x))^{2}}dx; and (b) writing the ellipse parametrically, finding the values of θ corresponding to the two points (0, 2) and (0.5, \sqrt{3}), and applying the trapezoid rule with n = 5 subdivisions to the integral for the arclength of a parametrized curve.

5. In Example 3 of the last section, suppose you wanted to find the distance traveled by the point P during the 5 seconds from when the wheel leaves the roof to when it hits the ground. (a) Explain why, even with the trapezoid rule, you cannot expect to get an accurate answer with a small value of n such as n = 10. Now suppose that we are interested only in the first 0.04 sec. Use the trapezoid rule with n = 4 subdivisions to find the distance traveled during the first 0.04 sec by (b) the center of the wheel, and (c) the point P.

6. The table below shows the horizontal and vertical velocities of a particle at 1-second intervals for the first 8 seconds. Use the trapezoid rule to find (a) the straight-line distance between the particle’s location at time t = 0 and at time t = 8, and (b) the actual distance traveled along its path between t = 0 and t = 8.

\begin{matrix}t & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ dx/dt & 7.02 & 6.59 & 5.96 & 5.23 & 4.40 & 3.67 & 2.74 & 1.81 & 1.08 \\ dy/dt & 4.12 & 5.21 & 5.84 & 6.15 & 6.24 & 6.23 & 6.09 & 5.80 & 5.45 \end{matrix}

32 5. NUMERICAL INTEGRATION: THE TRAPEZOID RULE

CHAPTER 6 Center of Mass

The seesaw principle. Suppose that a light-weight board 5 meters long is laid out between a point 10 meters from a reference point and a point 15 meters from the reference point. Suppose that someone weighing 50 kg is sitting at one end of the board (at 10 meters from the reference point), and someone weighing 75 kg is sitting at the other end. We want to know where to put the fulcrum so as to balance the board with the people on it. Here we will assume that the weight of the board is negligible, and that the people are point-masses concentrated at 10 and 15 on our axis:

0  10 50 kg  15 75 kg \overline{x}

The location of the balancing point \overline{x} is obtained by taking the ratio of the total moment M to the total mass m. Here m is simply 50 + 75 = 125 kg. The moment (in the x-direction, relative to the reference point 0) is the sum of (mass×x-coodinate), i.e., 50 · 10 + 75 · 15 = 1625 kg-m. (The units are kilogram-meters, i.e., mass times distance.) Then \overline{x} = M/m = 1625/125 = 13 meters. That is, the balancing point is 3/5 the way from the lighter person to the heavier person. (We would have gotten the same result if we had chosen our reference point differently, for example, if we had chosen one end of the board to be 0.)

If there are a number of point-masses located at various places on the x-axis, the procedure is the same. Suppose, for instance, that we have five masses: a mass m_{1} located at position x_{1}, a mass m_{2} located at x_{2}, and so on. Then

\overline{x} = \frac{M}{m} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5}} .

We shall call this the “seesaw principle:” the center of mass is equal to total moment divided by total mass, where moment is computed by adding up each piece of mass with the position of the piece of mass multiplied in. In reality, mass is usually not all concentrated in a few points, but rather is distributed continuously over some region. That is where integration (rather than taking a finite sum) comes in.

The center of mass is important for understanding the motion of large objects. This is because the vector sum of all external forces on the object is equal to the mass times the acceleration of the center of mass. For example, when a baton is twirled and thrown into the air (Example 11.1 in the Math 124 notes), its center travels along a simple parabolic falling body trajectory (but the path of the tip of the baton is more complicated).

There are many types of center of mass problems, involving three-dimensional shapes with varying densities. However, we shall deal only with problems of a particularly straightforward type: the center of mass of a flat uniform shape (also called its “ centroid”).

So suppose we have some shape in the xy-plane, for example, the region contained between two curves (a “top curve” and a “bottom curve”) between x = a and x = b. Suppose that this shape is made up of a uniform substance having a certain constant density D measured in mass per unit area — say, gm/cm^{2}. (If the substance has a known density in the sense of mass per volume — say, gm/cm^{3} — then D is equal to this volume density times the thickness of the plate.) The mass of the uniform plate is simply D times its area, i.e.,

m = D ∫_{a}^{b}(top function − bottom function)dx .

Let f(x) denote the distance across the plate from top to bottom measured at position x on the x-axis, i.e., f(x) is obtained by subtracting the bottom function from the top function. (In order to keep the notation simple, in this section we will not introduce separate symbols for the top function and the bottom function.) Then

m = D ∫_{a}^{b} f(x)dx .

\overline{y}  x  \overline{x}  ^{^{Δx} x  f(x)}

The uniform plate will balance at some point with coordinates (\overline{x}, \overline{y}). The practical meaning of \overline{x} and \overline{y} is as follows. If you take a razor-edge parallel to the y-axis and put it under the flat plate at a position x which is to the left of \overline{x}, then the plate will overbalance to the right; if you put the razor-edge under the plate at a position x > \overline{x}, then the plate will overbalance to the left. If you put the razor-edge exactly at the position \overline{x}, then the plate will balance. Similarly, if you take a razor-edge parallel to the x-axis and put it under the plate at a position y which is less than or greater than \overline{y}, then the plate will overbalance on one side or the other; but if you put it parallel to the x-axis exactly at \overline{y}, then the plate will balance. The intersection of the line parallel to the y-axis where the plate balances and the line parallel to the x-axis where it balances is the point (\overline{x}, \overline{y}). If you stuck a pencil under the plate at this point, the plate would spin around on the pencil point without falling to any side.

We first find \overline{x}, which is the ratio of total moment in the x-direction to the total mass. We already know how to find the total mass: m = D ∫_{a}^{b} f(x)dx. To find the moment, we use the same type of procedure as when we derived the integral formula for the area under a curve or the volume of a solid of revolution. Namely, we divide the interval from a to b into a large number of small subintervals of length Δx = (b − a)/n. The left endpoints of these subintervals are at a, a + Δx, a + 2Δx, … a + (n − 1)Δx. If we assume that the strip above each of the subintervals is a rectangle (this is approximately the case if Δx is small), then the total mass of all of the strips is obtained by summing

D f(a)Δx + D f(a + Δx)Δx + D f(a + 2Δx)Δx + ⋯ + D f(a + (n − 1)Δx)Δx.

However, we want the moment, which is equal to mass times location in the x-direction. We make the simplifying assumption that the entire strip above a subinterval has the same x-location as the left endpoint of the interval (this is approximately the case if Δx is small); so to get the moment we multiply each term in the above sum by the x-coordinate of the left endpoint. In other words, the moment is approximately

D a f(a)Δx + D(a + Δx)f(a + Δx)Δx + D(a + 2Δx)f(a + 2Δx)Δx +
⋯ + D(a + (n − 1)Δx)f(a + (n − 1)Δx)Δx,

which is the Riemann sum for the function D x f(x). Thus, in the limit we get the definite integral from a to b of D x f(x):

moment = D ∫_{a}^{b} xf(x)dx.

One way to think of this formula is to look at the contribution to the moment that comes from the part of the region that lies between x and x + Δx, where we suppose that Δx is very small. This is a thin strip Δx across whose height at the left is f(x). Since Δx is small, the strip is very close to being a rectangle, and so its area is very close to f(x)Δx, and its mass is very close to D f(x)Δx. Its contribution to moment is then very close to x times D f(x)Δx. Now to find the moment of the entire flat plate, we have to sum up the contributions as x goes from a to b. In the limit as Δx goes to zero this sum becomes an integral, and we write dx instead of Δx. That is, we get ∫_{a}^{b}(Dx f(x))dx, as before.

Thus, the total moment of a uniform flat plate is equal to the same type of formula as the total mass, except with x multiplied by f(x) inside the integral. Finally, to find \overline{x} we divide M = D ∫_{a}^{b} xf(x) dx by m = D ∫_{a}^{b} f(x) dx. The D’s cancel (that is, the location of the balancing point does not depend on whether the flat plate is made of a heavy or light substance), and we conclude that

\overline{x} = \frac{∫_{a}^{b} xf(x) dx}{∫_{a}^{b} f(x) dx}.

It remains to find a formula for \overline{y}. We shift perspectives, and look at the same shape from left to right as we move up the y-axis. Let g(y) denote the distance across the shape from left to right at the position y. To find g(y), we express the right boundary of the shape in the form x = g_{1}(y) and the left boundary in the form x = g_{2}(y); then g(y) = g_{1}(y)−g_{2}(y). We suppose that in the y-direction the shape is contained between y = c and y = d, which will be our limits of integration. (This is what we did when finding the area between two curves in cases when we integrated with respect to y rather than x.) The derivation of a formula for \overline{y} is exactly the same as for \overline{x}. The result is:

\overline{y} = \frac{∫_{c}^{d} yg(y) dy}{∫_{c}^{d} g(y) dy} .

Actually, the denominator in this formula has the same value as in the formula for \overline{x} — in both cases it is the area of the plate. The numerators, however, are different.

Example 6.1. A uniform flat shape has the form of a rectangular ruler of width w and length 2 feet, with the last foot tapered off to a point, i.e., it is in the shape of a rectangle of length 1 ft surmounted by an isosceles triangle of altitude 1 ft. Find the balancing point.

Our first task is to put this shape in the xy-plane in a convenient way. Let us take the x-axis as the axis of symmetry, and put the pointed end at the origin. (That way the slanted line has the simplest possible equation.) Thus, the shape is bounded by the two lines joining the origin to the points (1, ±w/2), the two horizontal lines joining (1, ±w/2) to (2, ±w/2), and the vertical line joining (2, −w/2) to (2, +w/2). See the drawing below, where the equations of these 5 lines are indicated:

[figure: 1; y = w/2; y = −w/2; y = \frac{w}{2} x; y = − \frac{w}{2} x; x = 2]

We first observe that, because of the symmetry of the shape, the balancing point is obviously on the x-axis, i.e., \overline{y} = 0. Thus, we need only find \overline{x}. In general, when finding the center of mass of a symmetrical shape you can reduce the amount of computation required by taking into account that the balancing point must be on the axis of symmetry.

To compute \overline{x}, we find the following formulas for the distance across from top to bottom (i.e., top function minus bottom function):

f(x) = \begin{cases}wx, & if 0 ≤ x ≤ 1; \\ w, & if 1 ≤ x ≤ 2. \end{cases}

Thus, the numerator (the moment integral) in the formula for \overline{x} is

∫_{0}^{1} wx^{2}dx + ∫_{1}^{2} wx dx = w/3 + 3w/2 = 11w/6 .

The denominator can be found either by integration or simply by using the formulas for the area of a rectangle and triangle: this gives us 3w/2. Thus, \overline{x} = (11w/6)/(3w/2) = 11/9 = 1.222. In other words, the balancing point is in the rectangular part of the shape a distance 2/9 the way from the base of the triangle to the opposite side of the rectangle.

Two observations should be made about this answer. (1) The width w canceled out; this means that the balancing point does not depend on whether the shape is wide or narrow (just as it does not depend on whether it is made of a heavy or light substance). (2) A rough check on the correctness of the answer is that we would expect an answer between 1 and 1.5, since a razor-edge placed at the base of the triangle would cause the shape to overbalance to the right, whereas a razor-edge place in the middle of the rectangular part would case the shape to overbalance to the left.

Example 6.2. Find the center of mass of a uniform flat plate in the shape of a quarter circle of radius 1 (say, the quarter of the unit circle that is in the first quadrant of the xy-plane) on top of a square of side 1 (i.e., the unit square in the corner of the fourth quadrant).

Here we do not have symmetry to help us, so we must compute both \overline{x} and \overline{y}. The denominator in both cases is the area of our shape, which is

(\frac{1}{4}π · 1^{2}) + (1^{2}) = (π/4) + 1 .

The distance from top to bottom is f(x) = 1 + \sqrt{1 − x^{2}}. Thus, in computing \overline{x}, we find that the moment integral is

∫_{0}^{1} x (1 + \sqrt{1 − x^{2}}) dx = ∫_{0}^{1} x dx + ∫_{0}^{1} x\sqrt{1 − x^{2}}dx .

The first of these two integrals is 0.5. The second integral can be evaluated by the method of substitution, since the derivative of what’s under the radical is essentially x dx. That is, we substitute u = 1 − x^{2}; then du = −2xdx, and so the integral becomes −0.5 ∫ \sqrt{u} du. As for the limits of integration, x = 0 implies u = 1 − 0^{2} = 1, and x = 1 implies u = 1 − 1^{2} = 0. Thus, the moment integral is

0.5 − 0.5 ∫_{1}^{0} \sqrt{u} du = 0.5 − 0.5(−2/3) = 5/6, so that \overline{x} = \frac{5/6}{(π/4) + 1} = 0.4667.

To compute \overline{y}, we see that the distance across our shape from the y-perspective (i.e., from left to right) is

g(y) = \begin{cases}1, & if −1 ≤ y ≤ 0; \\ \sqrt{1 − y^{2}}, & if 0 ≤ y ≤ 1. \end{cases}

Thus, the numerator of the \overline{y} formula is ∫_{−1}^{0} y dy+∫_{0}^{1} y\sqrt{1 − y^{2}}dy. The first of these integrals is −1/2, while the second integral happens to be exactly the same integral that we already computed in the last paragraph, except with a different letter as the variable of integration. That is, ∫_{0}^{1} y\sqrt{1 − y^{2}}dy has the same value as ∫_{0}^{1} x\sqrt{1 − x^{2}}dx, namely, 1/3. We conclude that

\overline{y} = \frac{−\frac{1}{2} + \frac{1}{3}}{\frac{π}{4} + 1} = −0.09335.

The reader should sketch a picture of the shape, and check that (0.4667, −0.09335) is a reasonable location for the balancing point.

Example 6.3. Find the centroid of Nicaragua (see the map on the next page). As an approximation to the actual shape of the country, suppose that it is a quadrilateral with vertices at the points

11^{∘} northern latitude & 86^{∘} western longitude, 11^{∘} northern latitude & 83^{∘} western longitude, 15^{∘} northern latitude & 83^{∘} western longitude, 13^{∘} northern latitude & 88^{∘} western longitude.

Also suppose that Nicaragua is a plane shape, i.e., disregard the mountains, curvature of the earth, etc.

Nicaragua

Here our first problem is to put the shape in the xy-plane in a convenient way. We could choose the conventional longitude/latitude coordinates as our xy-coordinates. But that would be a little cumbersome, because the origin of that coordinate system is at the Equator along the Greenwich Meridian, which is far from Nicaragua. Also note that “west longitude” means the negative direction from the Greenwich Meridian, so we would be using fairly large negative numbers for the x-coordinates of the corners (namely, −86, −83, −88). A better idea is to choose the origin at the intersection of the two slanted sides, which is (88^{∘} west, 13^{∘} north). Then the other three corners have the following coordinates relative to this origin: (2, −2), (5, −2), (5, 2). The two slanted sides have equations y = 0.4x (or x = 2.5y) and y = −x (or x = −y). The horizontal side is y = −2, and the vertical side is x = 5. Thus, the distance f(x) from top to bottom and the distance g(y) from left to right are:

f(x) = \begin{cases}1.4x, & if 0 ≤ x ≤ 2; \\ 0.4x + 2, & if 2 ≤ x ≤ 5; \end{cases} g(y) = \begin{cases}5 + y, & if −2 ≤ y ≤ 0; \\ 5 − 2.5y, & if 0 ≤ y ≤ 2. \end{cases}

The rest of this example is routine, and the details will be left to the reader. The answer is (3.10^{∘}, −0.31^{∘}). This must be added to the longitude −88^{∘} and the latitude 13^{∘} of the origin. Thus, the centroid of Nicaragua is at 12.7^{∘} northern latitude & 84.9^{∘} western longitude, i.e., about 70 miles east of Matagalpa, on the Rio Grande.

Homework 5.

1. Find the coordinates of the center of mass of a uniform flat plate that is bounded by:

(a) the line x + 2y = 6, the x-axis, and the y-axis; (b) the parabola y = x^{2}, the x-axis, and the line x = 2; (c) the right half of the circle x^{2} + y^{2} = r^{2} and the y-axis; (d) the top half of the ellipse x^{2} + 4y^{2} = 4 and the two lines joining the point (0, −4) to the points where the ellipse meets the x-axis.^{1}

2. Find the centroid of Nevada. Suppose that Nevada is a trapezoid with corners at the four points

42^{∘} northern latitude & 120^{∘} western longitude, 42^{∘} northern latitude & 114^{∘} western longitude, 39^{∘} northern latitude & 120^{∘} western longitude, 35^{∘} northern latitude & 114^{∘} western longitude.

3. Use the trapezoid rule with n = 6 subdivisions to find the center of mass of a uniform flat plate in the shape bounded by the curve y = tan x, the curve y = − tan x, and the line x = π/4.

1To compute the area of the plate in part (d), use the following general principle: If a shape such as the unit circle is expanded horizontally by a factor of a and vertically by a factor of b, then its area is multiplied by a factor of ab.

40 6. CENTER OF MASS

CHAPTER 7 Inverse Trig Functions

Given a function y = f(x), its inverse function is the function that “goes backwards.” We’ll say what we mean by this a little later. To find the inverse function, we (1) solve the equation y = f(x) for x in terms of y, obtaining x = g(y), and then (2) interchange the letters x and y, i.e., write y = g(x). Then we say that g(x) is the inverse function of f(x). We will illustrate inverse functions by a series of examples.

Example 7.1. y = f(x) = 2x + 1.

We first solve for x in terms of y: x = 0.5y − 0.5. Then we interchange x and y: y = g(x) = 0.5x − 0.5. At the right we show the graphs of both f(x) and g(x). Notice that if we imagine a mirror along the line y = x, then the two graphs are mirror images of one another. That is always the geometrical relationship between the graph of a function and the graph of its inverse function.

f(x) g(x)

When we say that the inverse function g(x) “ goes backwards,” we mean that if you compute f(x) and then apply g(x) to that value, i.e., if you compute g(f(x)), you will get x back again. The same thing happens if you evaluate the two functions in the other order, applying f to g(x), i.e., f(g(x)) is also equal to x:

If f(x) and g(x) are inverse functions of one another, then g(f(x)) = x and f(g(x)) = x.

For instance, in Example 7.1, we check that g(f(x)) = 0.5(2x + 1) − 0.5 = x and f(g(x)) = 2(0.5x − 0.5) + 1 = x.

Example 7.2. y = f(x) = x^{3}.

Solving for x, we obtain x = \sqrt[3]{y} = y^{1/3}, and so the

inverse function is y = g(x) = \sqrt[3]{x} = x^{1/3}. Again the graph of f(x) and the graph of g(x) are mirror images of one another with respect to the line y = x (see right). And again we find that applying g to f(x) or applying f to g(x) gets us back to x: g(f(x)) = \sqrt[3]{x^{3}} = x, f(g(x)) = (\sqrt[3]{x})^{3} = x. We say that taking the cube root is the inverse operation to cubing a number.

x^{3} ^{\sqrt[3]{x}}

Example 7.3. y = f(x) = x^{2}.

Here we have a situation that did not exist in the first two examples: solving for x gives two different functions, namely x = \sqrt{y} and x = −\sqrt{y}. We do not want this ambiguity, and so we make an agreement to take only the nonnegative square root: x = \sqrt{y}. This is equivalent to erasing the left half of the parabola y = x^{2}. Then the inverse function is y = \sqrt{x}. After crossing out the left half of the parabola, the resulting function y = f(x) = x^{2} for x ≥ 0 and the inverse function y = g(x) = \sqrt{x} have the same relation to one another as in Examples 7.1 and 7.2. Namely, (1) their graphs are mirror images of one another relative to the line y = x; and (2) each function “reverses” the other one. That is, if we take a nonnegative number, square it, and then take the nonnegative square root, we get back to where we started; and if we take a nonnegative number, take its square root, and then square it, we also get back to where we started.

Example 7.3 illustrates two things to think about when dealing with the inverse function of some y = f(x): (1) the possibility that you might have to select only part of y = f(x) (such as the right half of the parabola in Example 7.3) in order to obtain a well-defined inverse function y = g(x); (2) the possibility that the domain does not consist of all x, i.e., only x in a certain range can be substituted in g(x).

1. The Inverse Tangent Function

The inverse tangent function, which is called the arctangent, goes in the reverse direction from the tangent function. That is, the arctangent of x answers the question: What angle (in radians) has x as its tan? As in Example 7.3, there’s some ambiguity here. In fact, in the case of the arctangent, there are infinitely many possible angles whose tangent is x: if θ has the property that tan θ = x, then so does θ + π, θ + 2π, θ + 3π, θ − π, θ − 2π, etc., because the tangent function has period π. In other words, the part of the graph of y = tan x contained between the two asymptotes at x = −π/2 and x = π/2 is simply repeated between x = π/2 and x = 3π/2, between x = 3π/2 and x = 5π/2, and so on — infinitely often in both directions. (Recall that while sine and cosine each has period 2π, the tangent function has period π.) So in constructing the inverse tangent function the first thing we have to do is cross out all but one of the branches of the tangent function. The usual convention is to remove all but the branch near the y-axis, i.e., to look only at angles between −π/2 and π/2. After doing that, we obtain a well-defined arctangent function, which will be denoted y = arctan x. That is,

arctan x = \binom{the angle in radians in the interval from}{−π/2 to π/2 whose tangent is x}

Before going farther, we make three remarks about notation.

1. The function defined above is called the “ principal value” Arctangent (with a capital “A”). In some situations we might want to consider any angle whose tangent is x. In that case we’ll write arctan x with a small “a.”

2. In many textbooks and on most calculators, the function arctan x is denoted tan^{−1} x. We shall not use this notation, because it can be confusing. Namely, since tan^{2} x means (tan x)^{2}, you might erroneously think that tan^{−1} x is the same as (tan x)^{−1}, which is cot x. This is false. The functions arctan x and cot x are completely different: the first goes backwards from the tangent function, whereas the second is the reciprocal of the tangent function.

3. We are using arctan x to mean the angle in radians whose tangent is x. The angle in degrees would be (180/π) arctan x. Be careful when using calculators: most calculators will give you degrees unless you specifically change the mode to radians.

Just as the function y = tan x has vertical asymptotes at ±π/2, the function y = arctan x has horizontal asymptotes at ±π/2. As x goes from −∞ to +∞, its Arctan goes from −π/2 to π/2, never taking values below −π/2 or above π/2. As expected, the graph of y = tan x (more precisely, the branch of the graph that we’ve taken) and the graph of y = arctan x are mirror images of one another with respect to the line y = x.

arctan x tan x arctan x tan x^{π/2} ^{π/2}

2. The Derivative of y = Arctan x

To find dy/dx, where y = arctan x, we replace the relation y = arctan x with the equivalent relation expressing x as a function of y: x = tan y. In other words, the definition of Arctan tells us that when y = arctan x, then y is an angle whose tan is x. We then implicitly differentiate the relation x = tan y, i.e., we take d/dx of both sides. On the left we get 1. On the right we must use the chain rule, which says that the derivative of tan(some function of x) is sec^{2}(this function) times the derivative of this function. Thus, we obtain:

1 = (sec^{2} y)\frac{dy}{dx},  and so  \frac{dy}{dx} = \frac{1}{sec^{2} y} .

The only trouble with this is that, as generally happens with implicit differentiation, our formula for the derivative involves y. We would like to express \frac{dy}{dx} = \frac{d}{dx} arctan x in terms of x. Now using the trig identity sec^{2} y = 1 + tan^{2} y (this trig identity comes from the identity 1 = cos^{2} + sin^{2} if you divide through by cos^{2}), and also the fact that x = tan y, we can replace sec^{2} y in the denominator by 1 + x^{2}. Thus,

\frac{dy}{dx} = \frac{d}{dx} arctan x = \frac{1}{1 + x^{2}} .

This is a basic derivative formula.

If we have a function u(x) inside the Arctan, then we combine this formula with the chain rule, i.e., we have

\frac{d}{dx} Arctan(u) = \frac{1}{1 + u^{2}} \frac{du}{dx}.

Whenever we have a derivative formula, there’s a companion antiderivative formula:

∫ \frac{dx}{1 + x^{2}} = arctan x + C.

We shall later encounter some practical applications where this integration formula comes in handy.

Whenever you have a new function, such as arctan x, and a new derivative formula, you can combine it with other functions and solve the same type of problems as before — derivatives of products and of functions of functions, tangent line approximation, max/min, related rates, etc.

Example 7.4. Find the derivative of (x + 1) arctan(\sqrt{x}).

Using the product rule and the chain rule, we have

\frac{d}{dx} ((x + 1) arctan(\sqrt{x})) = (x + 1)\frac{1}{1 + (\sqrt{x})^{2}} · \frac{1}{2\sqrt{x}} + 1 · arctan(\sqrt{x})

= \frac{1}{2\sqrt{x}} + arctan(\sqrt{x}).

Example 7.5. Using the tangent line approximation and the fact that tan 45^{∘} = 1, find the angle in degrees whose tangent is 0.99.

We apply the tangent line approximation formula

f(x + Δx) ≈ f(x) + Δx f′(x)

to the situation: f(x) = \frac{180}{π} arctan x, x = 1, Δx = −0.01. We know that f(1) = 45, and we want to know: f(1 + (−0.01)) =? Here f′(x) = \frac{180}{π} · \frac{1}{1+x^{2}} = \frac{180}{π} · \frac{1}{1+1} = 90/π. Thus,

f(x + Δx) ≈ 45 + (−0.01)(90/π) = 44.71^{∘}.

Example 7.6. Suppose you are viewing a painting whose bottom is a distance a above eye level and whose top is a distance b above eye level. You can see detail in the painting best when it occupies the largest possible angle θ in your field of vision (see the diagram at the right below). How far should you stand back from the wall in order to get the best view?

Using this diagram, we cannot express θ directly as an inverse trig function, since it is not an angle in a right triangle. However, θ is the difference between the lower-left angle of the larger right triangle and the lower-left angle of the smaller right triangle. In both right triangles we have letters standing for the two legs. Thus, we can use the arctangent

ab θ x eye level

function. That is, θ is equal to the angle arctan(b/x) of the larger right triangle minus the angle arctan(a/x) of the smaller right triangle: θ = arctan(b/x) − arctan(a/x). We want to maximize θ as a function of x. Hence, we set dθ/dx equal to 0:

0 = \frac{dθ}{dx} = \frac{1}{1 + (b/x)^{2}} (−\frac{b}{x^{2}} ) − \frac{1}{1 + (a/x)^{2}} (−\frac{a}{x^{2}} ) ,

where we have used the formula for d/dx of arctan u. We now simplify, multiplying the numerators and denominators together in each term and bringing the first fraction (which has a negative sign) over to the left:

\frac{b}{b^{2} + x^{2}} = \frac{a}{a^{2} + x^{2}} .

To solve for x, we cross-multiply, then gather together the terms containing an x, and finally factor out common factors on both sides:

b(a^{2} + x^{2}) = a(b^{2} + x^{2}),  bx^{2} − ax^{2} = ab^{2} − ba^{2},  (b − a)x^{2} = ab(b − a).

Dividing by b − a and taking the positive square root gives x = \sqrt{ab}. From the practical setup it should be clear that this critical point really is a maximum. If you want to check by the second derivative method, just take the second derivative of θ with respect to x at the point x = \sqrt{ab}. You will get a negative value.

3. The Inverse Sine Function

The inverse sine function answers the question: what angle (in radians) has sine equal to x? There are two things to watch out for here:

1. As in the case of the tangent, there will be infinitely many angles whose sine is the same. So we must remove all of the sine curve except for one section before we construct the inverse function. The usual convention, as in the case of Arctan, is to take the part of the curve y = sin x between x = −π/2 and x = π/2, where y goes exactly once through all values from −1 to 1. That is, we define the principal value Arcsin of x as follows:

arcsin x = \binom{the angle in radians between}{−π/2 and π/2 whose sine is x.}

2. Unlike Arctan, the function Arcsin cannot be applied to all x, but rather only to x between −1 and 1. This is because, unlike the tangent function, which takes all values from −∞ to ∞, the sine function only takes values between −1 and 1, and for x outside of that interval there is no angle whose sine is x. Thus, the domain of arcsin x is the interval [−1, 1].

On the right is a graph of the function y = sin x (between x = −π/2 and x = π/2) and the function y = arcsin x (which has domain between x = −1 and x = 1). As usual, the two graphs are symmetric with respect to the line y = x.

arcsin x sin x arcsin x sin x  ^{−π/2}^{π/2}^{−1}  ^{1}

4. The Derivative of y = Arcsin x

The procedure is just like in the case of Arctan. We first replace the relation y =Arcsin x by the relation x = sin y, and then we take d/dx of both sides, obtaining 1 = (cos y)\frac{dy}{dx}, i.e., dy/dx = 1/ cos y. We can express the derivative in terms of x, which equals sin y, by means of the trig identity cos = \sqrt{1 − sin^{2}}. Thus, 1/ cos y = 1/\sqrt{1 − sin^{2} y} = 1/\sqrt{1 − x^{2}}. Hence we have

\frac{d}{dx} arcsin x = 1\sqrt{1 − x^{2}}.

(Note: When we take the square root in writing cos in terms of sin, in general we have to put a ± in front of the radical; however, in this case we know that we want the positive square root, because in our definition of the Arcsin function y has to be between −π/2 and π/2, and for these values cos y is positive.)

The companion antiderivative formula is

∫ \frac{dx}{\sqrt{1 − x^{2}}} = arcsin x + C.

Example 7.7. The tip of a 20-meter span of a drawbridge has been raised 8 meters above the horizontal, and it is being raised at 0.2 m/sec. How fast (in deg/sec) is the angle of the drawbridge increasing?

In this related rates problem, let θ be the angle in degrees, and let x be the height the tip has been raised. Then θ = \frac{180}{π} arcsin(x/20), and so

\frac{dθ}{dt} = \frac{dθ}{dx} · \frac{dx}{dt} = \frac{180}{π} \frac{1}{\sqrt{1 − (x/20)^{2}}} \frac{1}{20} · \frac{dx}{dt} = \frac{180}{π} \frac{0.05}{\sqrt{1 − (8/20)^{2}}} · 0.2 = 0.625 deg/sec.

Example 7.8. Evaluate: ∫ dx/\sqrt{r^{2} − x^{2}}, where r is a constant.

Here we’ll make a u-substitution to transform the integral into a form where we can use the above antiderivative formula, which we now restate using the variable u: ∫ du/\sqrt{1 − u^{2}} = arcsin u+C. The difficulty in our example is that we have r^{2} where we’d like to have 1. We can get a 1 there by factoring out r^{2} from r^{2} − x^{2}, i.e., by writing r^{2} − x^{2} = r^{2} (1 − (x/r)^{2}), and then pulling the r^{2} out of the radical. Thus,

∫ \frac{dx}{\sqrt{r^{2} − x^{2}}} = ∫ \frac{1}{r} · \frac{dx}{\sqrt{1 − (x/r)^{2}}} .

This suggests the u-substitution u = x/r. Then du = (1/r)dx, and so in our integral we can replace (1/r)dx by du and the denominator by \sqrt{1 − u^{2}}. So we obtain:

∫ \frac{du}{\sqrt{1 − u^{2}}} = arcsin u + C = arcsin(x/r) + C,

which is our final answer. As always in an antiderivative problem, we can check that this is right by taking the derivative of our answer and seeing that we do in fact get 1/\sqrt{r^{2} − x^{2}}.

5. The Inverse Cosine Function

So far we have dealt with the inverse functions of two of the six trig functions (sine, cosine, tangent, secant, cosecant, cotangent). The other four inverse trig functions are used less frequently, and in any case they can be expressed in a simple way in terms of Arcsin or Arctan. We illustrate this by discussing the function Arccosine.

A basic relationship between the cosine and the sine is that the cosine of an angle is the sine of the complementary angle. This means that if we want to know an angle whose cosine is x, we can take the complementary angle of the angle whose sine is x, i.e., we can take (π/2) − arcsin x. We shall therefore define the principal value arccosine, denoted arccos x, as follows:

arccos(x) = \frac{π}{2} − arcsin(x) for − 1 ≤ x ≤ 1.

Since Arcsin runs from −π/2 to π/2, it follows that arccos(x) runs from π to 0 as x runs from −1 to 1. The derivative of arccos(x) is easy to compute:

\frac{d}{dx} arccos(x) = 0 − \frac{d}{dx} Arcsin(x) = −\frac{1}{\sqrt{1 − x^{2}}} .

Homework 6.

1. Find d/dx of:

(a) x^{2} arctan(x/2),

(b) (x + 1)/ arcsin(x^{2}),

(c) arcsin(\sqrt{1 − x}),

(d) x arcsin(x) + \sqrt{1 − x^{2}}.

2. Find the anti-derivative of (a) (100 − 16x^{2})^{−1/2},

(b) (1 + 2x^{2})^{−1}.

3. Using the tangent line approximation and the fact that sin(30^{∘}) = 0.5, find the angle in degrees whose sine is 0.52.

4. Find the volume of the solid formed when the shape bounded by y = \sqrt{x^{−2} − 1}, the line y = 1, and the x- and y-axes is rotated around the y-axis.

5. Find the center of mass of the uniform flat plate bounded by the curve 1/\sqrt{1 − x^{2}}, the curve −1/\sqrt{1 − x^{2}}, the y-axis, and the line x = 0.5.

6. Graph and find max/min points of (a) y = (arctan x) − 0.5x,

(b) y = (arcsin(x/2)) − x.

7. Sketch and find the area between (a) y = 2/(1 + x^{2}) and y = x^{2},

(b) y = 1/\sqrt{1 − x^{2}} and y = 2.

8. A rocket launched vertically from a point 2 km from an observer is ascending at 0.20 km/sec at the instant when it is 1 km above the ground. Find the rate in degrees/sec at which its angle of inclination (as seen by the observer) is increasing at that moment.

9. A triangle has two fixed sides of length a and b with included angle θ that is free to widen. Let x be the side opposite θ. (a) Use the law of cosines to express θ as a function of x. (b) At the instant when x = c, what is the ratio of the rate at which the angle θ is widening (in rad/sec) to the rate at which the side x is increasing?

48 7. INVERSE TRIG FUNCTIONS

CHAPTER 8 The Natural Logarithm

The natural logarithm, denoted y = ln x, is the function defined as follows: ln x is the area under the reciprocal function 1/u from u = 1 to u = x. Using the integral notation, we write:

ln x = ∫_{1}^{x} \frac{du}{u}.

In the drawing to the right the shaded area is ln x. Here we are showing the case x > 1, when the area is positive. If x < 1, on the other hand, then the limits of integration are “backwards” (i.e., the top limit of integration is less than the bottom limit of integration). In other words, if x < 1, then we traverse the area going from right to left rather than from left to right. In that case the area in question is counted with a negative sign. If x = 1, then the area is 0. If x ≤ 0, then we cannot define

1 x y = \frac{1}{u}

ln x, because in that case the reciprocal function 1/u has an asymptote (i.e., it goes to infinity) between u = 1 and u = x. Thus, our first property of ln x is:

(Property 1) ln x \begin{cases}> 0 & if x > 1; \\ = 0 & if x = 1; \\ < 0 & if 0 < x < 1; \\ is undefined & if x ≤ 0. \end{cases}

We say that the domain of ln x is all positive x.

The next group of properties are the main reason for defining ln x.

(Property 2 a)  \frac{d}{dx} ln x = \frac{1}{x};  \frac{d}{dx} ln u = \frac{1}{u} \frac{du}{dx}.

The first of these equalities is a direct result of the “ fundamental theorem of calculus,” which says that the derivative of this type of area function is obtained by removing the integral sign and replacing the dummy variable by the variable in the upper limit of integration: \frac{d}{dx} ∫_{1}^{x} \frac{1}{u}du = \frac{1}{x}. The second equality follows from the chain rule: if y = ln u and u = u(x), then

\frac{dy}{dx} = \frac{dy}{du} · \frac{du}{dx} = \frac{1}{u} \frac{du}{dx}.

(Property 2 b) ∫ \frac{dx}{x} = ln x + C.

This means that ∫_{a}^{b} \frac{dx}{x} = ln b − ln a. Property 2(b) follows from 2(a), which says that ln x is an antiderivative of 1/x. We shall later see that if we want to allow the possibility of negative limits of integration, we must replace x by |x| inside the ln, i.e., we modify the antiderivative formula as follows:

∫ \frac{dx}{x} = ln |x| + C.

This antiderivative formula fills in a gap in the antiderivative formula we had before for x^{n}, namely, the case n = −1. For all n ≠ −1, the antiderivative of x^{n} is another power function; but the antiderivative of x^{−1} is a different type of function, namely, our “natural logarithm.” To summarize, we write:

∫ x^{n} dx = \begin{cases}\frac{1}{n+1}x^{n+1} + C & if n ≠ −1; \\ ln |x| + C & if n = −1. \end{cases}

Before discussing further properties of ln x, we give some examples of the use of Properties 2(a) and 2(b).

Example 8.1. Find the derivative of x ln^{2}(sin x).

By the product rule,

\frac{d}{dx}(x ln^{2}(sin x)) = x \frac{d}{dx}(ln^{2}(sin x)) + 1 · ln^{2}(sin x)
= x · 2 ln(sin x)\frac{d}{dx}(ln(sin x)) + ln^{2}(sin x)

= 2x ln(sin x)\frac{1}{sin x} cos x + ln^{2}(sin x) = 2x cot x ln(sin x) + ln^{2}(sin x).

Example 8.2. Find the definite integrals

(a) ∫_{0}^{2} \frac{dx}{3x + 1};  (b) ∫_{−2}^{−1} \frac{dx}{x};  (c) ∫_{0}^{π/4} tan x dx;  (d) ∫_{2}^{4} \frac{2x}{x^{2} − 2}dx.

In several of these integrals it is useful to rewrite Property 2(a) with x replaced by u, and then make a u-substitution to transform the integral with respect to x to an integral with respect to u having the form in Property 2(a), i.e., ∫ \frac{1}{u}du.

(a) We make the substitution u = 3x + 1, so that du = 3dx, and hence dx = \frac{1}{3}du. The limits of integration in (a) change as follows: x = 0 ⟹ u = 3 · 0 + 1 = 1; and x = 2 ⟹ u = 3 · 2 + 1 = 7. Thus, we obtain

∫_{0}^{2} \frac{dx}{3x + 1} = \frac{1}{3} ∫_{1}^{7} \frac{du}{u} = \frac{1}{3} ln 7.

(b) Since our limits of integration are negative, we use the version of the antiderivative formula with the absolute value inside the ln. We obtain:

ln |x| |_{−2}^{−1}= ln | − 1| − ln | − 2| = ln 1 − ln 2 = − ln 2.

We can verify directly that this is the right answer: the area in question is an integral of the negative branch of the reciprocal function (which is in the third quadrant). But the area there from −2 to −1 is exactly the same as the area under the positive branch of the reciprocal function from 1 to 2, except that it is counted negatively because of being below the axis. Thus, the answer − ln 2 agrees with the picture in the graph at the right.

−2  ^{−1}  ^{2}^{1}y = 1/x

(c) At first glance, this has nothing to do with ln. However, if we rewrite tan x as sin x/ cos x, and notice that the derivative of u = cos x is du = − sin xdx, we can transform this integral into ∫ du/u. More precisely, we obtain

∫_{0}^{π/4} \frac{sin x dx}{cos x} = ∫_{cos 0}^{cos(π/4)} \frac{−du}{u} = − ∫_{1}^{\sqrt{2}/2} \frac{du}{u} = − ln(\sqrt{2}/2).

(d) Here we notice that the expression in the denominator has its derivative sitting with dx in the numerator. This means that substituting u for the denominator will result in du/u. That is, we set u = x^{2} − 2, so that du = 2xdx. The limits of integration change from 2 to 2^{2} − 2 = 2 and from 4 to 4^{2} − 2 = 14. So we obtain ∫_{2}^{14} du/u = ln(14) − ln 2.

We will soon see that the answers to parts (c) and (d) can be simplified.

Before giving the next group of properties of ln, we make a useful observation. We defined ln a to be the area under y = 1/x from x = 1 to x = a. But it turns out that for any positive number b this value ln a is also the area under y = 1/x from x = b to x = ab. For example, ln 2 is the area from 1 to 2, and it is also the area from 3 to 6 (or from 1000 to 2000, etc.). Roughly speaking, this is because the latter area is obtained by taking the area from 1 to 2, expanding it horizontally, and then contracting it vertically by the same factor. More precisely, to show that ln a is equal to ∫_{b}^{ab} dx/x, we write ln a = ∫_{1}^{a} dx/x and make the u-substitution u = bx. Then du = b dx, and the limits of integration become b · 1 to b · a, i.e.,

ln a = ∫_{1}^{a} \frac{dx}{x} = ∫_{1}^{a} \frac{b dx}{b x} = ∫_{b}^{ab} \frac{du}{u}.

The name of the variable of integration is irrelevant, so that we can equally well write ln a = ∫_{b}^{ab} dx/x. It is now easy to obtain our next basic group of properties.

(Property 3 a) ln(ab) = ln a + ln b .

This follows because ln a = ∫_{b}^{ab} dx/x = ln(ab) − ln b.

(Property 3 b) ln(1/a) = − ln a.

This is because, by part (a), we have ln(1/a) + ln a = ln 1 = 0.

(Property 3 c) ln(a/b) = ln a − ln b .

This follows immediately from part (a) if we replace b by 1/b and use part (b).

(Property 3 d) ln(a^{n}) = n ln a .

First note that part (a) can be repeated when you have more than two factors: ln(abc) = ln a + ln(bc) = ln a + ln b + ln c, and so on. Thus, if n is a positive integer, we can write ln(a^{n}) = ln(a · a · a ⋯ a) = ln a + ln a + ln a + ⋯ + ln a = n ln a. If n is a negative integer, one uses a similar argument with a replaced by 1/a. The formula ln(a^{n}) = n ln a applies for any n (integer or not), but we shall omit the proof in the case when n is not an integer. (But see Problem 2 in the homework for a proof.)

Example 8.3. Simplify the answers to parts (c)–(d) of Example 2.

In part (c) we have − ln(\sqrt{2}/2) = −(ln(2^{1/2}) − ln 2) = − (\frac{1}{2} ln 2 − ln 2) = \frac{1}{2} ln 2. In part (d) we have ln(14) − ln 2 = ln(14/2) = ln 7.

1. Appearance of the Curve y = ln x.

Recall what Property 1 says: the function y = ln x has domain consisting of all positive x, its graph crosses the x-axis at x = 1, and the function is positive when x > 1 and negative when 0 < x < 1. It is an increasing function, because its derivative is 1/x, which is always positive for x > 0. However, when x is large this derivative is quite small. In other words, the slope of the curve tends to “level off.” On the other hand, when x is near 0 the slope is very large, i.e., the curve is steep. At the point x = 1 where the curve crosses the x-axis, the slope is 1, i.e., the curve is increasing at a 45^{∘} angle as it crosses the axis.

1  y = ln x

We now ask whether the curve y = ln x increases without bound as x increases, or whether it is bounded by a horizontal asymptote. It is possible for an increasing curve to be bounded — for example, the curve y =Arctan x. It is also possible for a curve whose slopes are becoming smaller and smaller (“leveling off”) to increase without bound — for example, the curve y = \sqrt{x}. So our question is: does y = ln x look more like the Arctan curve or the square root curve? The answer to this question is: more like the square root curve, i.e., there is no horizontal asymptote. To see this, we can use Property 3(d) with a very large value of n. For example,

ln (2^{1000}) = 1000 ln 2 = 693, ln (2^{1000000}) = 693147, ln (2^{10000000000}) = 693147181.

Thus, we can get ln x to be as large as we want by choosing x to be 2 to a large enough power.

Even though ln x and \sqrt{x} both approach infinity (fairly slowly) as x ⟶ ∞, the former function approaches infinity more slowly than the latter one. If we look at the ratio ln x/\sqrt{x}, and set x equal to an extremely large number (like 10^{1000}, for example), we find that

\frac{ln 10^{1000}}{\sqrt{10^{1000}}} = \frac{1000 ln 10}{10^{500}}.

Here the numerator is about 230, while the denominator is a 1 followed by 500 zeros! So the ratio is extremely close to 0, i.e., ln x, while fairly large, is much much smaller than \sqrt{x}. The same argument can be used to show that ln x approaches infinity slower than any n-th root function. For example, the ratio ln x/ \sqrt[10]{x} has limit 0 as x ⟶ ∞. To summarize: the natural logarithm function grows without bound as x increases, but very, very slowly.

At the other end, as x ⟶ 0, the function ln x approaches −∞. This is because a very small positive number is the reciprocal of a very large positive number, and so its logarithm is the negative of the logarithm of the very large number.

2. Relation with Logarithm-to-the-Base-a

Historically, the first logarithms used were “ logarithms to the base 10,” denoted log_{10}. The definition of the function y = log_{10} x is: y is the power to which 10 must be raised to get x, i.e., 10^{y} = x. More generally, if a is any fixed number greater than 1 (for example, log to the base a = 2 is often used in computer science), then we define log_{a} x as the power to which a is raised to get x. That is,

y = log_{a} x  means that  a^{y} = x.

It is not hard to relate this “old-fashioned” logarithm to the natural logarithm ln. Namely, if y = log_{a} x, then to find y in terms of ln we take ln of both sides of the relation a^{y} = x, obtaining ln(a^{y}) = y ln a = ln x, i.e., y = ln x/ ln a. Since y = log_{a} x, we conclude that

log_{a} x = \frac{ln x}{ln a}.

For example, log_{10} x = ln x/ ln(10) = 0.4343 ln x. The only difference between the different kinds of logarithm is a constant factor. Thus, there is no problem in taking the derivative of log_{a}: \frac{d}{dx} log_{a} x = \frac{1}{ln a} · \frac{1}{x}.

3. Logarithmic Differentiation

Although it may seem that we can already take the derivative of almost any function for which we can write down a formula, you may have noticed that there is one type of function which we have not been able to differentiate, namely, a function of the form y = f(x)^{g(x)}, where g(x) is not a constant. Of course, if g(x) is a constant, then we can use the rule for the derivative of f(x)^{n}. But this rule cannot be used if the exponent is a function of x rather than a constant. Instead we must use a procedure called logarithmic differentiation. As its name implies, it consists of two parts: (1) take the ln of both sides of the equation y = f(x)^{g(x)}, and (2) differentiate both sides of the resulting equation. In part (1), be sure to use Property 3(d) of ln, with a replaced by f(x) and n replaced by g(x). We illustrate with the following examples.

Example 8.4. Find the derivative of (a) (sin x)^{x}; (b) (ln x)^{ln x}.

(a) We first take the ln of both sides of y = (sin x)^{x}, obtaining ln y = ln ((sin x)^{x}) = x ln(sin x). When taking d/dx of both sides, on the left we use Property 2(a) of ln and on the right we use the product rule:

\frac{1}{y} · \frac{dy}{dx} = x\frac{1}{sin x} cos x + 1 · ln(sin x) = x cot x + ln(sin x).

Finally, we multiply both sides by y, and express dy/dx in terms of x by replacing y by (sin x)^{x}:

\frac{dy}{dx} = (x cot x + ln(sin x))y = (x cot x + ln(sin x))(sin x)^{x}.

(b) We write y = (ln x)^{ln x}, and ln y = ln ((ln x)^{ln x}) = ln x ln(ln x). Taking d/dx of both sides gives:

\frac{1}{y} \frac{dy}{dx} = ln x \frac{d}{dx}(ln(ln x)) + \frac{1}{x} ln(ln x) = ln x(\frac{1}{ln x} \frac{1}{x}) + \frac{1}{x} ln(ln x) = \frac{1 + ln(ln x)}{x},

and hence

\frac{dy}{dx} = \frac{1 + ln(ln x)}{x} y = \frac{1 + ln(ln x)}{x}(ln x)^{ln x}.

Homework 7.

1. Find the derivative of:

(a) ln(kx) (k is a constant); (d) ln(tan x);
(b) ln^{4} x;  (e) ln ((x^{2} + 16)^{2}) − x arctan(x/4).

(c) ln^{2}(ln x);

2. (a) Find the derivative of the functions ln(ax) − ln x − ln a (where a is any constant) and ln(x^{n}) − n ln x (where n is any constant).

(b) Use part (a) to give another derivation of Properties 3(a) and 3(d) of ln.

3. Simplify:

(a) ln 30 − ln 2 − ln 5;  (d) ln \sqrt{6.75} (in terms of ln 2 and ln 3);
(b) ln \sqrt{8} (in terms of ln 2); (e) ln(πr^{2}) − 2 ln r;
(c) ln \sqrt[3]{9} (in terms of ln 3); (f) ln \sqrt{r^{2} − x^{2}} − \frac{1}{2} ln(r − x).

4. Find:

(a) ∫_{−1}^{−0.5} \frac{dx}{x};  (d) ∫_{2}^{3} \frac{dx}{x ln^{2} x};
(b) ∫ cot x dx;  (e) ∫_{0}^{2π} \frac{sin(θ/2)dθ}{2 − cos(θ/2)}.

(c) ∫ \frac{dx}{x ln x} (hint: substitute u = ln x);

5. Find the center of mass of the uniform plate contained (a) between the graphs of y = 1/x, y = −1/x, x = 1 and x = 2;

(b) between the graphs of y = 1/x^{2}, y = −1/x^{2}, x = 1 and x = 2;

(c) between the graphs of y = 1/(1 + x^{2}), y = −1/(1 + x^{2}), x = 0 and x = 1.

6. Find the area (a) between the curves y = 5/(1 + x^{2}) and y = 2/x;

(b) between the curves y = 1/\sqrt{1 − x^{2}} and y = 2/(2 − x).

7. The region bounded by y = −1/x^{2}, y = −1/4 and y = −1 is revolved around the y-axis to form a container. Find its volume.

8. A bug slides down the curve y = ln x, moving leftward (i.e., its x-coordinate is decreasing) at 10 units/min. Find its speed along the curve when x = 2, when x = 1, and when x = 1/2.

9. By “experiment” (for example, substituting x = 1/10^{1000}), find the limit as x ⟶ 0 (where x is positive) of:

(a) x^{3} ln x, (b) x ln x, (c) \sqrt{x} ln x, (d) \sqrt[10]{x} ln x.

10. Find the derivative of:

(a) y = x^{x};  (d) y = x^{ln x};
(b) y = (sin x)^{\sqrt{x}}; (e) y = (cos x)^{Arctan x}.

(c) y = (ln x)^{x};

11. One morning in January it began to snow and kept on snowing at a constant rate. A snow plow began to plow at noon. It plowed clean and at a constant rate (volume per unit time). From one o’clock until two o’clock it went only half as far as it had gone between noon and one o’clock. When did the snow begin falling?

56 8. THE NATURAL LOGARITHM

CHAPTER 9 Improper Integrals

An improper integral is the area of a shape that extends out to infinity in some direction. In many cases, such a shape has infinite area, and we say that the improper integral is ∞. However, in certain cases the total area is finite, because the part extending out infinitely far is “pinched closely together,” so that the very narrow part going out infinitely far does not contribute much area. This all sounds rather vague. The best way to understand improper integrals is through examples.

Example 9.1. Find the area under the reciprocal cube curve y = 1/x^{3} to the right of the line x = 1.

First, suppose that we bound the area on the right at x = b, where b is a very large upper limit of integration. Then we compute

∫_{1}^{b} \frac{dx}{x^{3}} = −\frac{1}{2}x^{−2} |_{1}^{b}= \frac{(1 − b^{−2})}{2}.

For example, the area from x = 1 to x = 100 is exactly 0.49995, the area from x = 1 to x = 1000000 is exactly 0.4999999999995, and so on. Clearly, the limit as b ⟶ ∞ is 0.5. This is what we mean by the area out to infinity:

∫_{1}^{∞} \frac{dx}{x^{3}} = lim_{b⟶∞} ∫_{1}^{b} \frac{dx}{x^{3}} = lim_{b⟶∞} \frac{(1 − b^{−2})}{2} = 0.5.

Example 9.2. Find the area under the reciprocal curve y = 1/x to the right of the line x = 1.

Here we have

lim_{b⟶∞} ∫_{1}^{b} \frac{dx}{x} = lim_{b⟶∞} ln b = ∞.

Roughly speaking, the reciprocal cube function pinches together with the xaxis a lot faster than does the reciprocal function, and so we get finite area with 1/x^{3} but not with 1/x.

\begin{aligned}& y = 1/x \\ y = 1/x^{3} \end{aligned}

Example 9.3. Let y = 1/x^{n}, where n is a positive constant not equal to 1. Find the area under this curve to the right of the line x = 1.

Here we want the limit as b ⟶ ∞ of ∫_{1}^{b} x^{−n}dx = \frac{1}{−n+1}x^{−n+1} |_{1}^{b}= \frac{1}{n−1} (1 − b^{1−n}). What happens as b ⟶ ∞ depends on the exponent of b, which is 1 − n. If this is positive, i.e., if n < 1, then the limit is infinite. On the other hand, if n > 1, then b^{1−n} approaches 0, and so we obtain the finite area 1/(n − 1). We can summarize the situation as follows (note that in Example 9.2 we saw that the area is infinite in the special case n = 1):

∫_{1}^{∞} \frac{dx}{x^{n}} = \begin{cases}\frac{1}{n−1} & if n > 1; \\ ∞ & if n ≤ 1. \end{cases}

Example 9.4. Find the area bounded by the x- and y-axes, the line x = 1, and the reciprocal cube root curve y = 1/\sqrt[3]{x}.

Here the region extends out to infinity in the y-direction. But the curve comes very close to the y-axis (its asymptote), so there is very little area enclosed when y is large. For this reason, as we shall see, the total area is finite. The area is the improper integral ∫_{0}^{1} x^{−1/3}dx. To evaluate this, let us compute the ordinary (not improper) integral ∫_{a}^{1} x^{−1/3}dx for a > 0, and then see what happens when a ⟶ 0. That is,

y = 1/\sqrt[3]{x} y = 1/x^{3}

∫_{0}^{1} x^{−1/3}dx = lim_{a⟶0} ∫_{a}^{1} x^{−1/3}dx = lim_{a⟶0} (\frac{3}{2}x^{2/3} |_{a}^{1}) = lim_{a⟶0} (\frac{3}{2} (1 − a^{2/3})) = 3/2,

because a raised to any positive power approaches 0 as a ⟶ 0.

In Example 9.4, suppose we remove from our region the square of unit side that is bounded by the x- and y-axes, the line x = 1, and the line y = 1. The remaining area is the area to the left of the curve x = 1/y^{3} above the line y = 1. This is exactly the same area as in Example 9.1, except that the roles of x and y are reversed. Thus, we expect the area in Example 9.4 to equal 1 plus the area in Example 9.1. Thus, the value of 3/2 in the last paragraph agrees with the value of 1/2 in Example 9.1.

In the same way as in Example 9.4, we can compute the area bounded by the x- and y-axes, the line x = 1, and the curve y = 1/x^{n}. The computation is just like the special case when n = 1/3. The result is:

∫_{0}^{1} \frac{dx}{x^{n}} = \begin{cases}\frac{1}{1−n} & if n < 1; \\ ∞ & if n ≥ 1. \end{cases}

Example 9.5. Find ∫_{−2}^{2} dx/\sqrt{4 − x^{2}}.

Here the integrand (4−x^{2})^{−1/2} is infinite at both endpoints of the interval of integration — the region goes out to infinity along the line x = −2 and along the line x = 2. To get a region that does not extend to infinity, we integrate between −a and a, where a is a little less than 2. Then we take the limit as a ⟶ 2. In evaluating the integral, we use Example 7.8 (with r set equal to 2) of the section on the Arcsin. Thus,

∫_{−2}^{2} \frac{dx}{\sqrt{4 − x^{2}}} = lim_{a⟶2} ∫_{−a}^{a} \frac{dx}{\sqrt{4 − x^{2}}} = lim_{a⟶2} arcsin(x/2) |_{−a}^{a}
= arcsin(1) − arcsin(−1) = \frac{π}{2} − (−\frac{π}{2} ) = π.

Example 9.6. The region in the fourth quadrant bounded by the y-axis, the line y = −1, and the curve y = −1/x is rotated around the y-axis to obtain a container (an infinitely deep funnel). (a) Find the volume of the funnel. (b) Show that the funnel has infinite surface area. Thus, one could fill the funnel entirely with water, but if one cuts the side of the funnel into small pieces and spreads them out on a table, there is no table of finite size that would hold all of the pieces!

(a) We first write x as a function of y: x = −1/y. Then in the volume integral π ∫_{−∞}^{−1} dy/y^{2}, we make the change of variables u = −y, so that du = −dy. This transforms the volume to −π ∫_{∞}^{1} dy/y^{2} = π ∫_{1}^{∞} du/u^{2} (we used the fact that the sign of a definite integral changes when you reverse the upper and lower limits of integration). The last integral was computed in Example 9.3 (take n = 2): it has finite value 1. Thus, the volume of the funnel is π.

(b) We have not developed the formula for the surface area of a volume of revolution. However, we can answer the question in part (b) by observing that the surface area is greater than the area of the flattened surface contained in the third and fourth quadrants below y = −1 between the curves x = 1/y and x = −1/y. This flat area is equal to ∫_{−∞}^{−1} \frac{−2}{y}dy = 2 ∫_{1}^{∞} du/u (where we make the same substitution u = −y as before). But the latter improper integral is infinite, as we saw in Example 9.2. Therefore, the funnel has infinite surface area.

Example 9.7. Determine how the noise from a freeway drops off as you move farther away. Make the following simplifying assumptions: (1) the amount of noise generated by a segment of the freeway is proportional to the segment’s length (i.e., the traffic is uniformly distributed along the freeway); (2) the contribution of a small segment of the freeway to the amount of noise you hear is inversely proportional to the square of your distance from that segment; (3) the freeway is a straight line extending infinitely far in both directions. (We are also assuming that the noise from different parts of the freeway add up without “canceling each other,” and we are using a linear scale of noise, rather than a logarithmic scale such as the decibel system.)

[figure: 0; ⟵ −∞; ∞ ⟶; r; \sqrt{r^{2} + x^{2}}; x; x + Δx]

Let r be our perpendicular distance from the freeway. Take the freeway to be the x-axis, with 0 located at the point closest to us. We shall find the total noise by the same type of procedure that we used to find definite integral formulas for the area under a curve, the volume of a solid of revolution, and the total moment of a uniform plate. In those cases we had the integral from a to b of the contribution (to area, volume, moment, or whatever) coming from the part between x and x + Δx (where we replace Δx by dx when we go from the Riemann sum to the integral).

For the sake of review, the chart on the next page shows the contribution from the part between x and x + Δx and the integral that resulted.

TYPE OF PROBLEM CONTRIBUTION FORMULA
area under curve  ≈ f(x)Δx  ∫_{a}^{b} f(x)dx
vol of revolution  ≈ πf(x)^{2}Δx  π ∫_{a}^{b} f(x)^{2}dx
moment of plate  ≈ D x f(x)Δx  D ∫_{a}^{b} x f(x)dx

So in the freeway problem we want to find the contribution to noise that comes from the tiny segment [x, x+Δx]. According to assumptions (1) and (2), the noise you hear from this segment is proportional to the length Δx of the segment and inversely proportional to the square of your distance from the segment. In computing this distance, we take the distance from the left endpoint of the segment (since Δx is very small, this is very close to the distance from any other point on the segment); by the Pythagorean theorem, the square of this distance is r^{2} + x^{2}. Thus, if we let K denote the constant of proportionality, we find that the contribution is

≈ K \frac{Δx}{r^{2} + x^{2}} .

Thus, the noise coming from the entire highway between a and b is found by summing up these contributions — or rather (after taking the limit as Δx ⟶ 0) taking the integral with Δx replaced by dx: Noise= ∫_{a}^{b} \frac{K dx}{r^{2}+x^{2}} . Thus,

Noise = K ∫_{a}^{b} \frac{dx}{r^{2} + x^{2}} = \frac{K}{r^{2}} ∫_{a}^{b} \frac{dx}{1 + (x/r)^{2}} =

= \frac{K}{r^{2}} ∫_{a/r}^{b/r} \frac{r du}{1 + u^{2}} = \frac{K}{r} ∫_{a/r}^{b/r} \frac{du}{1 + u^{2}}

= \frac{K}{r} arctan u |_{a/r}^{b/r}= \frac{K}{r} (arctan(\frac{b}{r} ) − arctan(\frac{a}{r})).

If the freeway had finite length and were contained between a and b, then this would give us a rather complicated expression for the noise from the freeway. It would not be easy to say how the noise drops off as our distance r increases. However, at this point we use simplifying assumption (3), which tells us to take the limit as a ⟶ −∞ and b ⟶ ∞, i.e.,

Noise = \frac{K}{r} ( lim_{b⟶∞} arctan(\frac{b}{r} ) − lim_{a⟶−∞} arctan(\frac{a}{r}))

= \frac{K}{r} (\frac{π}{2} − (−\frac{π}{2} )) = \frac{Kπ}{r} .

Since K and π are constants, we end up with the simple conclusion: noise from the freeway drops off proportionally to distance. Thus, if you move twice as far away, the amount of noise drops only by 50%. (We say “only,” because in the case of a point source the noise would drop by a factor of 4, i.e., by 75%.)

Even though the simplifying assumptions might seem to detract from the realism of this example, our conclusion roughly agrees with what one observes in practice. For example, real estate agents know that noise from Interstate 5 has a bad effect on property values for quite a distance — much more so than would a “point source” such as a noisy factory.

Notice the role of simplifying assumption (3). Although it might seem odd at first, improper integrals have many practical applications: it often simplifies the calculations to suppose that something extends out to infinity.

Homework 8.

1. Find the following improper integrals (some of them might be ∞):

(a) ∫_{1}^{∞} dx/(x\sqrt{x}); (e) ∫_{0}^{∞} dx/(1 + x^{2}); (i) ∫_{−10}^{10} dx/\sqrt{100 − x^{2}};
(b) ∫_{0}^{1} dx/(x\sqrt{x}); (f) ∫_{−∞}^{∞} dx/(1 + x^{2}); (j) ∫_{0}^{1} x dx/\sqrt{1 − x^{2}},
(c) ∫_{1}^{∞} dx/\sqrt{x};  (g) ∫_{−∞}^{∞} dx/(9 + x^{2}); (k) ∫_{0}^{∞} x dx/(1 + x^{2});
(d) ∫_{0}^{1} dx/\sqrt{x};  (h) ∫_{0}^{1} dx/\sqrt{1 − x^{2}}; (l) ∫_{0}^{∞} x dx/(1 + x^{4}).

2. Let n be a positive number. Find (in terms of n) the volume of the solid obtained by rotating around the y-axis the part of y = −1/x^{n} below y = −1 in the fourth quadrant. For what values of n is this volume infinite?

3. Suppose that n > 2. Find (in terms of n) the balancing point of the uniform flat plate bounded by the graph of y = 1/x^{n}, the graph of y = −1/x^{n}, and the line x = 1. (The plate extends to the right all the way out to infinity.)

4. Show that the following uniform flat plates have finite mass but do not have center of mass at a finite distance:

(a) Problem 3 when 1 < n ≤ 2;

(b) the plate bounded by the graph of y = 1/(1 + x^{2}), the graph of y = −1/(1 + x^{2}), and the y-axis. This means that you could lift up the plate, but you could not balance it!

CHAPTER 10 The Number e and the Exponential Function

The number e is a fundamental constant of mathematics, every bit as important as π. It is defined as follows: e is the number whose natural logarithm is exactly 1. That is: ln e = 1. In geometric terms, this says that you look at the area under the reciprocal function y = 1/x between 1 and a, and move the upper limit of integration a to the right until you get an area of 1. (See the drawing below). It is not hard to see that e is somewhere between 2 and 3 (see Problem 1 at the end of the section). The value of e to several decimal places is: e = 2.71828 · · ·. (The exact value cannot be given in decimal form, since e, like π, is an irrational number.)

y = \frac{1}{x} Area = 1 e

In Chapter 8, we obtained the following relationship between the natural logarithm and the log-to-the-base-a: ln x = ln a · log_{a} x. In the special case a = e, since ln e = 1 this gives us

ln x = log_{e} x,

in other words, the natural logarithm is the same as the logarithm to the base e. That is, ln x is equal to the power to which e must be raised to get x. This can be expressed in the form e^{ln x} = x, and also in the form ln(e^{x}) = x. In other words, the functions ln x and e^{x} are inverse functions. We shall return to this later.

There is another way to come up with the number e, which we shall explain using bank interest. Suppose that you put an initial amount of money P (called the “principal”) in the bank at time t = 0. Suppose that the bank pays exactly 100% annual interest (this is not as unreasonable as it may seem, if you happen to live in a country with high inflation), compounded n times per year. If n = 1, then the money simply sits in the bank for one year without increasing, at which point you get your 100% interest; that is, you have 2P in the bank after the first year is up. If n = 2, then after 6 months the money in the bank increases from P to 1.5 P (that is, after half a year you get 50%), and after another 6 months the money in the bank increases from 1.5 P to 1.5 · 1.5 P = 2.25 P (that is, you get another 50%, this time of a greater amount). In general, if the interest is compounded n times in the year, then every 1/n years whatever sum you have in the bank is multiplied by (1 + \frac{1}{n}). After 1 year the amount in the bank is

(1 + \frac{1}{n})^{n} P.

For example, if n = 12 (monthly compounding), then you have 2.613 P after 1 year; if n = 365 (daily compounding), then you have 2.71457 P after 1 year; if n = 31536000 (compounding every second), then you have 2.71828 P after 1 year; and the larger n is (i.e., the more frequently the compounding occurs) the more money you have. However, this amount does not increase without bound. Rather, as n ⟶ ∞ the amount you have in the bank after 1 year becomes closer and closer to e P = 2.71828 ⋯ P. (In the next paragraph we explain why the value gets closer and closer to e P.) Taking the limit as n goes to infinity is called “continuous compounding.” For all practical purposes, compounding every second is equivalent to continuous compounding, and even daily compounding is close to continuous compounding.

To see why the limiting value as n ⟶ ∞ is eP, we compute the limit as n ⟶ ∞ of (1 + (1/n))^{n}. Instead of looking at this power, let us take its natural log and find the limit of that:

lim_{n⟶∞} ln ((1 + (1/n))^{n}) = lim_{n⟶∞} \frac{ln (1 + \frac{1}{n})}{1/n} .

But this is the slope of the chord on the graph of y = ln x joining the points (1, 0) and (1 + \frac{1}{n}, ln(1 + \frac{1}{n})). As n increases, this chord approaches the tangent line to the curve at x = 1, and so this slope approaches the derivative at x = 1, which is 1 (since \frac{d}{dx} ln x = \frac{1}{x}). Thus, as n ⟶ ∞, the ln of (1 + (1/n))^{n} approaches 1. Thus, lim_{n⟶∞}(1 + (1/n))^{n} is equal to the number whose ln is 1, namely e:

lim_{n⟶∞} (1 + \frac{1}{n})^{n} = e.

Let us return to the bank interest problem, and ask how much money is in the bank after t years. In t years the interest is compounded nt times, so that the amount in the bank is

(1 + \frac{1}{n})^{nt} P = ((1 + \frac{1}{n})^{n})^{t} P.

If the money is “compounded continuously,” i.e., we take the limit as n ⟶ ∞, then this amount is simply e^{t}P.

(In practice, most banks compound money daily: n = 365. We shall always assume, however, that the money is compounded “continuously,” i.e., n = ∞. As a practical matter, there is very little difference. For example, even at the high interest rate of 100%, with daily compounding $1000 grows to $2714.57 in a year, while with continuous compounding it grows to $2718.28. With lower interest rates, the difference between daily compounding and continuous compounding is even less.)

The function e^{x} is called the exponential function. It is the inverse function of log_{e}, which we saw is the natural log function ln. In other words, to say that y = e^{x} is equivalent to saying that x = ln y. In the graph at the right we show the graphs of e^{x} (solid line) and ln x (dotted line). As always with inverse functions, these two graphs are the mirror images of one another with respect to the line y = x.

The function y = f(x) = e^{x} has a very basic property, which we discover when we

y = e^{x}y = ln x

find its derivative y′. To find this derivative, we start with the relation x = ln y, and take d/dx of both sides:

1 = \frac{1}{y} \frac{dy}{dx},  so that  \frac{dy}{dx} = y,  i.e.,  \frac{d}{dx} e^{x} = e^{x}.

That is, the function y = e^{x} has the very special property that its derivative is the very same function. It turns out that there is no other function — except for constant multiples Ce^{x} — that has this property.

1. Some Properties of the Exponential Function

By the law of exponents, if A and B are any numbers or expressions, then

(Property 1)  e^{A} · e^{B} = e^{A+B};  e^{A}/e^{B} = e^{A−B};  (e^{A})^{B} = e^{AB}.

Because ln and the exponential function are inverse functions,

(Property 2)  ln(e^{A}) = A;  e^{ln A} = A.

By the chain rule,

(Property 3) \frac{d}{dx}e^{u(x)} = e^{u(x)}u′(x).

Since \frac{d}{dx}e^{kx} = ke^{kx} (by (3) with u(x) = kx), it follows that

(Property 4) ∫ e^{kx}dx = \frac{1}{k} e^{kx} + C.

Example 10.1. Find the center of mass of a uniform plate bounded by the lines x = 0, x = 2 and the curves y = 1/2^{x} and y = −1/2^{x}.

By symmetry, the center of mass is on the x-axis. Its x-coordinate is

\overline{x} = \frac{∫_{0}^{2} \frac{x}{2^{x}} dx}{∫{2 \atop 0} \frac{dx}{2^{x}}} = \frac{∫_{0}^{2} xe^{−(ln 2) x}dx}{∫_{0}^{2} e^{−(ln 2) x}dx},

where we have used properties (1) and (2) to write 1/2^{x} as e^{−(ln 2) x}. We next use property (4) with k = − ln 2 to find the bottom integral, and then the trapezoid rule (with n = 4) to find the top integral:

∫_{0}^{2} e^{−(ln 2) x}dx = −\frac{1}{ln 2}e^{−(ln 2) x}|_{0}^{2} = −\frac{1}{ln 2} \frac{1}{2^{x}} |_{0}^{2} = \frac{1 − (1/4)}{ln 2} = 1.08202;

∫_{0}^{2} \frac{x}{2^{x}} dx ≈ \frac{1}{2} (0 + \frac{0.5}{\sqrt{2}} + \frac{1}{2} + \frac{1.5}{2\sqrt{2}} + \frac{1}{2} · \frac{2}{2^{2}} ) = \frac{5}{16}\sqrt{2} + \frac{3}{8} = 0.81694.

(Note: this last integral can be evaluated precisely by the method of integration by parts, which we will get to in about a week; it turns out that the exact value is −1/(2 ln 2) + 3/(2 ln 2)^{2} = 0.83968.) Finally, \overline{x} = 0.83968/1.08202 = 0.77603.

Example 10.2. Sketch the curve f(t) = e^{−t/T} cos(ωt) for t > 0 (where ω and T are constants). For simplicity, take T = 1 sec.

To get an idea of what to expect, let’s sketch the two functions whose product is f(t). The graph of y = e^{−t} is obtained by reflecting the graph of y = e^{t} (shown above) about the y-axis. The graph of cos(ωt) is obtained by squeezing the graph of cos t by a factor of ω toward the y-axis (e.g., horizontal contraction by a factor of 5 if ω = 5, horizontal expansion by a factor of 5 if ω = 1/5). The graphs of y = e^{−t} and y = cos(5t) are shown below:

The effect of multiplying cos(ωt) (which oscillates between 1 and −1) by e^{−t} (which steadily decreases from 1 to 0) is to obtain a function which oscillates between maximum and minimum values (local maxima and minima) that are steadily decreasing. This type of function f(t), called “damped harmonic oscillation,” arises in practice, for example, if you want to describe the motion of an object at the end of a spring which is bouncing back and forth with amplitude that decreases because of friction and air resistance.

Now let us determine the t-values for which our function f(t) has max/min points and points of inflection. To find the max/min points, we set f′(t) = 0:

0 = \frac{d}{dt}e^{−t} cos(ωt) = −e^{−t} cos(ωt) + e^{−t}(−ω sin(ωt)) = −e^{−t}(cos(ωt) + ω sin(ωt)).

Since e^{−t} is never 0, this derivative vanishes when cos(ωt) + ω sin(ωt) = 0, i.e. (if we divide through by ω cos(ωt)), when tan(ωt) = −1/ω. Thus, our solutions t are all positive values (in radians) of arctan(−1/ω), divided by ω. For example, if ω = 5, then we successively obtain a minimum at t = 0.589, a maximum at t = 1.217, a minimum at t = 1.845, a maximum at t = 2.474, and so on. To find the inflection points, we set f′′(t) = 0, i.e. (after a little simplification):

e^{−t}((1 − ω^{2}) cos(ωt) + 2ω sin(ωt)) = 0.

Thus, the points of inflection occur when t = \frac{1}{ω} arctan(\frac{ω^{2}−1}{2ω}). For example, if ω = 5, then t = 0.2 arctan 2.4 = 0.235, 0.864, 1.492, 2.120, etc. The curve y = e^{−t} cos(5t) is sketched at the top of the next page.

y = e^{−t} cos(5t)

Homework 9.

1. (a) Let l_{1} be the tangent line to the reciprocal function at the point (2, 0.5). Find the area under l_{1} between x = 1 and x = 3.

(b) Let l_{2} be the chord joining (1, 1) to (2.4, 1/2.4) on the graph of the reciprocal function. Find the area under l_{2}.

(c) Explain how your answers to parts (a) and (b) imply that the number e must be between 2.4 and 3.

2. Simplify the following expressions:

(a) e^{ln 3+ln x}; (c) e^{ln x−2 ln y}; (e) ln (\sqrt{e^{x^{2}}}) .
(b) ln (1/e^{x^{2}}); (d) e^{− ln(1/x)};

3. Solve for y in terms of x:

(a) e^{0.1y} = x^{2};  (c) x = ln sin y; (e) ln(1 − y) − ln y = 0.01x.
(b) ln(y − x) = 2 ln x; (d) e^{y^{2}} = x^{2};

4. Find the derivative of

(a) e^{2x}Arctan x;  (b) \frac{x}{e^{sin(πx)}} ;  (c) e^{e^{x}};  (d) x^{2}e^{− cos x};  (e) e^{Arcsin x}.

5. The function obtained by averaging the function e^{x} and the function e^{−x}, i.e., (e^{x}+e^{−x})/2, comes up often enough in applications to have earned the right to have its own name: “hyperbolic cosine.” The function (e^{x} − e^{−x})/2 is called the “hyperbolic sine.” These are abbreviated “cosh” and “sinh” (“sinh” is pronounced “sinch”):

cosh x = \frac{e^{x} + e^{−x}}{2} ; sinh x = \frac{e^{x} − e^{−x}}{2} .

(a) Find derivative formulas for cosh and sinh similar to the derivative formulas for cos and sin.

(b) Simplify cosh^{2} x − sinh^{2} x. (c) Sketch the hyperbolic sine and cosine.

6. Evaluate:

(a) ∫_{ln 0.04}^{ln 4} e^{x/2}dx,  (b) ∫_{0}^{1} xe^{−x^{2}}dx,  (c) ∫_{0}^{π} e^{cos x} sin x dx,  (d) ∫_{0}^{∞} \frac{dt}{10^{t}} .

7. Find the volume of the solid obtained by rotating the shape in Example 10.1 in the text around the x-axis.

8. Find the x- and y-coordinates of the maxima/minima and points of inflection, and sketch:

(a) y = e^{\frac{2}{3} sin(3x)}, 0 ≤ x ≤ π/3; (b) y = x^{2} ln(1/x), 0 < x < 1.

CHAPTER 11 Exponential Growth and Decay

1. Continuously Compounded Interest

Let’s return to the bank interest problem and suppose a more realistic annual interest rate of k (this is a percent written in decimal form, for example 8%= 0.08 — in this problem and many that follow, the units of k are reciprocal time, i.e., 0.08\frac{1}{yr} annual interest rate). We again want to find out how our principal P grows in t years. We again assume continuous compounding. But this time we shall use a different method to derive the function y(t) that tells the amount of money in the bank after t years.

The method is to set up and solve a certain “ differential equation.” Recall that a differential equation is an equation for finding an unknown function in which the derivative of the unknown function (and perhaps its second derivative or higher derivatives) appears. For example, (1) the differential equation y′ = f(x), where f(x) is a given function, is satisfied by an antiderivative of f(x); (2) the differential equation ÿ = −g is satisfied by the falling body formula y(t) = −\frac{1}{2}gt^{2} + v_{0}t + y_{0}.

Differential equations are fundamental in applications, because often we have to get at a function y(t) indirectly, using some information about the rate of change of y(t) that comes from translating some basic principle (of physics or finances, for instance) into mathematics. For example, the differential equation ÿ = −g is simply the statement that gravity near the surface of the earth imparts a constant acceleration.

In the case of bank interest continuously compounded at rate k, the basic principle is that at any instant interest is accumulating at a rate in dollars per year equal to k times the current amount in the bank. For example, if k = 8% = 0.08 and you put $1000 in the bank, then at first your money increases at the rate of $80.00 per year; but a little while later, when you have (say) $1010 in the bank, at that instant it is increasing at the rate of $80.80 per year. That is, you are “earning interest on the interest” as well as on the original principal. The basic principle in italics can be written mathematically as follows:

\frac{dy}{dt} = ky,

i.e., the function y(t) telling the number of dollars in the bank at time t years increases at a rate equal to k times the present value of this function.

Our task now is to solve this equation to find an expression for y in terms of t. A useful technique in this and many other differential equations is called “ separation of variables.” This means that we get everything involving y on one side of the equation and everything involving t on the other. Here the dy and dt are handled just like algebraic quantities. (When discussing the chain rule and also the method of substitution in integrals, we saw that it often makes sense to split up expressions like dy/dx or dy/dt, treating dy, dx, dt separately.) In the above differential equation we separate variables by multiplying both sides by dt and dividing both sides by y. After that the second step is to put an integral sign on both sides:

∫ \frac{dy}{y} = ∫ k dt.

On the left we have an indefinite integral with respect to the variable of integration y, and on the right we have an indefinite integral with respect to t. On the left we get ln |y| + C_{1}. Since y(t) is a positive function, we drop the absolute value sign. Also, we move the constant of integration to the right of the equality, subtracting it from the constant of integration we get on the right from integrating ∫ k dt = kt + C_{2}. That is, we set C = C_{2} − C_{1}. Thus, we have:

ln y = kt + C.

Finally, to solve for y we exponentiate both sides (this means: raise e to the expression on each side of the equality). Since e^{ln y} = y, this gives us a formula for y:

y(t) = e^{kt+C} = e^{C}e^{kt}.

We are not done, however, because we must determine the constant e^{C} in front of e^{kt}. As in acceleration–velocity–distance problems, we do this by using initial information. In the bank interest problem, the initial information is that we had P dollars in the bank at time t = 0: y(0) = P. Substituting t = 0 in the equation y(t) = e^{C}e^{kt} gives the result: P = y(0) = e^{C}e^{0} = e^{C}. Thus, we conclude that

y(t) = P e^{kt}.

(The special case of this formula when k = 100% = 1 was obtained in the last section by a different method.)

A function of the form y = P e^{kt} is called exponential growth. Its graph is shown below (in the case k = 8% = 0.08).

y = P e^{0.08t}

Effective annual interest rate. By the effective interest rate we mean the percent by which the money actually increases if left in the bank 1 year. The increase is from P to P e^{k·1}, i.e., P (e^{k} − 1). Thus, the percent increase is e^{k} − 1 (which we usually convert to a percent by multiplying by 100):

effective annual interest rate = e^{k} − 1.

For example, if k = 0.08, then the effective rate is e^{0.08} − 1 = 0.0833 = 8.33%. If k = 100%, then the effective rate is e − 1 = 171.83%. In general, if k is small, then the effective rate is only slightly more than k; while if k is large, then the effective rate is much greater, because it takes into account the interest earned on the interest. Sometimes people say “annual interest rate” when they really mean “effective annual interest rate.” Sometimes k is called the instantaneous interest rate. Note. Whenever we say simply “interest rate” we shall mean the instantaneous rate k, not the effective rate e^{k} − 1.

Doubling time. At time t we have P e^{kt} in the bank. Let us ask how long it will take for the money to double. This time period, which we call D, must satisfy the equation P e^{k(t+D)} = 2P e^{kt}. Dividing both sides by P e^{kt} and then taking the ln of both sides, we obtain kD = ln 2, i.e., D = (ln 2)/k. Notice that the doubling time D does not depend on the starting time t or on the principal P: it depends only on the interest rate. For example, at 8% interest rate the doubling time is (ln 2)/0.08 = 8 2/3 years.

2. Other Examples of Exponential Growth.

Any time we have something whose rate of increase is proportional to its present value (i.e., \frac{dy}{dt} = ky), we obtain the exponential function y(t) = y_{0}e^{kt}, where k is the constant of proportionality between the rate of increase and the present value, and y_{0} is the initial value of our function. Among the examples of this situation are:

Inflation. Rate of increase in cost of living is measured as an annual percent of the current cost of living. If the rate of inflation remains constant, then this cost of living follows the law of exponential growth.

Example 11.1. In the six-month period from June to December 1990 in Nicaragua, the córdoba deflated from an exchange rate of about 200000 per dollar to 2000000 per dollar. Find both the effective annual rate and the instantaneous rate k of inflation of the price in córdobas of an item whose dollar cost remains constant.

Here the inflation was by a factor of 10 in 1/2 year, and hence by a factor of 100 in 1 year. So the cost increased by 99 = 9900%. Thus, the effective rate is 99 = e^{k} − 1. To find k, we add 1 to both sides and take ln, obtaining k = ln 100 = 4.605 = 460.5%.

Population. If the birth and death rates remain constant (and there is no immigration or emigration), then the population as a function of time follows the law of exponential growth, in which k is the difference: k = birth rate − death rate.

Example 11.2. If a bacteria culture has a 4% per hour instantaneous birth rate and a 2.5% instantaneous death rate, what is its doubling time?

Here k = 4% − 2.5% = 1.5%, and so D = (ln 2)/0.015 = 46.2 hours.

3. Exponential Decay

Here we have a function y(t) which decreases at a rate proportional to its current value, with proportionality constant k: \frac{dy}{dt} = −ky. Mathematically speaking, nothing much changes: to solve this differential equation we simply replace k by −k in our earlier derivation, thereby obtaining the function y(t) = y_{0}e^{−kt}. However, the graph of y_{0}e^{−kt} as t increases looks very different from the graph of y_{0}e^{kt}, namely, it ends up decreasing rapidly toward 0:

y_{0}  y = y_{0}e^{−0.08 t}

Half-life. An exponential decay function y(t) = y_{0}e^{−kt} has a “ half-life” rather than a doubling time. This is the length of time H that it takes to decrease to half its current value. To find H, we set y_{0}e^{−k(t+H)} equal to half of y_{0}e^{−kt}. Dividing this equation through by y_{0}e^{−kt} and then taking ln gives: −kH = ln(1/2) = − ln 2, and so H = (ln 2)/k. Note that this is the same as the doubling time formula in the case of exponential growth.

4. Examples of Exponential Decay.

Radioactive decay.

Example 11.3. Some future visitors to Earth find a planet that was rendered uninhabitable by a catastrophe that occurred many thousands of years before. By reconstructing the accident, they determine how much radioactive Plutonium 239 was released at that time. They also measure the amount of Plutonium 239 remaining, and find that only 1/500 of the radioactive Plutonium is still present. Using the value 24400 years for the half-life of Plutonium 239, determine how much time elapsed between the accident and the arrival of the visitors.

We first find k, using the relation kH = ln 2: k = ln 2/24400 = 0.0000284. Then we find t by setting y(t) = \frac{1}{500}y_{0}, i.e., \frac{1}{500}y_{0} = y_{0}e^{−kt}. Dividing by y_{0} and taking ln gives −kt = ln(1/500) = − ln 500, so that t = (ln 500)/0.0000284 = 218800 years.

Population when the death rate exceeds the birth rate.

“Real” salary (measured in inflation-adjusted dollars, i.e., buying power of what you earn) if the rate of salary increase is less than the inflation rate.

Finally, we should emphasize that real-world phenomena follow the law of exponential growth or decay only if the thing being measured satisfies the differential equation y′ = ky or y′ = −ky. In practice, when we suppose this to be the case we are making a simplifying assumption that may be quite inaccurate. For example, most human or animal populations are influenced by many factors and cannot be described at all accurately by the exponential growth model.

Homework 10.

1. Suppose that interest on money in the bank accumulates at an annual rate of 7% compounded continuously. How much should be invested today so that 20 years from now it will be worth $20000? B

2. Suppose that a sum of money draws interest compounded continuously. How long will it take to double if the annual rate is (a) 6%?, (b) 10%?, (c) 18%? How long will it take to double if the effective annual rate is (a) 6%?, (b) 10%?, (c) 18%?

3. In an experiment to study the growth of bacteria, a medical student measured 5000 bacteria at time 0 and 8000 at time 10 minutes. Assuming that the number of bacteria grows exponentially, how many bacteria will be present after 30 minutes?

4. Suppose that the population of a certain species grows at the instantaneous rate of 2% per year. If the present population is 1000000, what will the population be in 1 year? In 20 years? How long will the population take to double?

5. Assume that the population of the Earth changes at a rate proportional to its current population. Assume further that in A.D. 1650 its population was 250 million, and in A.D. 1950 its population was 2.5 billion. Find an expression giving the population of the Earth at any time. If the greatest population the Earth can support is 25 billion, when will this limit be reached?

6. If you hear on the news that prices are rising exponentially at the effective yearly rate of 12%, what is the instantaneous annual rate of inflation? How long does it take prices to double at this rate?

7. Suppose that the average yearly cost of college in the U.S. (tuition and expenses) is increasing at the instantaneous rate of 10% per year. By what factor will the cost of college increase for a child born now in the 18 years before she goes to college?

8. Suppose that bank interest is compounded continuously at 6%, and inflation on food has instantaneous rate 13%. (a) If you put $10000 in the bank, how much will you have 100 years later? (b) If it now costs $100/week to feed an average family, write a formula for weekly food cost t years later. (c) What is the total cost of feeding a family over the next 10 years? (d) What will the weekly food bill be in 100 years? (e) If your great grandchild takes your money in part (a) out of the bank at that time, how long will it feed his or her family?

9. The half-life of Carbon 14 is 5568 years. Archeologists find an ancient table made from a tree which had y_{0} grams of Carbon 14 when it was cut. Now the wood has only y_{1} grams of Carbon 14. Express the age of the table in terms of y_{0} and y_{1}.

10. A certain radioactive material has a half-life of 2 hours. Find the time needed for a given amount of this material to decay to one-tenth of its original mass.

11. A certain radioactive material is known to decay at a rate proportional to the amount present. A block of this material originally having a mass of 100 grams is observed after 20 years to have a mass of only 80 grams. Find an expression for the mass of the material at any time. Also find the half-life of the material.

12. Suppose that the average yearly real income (i.e., in inflation-adjusted dollars, say 1995-dollars) of a U.S. worker is now decreasing at the instantaneous rate of 1.5% per year, and that this trend continues indefinitely. If the current average salary is $18000, what will an average worker’s yearly income (in 1995-dollars) be in 30 years?

13. You are flying in a plane at constant altitude y_{0}, constant speed v_{0} and constant direction. You want to drop a container of relief supplies to a target that is straight ahead under your flight path. You will release the container at the instant when your line of sight to the target makes an angle α below the horizontal. The purpose of this problem is to determine α. Let v(t) denote the \underline{horizontal} component of the velocity of the container. Assume that air resistance causes a horizontal component of deceleration that is proportional to velocity, with proportionality constant k. Air resistance is the only force acting to change the velocity (since gravity does not affect horizontal motion).

(a) Write a differential equation for v(t), and find a formula for v(t).

(b) Find a formula for the horizontal distance traveled by the container between time 0 and time t.

(c) Suppose that after enough time has elapsed so that the exponential decay term in part (c) is negligible, a parachute opens and the container glides straight down until it hits the ground. Find α.

CHAPTER 12 Techniques of Integration I: Substitution, Powers of Trig Functions

1. Substitution

By a technique of integration we mean a method which can be used to find the antiderivative (indefinite integral) of certain types of functions. We first review u-substitutions. We shall use an example to illustrate the thought process that goes into deciding on a substitution of the form u = u(x).

Example 12.1. Find the x-coordinate of the center of mass of the uniform plate that is bounded by the x- and y-axes, the line x = 10, and the curve y = 1/(1 + 0.01x^{2}).

We must compute \overline{x} = ∫_{0}^{10} x dx/(1 + 0.01x^{2}) / ∫_{0}^{10} dx/(1 + 0.01x^{2}). At first the top

and bottom integrals look similar: both have (1 + 0.01x^{2}) in the denominator. However, they are actually quite different.

In the bottom we have something very close to ∫ du/(1 + u^{2}) =Arctan u + C. The only problem is the constant 0.01 in front of x^{2}. So we choose u so that u^{2} = 0.01x^{2}, i.e., u = 0.1x. Then x = 10u, dx = 10du. The limits of integration also change: x = 0 ⟹ u = 0, and x = 10 ⟹ u = 0.1 · 10 = 1. Hence, the bottom integral becomes ∫_{0}^{1} 10 du/(1 + u^{2}) = 10 Arctan u |_{0}^{1}= 10(π/4) = 2.5π.

The top integral, on the other hand, has nothing to do with Arctan. The x next to the dx changes everything. Namely, x is essentially the derivative of the denominator. Thus, a substitution u =denominator will lead to something like ∫ du/u = ln |u|+C. More precisely, we set u = 1+0.01x^{2}, so that du = 0.02x dx, and hence x dx = 50 du. This gives us 50 du/u inside the integral. The limits of integration change as follows: x = 0 ⟹ u = 1+0.01·0 = 1, and x = 10 ⟹ u = 1 + 0.01 · 10^{2} = 2. Thus, the top integral becomes 50 ∫_{1}^{2} du/u = 50 ln 2. Finally, we compute \overline{x} = 50 ln 2/2.5π = 4.4127.

2. Odd Powers of Sine and Cosine

The technique of u-substitution can be used to evaluate any integral of the form

∫ sin^{n}(ax + b) dx or ∫ cos^{n}(ax + b) dx, where n is a positive odd number.

The method works as follows. Say we want to integrate sin^{n}. First take out one sin and put it with the dx. What remains will be sin to an even power. Express this in terms of cos (using sin^{2} = 1 − cos^{2}). Now make a u-substitution for the cos. You will end up with only a polynomial to integrate. (If we had cos^{n} to integrate, we would do the same thing, with the roles of sin and cos reversed.) We illustrate with an example.

Example 12.2. Find ∫_{0}^{1} sin^{5}(πx)dx.

We first write

∫_{0}^{1} sin^{5}(πx)dx = ∫_{0}^{1} sin^{4}(πx) sin(πx)dx = ∫_{0}^{1}(1 − cos^{2}(πx))^{2} sin(πx)dx.

Next, we make the substitution u = cos(πx), so that du = −π sin(πx). Then sin(πx)dx = −\frac{1}{π} du, and so our integral becomes −\frac{1}{π} ∫ (1 − u^{2})^{2}du. The limits of integration change as follows: x = 0 ⟹ u = cos 0 = 1, and x = 1 ⟹ u = cos π = −1. To integrate the polynomial, we expand the square: (1 − u^{2})^{2} = 1 − 2u^{2} + u^{4}. So we obtain:

−\frac{1}{π} ∫_{1}^{−1} 1 − 2u^{2} + u^{4} du = −\frac{1}{π} (u − \frac{2}{3}u^{3} + \frac{1}{5}u^{5}) |_{1}^{−1}= \frac{16}{15π} .

It is also possible to use this method to integrate negative odd powers of sine and cosine. However, in that case instead of a polynomial in u you end up with a rational function (i.e., you have a denominator with a polynomial in it). We will treat some cases of this in part III of techniques of integration.

3. Even Powers of Sine and Cosine

The above technique does not work if you want to integrate something like cos^{2} or sin^{4}. To handle even powers one needs a trig identity, namely, the cos(2A) formula: cos(2A) = cos^{2} A − sin^{2} A. When combined with the trig identity 1 = cos^{2} A + sin^{2} A, this gives formulas for either cos^{2} A or sin^{2} A in terms of cos(2A). If we add these two identities and divide by 2, we get cos^{2} A; if we subtract them and divide by 2, we get sin^{2} A:

cos^{2} A + sin^{2} A = 1  cos^{2} A + sin^{2} A = 1
+ \underline{cos^{2} A − sin^{2} A} = \underline{cos(2A)}  − \underline{cos^{2} A − sin^{2} A} = \underline{cos(2A)}
2 cos^{2} A = 1 + cos(2A)  2 sin^{2} A = 1 − cos(2A) ,

i.e.,

cos^{2} A = \frac{1 + cos(2A)}{2},  sin^{2} A = \frac{1 − cos(2A)}{2}.

We now illustrate the use of these two identities in integration problems.

Example 12.3. Find the average value of sin^{2} x between 0 and π.

The average value of a function f(x) on the interval [a, b] is the integral of the function divided by the length of the interval: \frac{1}{b−a} ∫_{a}^{b} f(x)dx. So the average value of sin^{2} x is

\frac{1}{π} ∫_{0}^{π} sin^{2} x dx = \frac{1}{π} ∫_{0}^{π} (\frac{1 − cos(2x)}{2}) dx

= \frac{1}{π} (\frac{1}{2}x − \frac{1}{2} · \frac{1}{2} sin(2x)) |_{0}^{π}

= \frac{1}{π} (\frac{π}{2} − \frac{1}{4}(0 − 0)) = \frac{1}{2}.

Example 12.4. Find the antiderivative of cos^{4} x.

We have

∫ cos^{4} x dx = ∫ (cos^{2} x)^{2} dx = ∫ (\frac{1 + cos(2x)}{2})^{2} dx

= ∫ (\frac{1 + 2 cos(2x) + cos^{2}(2x)}{4}) dx.

We now use the identity for cos^{2}(A) once more, this time with A = 2x: cos^{2}(2x) = (1 + cos(4x))/2. In this way we end up replacing all powers of cosine with expressions of the form cos(ax), which are easy to integrate. After writing

1 + 2 cos(2x) + cos^{2}(2x) = 1 + 2 cos(2x) + (1 + cos(4x))/2

= \frac{3}{2} + 2 cos(2x) + \frac{1}{2} cos(4x)

and dividing this by 4, we obtain

∫ cos^{4} x dx = ∫ (\frac{3}{8} + \frac{1}{2} cos(2x) + \frac{1}{8} cos(4x)) dx

= \frac{3}{8}x + \frac{1}{4} sin(2x) + \frac{1}{32} sin(4x) + C.

Homework 11.

1. Find the x-coordinate of the center of mass of the flat plate bounded by the x- and y-axes, the line x = 5, and the curve y = 1/\sqrt{1 − 0.01x^{2}}.

2. Find the anti-derivative of (a) sin(2πt)/\sqrt{2 + cos(2πt)};

(b) cos(ax + b)/ sin^{8}(ax + b).

3. Find the volume of the solid obtained by revolving y = cos x around the x-axis between x = 0 and x = π/4.

4. Find the average value of sin^{n}x between 0 and π/2 when n is (a) 3, (b) 4, (c) 5, (d) 6, (e) 7.

CHAPTER 13 Techniques of Integration II: Integration by Parts

A general principle is that any derivative rule has a corresponding integration rule. For example, a derivative formula such as \frac{d}{dx} sin x = cos x has a corresponding antiderivative formula ∫ cos x dx = sin x + C. The u-substitution technique of integration comes from the chain rule for differentiation: if you check an antiderivative obtained by u-substitution in order to be sure that its derivative is the function you wanted to integrate, you use the chain rule. For example, in the top integral of Example 12.1 in the last section, to check that 50 ln u = 50 ln(1 + 0.01x^{2}) is the antiderivative of x/(1 + 0.01x^{2}), you use the chain rule:

\frac{d}{dx}50 ln u = \frac{50}{u} \frac{du}{dx} = \frac{50}{1 + 0.01x^{2}} (0.02x) = \frac{x}{1 + 0.01x^{2}} .

The product rule for derivatives also gives rise to a very useful integration rule, called integration by parts. To get this rule, we recall the product rule, which says that the derivative of u(x)v(x) is u(x)v′(x) + u′(x)v(x). We shall omit the (x), and write simply (uv)′ = uv′ + u′v. Now solve this equality for uv′, and then take the antiderivative of both sides. Since the antiderivative of (uv)′ is just uv, we obtain: ∫ uv′ = uv − ∫ u′v. This is the integration by parts rule in shorthand form. Written out completely, it is:

∫ u(x)v′(x)dx = u(x)v(x) − ∫ u′(x)v(x)dx (indefinite integral form).

For definite integrals, we put in the limits of integration:

∫_{a}^{b} u(x)v′(x)dx = u(x)v(x) |_{a}^{b} − ∫_{a}^{b} u′(x)v(x)dx (definite integral form).

At first, this might seem to be useless: the rule says that our integral is equal to uv minus some other integral, and so we still have an integral to compute. However, for an important class of functions f(x) it is possible to choose u(x) and v(x) in such a way that f(x) = u(x)v′(x) and the integral on the right — namely, ∫ u′(x)v(x)dx — is easier to evaluate than ∫ f(x)dx. This section is devoted to explaining how this is done. We first give a series of examples, and then we make up a table summarizing the main uses of integration by parts.

Example 13.1. Find ∫ x cos x dx, and then find the x-coordinate of the center of mass of the flat plate bounded by the x- and y-axes and the cosine curve in the first quadrant.

We choose u(x) = x and v′(x) = cos x. The first thing we must do is find v(x), which is an antiderivative of cos x. So we take v(x) = sin x. In an integration by parts problem, v′(x) must always be a function whose antiderivative you can find right away. (The integration by parts would work just as well if we took any other antiderivative of the form sin x + C; but to avoid unnecessary work we shall always choose v(x) by setting the constant of integration equal to 0.) Using u(x) = x and v(x) = sin x in the integration by parts rule, we obtain:

∫ x cos x dx = x sin x − ∫ 1 · sin x dx = x sin x − (− cos x) + C = x sin x + cos x + C.

It is instructive to check our answer by taking the derivative of x sin x + cos x. As always in an integration by parts problem, our answer has two parts: a product of two functions (namely x sin x) and another function which can be thought of as a “correction term” (namely, cos x). When we take the derivative of the product, the product rule gives us two terms. One of them is our original function inside the integral, and the other is a term we don’t want. The derivative of the “correction term” precisely cancels out the term we don’t want. We see how this works when we check our answer to Example 13.1:

\frac{d}{dx}(x sin x + cos x) = (x cos x + 1 · sin x) − sin x = x cos x.

Finally, \overline{x} in this example is equal to

∫_{0}^{π/2} x cos x dx / ∫_{0}^{π/2} cos x dx = (x sin x + cos x) |_{0}^{π/2}/ sin x |_{0}^{π/2}

= (π/2 − 1)/1 = \frac{π}{2} − 1 = 0.5708.

Example 13.2. Find ∫ xe^{5x}dx.

Again choose u(x) = x. Let v′(x) = e^{5x}, so that v(x) = \frac{1}{5}e^{5x}. Then

∫ xe^{5x}dx = x · \frac{1}{5}e^{5x} − ∫ 1 · \frac{1}{5}e^{5x}dx = \frac{1}{5}xe^{5x} − \frac{1}{5} ∫ e^{5x}dx = \frac{1}{5}xe^{5x} − \frac{1}{25}e^{5x} + C.

Example 13.3. Find the improper integral ∫_{0}^{∞} x^{10}e^{−x}dx.

We shall apply the definite integral version of integration by parts. Having upper limit of integration ∞ means that we can take a finite upper limit of integration b and then let b ⟶ ∞.

Before doing this problem, we shall digress for a minute, and show that x^{n}e^{−kx} always approaches 0 as x ⟶ ∞, for any n and any positive constant k. Look at the ln of this expression: ln(x^{n}e^{−kx}) = ln(x^{n}) + ln(e^{−kx}) = (n ln x) − kx. That is, we get a constant multiple of ln x minus a constant multiple of x. When x is large ln x is much smaller than x, and so eventually this difference must approach −∞. Hence, our original expression x^{n}e^{−kx}, which is e^{n ln x−kx}, is e to a power that approaches −∞, and this approaches zero, as claimed.

We shall apply this limit fact when we evaluate expressions of the form x^{n}e^{−x} at infinity (i.e., at b and then take b ⟶ ∞). That is, when we evaluate x^{n}e^{−x} at infinity we get 0. In our integral we take u(x) = x^{10} and v′(x) = e^{−x}, so that v(x) = −e^{−x}. Then

∫_{0}^{∞} x^{10}e^{−x} dx = −x^{10}e^{−x} |_{0}^{∞} − ∫_{0}^{∞} 10x^{9} (−e^{−x}) dx = 0 − 0 + 10 ∫_{0}^{∞} x^{9}e^{−x} dx.

The last integral is like the one we started with; however, the power of x has been reduced by 1. So we must apply integration by parts again, this time with u(x) = x^{9}. The result is:

∫_{0}^{∞} x^{9}e^{−x}dx = 9 ∫_{0}^{∞} x^{8}e^{−x}dx. Continuing in this way, we obtain

∫_{0}^{∞} x^{10}e^{−x}dx = 10 ∫_{0}^{∞} x^{9}e^{−x}dx = 10 · 9 ∫_{0}^{∞} x^{8}e^{−x}dx
= ⋯ = 10 · 9 · 8 ⋯ 2 · 1 · ∫_{0}^{∞} e^{−x}dx.

But ∫_{0}^{∞} e^{−x}dx = −e^{−x} |_{0}^{∞}= 1, so that our final answer is 10·9·8·7·6·5·4·3·2·1 = 3628800. (The product of all numbers from 1 to 10 is called “10 factorial,” and is written 10!.)

Example 13.4. Find ∫_{0}^{1} x^{3} ln x dx.

First note that the function x^{3} ln x is negative between 0 and 1 (because ln x is negative there), has value 0 when x = 1, and approaches the limit 0 as x ⟶ 0. (For the latter fact, see Problem 9(a) in the section on ln.) Thus, the function we’re integrating in Example 13.4 has the following appearance:

y = x^{3} ln x

We now use integration by parts. If we try to proceed as in Example 13.3, setting u(x) equal to x^{3}, we run into a problem: we would then have to take v′(x) equal to ln x, and we do not yet know an antiderivative of ln x. In an integration problem with ln x, it makes sense to reverse our choice of u and v. That is, we set u(x) = ln x and v′(x) = x^{3}, so that v(x) = x^{4}/4. This gives us:

∫_{0}^{1} x^{3} ln x dx = \frac{1}{4}x^{4} ln x |_{0}^{1} − ∫_{0}^{1} \frac{1}{4}x^{4} · \frac{1}{x} dx.

Notice what happened in the integral at the right: we have nothing more than a power of x to integrate. The point is that integration by parts can be viewed as a trade-off, where in the integral on the right you get to replace u by its derivative u′ (which should be a simpler function), but at the expense of replacing v′ by its antiderivative v. In this problem we had to replace x^{3} by a higher power x^{4}, but in return we got to replace ln x by the much simpler function 1/x which is its derivative. Returning to our computation, we find that x^{4} ln x gives 0 when evaluated at either 1 or 0, and hence

∫_{0}^{1} x^{3} ln x dx = −\frac{1}{4} ∫_{0}^{1} x^{3} dx = −\frac{1}{4} · \frac{1}{4} = −\frac{1}{16}.

Example 13.5. Find the antiderivative of ln x.

At first it might seem that the integral ∫ ln x dx should have nothing to do with integration by parts, since the function inside the integral is not a product. However, we can treat this integral in the same way as we treated ∫ x^{3} ln x dx in Example 13.4. Namely, we view ln x as the product 1 · ln x = x^{0} ln x, and we set u(x) = ln x and v′(x) = 1, so that v(x) = x. Then:

∫ ln x dx = x ln x − ∫ x · \frac{1}{x}dx = x ln x − ∫ 1 dx = x ln x − x + C.

Example 13.6. Find the antiderivative of arcsin x.

This antiderivative (and also the antiderivative of Arctan) can be computed in the same way as the antiderivative of ln x in Example 13.5. Namely, we take u(x) = arcsin x, v′(x) = 1, v(x) = x, so that u′(x) = 1/\sqrt{1 − x^{2}}, and

∫ arcsin x dx = x arcsin x − ∫ x\frac{1}{\sqrt{1 − x^{2}}} dx = x arcsin x − ∫ \frac{x dx}{\sqrt{1 − x^{2}}} .

The last integral can be evaluated using a u-substitution. That is, for the rest of the problem we forget about integration by parts, and use the letter u for an entirely different purpose. Namely, if we set u equal to the expression under the radical, we see that du is essentially what we have in the numerator. More precisely, u = 1 − x^{2}, du = −2xdx, so that

∫ arcsin x dx = x arcsin x + ∫ \frac{du}{2\sqrt{u}}
= x arcsin x + \sqrt{u} + C = x arcsin x + \sqrt{1 − x^{2}} + C.

The reader should check this answer by taking its derivative and seeing that, after cancelation, you get arcsin x.

Before doing more examples, we give a table that summarizes the most common uses of integration by parts. They all relate to functions which are the product of a power of x and either an exponential function, sine or cosine, or else one of the inverse functions ln, Arcsin, Arctan.

1. How to Choose u and v

I. In the following cases choose u(x) = x^{n} and v′(x) =the other function. Here n is a positive integer, and k, a, and b are constants.

∫ f(x)dx, where f(x) is a product x^{n} · \begin{cases}e^{kx}, \\ sin(ax + b), \\ cos(ax + b). \end{cases}

II. In the following cases choose v′(x) = x^{n} and u(x) = the other function. Here n is a positive integer or 0, m is a positive integer, and a and b are constants.

∫ f(x)dx, where f(x) is a product x^{n} · \begin{cases}ln(ax + b), \\ ln^{m}(ax + b), \\ arcsin(ax + b), \\ arctan(ax + b). \end{cases}

One type of problem in part II which has not yet occurred in the examples is x^{n} · ln^{m}(ax + b). We now give an example of this.

1. HOW TO CHOOSE u AND v 83

Example 13.7. Suppose that the shape in Example 13.4 is revolved around the x-axis to form a solid. Find its volume.

Since (x^{3} ln x)^{2} = x^{6} ln^{2} x, we choose u(x) = ln^{2} x and v′(x) = x^{6}. Then v(x) = x^{7}/7, and u′(x) = 2(ln x)/x. We have:

Vol = π ∫_{0}^{1} x^{6} ln^{2} x dx = π (\frac{1}{7}x^{7} ln^{2} x |_{0}^{1} − ∫_{0}^{1} \frac{1}{7}x^{7} · 2(ln x)\frac{1}{x} dx)

= π (0 − 0 − \frac{2}{7} ∫_{0}^{1} x^{6} ln x dx) .

We now have to use integration by parts a second time. (In general, to integrate x^{n} ln^{m} x, you’ll need to use integration by parts m times, each time lowering the power of ln x by one.) We obtain:

Vol = −\frac{2π}{7} ∫_{0}^{1} x^{6} ln x dx = −\frac{2π}{7} (\frac{1}{7}x^{7} ln x |_{0}^{1} − ∫_{0}^{1} \frac{1}{7}x^{7} · \frac{1}{x} dx)

= \frac{2π}{7} · \frac{1}{7} ∫_{0}^{1} x^{6} dx = \frac{2π}{343} = 0.01832.

Example 13.8. Find both coordinates of the center of mass of the flat plate bounded by the x- and y-axes and the curve y = e^{−x} in the first quadrant (out to infinity).

The denominator in \overline{x} and \overline{y} is the area, which is ∫_{0}^{∞} e^{−x}dx = 1, as we saw in Example 13.3. Thus,

\overline{x} = ∫_{0}^{∞} xe^{−x}dx = −xe^{−x} |_{0}^{∞} − ∫_{0}^{∞} (−e^{−x}) · 1 dx = 0 − 0 + ∫_{0}^{∞} e^{−x}dx = 1.

To compute \overline{y}, we must first write x in terms of y: y = e^{−x} ⟹ x = − ln y. In the y-direction, the shape is contained between y = 0 and y = 1. Thus,

\overline{y} = ∫_{0}^{1} y(− ln y)dy = − (\frac{1}{2}y^{2} ln y |_{0}^{1} − ∫_{0}^{1} \frac{1}{2}y^{2} · \frac{1}{y} dy) = \frac{1}{2} ∫_{0}^{1} y dy = \frac{1}{4}.

So the center of mass is at (1, 0.25).

Homework 12.

1. Find the anti-derivative of:

(a) x sin(ax + b); (b) x^{2} cos x; (c) arctan x; (d) x^{3}e^{x}; (e) x arctan x.

2. Find the center of mass of the uniform flat plate bounded by (a) the x- and y-axes and the curve e^{kx} in the second quadrant (out to −∞);

(b) the curves y = −x^{2} ln x and y = x^{2} ln x;

(c) the curves y = −x^{n} ln x and y = x^{n} ln x (express your answer in terms of n);

(d) the curves y = sin^{2}(πx) and y = − sin^{2}(πx) to the left of x = 1/4.

3. Find the volume of the container obtained by rotating x = y sin y around the y-axis between y = 0 and y = π/3.

CHAPTER 14 Techniques of Integration III: Rational Functions

In this section we treat integration of rational functions. A “rational function” is a fraction with polynomials in the numerator and denominator. For example,

\frac{x^{3}}{x^{2} + x − 6},  \frac{1}{(x − 3)^{2}} ,  \frac{x^{2} + 1}{x^{2} − 1}

are all rational functions of x. There is a general technique called “ partial fractions” for integrating a rational function. The algebraic steps in that technique are rather cumbersome if the polynomial in the denominator has degree more than 2 (except in the case when the denominator has the form (ax + b)^{n}, a case we shall treat below). However, in practice one does not often run across rational functions with high degree polynomials in the denominator for which one has to find the antiderivative function. So we shall explain how to find the antiderivative of a rational function only in the two cases which are most frequently encountered: (a) the denominator has the form (ax+b)^{n}; (b) the denominator is a quadratic polynomial ax^{2} + bx + c.

In the case when the denominator has the form (ax + b)^{n}, the u-substitution u = ax + b will always work. Just replace the denominator by u^{n}, replace x in the numerator by (u − b)/a, and replace dx by \frac{1}{a}du.

Example 14.1. Find

∫_{0}^{1} \frac{t^{3}dt}{(3 − 2t)^{2}} .

We set u = 3 − 2t, so that t = (3 − u)/2, dt = −\frac{1}{2}du. The limits of integration change: t = 0 ⟹ u = 3, and t = 1 ⟹ u = 1. So we obtain:

−\frac{1}{2} ∫_{3}^{1} (\frac{3 − u}{2})^{3} \frac{du}{u^{2}} = −\frac{1}{16} ∫_{3}^{1} \frac{(3 − u)^{3}}{u^{2}} du = \frac{1}{16} ∫_{1}^{3} \frac{27 − 27u + 9u^{2} − u^{3}}{u^{2}} du

= \frac{1}{16} ∫_{1}^{3} (\frac{27}{u^{2}} − \frac{27}{u} + 9 − u) du

= \frac{1}{16} (−\frac{27}{u} − 27 ln u + 9u − \frac{1}{2}u^{2}) |_{1}^{3}= 2 − \frac{27}{16} ln 3.

We now proceed to the case when the denominator is a quadratic polynomial. We can always factor out the coefficient of x^{2} and put it outside the integral. This will lead to a denominator in which x^{2} has coefficient 1, i.e., a denominator of the form x^{2} + bx + c. There are three possible cases, depending on how many real roots the quadratic has. Recall the quadratic formula used to solve ax^{2} + bx + c = 0, which when a = 1 takes the form r_{1}, r_{2} = (−b±\sqrt{b^{2} − 4c})/2. The sign of the discriminant b^{2} −4c tells us whether there is (a) 1 double root (when the discriminant is 0), (b) two distinct roots (when the discriminant is positive), or (c) no real roots (when the discriminant is negative). In the first case, the denominator takes the form (x − r)^{2}, and we can integrate the rational function by the method explained above (substituting u = x − r). The remainder of this section is concerned with the other two possibilities.

1. Quadratic Denominator That Factors

We first make an algebraic observation about what happens when we subtract as follows:

\frac{1}{x − r_{1}} − \frac{1}{x − r_{2}} = \frac{1}{x − r_{1}} · \frac{x − r_{2}}{x − r_{2}} − \frac{1}{x − r_{2}} · \frac{x − r_{1}}{x − r_{1}} = \frac{r_{1} − r_{2}}{(x − r_{1})(x − r_{2})}.

We shall use this in the reverse order: instead of combining 1/(x − r_{1}) and 1/(x − r_{2}) into a single fraction, we shall split up 1/(x−r_{1})(x−r_{2}) into a difference of two fractions. Namely, we divide through by r_{1} − r_{2} in the above equality:

\frac{1}{(x − r_{1})(x − r_{2})} = \frac{1}{r_{1} − r_{2}} (\frac{1}{x − r_{1}} − \frac{1}{x − r_{2}} ) .

This basic algebraic identity will allow us to split up our integral into two separate integrals, in each of which the denominator has the simpler form x − r. Here are some examples.

Example 14.2. Find

∫_{1}^{∞} \frac{dx}{2x^{2} + x}.

Since 2x^{2} + x = 2(x^{2} + 0.5) = 2(x − 0)(x − (−0.5)), we take r_{1} = 0, r_{2} = −0.5. Then

\frac{1}{2x^{2} + x} = \frac{1}{2} · \frac{1}{x^{2} + 0.5} = \frac{1}{2} · \frac{1}{0 − (−0.5)} (\frac{1}{x − 0} − \frac{1}{x − (−0.5)}) = (\frac{1}{x} − \frac{1}{x + 0.5}) .

Thus, our integral is

∫_{1}^{∞} (\frac{1}{x} − \frac{1}{x + 0.5}) dx = (ln x − ln(x + 0.5)) |_{1}^{∞} .

(We used the formula ∫ dx/(ax + b) = \frac{1}{a} ln |ax + b| + C, and we omitted the absolute value signs, because both x and x + 0.5 are positive between 1 and ∞.) To evaluate this antiderivative at ∞ means to take the limit of ln b − ln(b + 0.5) as b ⟶ ∞. Both ln b and ln(b + 0.5) approach ∞, so we have the indeterminate form ∞ − ∞. Write ln b − ln(b + 0.5) = ln (\frac{b}{b+0.5}). Notice that b/(b + 0.5) has limit 1 as b ⟶ ∞. Thus, its ln has limit ln 1 = 0. In other words, evaluating ln x − ln(x + 0.5) at ∞ gives 0. Thus, we conclude that ∫_{1}^{∞} dx/(2x^{2} + x) = −(ln 1 − ln 1.5) = ln 1.5.

Example 14.3. Find the x-coordinate of the center of mass of the uniform flat plate bounded by the x- and y-axes, the line x = 2, and the curve y = 1/(9 − x^{2}).

Here \overline{x} is the ratio of the integral ∫_{0}^{2} x dx/(9−x^{2}) to the integral ∫_{0}^{2} dx/(9−x^{2}). We first factor out the coefficient of x^{2}, which is −1, and write 9−x^{2} = −(x^{2} −9) = −(x+3)(x−3). Thus,

\frac{1}{9 − x^{2}} = −\frac{1}{(x − (−3))(x − 3)} = −\frac{1}{(−3) − 3} (\frac{1}{x − (−3)} − \frac{1}{x − 3})

= \frac{1}{6} (\frac{1}{x + 3} − \frac{1}{x − 3}) .

So the bottom integral is

\frac{1}{6} ∫_{0}^{2} (\frac{1}{x + 3} − \frac{1}{x − 3}) dx = \frac{1}{6} (ln(x + 3) − ln |x − 3|) |_{0}^{2} .

(We need absolute values because x − 3 is negative for x between 0 and 2.) The result is \frac{1}{6} ln 5.

The top integral is the same, except with an x, i.e.,

\frac{1}{6} ∫_{0}^{2} (\frac{x}{x + 3} − \frac{x}{x − 3}) dx,

and we could integrate each of these terms without difficulty. However, in this particular integral there happens to be an easier way: notice that in the original integral ∫_{0}^{2} x dx/(9 − x^{2}) the derivative of 9 − x^{2} is essentially what we have in the numerator. Hence, it is more efficient simply to substitute u = 9 − x^{2}, thereby transforming the top integral as follows (du = −2x dx, and the limits of integration become 9 − 0^{2} = 9 to 9 − 2^{2} = 5):

−\frac{1}{2} ∫_{9}^{5} \frac{du}{u} = −\frac{1}{2}(ln 5 − ln 9) = \frac{1}{2}(ln 9 − ln 5) = \frac{1}{2} ln 1.8.

Finally, \overline{x} = \frac{1}{2} ln 1.8 / \frac{1}{6} ln 5 = 3 ln 1.8/ ln 5 = 1.0956.

Example 14.4. Find the antiderivative of x^{4}/(x^{2} − x − 2).

If the numerator has degree 2 or more, then you can simplify the integral by first performing a long division of polynomials. In this case you get x^{4} divided by x^{2} − x − 2 is x^{2} + x + 3 with a remainder of 5x + 6. Thus, the integral is

∫ (x^{2} + x + 3 + \frac{5x + 6}{x^{2} − x − 2}) dx.

To handle the fraction, we factor x^{2}−x−2 = (x−2)(x+1). (If you can’t see the factorization by inspection, then use the quadratic formula to determine the roots, which in this case are 2 and −1.) Then

\frac{5x + 6}{x^{2} − x − 2} = (5x + 6) · \frac{1}{2 − (−1)} (\frac{1}{x − 2} − \frac{1}{x + 1}) = \frac{1}{3} (\frac{5x + 6}{x − 2} − \frac{5x + 6}{x + 1}) .

Thus, the problem boils down to integrating (5x + 6)/(x − 2) and (5x + 6)/(x + 1). These integrals are done by the method described at the beginning of this section. For example, substituting u = x − 2 in the first integral, we have:

∫ \frac{5x + 6}{x − 2} dx = ∫ \frac{5(u + 2) + 6}{u}du = ∫ (5 + \frac{16}{u}) du
= 5u + 16 ln |u| + C = 5(x − 2) + 16 ln |x − 2| + C.

The integral of (5x + 6)/(x + 1) is similar.

2. Quadratic Denominator That Does Not Factor

As a preliminary procedure, we first factor out the coefficient of x^{2}, so that the quadratic denominator has the form x^{2} + bx + c. If the numerator has degree 2 or greater, we then use long division of polynomials to write our rational function in the form

polynomial + \frac{αx + β}{x^{2} + bx + c}.

Next, we “complete the square” in the denominator: x^{2} + bx + c = (x + \frac{1}{2}b)^{2} + (c − \frac{1}{4}b^{2}),

and we set r = \sqrt{c − \frac{1}{4}b^{2}}, so that x^{2} + bx + c = (x + \frac{1}{2}b)^{2} + r^{2}. After a u-substitution of the form u = x + \frac{1}{2}b, we can transform the integral to a sum of two integrals of the form

∫ \frac{u du}{u^{2} + r^{2}} = \frac{1}{2} ln(u^{2} + r^{2}) + C,  and  ∫ \frac{du}{u^{2} + r^{2}} = \frac{1}{r}Arctan (\frac{u}{r}) + C.

We illustrate with two examples. In Example 14.5 we will already have the denominator in the form x^{2} + 1, so we do not make a u-substitution. In Example 14.6 we will have to do some algebra to get the integral in the two standard forms ∫ u du/(u^{2} + r^{2}) and ∫ du/(u^{2} + r^{2}).

Example 14.5. Integrating x^{n}Arctan x.

Applying integration by parts in accordance with the table in the last section (setting v′(x) = x^{n} and u(x) = Arctan x), we obtain

∫ x^{n}Arctan x dx = \frac{x^{n+1}}{n + 1}Arctan x − \frac{1}{n + 1} ∫ \frac{x^{n+1}}{1 + x^{2}} dx.

To evaluate the last integral, divide x^{2} + 1 into x^{n+1}, obtaining a polynomial quotient and a remainder. It turns out that the remainder will be x or −x if n is an even number, and 1 or −1 if n is an odd number. Thus, after integrating the polynomial quotient, you will be left with either ∫ x dx/(x^{2} + 1) = \frac{1}{2} ln(x^{2} + 1) + C, or else ∫ dx/(x^{2} + 1) =Arctan x + C.

Example 14.6. Find the antiderivative of x^{3}/(x^{2} + 2x + 5).

First, x^{3} divided by x^{2} + 2x + 5 is x − 2 with a remainder of −x + 10. Next, completing the square gives x^{2} + 2x + 5 = (x + 1)^{2} + 4 = (x + 1)^{2} + 2^{2}, and so we substitute u = x + 1 (then x = u − 1, dx = du). Thus,

∫ \frac{x^{3}dx}{x^{2} + 2x + 5} = ∫ (x − 2 + \frac{−x + 10}{x^{2} + 2x + 5}) dx

= \frac{1}{2}x^{2} − 2x + ∫ \frac{−x + 10}{(x + 1)^{2} + 4} dx

= \frac{1}{2}x^{2} − 2x + ∫ \frac{−(u − 1) + 10}{u^{2} + 4} du

= \frac{1}{2}x^{2} − 2x − ∫ \frac{u du}{u^{2} + 4} + 11 ∫ \frac{du}{u^{2} + 4}

= \frac{1}{2}x^{2} − 2x − \frac{1}{2} ln(u^{2} + 4) + \frac{11}{2}Arctan (\frac{u}{2}) + C

= \frac{1}{2}x^{2} − 2x − \frac{1}{2} ln(x^{2} + 2x + 5) + \frac{11}{2}Arctan (\frac{x + 1}{2}) + C.

Homework 13.

1. Find

(a) ∫_{0}^{1} dx/(4 − x^{2}); (e) ∫_{0}^{1} x^{4}dx/(4 − x^{2});  (h) ∫_{0}^{∞} dx/(x^{2} + 10x + 25);
(b) ∫_{0}^{1} x^{2}dx/(4 − x^{2}); (f) ∫_{0}^{1} x^{4}dx/(4 + x^{2});  (i) ∫_{0}^{∞} dx/(x^{2} + 10x + 29);
(c) ∫_{0}^{1} x^{3}dx/(4 + x^{2}); (g) ∫_{0}^{∞} dx/(x^{2} + 10x + 21); (j) ∫_{0}^{1} dx/(2x^{2} − x − 3).

(d) ∫_{1}^{∞} dx/(x^{2} + 3x);

2. Find the center of mass of a uniform flat plate bounded by:

(a) the lines x = 0 and x = 2 and the curves y = 1/(x^{2} + 3x + 2) and y = −1/(x^{2} + 3x + 2);

(b) the line x = 1, the curve y = x^{2}Arctan x, and the curve y = −x^{2} arctan x.

3. Find the anti-derivative of sec x = (cos x)^{−1} by the method used before for odd powers of cos x. That is, write (cos x)^{−1} = (cos x)^{−2} cos x = cos x/(1 − sin^{2} x). After making the appropriate u-substitution, you will have a rational function to integrate.

90 14. TECHNIQUES OF INTEGRATION III

CHAPTER 15 Techniques of Integration IV: Inverse Trig Substitution

The technique of “ inverse trig substitution” is used when you want to integrate a function involving the square root of a quadratic polynomial, such as \sqrt{25 − x^{2}}, \sqrt{x^{2} − 2x}, \sqrt{x^{2} + 2x + 2}, etc. It is an “inverse” substitution because, instead of writing a new variable u as a function of x (i.e., u = u(x)), you write x in terms of a new variable θ (i.e., x = x(θ)), where the new variable θ stands for an angle in a right triangle. Here is the procedure:

1. Take the quadratic which is under the radical and write it either as a sum of squares A^{2} + B^{2} or as a difference of squares A^{2} − B^{2}, where one of the terms A or B is a constant and the other is either x or a multiple of x (or at worst an expression of the form ax + b). In the three examples above, we select A and B as follows:

25 − x^{2} = 5^{2} − x^{2},  x^{2} − 2x = (x − 1)^{2} − 1^{2},  x^{2} + 2x + 2 = (x + 1)^{2} + 1^{2}.

2. Draw a right triangle, and label two of the sides A and B. If the quadratic under the radical is A^{2} + B^{2}, then take the two sides to be the legs of the right triangle. If the quadratic is A^{2} − B^{2}, then take A to be the hypotenuse and B to be one of the legs. In this way the radical itself (i.e., \sqrt{A^{2} + B^{2}} or else \sqrt{A^{2} − B^{2}}) will be the third side.

3. Let θ be one of the acute angles of the right triangle (either one will do). Then write everything in the integral — x, dx, and the square root — in terms of the new variable θ (also change the limits of integration, if you’re evaluating a definite integral). Hopefully, the new integral with θ as the variable of integration will be relatively easy to evaluate.

We now give some examples of the use of this method.

Example 15.1. Find the area in the first quadrant that is bounded by the x- and y-axes, the line x = 3, and the circle of radius 6 centered at the origin.

To evaluate ∫_{0}^{3} \sqrt{36 − x^{2}}dx, we go through

the three steps above. (1) Set A = 6 and B = x, so that the radical is \sqrt{A^{2} − B^{2}}. (2) Draw a right triangle with the sides and angle labeled as shown (see the figure at the right)

(3) Write x = 6 sin θ, so that dx = 6 cos θ dθ. In addition, the square root is the adjacent side, so that \sqrt{36 − x^{2}} = 6 cos θ. Since θ = arcsin(x/6), the limits of integration become arcsin(0) = 0 and arcsin(3/6) = π/6. Thus,

∫_{0}^{3} \sqrt{36 − x^{2}} dx = ∫_{0}^{π/6}(6 cos θ)6 cos θ dθ = 36 ∫_{0}^{π/6} cos^{2} θ dθ

= 36 ∫_{0}^{π/6} \frac{1 + cos(2θ)}{2} dθ = 36 (\frac{1}{2}θ + \frac{1}{4} sin(2θ)) |_{0}^{π/6}

= 3π + 9 sin(\frac{π}{3} ) = 3π + \frac{9}{2}\sqrt{3}.

We could have actually obtained the same result by some simple geometry. The region in question is made up of a 30^{∘} sector of a circle (i.e., 1/12 of a circle of radius 6) and a triangle of base 3 and altitude 3\sqrt{3} (see the figure at the right). So the area is

\frac{1}{12}π · 6^{2} + \frac{1}{2}3 · 3\sqrt{3} = 3π + \frac{9}{2}\sqrt{3} .

30^{∘}

Example 15.2. Find the center of mass of a uniform flat plate in the shape of one hump of the sine curve.

By considerations of symmetry, \overline{x} is clearly π/2. To find \overline{y}, we must find the distance across from left to right at height y. From the diagram at the right, it should be clear that this distance is π − 2 arcsin(y). The bottom integral in \overline{y} is just the area under y = sin x from 0 to π, which is

1 y  x arcsin y  arcsin y π

2. So it remains to compute the top integral

∫_{0}^{1} y(π − 2 arcsin y) dy = \frac{π}{2} − 2 ∫_{0}^{1} y arcsin y dy .

Using integration by parts, we transform the last integral as follows:

∫_{0}^{1} y arcsin y dy = \frac{1}{2}y^{2} arcsin y |_{0}^{1} − ∫_{0}^{1} \frac{1}{2}y^{2} \frac{1}{\sqrt{1 − y^{2}}} dy = \frac{π}{4} − \frac{1}{2} ∫_{0}^{1} \frac{y^{2}dy}{\sqrt{1 − y^{2}}} .

We now use an inverse trig substitution, drawing our triangle so that 1 is the hypotenuse and y is the leg opposite the angle θ. We have y = sin θ, dy = cos θ dθ, \sqrt{1 − y^{2}} = cos θ, and the limits of integration become 0 to π/2. Putting this all together, we find that the top integral in \overline{y} is equal to

\frac{π}{2} − 2 ∫_{0}^{1} arcsin y dy = \frac{π}{2} − 2(\frac{π}{4} − \frac{1}{2} ∫_{0}^{1}(y^{2}dy/\sqrt{1 − y^{2}}))
= ∫_{0}^{1}(y^{2}dy/\sqrt{1 − y^{2}}),

which with the substitution y = sin θ becomes

∫_{0}^{π/2} \frac{sin^{2} θ cos θ dθ}{cos θ} = ∫_{0}^{π/2} sin^{2} θ dθ

= ∫_{0}^{π/2} \frac{1 − cos(2θ)}{2}dθ = (\frac{1}{2}θ − \frac{1}{4} sin(2θ)) |_{0}^{π/2}= π/4 .

We conclude that \overline{y} = π/8 = 0.3927.

Using this type of procedure — integration by parts followed by inverse trig substitution — one can integrate x^{n} arcsin x for any positive integer n.

Example 15.3. Find the antiderivative of 1/\sqrt{x^{2} − 2x}.

As shown above, we can write x^{2} − 2x as a difference of two squares:

x^{2} − 2x = (x − 1)^{2} − 1^{2} .

So we choose the hypotenuse of the right triangle to be x − 1 and the leg opposite θ to be 1 (see the drawing at the right). Then sin θ = 1/(x − 1) and so x = 1+(1/ sin θ), dx = −(cos θ/ sin^{2} θ) dθ, and the radical is cot θ = cos θ/ sin θ. Hence,

∫ \frac{dx}{\sqrt{x^{2} − 2x}} = ∫ \frac{−(cos θ/ sin^{2} θ)dθ}{cos θ/ sin θ} = − ∫ \frac{dθ}{sin θ} .

This integral is very similar to Problem 3 in the last homework set (just reverse the roles of sin and cos). The result (taking into account the minus sign in front of the integral) is: \frac{1}{2} ln \frac{1+cos θ}{1−cos θ} + C. Finally, we can write the answer in terms of x by reading cos θ off the triangle: cos θ = \sqrt{x^{2} − 2x}/(x−1). (Note: The answer can be simplified algebraically, using the properties of ln; the result is: ln |x − 1 + \sqrt{x^{2} − 2x}| + C.)

Example 15.4. Find ∫ x dx/\sqrt{x^{2} + 2x + 2}.

As mentioned earlier, we can write x^{2} + 2x + 2 as a sum of two squares: x^{2} + 2x + 2 = (x + 1)^{2} + 1^{2}. So we choose x + 1 to be one leg (say, the leg opposite θ) and 1 to be the other leg of the right triangle. Then x + 1 = tan θ, and so x = tan θ − 1, dx = sec^{2} θ dθ. In addition, \sqrt{x^{2} + 2x + 2} is the hypotenuse of the triangle, which is sec θ. So our integral becomes

∫ \frac{(tan θ − 1) sec^{2} θ dθ}{sec θ} = ∫ tan θ sec θ dθ − ∫ sec θ dθ.

The function inside the first integral happens to be the derivative of sec θ. (If you didn’t notice this, you could convert the integral to sines and cosines, i.e., to ∫ sin θ dθ/ cos^{2} θ, and use the substitution u = cos θ.) The second integral is Problem 3 of the last homework set. Putting this together (and using the fact that sec θ = \sqrt{x^{2} + 2x + 2} and sin θ = (x + 1)/\sqrt{x^{2} + 2x + 2}), we obtain the following answer:

∫ \frac{x dx}{\sqrt{x^{2} + 2x + 2}} = \sqrt{x^{2} + 2x + 2} + \frac{1}{2} ln \frac{\sqrt{x^{2} + 2x + 2} + x + 1}{\sqrt{x^{2} + 2x + 2} − (x + 1)} + C.

Example 15.5. (Force of repulsion from a line source of electric charge.)

Suppose that we have a long straight wire with an electric charge uniformly distributed along it. Our purpose is to determine how the force of repulsion exerted on a similarly charged particle drops off as the particle moves away from the wire.

The basic physical law needed here is: the force of repulsion between two particles of charge q_{1} and q_{2} has magnitude k q_{1}q_{2}/distance^{2}, where k is a universal constant. Before doing this example, we should review Example 9.7 in the section on improper integrals (noise from the freeway). There also we had an inverse square law. As in the noise example, we set up our wire as the x-axis, taking 0 to be the point at the foot of the perpendicular drawn from the charged particle. If c stands for the amount of charge per unit distance along the wire, then the contribution to the force of repulsion coming from the small segment of wire between x and x + Δx is k q_{1}(cΔx)/distance^{2} = KΔx/(r^{2} + x^{2}), where we have set K = kq_{1}c. (See the diagram in the freeway example.)

However, there is one basic difference between the two examples. In the freeway problem we were dealing with a scalar (amount of noise) rather than force, which is a vector (it has both magnitude and direction). This means that the force contributed by the segment of the wire between x and x + Δx points at an angle away from the vertical. What is that angle? You should convince yourself that the angle is arctan(x/r). The second diagram to the right below shows the force vector contributed by the part of the wire between x and x + Δx.

The horizontal component of this force (i.e., the component parallel to the wire) is canceled by the force contributed from the opposite segment (the segment between −x and −x − Δx). So the contribution to the force of repulsion (the component pointing away from the wire) is only a portion of KΔx/(r^{2} + x^{2}). The portion can be determined by similar triangles (see the diagram at the right). Comparing the two triangles, we find that the side “?” which we want to find has the same ratio to KΔx/(r^{2} + x^{2}) as r does to \sqrt{r^{2} + x^{2}}. Thus, the contribution of the small segment is equal to:

\frac{r}{\sqrt{r^{2} + x^{2}}} · \frac{KΔx}{r^{2} + x^{2}} = \frac{KrΔx}{(r^{2} + x^{2})^{3/2}} . x r  ^{\sqrt{r^{2} + x^{2}}} \frac{KΔx}{r^{2} + x^{2}}  ?

Proceeding as in the freeway example and making the simplifying assumption that the wire extends from −∞ to ∞, we obtain the following definite integral for the force of repulsion: ∫_{−∞}^{∞} Kr dx/ (r^{2} + x^{2})^{3/2}. This is an integral that can be evaluated by an inverse trig substitution, taking x and r to be the legs of the right triangle (say, x is the leg opposite θ). Then x = r tan θ, dx = r sec^{2} θ dθ, \sqrt{r^{2} + x^{2}} = r sec θ. The limits of integration become −π/2 to π/2, so we obtain:

∫_{−∞}^{∞} \frac{Kr dx}{(\sqrt{r^{2} + x^{2}})^{3}} = ∫_{−π/2}^{π/2} \frac{Kr^{2} sec^{2} θ dθ}{(r sec θ)^{3}} = \frac{K}{r} ∫_{−π/2}^{π/2} cos θ dθ = \frac{2K}{r} .

Hence we conclude (as in the freeway example) that the force drops off proportionally to the distance (not the square of the distance).

1. A Final Word

Before leaving the subject of techniques of integration, we should point out that sometimes you have a choice of techniques, and your answer using one technique might at first glance seem to be different from your answer if you use a different technique.

Example 15.6. Find ∫ sin x cos x dx.

First method. Substitute u = sin x, so that du = cos x dx. We obtain: ∫ u du = \frac{1}{2}u^{2} + C = \frac{1}{2} sin^{2} x + C.

Another method. Substitute u = cos x, so that du = − sin x dx. Then we get: − ∫ u du = −\frac{1}{2}u^{2} + C = −\frac{1}{2} cos^{2} x + C.

A third method. Use the trig identity for the sine of twice an angle: sin(2x) = 2 sin x cos x. We can thus rewrite our integral as ∫ \frac{1}{2} sin(2x)dx = −\frac{1}{4} cos(2x) + C.

Why three different answers? The point is that the three answers are not really different. Namely, the second answer can be written −\frac{1}{2}(1−sin^{2} x)+C = \frac{1}{2} sin^{2} x+C− \frac{1}{2}; and the third answer (if we use the cosine double angle formula cos(2x) = cos^{2} x − sin^{2} x = 1 − 2 sin^{2} x) can be written as follows: −\frac{1}{4}(1 − 2 sin^{2} x) + C = \frac{1}{2} sin^{2} x + C − \frac{1}{4}. So the second and third answers differ from the first only by constants. But remember that the meaning of “+C” in the answer to an indefinite integral is that if you add any constant to the answer, you still have an antiderivative, i.e., the new function is still a correct answer.

Homework 14.

1. Use inverse trig substitution to find ∫ dx/\sqrt{r^{2} − x^{2}}, where r is a constant (we evaluated this earlier using a u-substitution).

2. Find the anti-derivative of (a) 1/\sqrt{x^{2} + r^{2}},

(b) 1/\sqrt{x^{2} − r^{2}}.

3. Find the center of mass of the uniform flat plate bounded by the curve y = x arcsin x, the curve y = −x arcsin x, and the line x = 0.5.

4. Find the x-coordinate of the center of mass of the uniform flat plate bounded by the xand y-axes, the line x = 2, and the curve y = 1/\sqrt{x^{2} + 6x + 13}.

5. Find ∫_{0}^{∞} dx/ ((x + 2)\sqrt{x^{2} + 4x + 5}).

CHAPTER 16 Differential Equations I: Separation of Variables

The differential equations in this section will be of a slightly more complicated form than in the section on exponential growth and decay, where we treated the differential equation y′ = ±ky (+k leading to exponential growth, and −k leading to exponential decay). Namely, we shall deal with differential equations of the form y′ = ±k(y−b), where b is some constant.

Before discussing the applications, let’s solve the differential equation \frac{dy}{dt} = k(y − b) by the same method — separation of variables — that we used before. We first multiply both sides by dt and divide through by y − b, and we then put an integral sign in front of both sides:

∫ \frac{dy}{y − b} = ∫ k dt
ln |y − b| + C_{1} = kt + C_{2}
ln |y − b| = kt + C (where C = C_{2} − C_{1})
|y − b| = e^{kt+C} = e^{C}e^{kt}
y − b = ±e^{C}e^{kt}.

To find the constant ±e^{C} in front of e^{kt}, we suppose that we have the initial information y(0) = y_{0}; then substituting t = 0 on both sides gives us: y_{0} − b = (±e^{C}) e^{0} = ±e^{C}, and so we can write y − b = (y_{0} − b)e^{kt}, i.e.,

The differential equation \frac{dy}{dt} = k(y − b) has solution y(t) = b + (y_{0} − b)e^{kt}.

If our differential equation is of the form \frac{dy}{dt} = −k(y − b) with a negative constant of proportionality −k, then the solution is exactly the same with k replaced by −k. Thus,

The differential equation \frac{dy}{dt} = −k(y − b) has solution y(t) = b + (y_{0} − b)e^{−kt}.

1. Installment Buying

Large purchases (cars, houses, furniture, etc.) are often made according to the following system. When you make the purchase (time t = 0), you pay only a small fraction of the price (called a “downpayment”) and agree to pay the remaining amount — which we shall denote y_{0} — over a certain period of time. In essence, you are borrowing the y_{0} dollars from the seller (or a mortgage company in the case of a house). We shall let T denote the term of this loan (in years), i.e., the time during which you continue to make payments, and we shall let a denote the rate in dollars per year at which you pay money during this time T. (If you make monthly payments, then a equals 12 times the amount you pay each month.) If the seller is willing to loan you the y_{0} interest free, then the situation is very simple: a = y_{0}/T. But in practice you must pay back not only y_{0} to make up the purchase price, but also interest on the money you owe. Let k be the instantaneous rate of interest.

In order to treat this problem by a differential equation, we shall make two simplifying assumptions:

1. The interest is compounded continuously. (This is the same simplifying assumption that we made in the section on exponential growth.)

2. Your payments are being made continuously (rather than at the end of the month, as usually happens in practice). For example, in the case of monthly payments of $300, this would mean, roughly speaking, that instead of waiting until the end of the month, you arrange to have your bank’s computer transfer one penny from your account to the seller’s account every 86.4 seconds.

Even though both of these simplifying assumptions seem to diminish the realism of our treatment of installment buying, in practice the inaccuracy introduced is very small — an error of considerably less than 1%.

From a practical point of view, perhaps the most useful thing to figure out is the monthly payment a/12, which depends on y_{0} (which is purchase price minus downpayment), the term T, and the interest rate k. But there are other questions which might also interest us. In any case, the way to approach the matter is to introduce the function

y(t) = the amount you still owe at time t.

This amount y(t) starts out at y(0) = y_{0}, and steadily decreases until at time t = T you’ve paid off the loan, i.e., y(T) = 0. This function y(t) satisfies a differential equation, which we first state in words: the instantaneous rate of change of the amount y(t) that you owe is the result of two contributions: (1) a positive contribution caused by interest, which accumulates at a rate equal to k times the current amount you owe, and (2) a negative contribution of a dollars/year, as you make payments to decrease your debt.

Written mathematically, the first contribution is ky and the second contribution is −a. Thus, the function y(t) satifies the differential equation

\frac{dy}{dt} = ky − a.

If we rewrite the right side in the form k(y − \frac{a}{k}), we see that this is the same type of differential equation that we already solved at the beginning of the section. We can simply take the solution we already found, where we substitute b = a/k. The result is:

y(t) = \frac{a}{k} + (y_{0} − \frac{a}{k}) e^{kt}.

We now want to interpret this result to obtain useful information. First, let us find the rate of payment a (or the monthly payment rate a/12) as a function of y_{0}, T, and k. The basic fact we use is that the amount you owe must be reduced to 0 when t = T. That is,

0 = y(T) = \frac{a}{k} + (y_{0} − \frac{a}{k}) e^{kT} .

To solve this for a, we do a little algebra. We start out by dividing through by e^{kT}, to make things look slightly simpler (no parentheses needed):

0 = \frac{a}{k}e^{−kT} + y_{0} − \frac{a}{k}.

(Recall that the reciprocal of e^{kT} is e^{−kT}.) Since a is the unknown, we move all the terms with a over to the left, and factor out a:

\frac{a}{k} (1 − e^{−kT} ) = y_{0}.

Now it is easy to solve, by multiplying both sides by k and dividing both sides by 1 − e^{−kT}:

a = \frac{ky_{0}}{1 − e^{−kT}} .

Example 16.1. A few years ago the State of Washington had a 12% ceiling on interest rates, while neighboring Oregon had no such limit. During this time an ad for a $450 dinette set running in Pacific Northwest newspapers said: no money down; residents of Washington take 32 months to pay at 12% interest; residents of Oregon make the same monthly payments as in Washington, although their interest rate will be 18%. Find the monthly payment that residents of both states make, and also find out how much beyond the 32 months must Oregonians continue to make payments because of their higher interest rate.

In Washington, T = 32/12 years, k = 0.12, and y_{0} = 450. Hence,

\frac{a}{12} = \frac{1}{12} · 0.12 · 450 / (1 − e^{−0.12·32/12}) = $16.43.

In Oregon, we now know the value of a (namely, 12 · 16.4323 =$197.19), and we want to find T. So we return to the equation y(T) = 0 above, and instead of solving for a we solve for T. Moving the term with T over to the left, we have

(\frac{a}{k} − y_{0}) e^{kT} = \frac{A}{k}.

We divide both sides by ((a/k) − y_{0}), simplify the fraction on the right by multiplying by k/k, and then take the ln of both sides. The result is

kT = ln (\frac{a}{a − ky_{0}} ) , i.e., T = \frac{1}{k} ln (\frac{a}{a − ky_{0}} ) .

Substituting k = 0.18, a = 197.19, and y_{0} = 450, we obtain: T = 2.9386 years = 35.26 months. That is, Oregonians must continue paying for about 3 1/4 months more.

The next example deals with home mortgages. By a “mortgage” we shall always mean a fixed-rate, fixed-payment mortgage, i.e., the standard kind. Adjustable-rate mortgages and other variants will not be examined here.

Example 16.2. To buy a house for $71,500.00, you make a downpayment of 20% and take a 30-year mortgage for the remaining 80% at 11% interest. Find your monthly payments a/12, your yearly rate of payment a, and the total amount you end up paying for the house. Also compute how much of the money paid in the first year is tax-deductible, and how much of the money paid in the 30th year is tax-deductible. (Note: money paid toward interest is deductible from U.S. federal income tax, while money that goes toward reducing the amount you owe is not.)

Here y_{0} = 0.8 · 71500 = 57200, and so

a = \frac{0.11 · 57200}{1 − e^{−0.11·30}} = $6532.96; \frac{a}{12} = $544.41.

The amount you end up paying for the house is equal to the downpayment plus aT, i.e., $14300 + 30 · $6532.96 = $210289, which is about three times the price of the house! (This isn’t as bad as it seems, for at least two reasons: (1) a significant part of the $138789 that goes for interest is being paid by the government, in the form of reduced income tax, and (2) the dollars that you pay 20 or 30 years later are dollars whose real value has been reduced by inflation.) Next, we want to find out how much of the $6532.96 paid between t = 0 and t = 1 is tax-deductible, and how much of the $6532.96 paid between t = 29 and t = 30 is tax-deductible. The simplest way to do this is to ask how much is not tax-deductible, and that is the amount which has gone to decrease your indebtedness. Thus, the amount in the first year that is not tax-deductible is

y(0) − y(1) = (\frac{a}{k} + (y_{0} − \frac{a}{k}) e^{k·0}) − (\frac{a}{k} + (y_{0} − \frac{a}{k}) e^{k·1})

= (y_{0} − \frac{a}{k}) (1 − e^{k}) = (\frac{a}{k} − y_{0}) (e^{k} − 1)

= (\frac{6532.96}{0.11} − 57200) (e^{0.11} − 1) = $254.71.

The remaining 6532.96−254.71 = $6278.25 is tax-deductible. In the 30th year, on the other hand, the non-tax-deductible amount is y(29) − y(30) = y(29) − 0 = y(29) = $6185.64, and so only $347.32 is deductible in the last year.

What is happening is that over the term of the mortgage the way your a dollars per year payment divides up is changing. At the beginning, when you still owe a lot of money, most of the a goes toward interest payment, and only very little goes toward reducing the principal. As time goes on, more and more of the a goes toward reducing the principal, and less and less of it goes toward interest payment. This can be represented visually by a graph (see the figure at the right) that

t_{1} t_{2}  T a ky_{0}  k y(t)

depicts what our original differential equation is saying. The graph shows the horizontal line a along with the curve ky(t). The differential equation says that at any instant, of the a dollars/year that you pay, ky(t) of it must go toward interest, and the remaining a − ky goes toward reducing the amount you owe. In symbols, dy/dt = −(a − ky) (the minus sign appears because we are decreasing y). Over any time interval from t_{1} to t_{2}, the amount you pay is a(t_{2} − t_{1}), which is the area under the horizontal line from t_{1} to t_{2}. Of that area, the area under the curve ky(t) is the amount that goes toward interest (the tax-deductible amount), and the area between the curve and the horizontal line is the amount that goes toward paying off the principal.

Recall that we made two simplifying assumptions at the very beginning. Since Example 16.2 is based on an actual purchase, we can show the actual values computed by the mortgage company, based on daily compounding and payments made at the end of each month. Our error turns out to be quite small: the true monthly payments for this mortgage were $545.12. We were off by $0.71, which is only 0.13% of the total.

Below we show the table summarizing a statement sent us by the mortgage company for the 8th year. This statement gives the same type of information as the above graph between t = 7 and t = 8. The first column has the constant value a/12 that is paid each month. The fourth column shows how much of this has gone toward the principal, and the sixth column shows the difference, which has gone toward interest. The sum of the amounts in the sixth column is the “interest total” given at the bottom — the amount that is tax-deductible.

   ANNUAL STATEMENT ON YOUR MORTAGAGE LOAN #5-03-00016965 INTEREST RATE 11.000%

    TOTAL     DATE     INTEREST      PRINCIPAL ACCOUNT     INTEREST
   PAYMEMT    PAID      PAID TO    Transactions   Balance   PAYMENT
            01-01-87                            54,542.50 ←BEGINNING BALANCE
    545.12  01-12-87   01-01-87         45.15-  54,497.35   499.97
    545.12  02-11-87   02-01-87         45.56-  54,451.79   499.56
    545.12  03-11-87   03-01-87         45.98-  54,405.81   499.14
    545.12  04-13-87   04-01-87         46.40-  54,359.41   498.72
    545.12  05-11-87   05-01-87         46.83-  54,312.58   498.29
    545.12  06-11-87   06-01-87         47.25-  54,265.33   497.87
    545.12  07-13-87   07-01-87         47.69-  54,217.64   497.43
    545.12  08-11-87   08-01-87         48.12-  54,169.52   497.00
    545.12  09-11-87   09-01-87         48.57-  54,120.95   496.55
    545.12  10-13-87   10-01-87         49.01-  54,071.94   496.11
    545.12  11-12-87   11-01-87         49.46-  54,022.48   495.66
    545.12  12-11-87   12-01-87         49.91-  53,972.57   495.21

   TOTAL AS OF DECEMBER 31, 1987                           5971.51

2. Newton’s Law of Cooling and Heating

Newton’s law of cooling states that if an object is brought into an environment, then the rate of cooling (or heating, if the object is colder than the environment) is proportional to the temperature difference between the object and the environment. Let y(t) be the temperature of the object at time t, let y_{0} = y(0) be the initial temperature of the object, and let b be the temperature of the environment. Then Newton’s law can be translated into the mathematical equation dy/dt = −k(y − b). Here k is the constant of proportionality, which in practice depends on the thermal properties of the object in the particular case at hand. k is positive, and the minus sign is necessary in front of k because dy/dt is negative when the object’s temperature y(t) is greater than b, and dy/dt is positive when y(t) is less than b. This differential equation is of the form that was solved at the beginning of the section. Thus, we have y(t) = b + (y_{0} − b)e^{−kt}. Notice that as t gets larger, the exponential term in this function gets smaller and smaller, so that y(t) approaches b, i.e., the temperature of the object gets closer and closer to that of the environment. (In this section we are assuming that the ambient temperature b is constant, i.e., that it is not heated up by the object and does not change with the changing time of day. In the final section on differential equations, we will treat an example where b changes with time.)

Example 16.3. A cup of coffee at 200^{∘}F is brought out into a room which is kept at 60^{∘}F. Two minutes later you measure the coffee’s temperature to be 180^{∘}. Find a formula for the coffee’s temperature at any time t.

Here we have y_{0} = 200 and b = 60, but we do not know what k is. In the formula for y(t), namely b + (y_{0} − b)e^{−kt} = 60 + 140e^{−kt}, we can find k by using the information given about y(2):

180 = y(2) = 60 + 140e^{−k·2}
120 = 140e^{−2k}
ln(120/140) = −2k
k = −\frac{1}{2} ln(120/140) = \frac{1}{2} ln(7/6) = 0.077.

Thus, y(t) = 60 + 140e^{−0.077t}.

3. Falling Bodies with Air Resistance

Now we shall find a formula for the velocity v(t) of a falling body, no longer assuming that air resistance is negligible.

The vertical motion of a falling object is affected by two things: gravity and air resistance. Near the surface of the earth, gravity contributes a constant acceleration g. (Let us take \underline{downward} to be the positive direction, in which case gravitational acceleration is g rather than −g.) The air resistance depends on the aerodynamic properties of the object and also on its speed. Generally, the faster the object is moving, the greater the force of air molecules pushing against it. In this section we shall assume that the force of air resistance is proportional to the velocity. Since acceleration is proportional to force, this also means that the acceleration (actually, deceleration) from air resistance is assumed to be proportional to v(t). Thus, there are two contributions to dv/dt: g and −kv(t) (here k is positive, and there’s a minus sign in front of k because the air resistance works in the opposite direction to velocity). In other words, the unknown function v(t) satisfies the differential equation dv/dt = g − kv, which we rewrite in the form

dv/dt = −k (v − \frac{g}{k} ) .

We see that this is in the form that was solved at the beginning of the section (with v instead of y and g/k in place of b). So our solution is the velocity function

v(t) = \frac{g}{k} + (v_{0} − \frac{g}{k} ) e^{−kt}.

Notice that as the time increases, the exponential term in the formula for v(t) approaches zero, and so the velocity approaches the value g/k. We call g/k the terminal velocity and denote it v_{∞}. The terminal velocity does not depend on the initial velocity: no matter what initial velocity we give to the object, eventually its velocity will become essentially equal to the value g/k (in the sense that the difference between v(t) and g/k cannot be detected in practice). In other words, the graph of v(t) has a horizontal asymptote at height g/k. Another way to find the terminal velocity is to ask when the two forces on the object — gravity and air resistance — balance, i.e., when the right side of the differential equation equals zero: g − kv = 0. This happens when v = g/k = v_{∞}.

Example 16.4. A parachutist drops from a plane at time t = 0. She freefalls for 10 seconds, during which time the constant of air resistance is k = 0.1. After her parachute opens, the air resistance becomes much greater: k = 1.4. Take downward to be the positive direction, and take g = 9.8 m/sec^{2}. Find formulas for her velocity for 0 ≤ t ≤ 20, and draw a graph of this velocity function. Since the parachutist dropped from the plane, rather than throwing herself downward, the initial velocity at time t = 0 is 0. Thus, for the first 10 seconds we have v(t) = \frac{g}{k} (1 − e^{−kt}) = 98 (1 − e^{−0.1t}) . Starting at t = 10 we have a new situation. The new initial velocity is v(10) = 98 (1 − e^{−0.1·10}) = 61.95 m/sec, and the new g/k is 9.8/1.4 = 7.0. In using our falling body formula, we must be careful, since the starting time is now 10 rather than 0. Thus, t − 10 is the time elapsed since the parachute opened, and so we use t − 10 in place of t in the formula. The result is:

v(t) = \begin{cases}98 (1 − e^{−0.1t}) , & for 0 ≤ t ≤ 10; \\ 7 + 54.95e^{−1.4(t−10)}, & for 10 ≤ t ≤ 20. \end{cases}

A graph of the velocity function is shown above.

4. Mixing Problems

Suppose we have a container which holds a certain volume V of water and at time t = 0 has a certain amount y_{0} of dissolved impurity (such as salt, pollutant, etc.). A pipe carries water into the container at a rate r, and another pipe carries water out of the container at the same rate r (so that the volume V of water in the container remains the same). The incoming water has a concentration k of impurity (measured in gm/liter, lb/ft^{3}, gm/m^{3}, etc.). Let y(t) be the amount of impurity in the water in the container at time t. We make the assumption that the impurity brought into the water by the first pipe immediately spreads uniformly throughout the container, so that the water in the container always has the same concentration of impurity throughout (i.e., there is no pileup of impurity near the mouth of the pipe). As a result, the concentration of impurity in the water taken up by the exit pipe is equal to the overall concentration in the container, namely, amount of impurity per volume = y(t)/V.

We are now ready to set up a differential equation for y(t). The basic principal is that the rate at which impurity enters the container is equal to the concentration of impurity (amount per volume) times the rate of the incoming water (volume per time), and the rate at which impurity leaves the container is equal to the concentration of impurity in the exit pipe times the rate of the water going out. Thus, dy/dt has a positive contribution of kr and a negative contribution of y/V times r, i.e.,

\frac{dy}{dt} = kr − \frac{y}{V} r = kV \frac{r}{V} − \frac{r}{V} y = −\frac{r}{V} (y − kV).

This differential equation is of the form that was solved at the beginning of the section, except that the k here has nothing to do with the −k that we had earlier. Now it is r/V that is playing the role of k, and kV that is playing the role of b. So we have the solution

y(t) = kV + (y_{0} − kV)e^{−rt/V} .

It is also sometimes useful to have a formula for the concentration in the container at time t. This is y(t)/V, which we get by dividing the above formula by V:

\frac{y(t)}{V} = k + (\frac{y_{0}}{V} − k) e^{−rt/V} .

Notice that as t ⟶ ∞, the concentration of impurity in the container approaches k, which is the concentration in the incoming water. This makes sense: roughly speaking, the incoming water with its concentration k eventually “displaces” the water that was originally there.

The special case k = 0. If k = 0, then our formula for y(t) becomes y(t) = y_{0}e^{−rt/V}, which is an exponential decay function. (In this case our original differential equation was simply dy/dt = −\frac{r}{V} y.) Thus, if you remove impurities from the water by pumping in pure water, you find that the amount of impurity decreases exponentially, so that eventually the amount of impurity is too small to be measured. What is the “half-life” of the impurities, i.e., the length of time required to cut the amount of impurity in half? We use the formula H = (ln 2)/k that we derived before, where k is replaced by r/V. The result is H = (ln 2)V/r.

This concludes our discussion of four applications of differential equations of the form y′ = ±k(y − b). We now make some final observations.

5. The Constant Solution

There is one and only one constant function y(t) =const that satisfies a differential equation of the form dy/dt = ±k(y − b). Namely, since a constant function has derivative zero, it follows that a constant function satisfies the differential equation if and only if 0 = ±k(y − b), i.e., if and only if y(t) = b. In all four of the practical applications discussed above, the constant solution has a special meaning.

Installment buying. The constant solution y(t) = a/k means that you have chosen your yearly rate of payment a to equal ky_{0} (the interest rate times the amount you borrowed). That is just enough to pay the interest, without diminishing the amount you owe. This constant solution separates the two cases: (1) a > ky_{0}, in which case the amount you owe y(t) = \frac{a}{k} +(y_{0} − \frac{a}{k}) e^{kt} = \frac{a}{k} −(\frac{a}{k} − y_{0}) e^{kt} decreases and becomes zero in a finite amount of time T; and (2) a < ky_{0}, in which case you are not even paying enough to take care of the interest, and the amount you owe grows without bound.

Law of cooling/heating. The constant solution y(t) = b means that you have brought an object into the room whose temperature is the same as that of the room, in which case, of course, the temperature of the object remains constant. This constant solution separates the two cases: (1) y_{0} > b, in which case you are bringing into the room a hotter object, which then cools, approaching the ambient temperature b as t ⟶ ∞; and (2) y_{0} < b, in which case you have a colder object, which heats up, approaching the ambient temperature as t ⟶ ∞. Note that in all cases the constant solution b is also the value that y(t) approaches as t ⟶ ∞.

Falling bodies with air resistance. The constant solution v(t) = g/k means that you have thrown the object downward at an initial velocity v_{0} exactly equal to the terminal velocity v_{∞} = g/k. In that case the object falls at constant speed. This constant solution separates the two cases: (1) v_{0} < g/k (for example, the object falls from rest), in which case the object accelerates downward, approaching v_{∞} = g/k as t ⟶ ∞; and (2) v_{0} > g/k (for example, a parachutist after the parachute opens), in which case the object decelerates, approaching v_{∞} = g/k as t ⟶ ∞. Again note that, whatever the initial velocity, the constant solution g/k is also the value that v(t) approaches as t ⟶ ∞.

Mixing problems. The constant solution y(t) = kV means that you start with a concentration y_{0}/V = k which is equal to the concentration of impurity in the incoming water; thus, the incoming water does not change the amount of impurity in the container. This constant solution separates the two cases: (1) y_{0} < kV, when the incoming water increases the amount of impurity until the concentration y(t)/V approaches k in the limit as t ⟶ ∞; and (2) y_{0} > kV, when the incoming water decreases the amount of impurity, and again the concentration approaches k as t ⟶ ∞. Once again, in all cases the constant solution kV is also the limiting value of y(t) as t ⟶ ∞.

6. A Final Remark

In all four of these applications, we made some simplifying assumptions, which introduce errors. That is, our formulas will not be an exact description of what happens in practice. The amount of error introduced depends on the particular situation. Generally speaking, in installment buying problems the error is very small, while in the other three applications our numerical values for y(t) or v(t) might be way off. However, even if we cannot always rely upon our quantitative results, we have obtained some useful qualitative information. For example, the role of the constant solution of the differential equation — as the limiting value of the function as t ⟶ ∞ — is actually observed in practice. For example, the terminal velocity of a falling body is the speed at which air resistance exactly balances (is equal to) the force of gravity.

Homework 15.

1. Suppose you buy an $800 stereo system, taking 3 years to pay at 16% interest with no money down. What are your monthly payments, and how much will you end up paying for the stereo in all?

2. Suppose you want to buy a $20000 boat but have only $5000 for a downpayment. You can get a 5-year loan for the rest at 18%.

(a) What are your monthly payments, and how much will you end up paying for the boat?

(b) If you can only afford $300 monthly payments, how long will it take you to pay off the loan?

(c) Suppose you arrange a loan where you are paying $300 per month as in part (b). How much of the $3600 that you pay in the first year is tax-deductible? How much of the $3600 that you pay in the fifth year is tax-deductible?

3. Suppose that a mortgage company will give you a 30-year mortgage at 12% interest, but they consider you an acceptable credit risk only if you make a 20% downpayment and if your monthly payments do not exceed one fourth of your $1600 monthly salary. What is the most expensive house you can buy, and how much cash on hand do you need to buy it?

4. You buy a house for $90000 and make a $20000 downpayment, using a 30-year mortgage for the rest.

(a) Suppose that between the time you agree to buy the house and the time the arrangements are finalized, the interest rate jumps from 11% to 12%. What increase in the price of the house would have been equivalent to the increase in interest rate (i.e., would have led to the same increase in your monthly payments, assuming your downpayment remains the same)?

(b) Suppose that the interest rate is 12%. How much more would you have to pay each month if you want to pay off the mortgage in 25 years rather than 30 years?

(c) What is the shortest term mortgage for which your monthly payments do not exceed $750?

5. An item is bought using a loan of y_{0} dollars, which is paid back at the steady rate of a dollars/year. The instantaneous interest rate is 10% (assume continuous compounding), and the term of the loan is 20 years. At what instant t is exactly half of the money going for interest (i.e., half of the a dollars per year being paid is going for interest at that instant) and half for principal?

6. Suppose that on your 16th birthday a wealthy distant relative dies and leaves you $30000 in a trust fund that works as follows. The money in the fund always receives 8% interest compounded continuously. You receive no money until you turn 21. Then for the next 30 years you receive a fixed amount each month, which is determined in such a way that the fund is precisely used up on your 51st birthday. (Make the simplifying assumption that you receive money continuously throughout the month rather than in a lump payment once each month.)

(a) How much money is in the trust fund when you turn 21?

(b) How much will you receive each month between your 21st and 51st birthdays?

(c) If you decide to take only $300 per month and leave the rest in the fund, how long will the trust fund last?

(d) How much can be withdrawn per month if you want the trust fund to last forever?

7. The table below is based on data published in the June 15, 1981 issue of The Seattle Times. It shows the monthly payments and total cost for a 42-month $6000 loan for a new car, depending on the interest rate demanded by various lending institutions.

Basic new car loan $6,000, 42 months

LENDER  INTEREST RATE MONTHLY PAYMENT TOTAL COST
Boeing Credit Union  13.75%  $180.78  $7,792.76
Auto Dealer No.1  16%  $187.50  $7,875.00
First Interstate Bank  18%  $193.59  $8,130.78
Auto Dealer No.2  18.25%  $194.35  $8,162.72
Rainer National Bank  19.25%  $197.45  $8,292.90
Seattle-First National Bank  20.5%  $201.35  $8,456.70

Use the procedure in this section to compute the monthly payments in each of the six cases, and see if your answers agree with the values given. They will not be exactly the same, because of our simplifying assumptions. However, you should not be more than about a dollar off.

8. You make a pudding for a dinner party, and put it in the refrigerator at 6 p.m. (t = 0). Your refrigerator maintains a constant temperature of 40^{∘}. The pudding will gel when it cools to 45^{∘}. When you put the pudding in the refrigerator you measure its temperature to be 190^{∘}, and when the first guest arrives at 7 p.m. you measure it again and get a temperature reading of 100^{∘}. When is the earliest you can serve dessert?

9. Water at 55^{∘} is pumped into a swimming pool on a 90^{∘} summer day. After 2 hours the temperature of the water is 60^{∘}. In how many hours (assuming the outside temperature remains 90^{∘}) will the water reach a comfortable swimming temperature of 70^{∘}?

10. A detective discovers a murder victim in a room at 8:00 a.m. one morning. The temperature of the body is 80.0^{∘}F. Two hours later (at 10:00 a.m.) the temperature of the body is 75.0^{∘}F. Assuming that the room is kept at the constant temperature 70.0^{∘}F, determine the time at which the murder was committed. (Assume that the murder victim had a normal temperature of 98.6^{∘}F at the time of death.)

11. The detective in Problem 10 discovers another murder victim at noon the next day in another 70.0^{∘}F room, and immediately has the body transported to a morgue, where the temperature is kept at 50.0^{∘}F. She measures the body’s temperature right before it is taken from the room, and again at 4 p.m., after it has been 4 hours in the morgue. The former temperature is 85.0^{∘}F, and the latter temperature is 65.0^{∘}F. When was the murder committed? (Assume that the same constant of proportionality in Newton’s law applies in the room and in the morgue.)

12. Suppose you drop an object from a great height, so that it reaches very nearly terminal velocity by time it hits the ground. By measuring the impact you determine that this terminal velocity is 49 m/sec.

(a) Find k.

(b) Graph the velocity function for the following values of v_{0}: 0 (dropped from rest), 20 m/sec downward, 100 m/sec downward, 49 m/sec downward, 20 m/sec upward.

(c) At what velocity must the object be thrown upward if you want it to reach its peak height after 3 sec?

(d) If v_{0} is 20 m/sec upward, how long does the object take to reach its peak height?

13. In Example 16.4, find formulas for the distance s(t) fallen from the plane at time t for 0 ≤ t ≤ 10 and for t > 10. Notice that the exponential decay term makes a negligible contribution when t is greater than about 15 sec. If the plane was at an altitude of 500 meters when the parachutist jumped, when does she reach the ground?

14. A dense heavy package is dropped from a plane. After 7 seconds the package releases a parachute that starts to slow its speed. Assume that the air resistance is negligible during the first 7 seconds, and that wind tunnel tests show that with the parachute open the air resistance balances gravity at a speed of 19.6 m/sec. Find formulas for the downward velocity v(t) for 0 ≤ t ≤ 15, and graph v(t).

15. Water is flowing at 3 m^{3}/min in one end and at the same rate out the other end of a vat containing 100 m^{3} of water. Let y(t) be the amount (in grams) of impurity in the vat of water at time t min. The water flowing in contains 0.1 gm of impurity per cubic meter. Find y(t) if the initial concentration of impurity in the container is

(a) 0.05 gm/m^{3}, (b) 0.20 gm/m^{3}, (c) 0.10 gm/m^{3}. Graph y(t) in each case (on the same xy-plane).

16. A tank originally contains 100 gal of fresh water. Water with 1/2 lb/gal of salt is poured into the tank at a rate of 2 gal/min, and the well-stirred mixture is allowed to leave at the same rate. After 10 minutes the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of 20 minutes.

17. Suppose that a room containing 1200 cubic feet of air is originally free of carbon monoxide. Beginning at time t = 0 cigarette smoke containing 4 percent carbon monoxide is introduced into the room at a rate of 0.1 ft^{3}/min, and the well-circulated mixture is allowed to leave the room at the same rate.

(a) Find an expression for the amount y(t) of carbon monoxide (in cubic feet) in the room at time t.

(b) Extended exposure to a carbon monoxide concentration as low as 0.012% is harmful to the human body. Find the time at which this concentration is reached.

18. Consider a lake of constant volume V which at time t contains an amount y(t) of pollutant, evenly distributed throughout the lake with a concentration c(t), where c(t) = y(t)/V. Assume that water containing a concentration k of pollutant (measured in tons/km^{3}) enters the lake at a rate r (measured in km^{3}/yr), and that water leaves the lake at the same rate. Suppose that pollutants are also added directly to the lake at a constant rate P (measured in tons/yr). Note that the given assumptions neglect a number of factors which may be important in some cases: the water added or lost by precipitation, absorption, and evaporation; the stratifying effect of temperature differences in a deep lake; the tendency of irregularities in the coastline to produce sheltered bays; and the fact that pollutants are not deposited evenly throughout the lake but rather at isolated points around its periphery. However, we shall neglect these factors in what follows. Thus, according to our assumptions, the rate of change of y(t) is the sum of three contributions: rk tons/yr in the incoming water, P tons/yr entering additionally, and minus r times y/V leaving in the outgoing water.

(a) Write a differential equation for y.

(b) Solve the differential equation to find an expression for y(t). The constants r, k, P, V, and y_{0} should appear in your expression.

(c) What is the limiting value of y(t) as t ⟶ ∞?

(d) If the addition of pollutants to the lake is terminated (i.e., k = 0 and P = 0 for t > 0), determine the time interval T which must elapse before the concentration of pollutants is reduced to 10% of its original value.

(e) The table below contains data for several of the Great Lakes. Use the data to determine the time T necessary to reduce the contamination of each of these lakes to 10% of the original value.

Lake Superior: V = 12200 km^{3}, r = 65.2 km^{3}/yr Lake Michigan: V = 4900 km^{3}, r = 158 km^{3}/yr Lake Erie: V = 460 km^{3}, r = 175 km^{3}/yr Lake Ontario: V = 1600 km^{3}, r = 209 km^{3}/yr

19. Suppose that at time t = 0 a country’s population is 300 million, and its death rate exceeds its birth rate by k. Normally, that would lead to exponential decay of the population given by the formula y(t) = y_{0}e^{−kt}. However, there is also a constant immigration of I million people per year.

(a) Write a differential equation for the population function.

(b) If k = 1% and I = 4 million people per year, find the population in 50 years, and also find the limiting value of the population.

20. The population of salmon in the Pacific Northwest, if not harvested, would grow by 10% per year (this is effective yearly rate, not instantaneous rate). Suppose that the current salmon population is 3 million, and that the salmon are being harvested at a constant rate of R per year. At what rate R should the salmon be harvested in order that the salmon population be 4 million in 4 years from now? What would happen if R were 300000 salmon per year?

110 16. SEPARATION OF VARIABLES

CHAPTER 17 Differential Equations II: More Separation of Variables

In certain practical situations we might want to make simplifying assumptions that lead to somewhat more complicated differential equations which are still not too complicated to be solved by the method of separation of variables. This section is devoted to several examples of this.

But first we want to illustrate how the general method of separation of variables can be used to solve differential equations that look quite different from those in the last chapter.

Example 17.1. Find an equation for the curve that passes through the point (2, 3) and satisfies the differential equation dy/dx = y^{2}/x^{3}.

We first separate variables in the differential equation, obtaining

\frac{dy}{y^{2}} = \frac{dx}{x^{3}}.

If we integrate both sides and combine the constants of integration into a single constant on the right, we get

−\frac{1}{y} = −\frac{1}{2} · \frac{1}{x^{2}} + C.

To find C, we use the “initial data” that the point (2, 3) is on the curve. Thus, −\frac{1}{3} = −\frac{1}{2} · \frac{1}{2^{2}} + C, from which we get C = \frac{1}{8} − \frac{1}{3} = −\frac{5}{24}. Finally, we solve for y by taking the negative reciprocal of both sides. The result is:

y = \frac{1}{\frac{1}{2x^{2}} + \frac{5}{24}} = \frac{24x^{2}}{12 + 5x^{2}} .

1. Spread of an Epidemic

Suppose that y(t) is the number of people (or some type of animal) who have a certain contagious disease at time t. We are interested in the rate of change of y(t), and eventually in a formula for y(t). Since the spread of the disease depends on the number of people from whom one can catch it, we might try supposing that the rate of spread of the disease is simply proportional to the number who have it, i.e., dy/dt = ky. This would lead to the exponential growth function y(t) = y_{0}e^{kt}, a function which might be a reasonable approximation for some small time interval, but which clearly is far off for large t: the function y_{0}e^{kt} rapidly grows without bound, while the total population has some finite value P. The number of sick people cannot be greater than the number of people.

So let’s try another type of simplifying assumption: that the spread of the disease is proportional to the number of contacts between a sick person and a healthy person, and that this, in turn, is proportional to the product of the number of people with the disease and the number of people without the disease. That is, at any time t, dy/dt is proportional to the product of y(t) and P − y(t):

\frac{dy}{dt} = ky(P − y).

It must be admitted that this model for the spread of an epidemic is still very simplistic. It assumes, for instance, that no attempt is made to quarantine sick people, that everyone is susceptible to the disease, that someone who gets the disease remains sick (and a carrier) indefinitely, and so on. But in some situations this differential equation is a reasonable approximation to reality.

We now solve the above differential equation by separation of variables. The first step is to move the dt to the right, to move the part on the right with y over to the left, and then to insert an integral sign on both sides:

∫ \frac{dy}{y(P − y)} = ∫ kdt = kt + C_{2}.

The anti-derivative on the left can be found by the method of partial fractions, i.e., we write:

\frac{1}{y(P − y)} = −\frac{1}{(y − 0)(y − P)} = −\frac{1}{0 − P} (\frac{1}{y} − \frac{1}{y − P} ) = \frac{1}{P} (\frac{1}{y} + \frac{1}{P − y} ) ,

after which we see that the above indefinite integral with respect to y is

\frac{1}{P} (ln |y| − ln |P − y|) + C_{1} = \frac{1}{P} (ln y − ln(P − y)) + C_{1}

(since both y and P − y are positive)

= \frac{1}{P} ln (\frac{y}{P − y} ) + C_{1}.

We next move the C_{1} to the right of the equality, multiply through by P, and set C = P (C_{2} − C_{1}). We then exponentiate both sides. The result is

\frac{y}{P − y} = e^{P kt+C}.

To solve this for y, the simplest thing is to take the reciprocal of both sides, getting

\frac{P − y}{y} = \frac{P}{y} − 1 = e^{−C}e^{−P kt}.

To find y we add 1 to both sides, take the reciprocal again and multiply through by P, obtaining:

y(t) = \frac{P}{1 + e^{−C}e^{−P kt}} .

To find C we need some initial information, such as y_{0} = y(0). Substituting t = 0 (it’s simplest to substitute into our earlier equation which had (P/y) − 1 on the left) gives: e^{−C} = (P/y_{0}) − 1. In a practical situation, if you’re not given the value of k, you can find it if you know the value of y(t) at some other particular time.

Cautionary note: The above solution technique works only when k and P are constant with respect to time t.

Example 17.2. Let y(t) denote the fraction of the population having the disease at time t months (in other words, we are measuring population in units of fraction of the total, rather than in units of individuals; in that way the total population P = 1). At time t = 0 suppose that 1% of the population are sick, and after a month 10% are sick. Find a formula for y(t), graph y(t), and find out at what time the rate of spread of the disease starts to slow down.

As shown above, we have

y(t) = \frac{1}{1 + e^{−C}e^{−kt}} .

Substituting t = 0 gives e^{−C} = (1/0.01) − 1 = 99. To find k, we substitute t = 1 month: (1/y(1)) − 1 = e^{−C}e^{−k·1} = 99e^{−k}, so that 9 = 99e^{−k}, which leads to −k = ln(1/11), i.e., k = ln 11 = 2.4. At the right, we show the graph of our solution y(t) = 1 / (1 + 99e^{−2.4t}).

To ask when the rate of spread of the disease starts to slow down is the same as asking when y′ = dy/dt reaches its maximum. This is the point of inflection in the above graph, which we find by setting y′′ = 0. We could take the second derivative of our y(t) formula. But an easier way is to take d/dt of both sides of the original differential equation y′ = ky(1 − y). Using the product rule on the right, we obtain

y′′ = ky(−y′) + ky′(1 − y) = ky′(−y + 1 − y) = ky′(1 − 2y).

This vanishes when y = 1/2. In other words, the spread of the disease starts to ease off as soon as more than 50% of the population has the disease. To find the time when this occurs, we can set the y(t) formula equal to 1/2 and solve for t. The result is: approximately 2 months.

2. More Mixing Problems

Suppose that water is being pumped into our container at a constant rate r_{1} that is faster than the constant rate r at which water is leaving. This means that the volume V is no longer constant: it changes at rate r_{1} − r, and so is given by the formula V = V_{0} + (r_{1} − r)t. On the right side rk − \frac{r}{V} y of the differential equation in the last section the effect is to change the first r to r_{1} and replace the constant volume V by a function of t:

\frac{dy}{dt} = r_{1}k − \frac{ry}{V_{0} + (r_{1} − r)t}.

If k ≠ 0 and we try to solve this by separation of variables, we find that we are stymied. This problem can be solved by a different technique that will not be covered in this course, but it cannot be solved by separation of variables. So for now let us assume that k = 0, i.e., that fresh water is being pumped into the container. In that case, we can use separation of variables: we divide both sides by y, multiply both sides by dt, and put an integral sign in front of both sides:

∫ \frac{dy}{y} = ∫ −\frac{r dt}{V_{0} + (r_{1} − r)t} = −\frac{r}{V_{0}} ∫ \frac{dt}{1 + \frac{r_{1}−r}{V_{0}} t}.

Performing both integrations, we obtain

ln |y| + C_{1} = −\frac{r}{r_{1} − r} ln|1 + \frac{r_{1} − r}{V_{0}} t| + C_{2}.

Since the expressions in the absolute value signs are both positive, we can drop the absolute value. We also set C = C_{2} − C_{1} and exponentiate both sides. The result is:

y(t) = e^{C} (1 + \frac{r_{1} − r}{V_{0}} t)^{−r/(r_{1}−r)} .

Finally, by substituting t = 0 on both sides, we find that e^{C} = y(0) = y_{0}.

Notice that this variant on the mixing problem has a very different solution from the variant we treated in the last section, where we had r_{1} = r. Finally, be careful to note that the above solution is correct only when r and r_{1} are constants (do not change with t).

Example 17.3. At time t = 0 a container has y_{0} gm of salt dissolved in 100 liters of water. Fresh water is pumped into the container at 5 liters/min, and the well-mixed solution leaves the container at 3 liters/min. Find a formula for the amount of salt in the water at time t.

In this problem V_{0} = 100, r_{1} = 5, and r = 3. Hence, by the formula just derived, y(t) = y_{0}(1 + 0.02t)^{−1.5}.

3. Harmonic Oscillation

Suppose that we have an object of mass m at the end of a spring. Let y(t) be the distance of the object from the equilibrium point at time t (i.e., the distance through which the spring is extended). We shall assume that the only force on the object is that of the spring (i.e., we neglect friction, air resistance, gravity, etc.). We shall also assume the following basic physical relationship: the force of the spring is proportional to the amount by which the string is extended. Since acceleration is proportional to force (it is equal to force divided by m), it follows that the acceleration ÿ(t) is proportional to y(t). Note that the direction of pull is opposite to the direction of extension — that is, if the string is extended, then the pull is inward, i.e., negative; while if the string is compressed, then the pull is outward, i.e., positive. Thus, the proportionality constant between ÿ(t) and y(t) is negative; we shall let −k be this constant. Thus, the distance function y(t) satisfies the differential equation

ÿ = −ky.

This is called a second order differential equation, because it involves the second derivative. Before solving it, we rewrite it in the form

ÿ + ky = 0.

Instead of worrying about the differential equation in its present form, we instead look at a different expression involving only y(t) and its first derivative ẏ(t):

ẏ^{2} + ky^{2}.

This is basically the sum of the squares of the function and its first derivative, with the constant k thrown in. Why would we want to look at that? In the next paragraph we’ll say something about where this expression comes from. For now what we want to do is just take its derivative:

\frac{d}{dt}(ẏ^{2} + ky^{2}) = 2ẏ ÿ + 2ky ẏ = 2ẏ(ÿ + ky).

But according to our differential equation, ÿ + ky = 0, so the derivative of our sum-of-squares expression is zero. The only function with zero derivative is a constant. What is the value of this constant? We use our initial conditions to answer this question: at time t = 0 the sum-of-squares expression has value ẏ(0)^{2} + ky(0)^{2} = v_{0}^{2} + ky_{0}^{2}. Let us suppose that the object is at location 0 at time t = 0; then the constant is simply v_{0}^{2}, where v_{0} is the object’s initial velocity. We conclude that ẏ^{2} + ky^{2} = v_{0}^{2}.

Before solving this last differential equation, we make a remark about where the sumof-squares expression comes from. It is essentially what physicists call the energy of the object. More precisely, if we multiply it by the constant \frac{1}{2}m (where m is the mass), we get the energy. The conclusion of the last paragraph — that this is constant — is an example of another fundamental principle of physics, the law of conservation of energy. If we work only with the mathematics, then admittedly the expression ẏ^{2} + ky^{2} seems to be pulled out of a hat. However, if we know a little physics, then the method of solving this differential equation seems quite natural.

Anyway, we now have the differential equation

ẏ^{2} + ky^{2} = v_{0}^{2},

which we shall see can be solved by separation of variables. If we move ky^{2} to the right and take square roots, we get (using the notation dy/dt instead of ẏ)

\frac{dy}{dt} = \sqrt{v_{0}^{2} − ky^{2}},

and then, separating variables,

∫ \frac{dy}{\sqrt{v_{0}^{2} − ky^{2}}} = ∫ dt.

On the left, if we pull the k out of the radical (so that it becomes \sqrt{k}) and set r^{2} = (v_{0}/\sqrt{k})^{2}, then we have Example 7.8 of the chapter on inverse trig functions (with y in place of x and with r replaced by v_{0}/\sqrt{k}). Using the answer to Example 7.8, we find that the anti-derivative on the left is

\frac{1}{\sqrt{k}} Arcsin (\frac{y}{v_{0}/\sqrt{k}} ) + C_{1}.

Thus, we can write

\frac{1}{\sqrt{k}} Arcsin (\frac{y}{v_{0}/\sqrt{k}} ) = t + C

(where C = C_{2} − C_{1} is the difference of the constants of integration that one gets on the right and on the left). To find C we again use the initial information that y(0) = 0 when t = 0; since Arcsin(0) = 0, we find that C = 0. Finally, we solve for y by multiplying through by \sqrt{k} and then taking the sine of both sides:

\frac{y}{v_{0}/\sqrt{k}} = sin(\sqrt{k} t),  so that  y(t) = \frac{v_{0}}{\sqrt{k}} sin(\sqrt{k} t) = \frac{v_{0}}{ω} sin(ωt),

where we have set ω = \sqrt{k}, which is the angular frequency.

Thus, the object oscillates back and forth with sinusoidal motion having period 2π/ω. This motion is called harmonic oscillation.

Notice that once we have found the function y(t) = \frac{v_{0}}{ω} sin(ωt), it is very easy to check that it really does satisfy the original differential equation ÿ = −ky = −ω^{2}y; namely, just take the second derivative of y(t), and see that both sides of the equality are equal to −v_{0}ω sin(ωt). It is also clear that y(0) = 0 and ẏ(0) = \frac{v_{0}}{ω} · ω · cos(ω · 0) = v_{0}.

What would happen if we had to take into account friction or air resistance? Suppose that this contributed a force proportional to velocity (as we assumed in Section 3 of Chapter 16). Then our original differential equation would have another term equal to a constant times ẏ. One can show that the solution is a function similar to Example 10.2 in the chapter on the exponential function — it has an exponential decay factor. This is called damped harmonic oscillation.

Homework 16.

1. Find an equation in the form y = f(x) for the curve that passes through the point (3, 1) and satisfies the differential equation

(a) \frac{dy}{dx} = \frac{y^{3}}{x^{2}} ; (b) \frac{dy}{dx} = x^{2}y^{2}; (c) \frac{dy}{dx} = \frac{cos(πx)}{y}.

2. In modeling population growth one might take into account that the reproduction rate (birth rate minus death rate) is not constant, but rather diminishes as the population grows — because the food supply is not sufficient to sustain growth, crowded conditions cause disease and death, etc. The logistic model of population growth is the simplest model in which, instead of having a constant k in dy/dt = ky, we have a reproduction rate which decreases as the population increases. In the logistic model we insert a factor of (1 − y/M) on the right:

\frac{dy}{dt} = k (1 − \frac{y}{M} ) y,

where M is a constant which has the practical meaning of the maximum possible population. (Note that the population would stop growing — i.e., dy/dt would become zero — if y were to become equal to M.)

(a) Derive a general formula for y(t) which will involve the constants k and M and another constant whose value depends on y_{0}.

(b) Suppose that a beetle population on an island obeys the logistic equation. The maximum beetle population that the island can support is 100 million. At present there are 5 million beetles on the island, and last year at this time there were 2 million beetles. Find a formula for the beetle population at time t. How long will it be before the population reaches 95 million?

3. In Example 17.3, what function do you get for the amount y(t) of salt at time t if the rate of the outgoing water is 5 liters/min and the rate of the incoming fresh water is 3 liters/min? What is the domain of this function (until the container is empty)?

4. When an object falls, the deceleration caused by air resistance is some function r(v) of its velocity, and so v(t) satisfies the differential equation dv/dt = g−r(v), where g = 9.8 m/sec^{2} (if we choose downward as the positive direction). In the last section we assumed that r(v) was a linear function of v, namely, we took r(v) ≈ kv. For example, if k = 0.102, we have the differential equation dv/dt = 9.8 − 0.102v. In many cases using a linear approximation kv to r(v) is accurate enough — that way we get a differential equation that is particularly easy to solve. (Using kv instead of a more complicated r(v) is called “linearizing” the differential equation.) Now suppose that the linear approximation isn’t good enough, and you need a quadratic correction term 0.002v^{2}, i.e., r(v) = 0.102v + 0.002v^{2}.

(a) Graph the right side of the resulting differential equation, and find the terminal velocity.

(b) Solve the differential equation to find a formula for v(t) if the object is dropped from rest.

(c) Write a definite integral for the distance the object falls in the first 3 seconds (do not bother to evaluate the integral).

(d) Now suppose that you’re interested in distance fallen not as a function of time as in part (c), but rather as a function of velocity. Namely, you want to answer the question: How far does the object fall (from rest) before reaching a speed of 48 m/sec? Thus, consider distance s as a function of velocity, and use the chain rule to express ds/dv in terms of ds/dt (which is v) and dv/dt, thereby obtaining a differential equation for s(v). Now write an integral for the distance fallen as v increases from v_{1} to v_{2}, and answer the question in italics.

5. Answer the same question as in Problem 4(d) under the assumption that the air resistance r(v) is not given by a formula at all, but rather by experimental readings. That is, using a wind tunnel, you let air rush past the object at velocities 8, 16, 24, 32, 40, and 48 m/sec, measuring the force of air resistance, which you then divide by the mass of the object to obtain r(v). The values you obtain are:

\begin{matrix}v & 0 & 8 & 16 & 24 & 32 & 40 & 48 \\ r(v) & 0 & 0.91 & 1.82 & 2.75 & 3.75 & 4.80 & 5.84 \end{matrix}

Use the trapezoid rule to evaluate the integral that represents the distance the object falls before reaching a speed of 48 m/sec.

6. A vat contains V liters of water with y_{0} grams of salt at time t = 0. Fresh water enters the vat at a rate r(t) that changes with time, and the well-stirred solution leaves the vat at the same rate.

(a) Find a differential equation for the amount of salt at time t.

(b) If V = 100 liters, and the rate r(t) starts at 2 liters/min and increases steadily at 0.5 liters/min each min, derive a formula for y(t).

(c) If V = 100 liters, and the rate r(t) varies sinusoidally with a minimum value of 0 at t = 0 and then a maximum of 2 liters/min after 1 minute (here r(t) goes from 0 to 2 and back to 0 once every 2 minutes), derive a formula for y(t).

7. Answer Problem 6(b) under the assumption that the rate at which water leaves the vat remains at 2 liter/min, and only the incoming rate increases. (Note that the volume is no longer constant.) Also find (in terms of y_{0}) the limiting value of the amount of salt as t ⟶ ∞.

8. A vat contains 100 liters of water with y_{0} grams of salt at time t = 0. Water with k gm/liter of salt enters the vat at a rate that changes gradually over 10 minutes, as measured at 1-minute intervals and tabulated below:

t  0  1  2  3  4  5  6  7  8  9 10 r(t) 2.50 2.56 2.62 2.67 2.71 2.73 2.74 2.74 2.72 2.68 2.62

The well-stirred solution leaves the vat at the same rate.

(a) Find a differential equation for the amount of salt at time t.

(b) Find the solution y(t) of the differential equation. Your solution should contain the constants k and y_{0}, and should include an integral involving r(t).

(c) If we flush out the salt by pumping in pure water (i.e., if k = 0), find the percent by which the amount of salt decreases during the 10 minutes. (Use the trapezoid rule to evaluate the integral.)

(d) If k = 0.40 gm/liter and y_{0} = 30 gm, find y(10).

9. At time t = 0 a car is traveling at 30 m/sec. From time t = 0 until the car stops, a constant breaking force is applied which causes a deceleration of 8 m/sec^{2}. The other force acting on the car is air resistance (and friction), causing a deceleration which for the range of velocities we’re dealing with is approximately equal to 0.4v + 0.01v^{2}. Find

(a) the differential equation for v(t);

(b) the velocity function v(t) for 0 ≤ t ≤ T; and

(c) the time T when the car stops.

10. A rocket sled starts from rest and is propelled by a constant thrust (force) F in a horizontal path. Suppose that, as the propellant is burned, the mass of the rocket, which initially is m, decreases at the constant rate α. Also assume that the force of air resistance and friction is proportional to the velocity with proportionality constant K.

(a) Using the basic physical principle that

(mass of rocket at time t) · (acceleration)=total of forces acting on rocket,

find a differential equation for v(t).

(b) Solve the differential equation in part (a) to find a formula for v(t) in terms of t and the constants F, m, α, K, and v_{0}.

11. In Section 3 we derived the formula y(t) = \frac{v_{0}}{ω} sin(ωt) for harmonic oscillation, assuming zero initial position. Find the formula for harmonic oscillation if the initial velocity is zero, but the initial position is y_{0} ≠ 0.

12. Find the formula for harmonic oscillation if y(0) = y_{0} and ẏ(0) = v_{0}, where we no longer suppose that either of these initial values is zero.

120 17. MORE SEPARATION OF VARIABLES

CHAPTER 18 More Differential Equations

A very common type of differential equation is the differential equations of the form \frac{dy}{dt} = ky + l, where both k and l are allowed to be functions of t. Such a differential equation is called linear (because y occurs only to the first power) and, if l ≠ 0, nonhomogeneous. If l = 0, then it is called a linear homogeneous differential equation.

The homogeneous differential equation \frac{dy}{dt} = k(t)y is of the type solved in the last section by separation of variables. However, if l(t) is not zero (and not a constant multiple

of k(t)), then the nonhomogeneous differential equation \frac{dy}{dt} = k(t)y +l(t)y cannot be solved by separation of variables. The method used to solve such a differential equation will not be covered in Math 125 (but you will get it in the next course that you take on differential equations). In this chapter our interest is not in solving these differential equations, but rather in setting them up correctly.

Example 18.1. In Example 18.2 of the last section, suppose that the water pumped into the container is not fresh, but rather has a concentration of 0.4 gm/liter. Find a differential equation for the amount of salt in the water at time t.

As we saw in the last section, the differential equation for the amount of salt y(t) is

\frac{dy}{dt} = r_{1}k − \frac{ry}{V_{0} + (r_{1} − r)t} = 5 · 0.4 − \frac{3y}{100 + 2t} = −\frac{3}{100 + 2t}y + 2,

i.e., our function k(t) is −3/(100 + 2t) and our function l(t) is 2.

Example 18.2. (Heating/cooling with sinusoidal ambient temperature.)

In Newton’s law of cooling we no longer assume that the ambient temperature is constant. Suppose we are interested in the temperature y(t) inside an unheated building under the assumption that the outside temperature is varying sinusoidally between its warmest value in mid-afternoon and its coldest value some time after midnight. For simplicity, we shall choose t = 0 to be the time at which the temperature reaches its average value D (some time in mid-morning). Then when t = 6 it reaches its maximum D + A, when t = 12 it’s back to D, when t = 18 it reaches its minimum D − A, and finally when t = 24 it comes back to D and repeats the temperature values of the day before. The formula for such a sinusoidal function is b(t) = A sin(αt) + D, where in this case α = 2π/24 = π/12 (24 is the period of the sinusoidal function). According to Newton’s law of cooling, we have

\frac{dy}{dt} = −k(y − b) = −k (y − (A sin(αt) + D)) ,

where the constants A and D depend on the range of oscillation of the outside temperature, α = π/12, and k depends on the particular building (the better insulated the building is, the smaller k is). This is a differential equation of the form dy/dt = k(t)y + l(t), where the constant −k is playing the role of k(t), and the function k (A sin(αt) + D) is playing the role of l(t).

Without going through the method of solution, we will give the result — the function y(t) that satisfies this differential equation. Here y(t) is written in terms of t and the constants k, A, D, and y_{0}. It turns out to be convenient to introduce a new constant θ, which is the angle constructed using α as the opposite side and k as the adjacent side (i.e., θ =Arctan(α/k)). See the diagram at the right.

θ k α

It can be shown that

y(t) = D + (y_{0} − D + A sin θ cos θ)e^{−kt} + A cos θ sin(αt − θ).

For large t, the exponential decay term in the formula for y(t) becomes negligible. If we drop that term we obtain the “steady state” solution, which does not depend on our initial value for the building’s temperature. Thus, we arrive at the following formula for the steady state temperature function of the building:

y(t) = D + (A cos θ) sin(αt − θ), where θ = Arctan(α/k) and α = π/12.

The practical interpretation of this formula is that the building’s insulation has two effects:

(1) to decrease the magnitude of temperature oscillation around D, namely by a factor of cos θ = k/\sqrt{k^{2} + α^{2}}, and (2) to cause a “phase shift” of θ (equivalent to a “delay” of 12θ/π hours if θ is measured in radians, or θ/15 hours if θ is measured in degrees).

The practical meaning of the steady state solution is as follows. Suppose that the house is heated to a high temperature y_{0}, and then at time t = 0 the heat is turned off. For a while, the house cools off to the range in which the outside temperature is oscillating (around D). If t is small, then the term

(y_{0} − D + \frac{Akα}{k^{2} + α^{2}} ) e^{−kt}

is important, since it describes this cooling process. But after a while the outside temperature “takes over,” and the inside temperature no longer depends to any significant extent on what the initial inside temperature y_{0} was. As the exponential decay (“cooling”) term becomes negligible, the inside temperature can be described by the steady state solution, which is a purely sinusoidal function, as we have seen.

In Example 2 we admittedly adopted a somewhat unrealistic simplifying assumption to the effect that the outside temperature varies sinusoidally day in and day out. Nevertheless, from a qualitative point of view our solution reveals phenomena that do occur in practice, namely, the decrease in magnitude of oscillation and the phase shift which are caused by the building’s insulation. This explains why on a hot summer day the inside of a house remains hot long after the outside temperature has dropped to a comfortable level.

Homework 17.

1. In Example 18.2, show that the solution given satisfies the initial condition y(0) = y_{0} and also the differential equation. You will have to use the trig formulas for sin(A ± B), cos(A ± B), and also the triangle relating α, k, and θ.

2. In Example 18.2, (a) find a formula for the initial temperature which would cause the building’s temper-

ature function y(t) immediately to equal the steady state function;

(b) explain what happens to the phase shift (time delay) and the oscillation amplitude of the steady state function if the building is very well-insulated (k is near 0) and if the building is very poorly insulated (k is large);

(c) find an approximate formula in terms of k for the time delay in the steady state function if the building is very well-insulated and if it is very poorly insulated (note: the units of k are time^{−1}); (d) if the outside temperature oscillates between 70^{∘}F and 90^{∘}F, reaching 90^{∘}F at 3 p.m., while the inside temperature oscillates between 75^{∘}F and 85^{∘}F, determine when the inside of the building is hottest, and draw a graph of both the outside and inside steady state temperature functions.

3. In Problem 18(b) of the first problem set on differential equations, suppose that the immigration I is only 2 million people/year at time t = 0, but that it increases steadily each year by 250000 people/year. As before, suppose that the country’s population is 300 million at time t = 0, and the death rate exceeds the birth rate by 1%. Write a differential equation for the population function.

4. Suppose that a cup of coffee at temperature y_{0} is brought into a room whose temperature is b at time t = 0 but is increasing at the rate a (in ^{∘}F/min). Let k denote the constant of proportionality in Newton’s law. Find a differential equation for the temperature function y(t).

5. In Problem 6 of the last problem set, suppose that, instead of fresh water, water with 2 gm/liter of salt is entering the vat. As before, the vat contains 100 liters with y_{0} grams of salt at time t = 0, water comes into the vat at 2 + 0.5t liters/min and leaves the vat at 2 liters/min. Give the differential equation satisfied by y(t).

6. Suppose that a vat contains V liters of water and y(t) grams of salt. Water is coming into the vat at constant rate r liters/min and leaving at the same rate. The concentration of salt at time t = 0 is k gm/liters, but this concentration is increasing steadily at α gm/liter each minute. Write a differential equation for y(t).

7. The normal mean temperature in Fairbanks, Alaska is closely approximated by the sinusoidal function

37 sin (\frac{2π}{365}(t − 101)) + 25,

where t is the number of days from the beginning of the year. This formula gives the expected average daily temperature during the t-th day (obtained by averaging the temperature on the t-th day over a 30-year period).^{1}

Suppose that on April 11 you sell a building in Fairbanks (so take April 11, rather than the beginning of the year, to be t = 0). As part of the purchase agreement, you are required to pay for electricity to keep the building heated to 65^{∘}F for the first 4 years. The cost of heating the building is known to be $0.50 for each degree-day below 65^{∘}, and this cost is not expected to change during the 4-year period. That is, at any time t the rate at which the heating bill accumulates is equal to $0.50 times (65 − current temperature). To cover your obligation to pay for heating you leave an amount y_{0} of money with the electric company. The electric company agrees to pay interest k (compounded continuously) on the money, and also each day deduct the heating cost for that day. (We make the simplifying assumption that this is a continuous process, rather than something that happens once a day. In the same way, when writing the above formula for the temperature function in Fairbanks we suppose that t is a continuous variable.)

1This formula for the temperature in Fairbanks is taken from an article by B. M. Lando and C. A. Lando in the September 1977 issue of The Mathematics Teacher.

(a) Write a differential equation for the amount of your money the electric company has at time t.

(b) If k = 0, i.e., if the electric company pays no interest on the money you deposited with them, how much should you leave with them to cover the 4 years’ heating cost?

(c) Now suppose that k is 9\frac{1}{8}% annual interest (i.e., k = 0.09125 yr^{−1} — but be careful to use consistent time units). Find a formula for y(t) in terms of t and the constant y_{0}.

(d) In the situation of part (c), how much money should you leave with the electric company to exactly cover the heating cost for 4 years?

8. At time t = 0 a vat contains 100 liters of water with y_{0} grams of salt. At that time you start letting water flow out of the vat at a rate r(t) that depends on time: the rate starts at 12 liters/min at time t = 0, and it decreases at a liters/min^{2} (until it reaches 0, at which point the experiment stops). There is no water coming into the vat, but someone is sprinkling salt into the water at the rate l gm/min.

(a) Write formulas for r(t) and the volume V (t).

(b) Find the differential equation satisfied by the amount y(t) of salt at time t. (The constants a and l should appear in your differential equation.)

9. At time t = 0 we have x_{0} grams of radioactive substance A and y_{0} grams of radioactive substance B. The former has half-life H_{1}, and so has radioactive decay constant k_{1} = (ln 2)/H_{1}; the latter has a different half-life H_{2} and decay constant k_{2} = (ln 2)/H_{2}. The substance A decays into the substance B, which then, in turn, decays into some nonradioactive substance.

(a) Find a formula for the amount x(t) of A at time t.

(b) Find a differential equation satisfied by the amount y(t) of B at time t.

10. The diagram at the right shows two vats, the first containing V liters of water and the second containing W liters of water. Water with k gm/liter of impurity is entering the first vat at constant rate r, water is passing from the first vat to the second vat at the same rate, and water is leaving the second vat also at rate r. Let x(t) be the amount of impurity in the first vat at time t, and let y(t) be the amount in the second vat at time t.

(a) Find a formula for x(t).

(b) Find a differential equation satisfied by y(t).

CHAPTER 19 Taylor’s Theorem

The dominant theme in this chapter is approximation and the main result, an extension of the Mean Value Theorem called Taylor’s Theorem, is of fundamental importance because it provided a method of approximating functions by polynomials. Thus, such problems as computing e^{x}, sin x, or ln x to a desired degree of accuracy can be solved by simply evaluating a polynomial. The idea behind Taylor’s theorem also explains the technique for evaluating certain limits called indeterminate forms and provides a way of estimating the rate of convergence of Newton’s method for approximating roots of equations.

1. Taylor Polynomials

We know that if the function f is differentiable (has a derivative) at x = a, then the tangent line

P_{1}(x) = f(a) + f′(a)(x − a)

at the point (a, f(a)) is a good local approximation to the graph of f, that is, P_{1}(x) ≈ f(x) if x ≈ a. For example, the tangent to the function f(x) = \sqrt{x} at x = 4 is

P_{1}(x) = 2 + 0.25(x − 4).

Now choose a value of x reasonably near 4, say 4.3. Then P_{1}(4.3) = 2.075 and \sqrt{4.3} = 2.0736... so that the error in the approximation \sqrt{4.3} ≈ 2.075 is less than 0.0014 in absolute value.

The proper generalization of this notion of approximation by the tangent stems from the observation that P_{1}(x) is a polynomial of degree one that agrees with f and f′ at x = a:

P_{1}(a) = f(a), P1^{′}(a) = f′(a).

The idea is to construct polynomials P_{n}(x) of higher degree whose value and first n derivatives match those of f at x. Thus, for n = 2, we seek a polynomial P_{2}(x) such that

P_{2}(a) = f(a), P2^{′}(a) = f′(a), P2^{′′}(a) = f′′(a). (1)

Since P_{1}(x) has the form c_{0} + c_{1}(x − a), it is reasonable to expect that P_{2}(x) will have the form

P_{2}(x) = c_{0} + c_{1}(x − a) + c_{2}(x − a)^{2}

for some choice of the coefficients c_{0}, c_{1} and c_{2}. The condition P_{2}(a) = f(a) is satisfied by choosing c_{0} = f(a) and since P2^{′}(x) = c_{1} +2c_{2}(x−a), the condition P2^{′}(a) = f′(a) is satisfied if c_{1} = f′(a). And, finally, from P2^{′′}(x) = 2c_{2} it follows that P2^{′′}(a) = f′′(a) if c_{2} = \frac{1}{2}f′′(a). Hence

P_{2}(x) = f(a) + f′(a)(x − a) + \frac{f′′(a)}{2}(x − a)^{2} (2)

is a polynomial that satisfies the conditions in (1). And, as we shall see in the next section, there is no other choice, that is, there is exactly one polynomial of degree two such that the value of the polynomial and its first two derivatives agree with the value of f and its first two derivatives at x = a. Similarly,

P_{3}(x) = f(a) + f′(a)(x − a) + \frac{f′′(a)}{2!}(x − a)^{2} + \frac{f′′′(a)}{3!}(x − a)^{3}

is the unique polynomial of degree three such that

P_{3}(a) = f(a), P3^{′}(a) = f′(a), P3^{′′}(a) = f′′(a), P3^{′′′}(a) = f′′′(a).

The following definition extends this idea to polynomials of degree n. We remind you that 0! = 1, 1! = 1, n! = 1 · 2 · 3 ⋯ n, f^{(0)} = f, and f^{(n)} denotes the nth derivative of f.

Definition of Taylor Polynomials. If the first n derivatives of f exist at a, then the nth degree Taylor polynomial^{1} for f at a is defined by

P_{n}(x) = f(a) + f′(a)(x − a) + \frac{f′′(a)}{2!}(x − a)^{2} + ⋯ + \frac{f^{(n)}(a)}{n!}(x − a)^{n}, (2)

or, in summation notation,

P_{n}(x) = ∑_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x − a)^{k}. (3)

The nth-degree Taylor polynomials is structured so that the value of P_{n}(x) and its first n derivatives match those of f at a. Having thus constructed P_{n}(x) like f at a, we expect P_{n}(x) to closely approximate f(x) if x is near a. It does, and in the next section we will estimate the accuracy of that approximation.

Example 1. Write the third degree Taylor polynomial for f(x) = \sqrt{x} at a = 81 and use it to approximate \sqrt{85}.

Solution. According to formula (2)

P_{3}(x) = f(81) + f′(81)(x − 81) + \frac{f^{(2)}(81)}{2!}(x − 81)^{2} + \frac{f^{(3)}(81)}{3!}(x − 81)^{3}

with f(x) = \sqrt{x}. From the computations

f(x) = \sqrt{x}, f′(x) = \frac{1}{2}x^{−1/2}, f^{(2)}(x) = −\frac{1}{4}x^{−3/2}, f^{(3)}(x) = \frac{3}{8}x^{−5/2},

f(81) = 9, f′(81) = \frac{1}{2 · 9}x^{−1/2}, f^{(2)}(81) = −\frac{1}{4 · 9^{3}} , f^{(3)}(81) = \frac{3}{8 · 9^{5}} ,

we obtain the Taylor polynomial

P_{3}(x) = 9 + \frac{1}{2 · 9}(x − 81) − \frac{1}{2 · 4 · 9^{3}} (x − 81)^{2} + \frac{1}{2 · 8 · 9^{5}} (x − 81)^{3}.

Notice that the linear part, the first two terms on the right, correspond to the tangent to f(x) = \sqrt{x} at the point (81, 9). At the x = 85,

P_{3}(85) = 9 + \frac{1}{2 · 9}(4) − \frac{1}{2 · 4 · 9^{3}} (4)^{2} + \frac{1}{2 · 8 · 9^{5}} (4)^{3} = 9 + \frac{2}{9} − \frac{2}{9^{3}} + \frac{4}{9^{5}} = 9.2195464... .

Clearly, this is a good approximation to \sqrt{85} = 9.2195444... . Actually, |P_{3}(85) − \sqrt{85} < .21 × 10^{−5}. ‡ Example 2. Write the 3rd-degree Taylor polynomial for f(x) = x^{3} − x^{2} + x − 1 at a = 2.

1After one of the contributors to the subject, English mathematician Brook Taylor (1685-1731)

Solution. First, f(2) = 5, and from the derivatives

f′(x) = 3x^{2} − 2x + 1, f′′(x) = 6x − 2, f^{(3)}(x) = 6

we have

f′(2) = 9, f′′(2) = 10, f^{(3)}(2) = 6.

Hence,

\begin{aligned}P_{3}(x) & = f(2) + f′(2)(x − 2) + \frac{f′′(2)}{2!}(x − 2)^{2} + \frac{f^{(3)}(2)}{3!}(x − 2)^{3} \\ & = 5 + 9(x − 2) + 5(x − 2)^{2} + (x − 2)^{3}. ‡ \end{aligned}

In Example 2, it important to observe that P_{3}(x) is more than just a good approximation to f(x) near a = 2; indeed, P_{3}(x) = f(x) for all x. In other words, P_{3}(x) = 5 + 9(x − 2) + 5(x − 2)^{2} + (x − 2)^{3} is simply another way of writing f(x) = x^{3} − x^{2} + x − 1. One way to see this is to expand and collect terms in P_{3}(x). In the next section we shall establish the general fact that if f(x) = c_{0} + c_{1}x + ⋯ c_{n}x^{n} is a polynomial of degree n and P_{n}(x) is the nth-degree Taylor polynomial for f at a, then P_{n}(x) = f(x) for all x.

2. Taylor’s Theorem

Now we turn to the problem of estimating the accuracy of the approximation f(x) ≈ P_{n}(x). Our method will be to find a bound for the quantity

R_{n}(x) = f(x) − P_{n}(x),

called the error of the nth remainder term. The key step is the following important result known as Taylor’s Theorem.

Theorem 1. Taylor’s Theorem If the function f has derivatives of all orders up to and including n + 1 on an open interval (a − r, a + r) about a and x is any number between a − r and a + r other than a, then there is a number c between a and x such that

f(x) = f(a) + f′(x − a) + ⋯ + \frac{f^{(n)}(a)}{n!} + \frac{f^{(n+1)}(a)}{(n + 1)!}(x − a)^{n+1}, (1)

that is, f(x) = P_{n}(x) + R_{n}(x) and

R_{n}(x) = \frac{f^{(n+1)}(c)}{(n + 1)!}(x − a)^{n+1}. (2)

Further, if |f^{(n+1)}(t)| ≤ M on the interval from a to x,

|R_{n}(x)| ≤ \frac{M}{(n + 1)!}|x − a|^{n+1}. (3)

We shall refer to formula (1) as Taylor’s formula with remainder. Once a and n are chosen, it expresses f(x) as the sum of a Taylor polynomial of degree n in powers of x − a plus a remainder (2) which we will routinely estimate by (3).

Proof. The proof rests on a shrewd choice of a constant K followed by successive applications of Rolle’s Theorem to the function g defined by

g(t) = f(t) − P_{n}(t) − K(t − a)^{n+1} (4) in which t varies over the closed interval from a to x. First, because f(a) = P_{n}(a), g(a) = f(a) − P_{n}(a) − K(a − a)^{n+1} = 0 and, also,

g(x) = f(x) − P_{n}(x) − K(x − a)^{n+1} = 0

provided

K = \frac{f(x) − P_{n}(x)}{(x − a)^{n+1}} .

Remember: x is fixed, t is the variable. Using this value of K, we may apply Rolle’s Theorem on the interval from a to x to conclude that g′(c_{1}) = 0 for at least one choice of c_{1} between a and x. f′(a) = Pn^{′}(a), the derivative g′(t) = f′(t) − Pn^{′}(t) − (n + 1)(t − a)^{n} is also zero at t = a. Hence g′(a) = g′(c_{1}) = 0, and a second application of Rolle’s Theorem shows that there is a number c_{2} between c_{1} and a at which g′′(c_{2}) = 0.

The idea is to continue attacking with Rolle’s Theorem, obtaining numbers c_{1} ≤ c_{2} < ⋯ c_{k} at which g′(c_{1}) = 0, g′(c_{2}) = 0, ..., g′(c_{k}) = 0. This is possible so long as g^{(k)}(a) = 0. Since f^{(k)}(a) = P_{n}^{(k)}(a) for k = 0, 1, ..., n and

g^{(k)}(t) = f^{(k)}(t) − P_{n}^{(k)}(t) − (n + 1)(n)(n − 1) ⋯ (n − k + 2)K(t − a)^{n+1−k},

it follows that g^{(k)}(a) = 0 if k = 0, 1, ..., n but

\begin{aligned}g^{(n+1)}(t) & = f^{(n+1)}(t) − 0 − (n + 1)!K \\ & = f^{(n+1)}(t) − (n + 1)!K \end{aligned}

need not be zero at t = a. Therefore, after n applications, we find that g^{(n)}(a) = g^{(n)}(c_{n}) = 0. A final approximation of Rolle’s Theorem yields a number c_{n+1} = c between c_{n} and a such that g^{(n+1)}(c) = 0, that is

\begin{aligned}g^{(n+1)}(c) & = f^{(n+1)}(c) − (n + 1)!K = 0 \\ K & = \frac{f^{(n+1)}(c)}{(n + 1)!} . \end{aligned}

Substitution of this new expression for K in (4) gives

g(t) = f(t) − P_{n}(t) − \frac{f^{(n+1)}(c)}{(n + 1)!}(t − a)^{n+1}.

But g(x) = 0. Hence

f(x) = P_{n}(x) + \frac{f^{(n+1)}(c)}{(n + 1)!}(x − a)^{n+1}.

This establishes formula (1).

Estimate (3) follows directly from the remainder formula (2). For if |f^{(n+1)}(t)| ≤ M everywhere between a and x, then, in particular |f^{(n+1)}(c)| ≤ M and

|R_{n}(x)| = |\frac{f^{(n+1)}(c)}{(n + 1)!}(x − a)^{n+1}| ≤ \frac{M}{(m + 1)!}|x − a|^{n+1}. ‡

Example 1. (a) Write Taylor’s formula with remainder for n = 3, a = 0 and f(x) = \sqrt{1 + x}. (b) Use it to approximate \sqrt{1.15} and estimate the error.

Solution. (a) The third degree Taylor polynomial for f(x) = \sqrt{1 + x} at a = 0 is

P_{3}(x) = 1 + \frac{1}{2}x − \frac{1}{8}x^{2} + \frac{1}{16}x^{3},

by Example 3 of the last section and, since f^{(4)}(x) = \frac{−3·5}{2^{4}}(1 + x)^{−7/2},

R_{3}(x) = −\frac{3 · 5}{2^{4} · 4!}(1 + c)^{−7/2}x^{4}

for some c between 0 and x. Hence Taylor’s formula (1) is

\sqrt{1 + x} = 1 + \frac{1}{2}x − \frac{1}{8}x^{2} + \frac{1}{16}x^{3} − \frac{3 · 5}{2^{4} · 4!}(1 + c)^{−7/2}x^{4}.

(b) At x = 0.15, this formula gives the approximation

\sqrt{1.15} ≈ 1 + \frac{1}{2}(0.15) − \frac{1}{8}(0.15)^{2} + \frac{1}{16}(0.15)^{3}

rounded to six places.

Estimation of the error

R_{3}(0.15) = −\frac{15}{384}(1 + c)^{−7/2}(0.15)^{4}

begins with the admission that we simply don’t know the value of c; the theory only tells us that 0 < c < 0.15. The best we can hope for is to put an upper bound on the error. To

do this, we argue that, since 0 < c, 1 < 1 + c and, therefore, (1 + c)^{−7/2} = \frac{1}{(1 + c)^{−7/2}} < 1.

This gives the bound

\begin{aligned}|R_{3}(0.15)| & = |−\frac{15}{384}(1 + c)^{−7/2}(0.15)^{4}| ≤ \frac{15}{384}(0.15)^{4} \\ & = 0.00001977... . \end{aligned}

Consequently, |R_{3}(0.15)| < 0.00002 and the approximation \sqrt{1.15} ≈ 1.072398 is off by no more than 2 units in the fifth decimal place. Thus

\sqrt{1.15} ≈ 1.0724 ,

meaning that 1.0724 is the best 4 decimal place approximation to \sqrt{1.15}. ‡

Some remarks on procedure are now in order. In the initial approximation \sqrt{1.15} ≈ 1.072398, we rounded to six decimal places, our strategy being to keep more digits than would be justified in the final result because we were uncertain as to the magnitude of the remainder term. Then, once we obtained a bound for the error, we rounded further to reflect the accuracy justified. Also, we should observe that the process of bounding the remainder usually amounts to placing c at one end or the other of the interval connecting a to x. Although useful, the resulting estimate may be too great by several factors.

The use of absolute values in the error estimates (3) is natural because usually we are only interested in the magnitude of the error, not whether it is positive or negative. Nonetheless, it is frequently worthwhile to look at the sign of R_{n}(x) when making the approximation f(x) ≈ P_{n}(x). Thus, in the preceding example, we can see that \sqrt{1.15} < 1.072398 because R_{3}(0.15) < 0.

Example 2. In applying Taylor’s formula (1) as an approximation tool, it is essential that x, the point at which the approximation is desired, be as near as possible to the base point a, where the derivatives are to be computed. On the other hand, you want the derivatives to be easily computable. In practice, the choice of a a represents a compromise between these two goals. Take f(x) = x^{x} so that f^{(k)}(x) = e^{x}; hence the only reasonable choice is a = 0 because f^{(k)}(0) = 1 (easy to compute) for k = 0, 2, 3, .... Taylor’s formula with remainder at a = 0 for f(x) = e^{x} is

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ⋯ + \frac{x^{n}}{n!} + \frac{e^{c}}{(n + 1)!}x^{n+1},

where c is between 0 and x. At x = \frac{1}{2} = 0.5, this formula reduces to

\sqrt{e} = e^{0.5} = 1 + (0.5) + \frac{(0.5)^{2}}{2!} + ⋯ + \frac{(0.5)^{n}}{n!} + \frac{e^{c}}{(n + 1)!}(0.5)^{n+1}

and the challenge is to find a small positive integer n such that

R_{n}(0.5) = \frac{e^{c}}{(n + 1)!}(0.5)^{n+1} < 0.5 × 10^{−6}.

First we find a bound for e^{c}. Since e < 3 and 0 < c < 0.5, it follows that e^{c} < 3^{0.5} < 2. Therefore

R_{n}(0.5) < \frac{2}{(n + 1)!}(0.5)^{n+1}.

If, now, n is chosen so that the right side is less than 0.5×10^{−6}. The smallest such n, found by examining the following table, is n = 7 where R_{7}(0.5) < 0.2 × 10^{−6}.

TABLE 13.1

\begin{matrix}n & \frac{2}{(n + 1)!}(0.5)^{n+1} \\ 1 & 0.25 \\ 2 & 0.04166667 \\ 3 & 0.00520833 \\ 4 & 0.00052083 \\ 5 & 0.00004340 \\ 6 & 0.00000310 \\ 7 & 0.00000019 \end{matrix}

Hence the approximation

\begin{aligned}\sqrt{e} & ≈ 1 + (0.5) + \frac{(0.5)^{2}}{2} + ⋯ + \frac{(0.5)^{7}}{7!} \\ & ≈ 1.6487212 \end{aligned}

has an error of at most 2 units in the seventh decimal place and, therefore, \sqrt{e} ≈ 1.648721. ‡

In this example we obtained the best (rounded) six decimal place approximation by forcing the error under 0.5 × 10^{−6}. This suggests that five place accuracy will be obtained by forcing the error under 0.5×10^{−5}, four place accuracy if the error is less than 0.5×10^{−4}, and so on. This often is the case and the rule is worth using, but it does not always work as may be seen in the problem of approximating

ln 1.25 = 0.2231435...

to four place accuracy. Suppose we obtain the approximation 0.22314 with a maximum error of 0.5 × 10^{−4}, meaning that ln 1.25 is greater than or equal to 0.22314 − 0.00005 = 0.22309 and less than or equal to 0.22134 + 0.00005 = 0.22319. However, approximation is obtained by rounding down to 0.2231 or up to 0.2232.

Example 3. (a) If f(x) = ln x, write Taylor’s formula with remainder for n = 5 at a = 1.

(b) Use the formula to approximate ln 0.8 and estimate the error.

Solution. (a) Based on the derivatives

f(x) = ln , f(1) = 0  f^{(4)}(x) = −\frac{3!}{x^{4}} , f^{(4)}(1) = −3!

f′(x) = \frac{1}{x}, f′(1) = 1  f^{(5)}(x) = \frac{4!}{x^{5}} , f^{(5)}(1) = 4!

f^{(2)}(x) = \frac{1}{x^{2}} , f^{(2)}(1) = −1 f^{(6)}(x) = −\frac{5!}{x^{6}} , f^{(6)}(c) = −\frac{5!}{c^{6}}

f^{(3)}(x) = \frac{2}{x^{3}} , f^{(3)}(1) = 2

the fifth degree Taylor’s formula is

ln x = (x − 1) − \frac{1}{2}(x − 1)^{2} + \frac{1}{3}(x − 1)^{3} − \frac{1}{4}(x − 1)^{4} + \frac{1}{5}(x − 1)^{5} − \frac{1}{6c^{6}} (x − 1)^{6}.

In the remainder

R_{6}(x) = −\frac{1}{6c^{6}} (x − 1)^{6},

c denotes a number between 1 and x.

(b) The error in the approximation

\begin{aligned}ln 0.8 & ≈ −0.2 − \frac{1}{2}(−0.2)^{2} + \frac{1}{3}(−0.2)^{3} − \frac{1}{4}(−0.2)^{4} + \frac{1}{5}(−0.2)^{5} \\ & ≈ −0.223131 \end{aligned}

is

R_{6}(0.8) = −\frac{1}{6c^{6}} (−0.2)^{6}.

It is worth recording that R_{6}(0.8) < 0 so that ln 0.8 < −0.223131. Because 0.8 < c <

1, \frac{1}{c} < \frac{1}{0.8} and \frac{1}{c^{6}} < \frac{1}{(0.8)^{6}} ; therefore,

|R_{6}(0.8)| = −\frac{1}{6c^{6}} (−0.2)^{6}| = \frac{1}{6c^{6}} (0.2)^{6} < \frac{1}{6} (\frac{0.2}{0.8})^{6} = 0.0000406... .

Hence −0.000041 < R_{6}(0.8) < 0 and

−0.223172 < ln 0.8 < −0.223131.

From this we can conclude that ln 0.8 = −0.2231..., meaning that the first four decimal digits are the actual digits in the complete representation of ln 0.8, but only that ln 0.8 ≈ −0.223, not knowing whether the fifth digit is less than or greater than or equal to 5. Note that the error is less than 0.5 × 10^{−4}. ‡ In the Taylor’s formula expansion of ln x in powers of x − 1 plus a remainder,

\begin{aligned}ln x & = (x − 1) − \frac{1}{2}(x − 1)^{2} + ⋯ + \frac{(−1)^{n−1}}{n} (x − 1)^{n} + \frac{(−1)^{n+1}}{n + 1} \frac{1}{c^{n+1}} (x − 1)^{n+1} \\ & = P_{n}(x) + R_{n}(x), \end{aligned}

the approximation ln x ≈ P_{n}(x) is good for small n only so long as x is near 1. The following table illustrates this dependence for two values of x. The actual value ln 1.2 = 0.1823215... is approximated well by P_{n}(1.2).

Table 13.2

n P_{n}(1.2) ACTUAL ERROR P_{n}(2)  ACTUAL ERROR
1 0.2  − 0.017678...  1  −0.306852...
2 0.18  0.002321...  0.5  0.193147...
3 0.182266... −0.000345...  0.833333... −0.140186...
4 0.182266... 0.000054...  0.583333... 0.1098138...
5 0.182330... −0.000009...  0.783333... −0.090186...
6 0.1823200 0.000001...  0.616666... 0.076480...

for n = 2, , 3, 4, 5, 6. In fact P_{6}(0.8) = 0.1823200 is accurate through 5 decimal places. But at x = 2, the approximation of the actual value ln 2 = 0.693147... by P_{n}(2) is worthless for small n. At n = 6 the best estimate we can make for the remainder R_{6}(2) = \frac{1}{7}c^{−7} = \frac{1}{7}c^{−7} is R_{6}(2) < \frac{1}{7} ≈ 0.14, while the absolute value of the actual error in the approximation

P_{n}(2) ≈ ln 2 is about half the remainder, that is, |tlj 2 − P_{n}(2)| ≈ \frac{1}{2(n + 1)}. This means

it would take an n of about 10,000 to reduce the error to 0.5 × 10^{−4}. Such computations dramatically illustrate why we insist that x − a be small in Taylor’s formula when the goal is the approximation of f(x). In general, x is given and the initial problem is to find a nearby point a at which f and its derivatives can be easily calculated.

Example 4. Approximate cos 68^{∘} to four decimal place accuracy.

Solution. The point nearest to 68^{∘} at which f(x) = cos x and its derivatives are easily evaluated is a = 60^{∘} = \frac{π}{3} radians. For this reason our decision is to expand cos x in powers of x − \frac{π}{3} plus a remainder. Since | sin c| ≤ 1 and | cos c| ≤ 1, we have the estimate

|rn(x)| = |\frac{f^{n+1}(c)}{(n + 1)!} (x − \frac{π}{3} )^{n+1}| ≤ \frac{1}{(n + 1)!} |x − \frac{π}{3} |^{n+1} .

To obtain the desired accuracy, we let x = \frac{68π}{180} in the remainder and increase n until

\frac{1}{(n + 1)!} |\frac{68π}{180} − \frac{π}{3} |^{n+1} < 0.5 × 10^{−4}.

This inequality is first satisfied at n = 3 where R_{3} (\frac{68π}{180} ) < 0.000012. From the derivatives

f(x) = cos x, f (\frac{π}{3} ) = \frac{1}{2},  f^{(3)}(x) = sin x, f^{(3)} (\frac{π}{3} ) = \frac{\sqrt{3}}{2},
f′(x) = − sin x, f′ (\frac{π}{3} ) = −\frac{\sqrt{3}}{2}, f^{(4)}(x) = cos x, f^{(4)}(c) = cos c;
f^{(2)}(x) = − cos x, f^{(2)} (\frac{π}{3} ) = −\frac{1}{2},

it follows that Taylor’s formula with remainder at a = \frac{π}{3} and with n = 3 is

cos x = \frac{1}{2} − \frac{\sqrt{3}}{2} (x − \frac{π}{3} ) − \frac{1}{2} \frac{1}{2!} (x − \frac{π}{3} )^{2} + \frac{\sqrt{3}}{2} \frac{1}{3!} (x − \frac{x}{3})^{3} + \frac{cos c}{4!} (x − \frac{x}{3})^{3} .

Hence, in the approximation

cos 68^{∘} = cos \frac{68π}{180} ≈ \frac{1}{2} − \frac{\sqrt{3}}{2} (\frac{68π}{180} ) − \frac{1}{2} \frac{1}{2!} (\frac{68π}{180} )^{2} + \frac{\sqrt{3}}{2} \frac{1}{3!} (\frac{68π}{180} )^{3}

the error is no more than two units in the fifth decimal digit and, therefore, cos 68^{∘} ≈ 0.3746 rounded to four decimal places. ‡

Our final example on approximation illustrates the power of Taylor’s formula in a different setting.

Example 5. Approximate the integral ∫_{0}^{0·2} \sqrt{x} sin x dx by making a substitution based

on the approximation sin x ≈ x − \frac{x^{3}}{3!} . Estimate the error.

Solution. Since the integral is one which cannot be evaluated by finding an antiderivative,

we simply replace sin x by x − \frac{x^{3}}{3!} and write

∫_{0}^{0·2} \sqrt{x} sin x dx ≈ ∫_{0}^{0·2} \sqrt{x} (x − \frac{x^{3}}{3!} ) dx ∫_{0}^{0·2} (x^{3/2} − \frac{1}{6}x^{7/2}) dx
= [\frac{2}{5}x^{5/2} − \frac{1}{27}x^{9/2}]_{0}^{0·2} = \frac{2}{5}(0.2)^{5/2} − \frac{1}{27}(0.2)^{9/2}
≈ 0.00712892 .

Of course, the error committed depends on the size of the remainder term in the expansion

sin x = x = 0 + x − 0 · \frac{x^{2}}{2!} − \frac{x^{3}}{3!} + 0 · \frac{x^{4}}{4!} + cos c · \frac{x^{5}}{5!}
= x − \frac{x^{3}}{3!} + + cos c · \frac{x^{5}}{5!} .

In the remainder terms R_{4}(x), c is between 0 and x. (We could have chosen n = 3, but the term involving x^{4} is absent and R_{4}(x) is smaller than R_{3}(x) in absolute value on the interval [0, 0.2].) It follows from the last equation that the difference between the function and its Taylor approximation is given by

\sqrt{x} sin x − \sqrt{x} (x − \frac{x^{3}}{3!} ) = \sqrt{x} cos c

and hence that

|\sqrt{x} sin x − \sqrt{x} (x − \frac{x^{3}}{3!} ) | = |\sqrt{x} cos c \frac{x^{5}}{5!} | ≤ \frac{x^{11/2}}{5!}

on the interval [0, 0.2]. This inequality provides a concrete statement about how well

\sqrt{x} (x − \frac{x^{3}}{3!} ) approximates \sqrt{x} sin x. Next we use the integral inequality |∫_{a}^{b} ((u(x) − v(x)) dx| ≤

∫_{a}^{b} |u(x) − v(x)| dx to obtain the following estimate:

\begin{matrix}| ∫_{0}^{0·2} \sqrt{x} sin x dx − ∫_{0}^{0·2} \sqrt{x} x − \frac{x^{3}}{3!} dx| ≤ ∫_{0}^{0·2} |\sqrt{x} sin x − \sqrt{x} (x − \frac{x^{3}}{3!} ) | dx ≤ ∫_{0}^{0·2} \frac{x^{11/2}}{5!} dx \\ = \frac{1}{120} \frac{2}{13} (0.2)^{13/2} = 0.000000036... . \end{matrix}

To sum up, the error in the approximation ∫_{0}^{0·2} \sqrt{x} sin x dx ≈ 0.00712892 is no more than

0.00000004, accordingly, ∫_{0}^{0·2} \sqrt{x} sin x dx ≈ 0.007129” rounded to six decimal places. ‡

3. L’Hospital’s Rule: The Indeterminate Form \frac{0}{0}

In our previous work we have been frequently asked to find the limit of a quotient f(x)/g(x). The method introduced in Section 2.5 asserted that if lim_{x→a} f(x) = A and lim_{x→a} g(x) = B, and B ≠ 0, then

lim_{x→a} \frac{f(x)}{g(x)} = \frac{A}{B} .

However, nothing was said there about the case B = 0. Of course, one possibility is B = 0 and A ≠ 0. If, for instance, f(x) = x, g(x) = x − 4, and a = 4, then

A = lim_{x→4} f(x) = 4, B = lim_{x→4} g(x) = 0

and

lim_{x→4} \frac{f(x)}{g(x)} = lim_{x→4} \frac{x}{x − 4}

fails to exist because the ratio increases to ∞ as x → 4^{+} and decreases to −∞ as x → 4^{−}. The more interesting situation is B = 0 and A = 0. In this case the limit of the quotient f(x)/g(x) is said to be “indeterminate at a” because it assumes the form “0/0” as x approaches a. The actual limit may or may not exist and, if it does any real number is a possible limit. Consider f(x) = x and g(x) = x^{2}. Then lim_{x→0} f(x) = 0, lim_{x→0} g(x) = 0 and

lim_{x→0} \frac{f(x)}{g(x)} = lim_{x→0} \frac{x}{x^{2}} = lim_{x→0} \frac{1}{x} does not exist);

but the functions f(x) = k sin x (k a real number) and g(x) = x both have limit zero at x = 0 and, as we saw in Section ,

lim_{x→0} \frac{f(x)}{g(x)} = lim_{x→0} \frac{k sin x}{x} = k lim_{x→0} \frac{sin x}{x} = k · 1 = k.

3. L’HOSPITAL’S RULE: THE INDETERMINATE FORM \frac{0}{0} 135

Taylor’s formula suggests a method for evaluation limits which have the indeterminate form 0/0.

Let f and g be two functions which have continuous second derivatives in some open interval about a. Then, using Taylor’s formula, we may write

f(x) = f(a) + f′(a)(x − a) + \frac{f′′(z_{1})}{2}(x − a)^{2}

for some z_{1} between a and x, and

g(x) = g(a) + g′(a)(x − a) + \frac{g′′(z_{2})}{2}(x − a)^{2}

for some z_{2} between a and x. Now assume that

lim_{x→a} f(x) = f(a) = 0 and lim_{x→a} g(x) = g(a) = 0.

Thus for x in the indicated interval and x ≠ a,

\begin{aligned}\frac{f(x)}{g(x)} & = \frac{f(a) + f′(a)(x − a) + \frac{f′′(z_{1})}{2}(x − a)^{2}}{g(a) + g′(a)(x − a) + \frac{g′′(z_{2})}{2}(x − a)^{2}} = \frac{f′(a)(x − a) + \frac{f′′(z_{1})}{2}(x − a)^{2}}{g′(a)(x − a) + \frac{g′′(z_{2})}{2}(x − a)^{2}} \\ & = \frac{f′(a) + \frac{f′′(z_{1})}{2}(x − a)}{g′(a) + \frac{g′′(z_{2})}{2}(x − a)}. \end{aligned}

As x → a, z_{1} and z_{2} must also tend to a and the second term in both the numerator and denominator of this last quotient tend to zero. Hence

lim_{x→a} \frac{f(x)}{g(x)} = \frac{f′(a)}{g′(a)}

provided that g′(a) ≠ 0. This analysis suggests that, under certain conditions, a quotient f(x)/g(x) that has the indeterminate form 0/0 at x = a and the quotient of derivatives f′(x)/g′(x) have the same limit as x tends to a. The various conditions that permit this procedure are commonly known as L’Hospital’s rules after the 17th century French mathematician who popularized them.

Theorem 2. Suppose that f and g are continuous functions on the closed interval [a, b], differentiable on the open interval (a, b), and that

lim_{x→a^{+}} f(x) = f(a) = 0, lim_{x→a^{+}} g(x) = g(a) = 0. (1)

Suppose also that

g′(x) ≠ 0 for a < x < b. (2)

Then,

(3) if lim_{x→a^{+}} \frac{f′(x)}{g′(x)} = ∞ [or −∞], it follows that lim_{x→a^{+}} \frac{f(x)}{g(x)} = ∞ [or −∞].

Proof. Our proof skirts Taylor’s formula, depending rather on an approach which avoids assuming that f is differentiable at x = a. To begin, we choose a number x, a < x < b, and consider the new function

H(t) = f(x)g(t) − g(x)f(t), a ≤ t ≤ x. (5)

As a function of t, H is continuous on [a, x] and differentiable for a < t < x. Further, H(x) = f(x)g(x) − g(x)f(x) = 0, H(a) = f(x)g(a) − g(x)f(a) = 0 by (1) and, since x is fixed,

H′(t) = f(x)g′(t) − g(x)f′(t).

According to Rolle’s theorem this derivative must be zero at least once between 0 and x. Hence there is a number z, a < z < x, such that H′(z) = f(x)g′(z) − g(x)f′(z) = 0 or, since g′(z) ≠ 0 by (2),

\frac{f(x)}{g(x)} = \frac{f′(z)}{g′(z)}.

Because a < z < x, we see that z → a^{+} if x → a^{+} and, therefore,

lim_{x→a^{+}} \frac{f(x)}{g(x)} = lim_{x→a^{+}} \frac{f′(z)}{g′(z)} = lim_{z→a^{+}} \frac{f′(z)}{g′(z)}.

Clearly, if lim_{z→a^{+}} \frac{f′(z)}{g′(z)} = L, then lim_{x→a^{+}} \frac{f(x)}{g(x)} = L. On the other hand, if \frac{f′(z)}{g′(z)} increases

without bound as z → a^{+}, that is if lim_{z→a^{+}} \frac{f′(z)}{g′(z)} = ∞, then lim_{x→a^{+}} \frac{f(x)}{g(x)} = ∞ also. The same

analysis applies if lim_{z→a^{+}} \frac{f′(z)}{g′(z)} = −∞. ‡

Although our result applies only to right-hand limits, the same argument establishes the result for left-hand limits. In combination, the result for right- and left-hand limits establishes the same result for two-sided limits. Thus the results of Theorem 2 are valid if lim_{x→a^{+}} is replaced everywhere by lim_{x→a^{−}} or everywhere by lim_{x→a} and the assumptions on f and

g are modified accordingly.

Example 1. Find lim_{x→0^{+}} \frac{e^{x} − 1}{\sqrt{x}}.

Solution. Since lim_{x→0^{+}} (e^{x} − 1) = lim_{x→0^{+}} \sqrt{x} = 0, the given limit has the indeterminate form

0/0. Further, both numerator and denominator are continuous for 0 ≤ x, differentiable for 0 < x and \frac{d}{dx}(\sqrt{x}) is never 0. Thus, L’Hospital’s rule,

lim_{x→0^{+}} \frac{e^{x} − 1}{\sqrt{x}} = lim_{x→0^{+}} \frac{\frac{d}{dx}[e^{x} − 1]}{\frac{d}{dx}[\sqrt{x}]} = lim_{x→0^{+}} \frac{e^{x}}{\frac{1}{2\sqrt{x}}} = lim_{x→0^{+}} \frac{1}{2}\sqrt{x}e^{x} = 0.

Note that \sqrt{x} does not have a derivative at x = 0. ‡

Remark. Strictly speaking, the first equality in the calculation is not justified until the limit on the right is shown to exist or to tend to ∞ or −∞. However, it is common practice in applying L’Hospital’s rule to streaming the computations as shown.

Example 2. Find lim_{x→0} \frac{sin 4x}{3x}. Solution. This quotient also has the indeterminate form

0/0 because lim_{x→0} sin 4x = lim_{x→0} sin 3x = 0. Hence we may apply L’Hospital’s rule to find that

lim_{x→0} \frac{sin 4x}{sin 3x} = lim_{x→0} \frac{\frac{d}{dx}[sin 4x]}{\frac{d}{dx}[sin 3x]} = lim_{x→0} \frac{4 cos 4x}{3 cos 3x} = \frac{4}{3}. ‡

Example 3. Find lim_{x→0} \frac{xe^{x} − sin x}{sin (x^{2})}.

Solution. First we check to see that lim_{x→0} (xe^{x} − sin x) = lim_{x→0} sin (x^{2}) = 0. Then, by L’Hospital’s rule,

lim_{x→0} \frac{xe^{x} − sin x}{sin (x^{2})} = lim_{x→0} \frac{\frac{d}{dx}[xe^{x} − sin x]}{\frac{d}{dx}[sin (x^{2})]} = lim_{x→0} \frac{xe^{x} + e^{x} − cos x}{2x cos (x^{2})}.

But the last limit tells us nothing because lim_{x→0} (xe^{x} + e^{x} − cos x) = lim_{x→0} 2x cos (x^{2}) = 0, that is, we are again faced with an indeterminate form 0/0. Our response is to apply L’Hospital’s rule a second time:

lim_{x→0} \frac{xe^{x} + e^{x} − cos x}{2x cos (x^{2})} = lim_{x→0} \frac{\frac{d}{dx}[xe^{x} + e^{x} − cos x]}{\frac{d}{dx}[2x cos (x^{2})]} = lim_{x→0} \frac{2e^{x} + xe^{x} + sin x}{2 cos (x^{2}) − 4x^{2} sin (x^{2})} = \frac{2}{2} = 1.

Therefore

lim_{x→0} \frac{xe^{x} − sin x}{sin (x^{2})} = 1. ‡

Example 4. Find lim_{θ→0} \frac{2 sin θ − sin 2θ}{θ − sin θ}.

Solution. This limit may be found by three applications of L’Hospital’s rule:

lim_{θ→0} \frac{2 sin θ − sin 2θ}{θ − sin θ} = lim_{θ→0} \frac{2 cos θ − cos 2 cos 2θ}{1 − cos θ} = lim_{θ→0} \frac{−2 sin θ + 4 sin 2θ}{sin θ}
= lim_{θ→0} \frac{−2 cos θ + 8 cos 2θ}{cos θ} = \frac{6}{1} = 6. ‡

CAUTION. Never apply L’Hospital’s rule lim_{x→a} \frac{f(x)}{g(x)} = lim_{x→a} \frac{f′(x)}{g′(x)} without first verifying

that lim_{x→a} f(x) = lim_{x→a} g(x) = 0. For example, lim_{x→1} \frac{sin \frac{π}{6} x}{1 + ln x} = \frac{1}{2} because lim_{x→1} (1 + ln x) = 1 and lim_{x→1} sin \frac{π}{6} x = \frac{1}{2}. Clearly, it is not legitimate to apply L’Hospital’s rule. Also, keep in mind that L’Hospital’s rule replaces the quotient f(x)/g(x) with the quotient of derivatives f′(x)/g′(x), which is not at all the same as differentiating f(x)/g(x) according to the quotient rule.

4. Other Indeterminate Forms A second version of L’Hospital’s rule applies to quotients f(x)/g(x) in which both f(x)

and g(x) tend to one of ∞ or −∞ as x → a^{+}, x → a^{−}, x → a, x → ∞, or x → −∞. For example, it may be that f(x) → −∞ and g(x) → ∞ as in

lim_{x→0^{+}} \frac{ln x}{csc x},

or that f(x) → ∞ and g(x) → ∞ as in

lim_{x→∞} \frac{x^{2}}{e^{x}} .

We cover all possibilities by saying that f(x)/g(x) has the indeterminate form \frac{∞}{∞}.

Theorem 3. L’Hospital’s rule \frac{∞}{∞}.

Let f and g be differentiable functions in the open interval (a, b) and suppose that (a) lim_{x→a^{+}} f(x) = ∞(or − ∞) and lim_{x→a^{+}} g(x) = ∞(or − ∞).

Suppose also that

(b) g′(x) ≠ 0 for a < x < b. Then, if

(c) lim_{x→a^{+}} \frac{f′(x)}{g′(x)} = L or is ∞ or −∞,

it follows that

lim_{x→a^{+}} \frac{f(x)}{g(x)} = lim_{x→a^{+}} \frac{f′(x)}{g′(x)}.

Exactly as in Theorem 2, this result remains valid if the limit x → a^{+} is replaced by any one of x → a^{−}, x → a, x → ∞, or x → −∞. The proof, while following the pattern set in proving L’Hospital’s rule for the case 0/0, is tedious and we omit it. It can be found in any advanced calculus text.

Example 1. Find (a) lim_{x→∞} \frac{x}{e^{x}} and (b) lim_{x→∞} \frac{x^{2}}{e^{x}}.

Solution. Since lim_{x→∞} x = lim_{x→∞} x^{2} = ∞ and lim_{x→∞} e^{x} = ∞, both quotients have the indeterminate form ∞/∞. Thus, by L’Hospital’s rule,

(a)  lim_{x→∞} \frac{x}{e^{x}} = lim_{x→∞} \frac{\frac{d}{dx}[x]}{\frac{d}{dx}[e^{x}]} = lim_{x→∞} \frac{1}{e^{x}} = 0

and, applying the rule twice,

(b)  lim_{x→∞} \frac{x^{2}}{e^{x}} = lim_{x→∞} \frac{\frac{d}{dx}[x^{2}]}{\frac{d}{dx}[e^{x}]} = lim_{x→∞} \frac{2x}{e^{x}} = lim_{x→∞} \frac{\frac{d}{dx}[2x]}{\frac{d}{dx}[e^{x}]} = lim_{x→∞} \frac{2}{e^{x}} = 0. ‡

More generally, we may choose any positive integer k and apply L’Hospital’s rule k times to find that

lim_{x→∞} \frac{x^{k}}{e^{x}} = lim_{x→∞} \frac{k!}{e^{x}} = 0,

thereby showing that e^{x} increases much faster than x^{k}, for any positive integer k.

Example 2. Find lim_{x→∞} \frac{ln x}{x^{2}}.

Solution. Both numerator and denominator tend to ∞ as x → ∞. Hence

lim_{x→∞} \frac{ln x}{x^{2}} = lim_{x→∞} \frac{\frac{1}{x}}{2x} = lim_{x→∞} \frac{1}{2x^{2}} = 0.

Thus, ln x increases at a much slower rate than x^{2}. ‡ Example 3. Find lim_{x→0^{+}} \frac{ln (sin x)}{ln x}.

Solution. Since both numerator and denominator tend to −∞ as x → 0^{+},

lim_{x→0^{+}} \frac{ln (sin x)}{ln x} = lim_{x→0^{+}} \frac{\frac{cos x}{sin x}}{\frac{1}{x}} = lim_{x→0^{+}} \frac{x cos x}{sin x}.

To proceed, note that the last quotient has the indeterminate form 0/0. Hence

lim_{x→0^{+}} \frac{ln (sin x)}{ln x} = lim_{x→0^{+}} \frac{x cos x}{sin x} = lim_{x→0^{+}} \frac{cos x − x sin x}{cos x} = 1.

or

lim_{x→0^{+}} \frac{x cos x}{sin x} = lim_{x→0^{+}} \frac{x}{sin x} · lim_{x→0^{+}} cos x = lim_{x→0^{+}} \frac{1}{cos x} · lim_{x→0^{+}} cos x = 1 · 1 = 1 ‡

We now turn our attention to some limit problems which can be treated by our technique only after some algebraic manipulations. Consider, for example, the product f(x)g(x) in which f(x) → 0 which g(x) → ∞ or −∞. We say that the resulting indeterminate form is of type 0.∞ and we handle it by rearranging f(x)g(x) to

\frac{\frac{f(x)}{1}}{g(x)} or \frac{\frac{g(x)}{1}}{f(x)},

thus reducing the product to one of the forms \frac{0}{0} or \frac{∞}{∞}.

Example 4. Find lim_{x→0^{+}} x ln x.

Solution. This product is indeterminate of type 0·∞ since lim_{x→0^{+}}x = 0 and lim_{x→0^{+}} ln x = −∞.

We change it into the form ∞/∞ and apply L’Hospital’s rule as follows:

lim_{x→0^{+}} x ln x = lim_{x→0^{+}} \frac{\frac{ln x}{1}}{\frac{1}{x}} = lim_{x→0^{+}} \frac{\frac{1}{x}}{−\frac{1}{x^{2}}} = lim_{x→0^{+}} (−x) = 0. ‡

PROBLEM SET 19.1

For each of the polynomials in Problems 1–3, write the Taylor polynomial of the same degree at the given value of a and show algebraically that the two polynomials are identical.

\begin{matrix}1. p(x) = 3x + 5, a = 1 & 2. p(x) = x^{5} − 1, a = 1 \\ 3. p(x) = (2x − 3)^{3}, a = −2 \end{matrix}

4. Write the third degree Taylor polynomial for f(x) = 1/\sqrt{x} at x = 81 and use it to approximate \sqrt{83}. Compared with the calculator value of \sqrt{83}; what is the error?

5. Write the fifth degree Taylor polynomial for f(x) = ln (1 + x) at x = 1. Use it to approximate 1/\sqrt{83}. What is the error?

In Problems 6–9, calculate the Taylor polynomial of the indicated degree at a.

6. f(x) = x ln x; n = 3, a = 1  7. h(t) = sin t; n = 4, a = \frac{π}{3}
8. f(t) = \sqrt{\frac{1+t}{1−t}}; n = 2, a = 0  9. f(x) = sin^{2} x; n = 5, a = π

10. If p(x) = 3x^{3} + 7x^{2} + 10x + 2 is the third degree Taylor polynomial at a = −1 for f, find f′′(−1).

11. Show that e^{x} > 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} if x > 0. [Hint: Show that the function F, obtained by subtracting the polynomial from e^{x}, and its first three derivatives are zero at x = 0 and that the fourth derivative is positive. Then integrate.]

12. On a single coordinate system plot sin x and its Taylor polynomials P_{1}, P_{3}, and P_{5} at a = 0.

PROBLEM SET 19.2

For each of the functions in Problems 1–5, write Taylor’s formula with remainder for the given values of a and n.

1. \frac{1}{1 − x}; a = 0, n = 6  2. (+x)^{−1/2}; a = 0, n = 4

3. ln \frac{1 + x}{1 − x}; a = 0, n = 5  4. sin x; a = 0, n = 6

5. cos x3x; a = \frac{π}{2} , n = 4

In Problems 6–8, use Taylor’s formula with remainder at a = 0 with n = 3 to approximate the indicated quantity. Estimate the error and round your approximation to the number of decimal places justified.

\begin{matrix}6. \sqrt{1.13} & 7. e^{0.2} \\ 8. ln 0.98 \end{matrix}

In Problems 9–11, use Taylor’s formula with remainder to find the best (rounded) four decimal place approximation to the given quantity. As a first step, choose n so that |R_{n}(x)| < 0.5 × 10^{−4}.

\begin{matrix}9. \sqrt{1.2} & 10. cos 29^{∘} \\ 11. ln 1.25 \end{matrix}

12. The problem is to find the best four decimal place approximation to ∫_{0}^{0.3} sin^{4} θ dθ. Compare the effort in (a) using the approximation sin θ ≈ θ, (b) approximating with Simpson’s method based on two parabolas, and (c) evaluating the integral analytically.

13. Approximate the integral ∫_{0}^{0·4} x\sqrt{cos x} dx in two ways, first by using cos x ≈ 1 − \frac{x^{2}}{2} in the integrand and then with Simpson’s method based on four parabolas.

142 19. APPROXIMATING FUNCTIONS

PROBLEM SET 19.3

For Problems 1-9, write L’Hopital’s Rule to find the indicated limits.

1. lim_{x→2} \frac{x^{2} + x − 6}{2x^{3} − 3x^{2} − 4x + 4}  2. lim_{x→0} \frac{e^{x} − e^{−x}}{sin x}
3. lim_{x→π} \frac{sin mx}{sin nx}; m, n integers  4. lim_{x→0} \frac{1 − cos 5x}{x}
5. lim_{x→0} \frac{e^{x} − e^{−x} − 2x}{x − sin x}  6. lim_{x→∞} \frac{\frac{π}{2} − Arctan x}{ln (1 + \frac{1}{x})}
7. lim_{α→π} \frac{1 + tan \frac{3α}{4}}{cot \frac{α}{2}}  8. lim_{α→ \frac{π}{2}} \frac{sin^{2} 2α}{1 + cos 2α}

9. lim_{θ→0} \frac{θ − sin (sin θ)}{θ − sin θ}

In Problems 10 and 11, formulate and solve a limit problem that, to your mind, substantiates the claim.

10. Arctan ≈ x for small values of x.

11. The area of the region bounded by the tangent curve and the sine curve from 0 to a small positive number x is about three times the corresponding area of the region between the line y = x and the sine curve.

12. Show that lim_{x→0} \frac{∫_{0}^{x} \frac{sin t}{t} dt}{x} = 1.

4. OTHER INDETERMINATE FORMS 143

PROBLEM SET 19.4

For Problems 1-10, find the limits.

1. lim_{x→∞} \frac{x^{4}}{e^{x}}  2. lim_{x→∞} \frac{ln (1 + e^{−3x})}{x}
3. lim_{x→∞} x^{3} (\frac{π}{2} − Arctan x − \frac{1}{x})  4. lim_{x→0^{+}} (1 + x)^{ln x}
5. lim_{x→∞} x^{3} (\frac{1}{2} ln \frac{x+1}{x−1} − −\frac{1}{x})  6. lim_{x→∞} (1 + \frac{t}{x})^{x}
7. lim_{x→∞} (\frac{1 + \frac{1}{2x}}{1 − \frac{1}{2x}} )^{x}  8. lim_{x→∞} (\frac{1}{x−1} − \frac{1}{ln x})
9. lim_{x→∞} (\frac{x+4}{x−4})^{x}  10. lim_{x→0} [\frac{1}{sin^{2} x} − \frac{1}{x^{2}} ]

11. Translate each of the following statements into a limit problem involving ratios. Solve the problems. (a) As x → ∞, ln x increases at a much slower rate than any positive power of x. (b) As x → ∞, e^{x} increases at a much faster rate than any positive power of x.

144 19. APPROXIMATING FUNCTIONS

CHAPTER 20 Infinite Series

As we saw in the preceding chapter, a smooth function can be approximated by its Taylor series expansion. And our experience, based on examples, is that the accuracy of such approximations improves as the number of terms increase. The idea of approximation by finite series is a truncated version of the more general idea of representation by an infinite series. In this chapter we shall consider the elementary theory of infinite series in some detail and, in the later sections, discuss the representation of functions by special infinite series called power series.

1. Infinite Sequences

At the intuitive level a sequence is an ordered succession of real numbers such as

1, 3, 5, 7 1, 5, 14, 30, 55

and

1, 2, 3, ..., n...  or  1^{2}, (1/2)^{2}, (1/3)^{2}, ..., (1/n)^{2}, ...

The first two sequences stop; they are finite sequences. The second two never stop; they are called infinite sequences. Only infinite sequences are of interest to us and therefore we shall use “sequence” to mean “infinite sequence”. The members of a sequence are called terms and they are ordered in the sense that there is a first term, a second term, a third term, and so forth. It is customary to denote the nth term of a sequence by a subscripted symbol such as a_{n}, b_{n}, x_{n}, or y_{n} and indicate the sequence by writing

(1) a_{1}, a_{2}, a_{3}, ..., a_{n}... .

The sequence (1) can be defined by the compact notation {a_{n}}. Thus {n^{2}} denotes the sequence

1^{2}, 2^{2}, 3^{2}, ..., n^{2}, ...,

and {sin nπ/2} denotes the sequence

sin \frac{π}{2} , sin \frac{2π}{2} , sin \frac{3π}{2} , ..., \frac{nπ}{2}

In special situations we may find it convenient not to label the first term of a sequence with the first counting number. In such instances we adopt the following notation:

{a_{n}}_{n=0}^{∞} to mean a_{0}, a_{1}, a_{2}, ..., a_{n}, ...
{a_{n}}_{n=2}^{∞} to mean a_{2}, a_{3}, a_{4}, ..., a_{n}, ...

and

{a_{n}}_{n=k}^{∞}to mean a_{k}, a_{k+1}, a_{k+2}, ..., a_{n}, ...

From the notation it is clear that a sequence is a special type of function in which the independent variable ranges over an infinite block of integers. This observation prompts the following definition.

Definition

A sequence is a function whose domain is the set of all integers greater than or equal to some fixed integer k. The value of a sequence at n is called the nth term.

Sequences are introduced in a variety of ways. Usually the terms are computed from a formula for a_{n}. The following table illustrates this situation for a number of simple examples:

nth term  first five terms  100th term
a_{n} = 2n + 1  3,5,7,9,11  201
a_{n} = 1 + (−1)^{n}  0,2,0,2,0  2

a_{n} = \frac{(−1)^{n}n^{2}}{4n − 1}  −1/3, 4/7, −9/11, 16/15, −25/19  10, 000/399
a_{n} = n/(n + 1)  1/2, 2/3, 3/4, 4/5, 5/6  100/101

In each case the terms are obtained by substituting the values n = 1, 2, 3, 4, 5, 100. Of course, a_{n} need not be an integer or a simple fraction. And in many instances, such as

a_{n} = n ∫_{0}^{n} e^{t^{2}} dt or a_{n} = sin (e^{cos n}),

it may not be possible to calculate each term exactly. Occasionally a_{n} is not given by a single formula, as in the rule

a_{n} = \{ \begin{matrix}n^{2} & if n is even \\ π & if n is odd \end{matrix}

which defines the sequence

π, 4, π, 16, π, 36, ...

And there is the possibility of not being able to write any kind of formula for a_{n}. This is certainly the case for the sequence {a_{n}} in which a_{n} is the nth digit in the decimal expansion of the irrational number \sqrt{2}. Finally, it may be that each term of a sequence is defined in terms of its predecessors as, for example,

a_{n+1} = a_{n} + \frac{1}{a_{n}} , a_{1} = 1.

The first few terms are

a_{1} = 1, a_{2} = 2, a_{3} = 5/2, a_{4} = 20/10, a_{5} = 941/290, a_{6} = 969, 581/272, 890

Given in this manner, a sequence is said to be defined recursively. Newton’s formula x_{n+1} = x_{n} − f(x_{n})/f′(x_{n}), n = 0, 1, 2, ..., for finding a root of f(x) = 0 is a recursively defined sequence. For example, using this formula to find the root of x^{3}−4 = 0 leads to the recursive sequence x_{n+1} = \frac{2}{3}[x_{n} + 2/x_{n}^{2}], n = 0, 1, 2, ... whose terms, starting with, say, x_{0} = 2, will approximate the cube root of 4.

We are interested in studying sequences {a_{n}} with the property that the nth term a_{n} approaches some real number as n increases. If, for example,

a_{n} = 1 − \frac{1}{n},

then a_{n} is clearly near 1 for large n, A fact we indicate by writing

a_{n} → 1 as n → ∞ or lim_{n→∞} a_{n} = 1

or, in words, we say that the sequence {1 − 1/n} has limit 1. More formally, we have the following definition.

Definition A sequence {a_{n}} has the number L as a limit, written

(2)  lim_{n→∞} a_{n} = L or a_{n} → L as n → ∞,

if, for every positive number ϵ, there exists a positive integer N such that

(3) |a_{n} − L| < ϵ if n ≥ N.

Whenever a sequence has a limit, we say that the sequence is convergent and that it converges to the limit L. A sequence that fails to have a limit is termed divergent.

You should think of the above definition as a technical way of saying that as n increases the terms a_{n} get close and stay close to some number L. One way to obtain a visual image is to associate the numbers a_{n} and L with points on a coordinate line. According to the definition of limit, we may choose any positive number ϵ we please −1, 1/2, or 10^{−10}, it doesn’t matter – and be certain that there is some index N such that each term a_{n} with index n ≥ N lies between L − ϵ and L + ϵ. In effect, then, once a segment is drawn with center at L, only a finite number of the terms are outside that segment. From this visualization, that the terms of a sequence with limit L must cluster about L, it is clear that a sequence cannot have two limits.

Example 1. Show that lim_{n→∞} \frac{n + 1}{n} = 1.

Solution. Here a_{n} = \frac{n + 1}{n}, L = 1 and a_{n} − L = a_{n} − 1 = \frac{n + 1}{n} − 1 = \frac{1}{n}.

At the intuitive level, it is clear that a_{n} → 1 as n → ∞ because a_{n} − 1 = \frac{1}{n} → 0 as n → ∞. More technically, we must follow the procedures set forth in the definition of a limit. Let ϵ be a positive number and pose the following question: How large must we choose n in order that

|a_{n} − 1| = \frac{1}{n} < ϵ?

Obviously, \frac{1}{n} < ϵ if \frac{1}{ϵ} < n. So if we let N be the smallest integer such that \frac{1}{ϵ} < N, then n ≥ N implies that n > \frac{1}{ϵ} and, hence, that

|a_{n} − 1| < ϵ.

This analysis amounts to a strategy for finding N once ϵ is chosen. For example, if ϵ = 0.5, \frac{1}{ϵ} = 2 and we choose N = 3; if ϵ = 0.025, \frac{1}{ϵ} = 40 and N = 41; and if ϵ = 0.00001, \frac{1}{ϵ} = 100, 000 and N = 100, 001. Two things are obvious: first, the choice of N depends on the choice of ϵ and, second, making ϵ small forces N to be large. ‡ Example 2. Show that the sequence a_{n} with a_{n} = (−1)^{n}(1 + 1/n) is divergent.

Solution. Since (−1)^{n} is −1 if n is odd and 1 is odd and 1 if n is even, the terms a_{n} oscillate from slightly larger than 1 for even n to slightly less than −1 for odd n. The terms of this sequence clearly can’t get close and stay close to any number. In fact any two consecutive terms differ by more than 2 in absolute value. ‡

Example 3. Show that lim_{n→∞} \frac{3n^{2} − n + 1}{2n^{2} + 5n} = 3/2.

Solution. The nth term of this sequence is a_{n} = \frac{3n^{2} − n + 1}{2n^{2} + 5n} and at the intuitive level it

is fairly clear that the limit is 3/2. For if we divide both numerator and denominator by n^{2}, we have

a_{n} = \frac{3 − 1/n + 1/n^{2}}{2 + 5/n}

and since 3 − 1/n + 1/n^{2} → 3 and 2 + 5/n → 2 as n → ∞, it follows that

lim_{n→∞} a_{n} = lim_{n→∞} \frac{3 − 1/n + 1/n^{2}}{2 + 5/n}

and since 3 − 1/n + 1/n^{2} → 3 and 2 + 5/n → 2 as n → ∞, it follows that

lim_{n→∞} a_{n} = lim_{n→∞} \frac{3 − 1/n + 1/n^{2}}{2 + 5/n} = \frac{3}{2}.

To follow a more rigorous approach, we examine the difference

a_{n} − \frac{3}{2} = \frac{3n^{2} − n + 1}{2n^{2} + 5n} − \frac{3}{2} = \frac{−17n + 2}{2(2n^{2} + 5n)}

and show that the expression on the right tends to 0 as n → ∞. First, if 2 ≤ n,

\begin{aligned}|a_{n} − \frac{3}{2}| & = |\frac{−17n + 2}{|4n^{2} + 10n|}| = \frac{| − 17n + 2|}{|4n^{2} + 10n|} ≤ \frac{| − 17n| + 2}{|4n^{2} + 10n|} \\ & = \frac{17n + 2}{4n^{2} + 10n} ≤ \frac{17n + n}{4n^{2} + 0} = \frac{18n}{4n^{2}} = \frac{9}{2n}. \end{aligned}

Given a positive number ϵ, we want to be certain than 9/2n < ϵ for n sufficiently large. Following the technical of Example 1, we choose N as the smallest integer such that 9/2ϵ < N. Then, if n ≥ N if follows that 9/2n ≤ 9/2N ≤ 9/2N < ϵ and hence that

|a_{n} − \frac{3}{2}| < ϵ.

Again, we have a rule that computes N in terms of ϵ. For example, if ϵ = 0.0017, 9/2ϵ = 2647.05... and N = 2648. ‡

This example illustrates both the intuitive approach and a strict application of the limit definition. In practice, we shall not insist on a strict or rigorous approach, rather preferring to attack limit problems with a blend of common sense and calculus.

If f is a function whose domain includes the integer k and all x ≥ k, then we can define a sequence by the formula

a_{n} = f(n), n = k, k + 1, k + 2, ... .

and if it turns out that f has a horizontal asymptote,

lim_{n→∞} f(x) = L,

then the sequence must do the same, that is, we must also have lim_{n→∞} a_{n} = L. Thus, from the function f(x) = (1 + 2/x)^{x} comes the sequence a_{n} = (1 + 2/n)^{n}. This correspondence together with our previous work on indeterminate forms provides a useful method for finding limits.

Example 4. Find lim_{n→∞} \frac{ln n}{n}.

Solution. The sequence a_{n} = \frac{ln n}{n} is associated with f(x) = \frac{ln x}{x}, a function that assumes the indeterminate form ∞/∞ as x → ∞. Hence we may apply L’Hospital’s rule to obtain

lim_{n→∞} \frac{l^{x}}{x} = lim_{n→∞} \frac{1/x}{1} = lim_{n→∞} \frac{1}{x} = 0.

Therefore,

lim_{n→∞} \frac{ln n}{n} = 0. ‡

As we saw in Example 2, a sequence may diverge because its terms oscillate, never clustering around a fixed value. A sequence also may diverge because its terms increase or decrease without bound. This is certainly the case for the sequences

1^{2}, 2^{2}, 3^{2}, ..., n^{2}, ...,
ln 1, ln 2, ln 3, ..., ln n, ... ;

situations which we describe by writing

lim_{n→∞} n^{2} = ∞, lim_{n→∞} ln n = ∞, lim_{n→∞} − \sqrt{n} = −∞

In general we write

(4)  lim_{n→∞} a_{n} = ∞

to mean that given a number M there is an integer N such that

a_{n} > M for n ≥ N.

A similar meaning is attached to

(5)  lim_{n→∞} a_{n} = −∞.

Such sequences are divergent; they have no limit. Nonetheless, we describe (4) by saying that {a_{n}} diverges to ∞; the sequence in (5) diverges to −∞. It is worth noting that the reciprocal of a sequence that diverges to ∞ or −∞ is a convergent sequence with limit 0. Thus,

lim_{n→∞} ln n = ∞ but lim_{n→∞} \frac{1}{ln n} = 0

and

lim_{n→∞} n^{3} = ∞ but lim_{n→∞} \frac{1}{n^{3}} = 0.

The following theorem catalogues some useful limits.

Theorem 1.

(6)  lim_{n→∞} r^{n} = 0 if |r| < 1.
(7)  lim_{n→∞} \frac{1}{n^{k}} = 0 (k > 0).
(8)  lim_{n→∞} a^{1/n} = 1 (a > 0).
(9)  lim_{n→∞} n^{1/n} = 1.
(10)  lim_{n→∞} (1 + \frac{x}{n})^{n} = e^{x} (all x)

Proof.

(6) We want to show that |r^{n} − 0| = |r^{n}| = |r|^{n} is small for large n. Since |r| < 1, 1/|r| > 1 so 1/|r| = 1 + h for some positive h. An induction argument based on

(1 + h)^{2} = 1 + 2h + h^{2} > 1 + 2h,
(1 + h)^{3} = (1 + h)(1 + h)^{2} > (1 + h)(1 + 2h)
= 1 = 3h + 2h^{2} > 1 + 3h,
(1 + h)^{4} = (1 + h)(1 + h)^{3} > (1 + h)(1 + 3h) = 1 + 4h + 3h^{2} > 1 + 4h,

shows that (1 + h)^{n} > 1 + nh for every positive integer n. Hence

|r|^{n} = \frac{1}{(1 + h)^{n}} < \frac{1}{1 + nh} < \frac{1}{nh}.

But h is fixed and 1/n → 0 as n → ∞. Therefore 1/nh → 0 and, in turn, |r|^{n} → 0 as n → ∞.

(7) By the definition of exponents, n^{k} = e^{k ln n}. Now, if k > 0, k ln n → ∞ as n → ∞ and e^{x} → ∞ as x → ∞. Hence n^{k} → ∞ as n → ∞ and, therefore, its reciprocal 1/n^{k} must tend to 0.

(8) Let b_{n} = a^{1/n}, a > 0. Then ln b_{n} = \frac{1}{n} ln a → 0 as n → ∞. Hence

b_{n} = e^{ln b_{n}} → e^{0} = 1 as n → ∞.

Here we have used the fact that the function e^{x} is continuous at x = 0.

(9) Let a_{n} = n^{1/n}. Then ln a_{n} = \frac{1}{n} ln n → 0 as n → ∞ by Example 4. Hence

a_{n} = e^{ln a_{n}} → e^{0} = 1 as n → f.

(10) Fix x and consider f(t) = (1 + \frac{x}{t})^{t} as t → ∞. Since y = ln f(t) = t ln (1 + \frac{x}{t}) has the indeterminate form ∞ · 0 as t → ∞, we may apply L’Hospital’s rule to see that

lim_{t→∞} y = lim_{t→∞} \frac{ln (1 + \frac{x}{t})}{\frac{1}{t}} = lim_{t→∞} \frac{\frac{1}{1+ \frac{x}{t}} · \frac{−x}{t^{2}}}{\frac{−1}{t^{2}}} = lim_{t→∞} \frac{x}{1 + \frac{x}{t}} = x

and hence that f(t) = e^{y} → e^{x} as t → ∞. The result now follows in view of the correspondence (1 + \frac{x}{n})^{n} = f(n). ‡

Sometimes it is possible to determine the limit of a sequence by comparing its terms with the terms of sequences having known limits.

Theorem 2. Pinching Principle for Sequences

If the sequences {a_{n}} and {b_{n}} have a common limit L and {b_{n}} is a sequence such that (11) a_{n} ≤ b_{n} ≤ c_{n},

then {b_{n}} also has limit L.

Proof. Choose a positive number ϵ and draw an open interval about L running from L − ϵ to L + ϵ. Since both {a_{n}} and {c_{n}} converge to L, it is possible to find an N such that a_{n} and c_{n} belong to (L − ϵ, L + ϵ) for n ≥ N. But b_{n} is between a_{n} and c_{n}. Therefore,

|b_{n} − L| < ϵ for n ≥ N.

This proves that lim_{n→∞} b_{n} = L. ‡

This result is especially useful in the case a_{n} = 0 for all n. The assertion then amounts to saying that

(12)  if 0 ≤ b_{n} ≤ c_{n}

and lim_{n→∞} c_{n} = 0, then lim_{n→∞} b_{n} = 0.

Computations with limits is sometimes aided by the application of some almost selfevident rules. We omit the formal proofs.

Theorem 3. If {a_{n}} has limit A and {b_{n}} has limit B, then

(13)  lim_{n→∞} (a_{n} + b_{n}) = A + B
(14)  lim_{n→∞} a_{n}b_{n} = AB,
(15)  lim_{n→∞} ka_{n} = kA (k = constant),

and

(16)  lim_{n→∞} \frac{a_{n}}{b_{n}} = \frac{A}{B} if B ≠ 0, b_{n} ≠ 0.

A straightforward application of (13), (15), and (16) is given by

lim_{n→∞} \frac{3n^{2} − n + 1}{2n^{2} + 5} = lim_{n→∞} \frac{3 − 1/n + 1/n^{2}}{2 + 5/n^{2}} = \frac{lim_{n→∞} (3 − 1/n + 1/n^{2}}{lim_{n→∞} (2 + 5/n^{2}}

= \frac{lim_{n→∞} 3 − lim_{n→∞} (1/n) + lim_{n→∞} (1/n^{2})}{lim_{n→∞} 2 + 5 lim_{n→∞} 1/n^{2}} = \frac{3 − 0 + 0}{2 + 5.0} = 3/2.

Theorem 4. If lim_{n→∞} a_{n} = L and f is a continuous function, then

(17)  lim_{n→∞} f(a_{n}) = f(L).

Thus the composition of a continuous function f and a convergent sequence {a_{n}} is a new sequence {f(a_{n})} whose limit is given by (17).

Example 5. Find lim_{n→∞} cos (\frac{2n − 3}{3n + 4}).

Solution. Since cos x is continuous and lim_{n→∞} \frac{2n − 3}{3n + 4} = \frac{2}{3}, it follows from (17) that

lim_{n→∞} cos (\frac{2n − 3}{3n + 4}) = cos ( lim_{n→∞} \frac{2n − 3}{3n + 4}) = cos (\frac{2}{3}) . ‡

2. Monotonic and Bounded Sequences

In the preceding section sequences were shown to be convergent by actually finding the limit. A far more common occurrence is that a given sequence converges, that is, it has a limit, but it is quite impossible to exhibit that limit. Hence it is important to develop methods that guarantee the existence of a limit without actually locating it. For the sequences of concern to us in this section – the monotonic sequences – there is a simple convergence test.

Definition. Monotonic Sequence A sequence {a_{n}} is

increasing if a_{1} ≤ a_{2} ≤ a_{3} ≤ ⋯ ≤ a_{n} ≤ a_{n+1} ≤ ⋯;

strictly increasing if a_{1} < a_{2} < a_{3} < ⋯ < ⋯ < a_{n} < a_{n+1} < ⋯;

decreasing if a_{1} ≥ a_{2} ≥ a_{3} ≥ ⋯ ≥ a_{n} ≥ a_{n+1}

strictly decreasing if a_{1} > a_{2} > a_{3} > ⋯ > ⋯ > a_{n} > a_{n+1} > · · ·.

A sequence is termed monotonic if any one of these conditions hold.

Technically speaking, the monotonicity of a sequence cannot be determined except by examining the formula for the nth term. Nonetheless, a possible permanent trend can often be detected by examining the first few terms.

Example 1.

(a) sequence: 0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, ... type: strictly increasing

(b) sequence: 1, 1, \frac{3}{2}, \frac{3}{2}, 2, 2, \frac{5}{4}, \frac{5}{4}, ... type: increasing
(c) sequence: \frac{2^{2}}{2!} , \frac{2^{3}}{3!} , \frac{2^{4}}{4!} , \frac{2^{5}}{5!} , \frac{2^{6}}{6!} , ... type: strictly decreasing
(d) sequence: 0, −1, −1, −2, −2, −3, −3, −4, ... type: decreasing ‡

Each of the sequences of this example are monotonic, but the sequence

1, \frac{1}{2}, 2, \frac{1}{3}, 3, \frac{1}{4}, ...

is not.

Example 2. Show that the following sequences are monotonic.

(a) a_{n} = ln n

(b) a_{n} = \frac{n − 1}{n + 1}

(c) a_{n} = n^{1/n} with domain n = 3, 4, 5, ...

(d) a_{n+1}a_{n} + 1/a_{n}, a_{1} = 1

(e) a_{n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + ⋯ + \frac{1}{2n}

(f) a_{n} = \frac{n^{2}}{2^{n}} with domain n = 3, 4, 5, ...

Solutions. (a) ln n < ln (n + 1) because ln x has a positive derivative for all positive x.

(b) a_{n+1} − a_{n} = \frac{n}{n + 2} − \frac{n − 1}{n + 1} = \frac{2}{(n + 2)(n + 1)} > 0 for all n so \{a_{n}\} is strictly increasing.

(c) Write ln a_{n} = ln n^{1/n} = \frac{ln n}{n}. Let f(x) = \frac{ln x}{x} and note that f′(x) = \frac{1 − ln x}{x^{2}} < 0 if

ln x > 1 or if x > e. Hence \frac{ln x}{x} is decreasing if x > e. Therefore ln a_{n} > ln a_{n+1} if n ≥ 3 or, finally, a_{n} > a_{n+1} if n ≥ 3 because ln x increases with x. Consequently the sequence {n^{1/n}}_{n=3}^{∞} is strictly decreasing.

(d) a_{1} is positive, and if a_{n} is positive, then a_{n+1} = a_{n} +1/a_{n} > 0. By induction every term is positive. Hence a_{n+1} = a_{n} + 1/a_{n} > a_{n} for all n. The sequence is strictly increasing.

(e) The sequence is strictly increasing since

a_{n+1} − a_{n} = \frac{1}{2n + 2} + \frac{1}{2n + 1} − \frac{1}{n + 1} = \frac{−1}{2(n + 1)} + \frac{1}{2n + 1} = \frac{1}{2(n + 1)(2n + 1)} > 0.

(f) The sequence is strictly decreasing because

\frac{a_{n}}{a_{n+1}} = \frac{n^{2}}{2^{n}} · \frac{2^{n+1}}{(n + 1)^{2}} = \frac{2}{(1 + 1/n)^{2}} > 1

if n ≥ 3, that is, (1 + 1/n)^{2} < 2 if n ≥ 3. ‡

The question of convergence for monotonic sequences is especially easy to answer. For example, the strictly increasing sequence {n^{2}} diverges to infinity: its terms do not lie below some fixed value. On the other hand the sequence \{\frac{n−1}{n}\} is strictly increasing with limit 1: its values do lie below some fixed number, in fact, \frac{n−1}{n} = 1 − \frac{1}{n} < 1 for all n. The difference is that one sequence is bounded above, the other is not. The following definition formalizes this key observation.

Definition. Bounded Sequence A sequence {a_{n}} is bounded above if there is a constant B such that

a_{n} ≤ B for all n;

it is bound below if there is a constant A such that

A ≤ a_{n} for all n.

A sequence that is both bounded above and below is called a bounded sequence.

Not every bounded sequence converges, as one sees by examining the oscillatory sequence {(−1)^{n}}, but as the above examples suggest, a bounded and monotonic sequence must converge.

Theorem 5. Every increasing sequence that is bounded above has a limit; and every decreasing sequence that is bounded below has a limit.

This theorem offers no hint as how to find the limit; it merely asserts that under certain conditions the limit will exist. Once it is shown that a_{n} ≤ B for all n, the theorem tells us that lim_{n→∞} a_{n} = L. The bound B need not be the limit: all we know is that L ≤ B and a_{n} ≤ L for all n. If {a_{n}} is strictly increasing, then a_{n} < L for all n.

We shall regard Theorem 5 as being geometrically obvious. However, it should be noted that its proof rests ultimately on the Least Upper Bound Axiom, an assumption about real numbers that undergirds all of calculus.

LEAST UPPER BOUND AXIOM Every set of real numbers that is bounded above has a least upper bound; and every set of real numbers that is bounded below has a greatest lower bound.

For example, the number 2 is an upper bound for the open interval 0 < x < 1, so is \frac{3}{2}; but 1 is at least upper bound. The limit L of a bound and increasing sequence is the least number which is an upper bound for all its terms.

Example 3. Use Theorem 5 to analyze the convergence of divergence of each sequence in Example 2.

Solution. (a) The sequence {ln n} is divergent; it is increasing but not bounded above.

(b) The sequence \{\frac{n−1}{n+1}\} is increasing and bounded above by 1, that is, \frac{n−1}{n+1} ≤ 1 for all n. Hence the sequence has a limit, by Theorem 5.

(c) {n^{1/n}}_{n=3}^{∞} is a decreasing sequence of positive numbers. Hence 0 is a lower bound. Therefore Theorem 5 applies and the sequence has a limit.

(d) The recursive sequence a_{n+1} = a_{n}+1/a_{n}, a_{1} = 1 is increasing and, according to Problem Set 14.1, a_{n+1} > \sqrt{2n + 1} for all n. since \sqrt{2n + 1} → ∞ as n → ∞, \{a_{n}\} is not bounded above and a_{n} → ∞ as n → ∞. There is no limit.

(e) The sequence is increasing, and bounded above because

a_{n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + ⋯ + \frac{1}{2n} < \frac{n}{n + 1} < 1

for all n. By Theorem 5, the sequence has a limit.

(f) \{\frac{n^{2}}{2^{n}} \}_{n=3}^{∞} is a decreasing sequence of positive numbers, hence 0 is a lower bound. Again Theorem 5 tells us that the sequence has a limit. ‡

Whether a sequence {a_{n}} has a limit depends on the behavior of a_{n} for large n; terms at the beginning have no effect. In practice this means you may ignore any initial block. Put technically, if {a_{n}}_{n=k}^{∞} has limit L, then {a_{n}}_{n=1}^{∞} will also have limit L. Thus, in reference to Theorem 5, a sequence that is eventually increasing (increasing from some term on) and bounded above has a limit; a sequence that is eventually decreasing and bound below has a limit.

Example 4. Use Theorem 5 to establish that the sequence \{\frac{n^{10}}{e^{n}} \}_{n=1}^{∞} has a limit.

Solution. The sequence a_{n} = n^{10}/e^{n} increases through the tenth term a_{10} and thereafter decreasing, as may be seen by differentiating f(x) = x^{10}/e^{x}. Hence Theorem 5 applies and a limit exists. The limit is 0 as may be seen by successive applications of L’Hospital’s rule. ‡

3. Convergent and Divergent Series, Partial Sums

The most familiar finite series is the finite geometric series

3. CONVERGENT AND DIVERGENT SERIES, PARTIAL SUMS 155

(1) 1 + x + x^{2} ⋯ + x^{n−1} whose sum s_{n} is given by

(2) s_{n} = \frac{1 − x^{n}}{1 − x} = \frac{1}{1 − x} − \frac{x^{n}}{1 − x} if x ≠ 1.

Of the two terms on the right in (2), one involves n, the other does not; and in the second term, if |x| < 1, x^{n} → 0 as n → ∞, by (6) of Theorem 1. Hence

s_{n} → \frac{1}{1 − x} as n → ∞ if |x| < 1.

Or, saying the same thing,

1 + x + x^{2} + ⋯ + x^{n−1} → \frac{1}{1 − x} as n → ∞ if |x| < 1.

We describe this limit by writing an infinite series:

1 + x + x^{2} + ⋯ = \frac{1}{1 − x} if |x| < 1.

The three dots are important; they serve to indicate that the established pattern continues without end. Thus, if x = 1/2,

+\frac{1}{2} + \frac{1}{2^{2}} + + ⋯ = \frac{1}{1 + \frac{1}{2}} = 2,

meaning that 2 is the limit of the partial sums

s_{1} = 1
s_{2} = 1 + \frac{1}{2}
s_{3} = 1 + \frac{1}{2} + \frac{1}{2^{2}}
⋮
s_{n} = 1 + \frac{1}{2} + \frac{1}{2^{2}} + ⋯ + \frac{1}{2^{n}}
⋮

This example illustrates the basic concept of an infinite series. Definition. Any sequence {a_{n}}_{n=1}^{∞} of real numbers can be used to form an infinite series (3) a_{1} + a_{2} + a_{3} + ⋯, which is said to be convergent if its sequence of partial sums

(4) s_{n} = a_{1} + a_{2} + a_{3} + ⋯ + a_{n} converges to a limit S. The limit S is called the sum of the infinite series (3) and we write (5) S = a_{1} + a_{2} + a_{3} + · · · . In case the sequence of partial sums (4) fails to converge, we say that the infinite series (3) is divergent.

The numbers a_{n} are the terms of the series (3); a_{1} is the first term, a_{2} is the second term, etc. They can be computed from the partial sums s_{n} by the formula

a_{1} = s_{1}, a_{2} = s_{2} − s_{1}, a_{3} = s_{3} − s_{2}, · · · .

For clarity, we shall often expand the notation in (3) to read

a_{1} + a_{2} + a_{3} + ⋯ + ⋯,

especially if we wish to remove any doubt as to the formula for a_{n}. And if brevity dictates, we shall adopt the summation notation

a_{1} + a_{2} + a_{3} + ⋯ + a_{n} + ⋯ = ∑_{n=1}^{∞} a_{n}.

In many problems it is convenient to begin indexing an infinite series at 0 instead of 1. We then write

a_{0} + a_{1} + a_{2} + ⋯ + a_{n} + ⋯ = ∑_{n=0}^{∞} a_{n}.

In this setting the partial sums are s_{0} = a_{0}, s_{1} = a_{0} + a_{1}, etc.

Example 1. Find the sum S of the series 1 + \frac{5}{8} + (\frac{5}{8})^{2} + (\frac{5}{8})^{3} + · · ·.

Solution. The partial sums are

s_{1} = 1,
s_{2} = 1 + \frac{5}{8},
⋮
s_{n} = 1 + \frac{5}{8} + (\frac{5}{8})^{2} + ⋯ + (\frac{5}{8})^{n} = \frac{1 − (\frac{5}{8})^{n+1}}{1 − \frac{5}{8}}

and, since lim_{n→∞} (\frac{5}{8})^{n+1} = 0, it follows that lim_{n→∞}s_{n} = 8/3. Hence the series has sum S = 8/3, that is,

\frac{8}{3} = 1 + \frac{5}{8} + (\frac{5}{8})^{2} + (\frac{5}{8})^{3} + · · · . ‡

Example 2. Show that the geometric series 1 + x + x^{2} + ⋯ = ∑_{n=0}^{∞} x^{n} diverges if |x| ≥ 1.

Solution. At x = 1 the geometric series reduces to

1 + 1 + 1 + ⋯,

a divergent series because its sequence of partial sums

s_{1} = 1, s_{2} = 2, s_{3} = 3, ⋯, s_{n} = n ⋯

diverge to infinity. At x = −1, the geometric series reduces to

1 − 1 + 1 − 1 + ⋯,

and this series diverges because its partial sums

s_{1} = 1, s_{2} = 0, s_{3} = 1, s_{4} = 0, ⋯

oscillate between 0 and 1 and therefore fail to have a limit.

If x > 1 the geometric series diverges because, by (2),

s_{n} = \frac{1 − x^{n}}{1 − x} = \frac{x^{n} = 1}{x − 1} → ∞ as n → ∞.

And if x < −1, the partial sums oscillate wildly with s_{2n} → ∞ and s_{2n+1} → −∞ as n → ∞. ‡

This example and our opening remarks establish the following important theorem.

Theorem 6. (Geometric Series). The geometric series

1 + x + x^{2} + ⋯ = ∑_{n=0}^{∞} x^{n}

converges to \frac{1}{1 − x} if |x| < 1; that is,

(6) 1 + x + x^{2} + ⋯ = \frac{1}{1 − x} if |x| < 1;

and the geometric series diverges if |x| ≥ 1.

In (6) the number x is called the common ratio.

Example 3. As a consequence of Theorem 6:

(a) 1 − \frac{2}{3} + (\frac{2}{3})^{2} − (\frac{2}{3})^{3} + ⋯ = _{\frac{1}{1−(\frac{2}{3})}} = \frac{3}{5} because |−\frac{2}{3} | < 1;

(b) 1 + e^{−1} + e^{−2} + e^{−3} + ⋯ = _{\frac{1}{1− \frac{1}{3}}} = \frac{e}{e−1} because e^{−1} < 1;

(c) 1 + e + e^{2} + e^{3} + ⋯ diverges because e > 1. ‡

One of the most illuminating of the many applications of the geometric series is to the study of decimal expansions of rational numbers. The connection arises from the observation that since .9 means 9/10 and .99 means 9/10 + 9/10^{2}, then the only logical meaning for the infinite decimal .999... is the infinite series 9/10 + 9/10^{2} + 9/10^{3} + ⋯ + 9/10^{n} + · · ·. The sum of this series is, as always, the limit of its partial sums, of which the nth is

s_{n} = \frac{9}{10} + \frac{9}{10^{2}} + \frac{9}{10^{3}} + \frac{9}{10^{n}}
= \frac{9}{10} [1 + \frac{1}{10} + \frac{1}{10^{2}} + ⋯ + \frac{1}{10^{n−1}} ]
= \frac{9}{10} = \frac{1−1/10^{n}}{1−1/10} = 1 − 1/10^{n}.

Since lim_{n→∞} s_{n} = 1, we conclude that

.999... = \frac{9}{10} + \frac{9}{10^{2}} + \frac{9}{10^{3}} + ⋯ = 1.

In general, if {a_{n}}_{n=0}^{∞} is a sequence of integers taken from 0 through 9, the infinite decimal .a_{1}, a_{2}a_{3} ⋯ denotes the infinite series

∑_{n=1}^{∞} \frac{a_{n}}{10^{n}} = \frac{a_{1}}{10} + \frac{a_{2}}{10^{2}} + +\frac{a_{3}}{10^{3}} + + ⋯ + \frac{a_{n}}{10^{n}} + ⋯

And if the a_{n} are periodic, in the sense that a certain pattern is repeated, then the decimal represents a rational number which can be found by a procedure similar to that just illustrated.

Example 4. Find the rational number represented by .444....

Solution. This decimal corresponds to the series

∑_{n=1}^{∞} \frac{4}{10^{n}} = \frac{4}{10} + \frac{4}{10^{2}} + \frac{4}{10^{3}} + ⋯

which has as its partial sum

\begin{aligned}s_{n} & = \frac{4}{10} + \frac{4}{10^{2}} + ⋯ \frac{4}{10^{n}} \\ & = \frac{4}{10} [1 + \frac{1}{10} + ⋯ + \frac{1}{10^{n−1}} ] = \frac{4}{10} \frac{1−1/10^{n}}{1−1/10} = \frac{4}{9}(1 − 1/10^{n}). \end{aligned}

Since lim_{n→∞} s_{n} = 4/9, it follows that .444... = 4/9. ‡

You should have observed by now that the terms of a convergent series are ultimately very small.

Theorem 7. If the series ∑_{k=1}^{∞} a_{k} tend to a finite limit as n → ∞. Therefore, a_{n} =

s_{n+1} − s_{n} → S − S = 0 as n → ∞. ‡ The content of this theorem is of more immediate use when stated as a

SUFFICIENT CONDITION FOR DIVERGENCE

(7) If the sequence {a_{n}} either fails to have a limit or has a limit different from 0, then the series ∑_{n=1}^{∞} a_{n} diverges.

Example 5. In consequence of (7), the following series are divergent.

(a)  ∑_{n=1}^{∞} \frac{n}{n+1} = \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + ⋯, a_{n} → 1 as n → ∞

(b)  ∑_{n=1}^{∞} \frac{(−1)^{n}n}{n+1} = −\frac{1}{2} + \frac{2}{3} − \frac{3}{4} + ⋯, a_{n} → 1 as n → ∞

(c)  ∑_{k=1}^{∞} ln k = ln 1 + ln 2 + ln 3 + ⋯, a_{n} → ∞ as n → ∞

(d)  ∑_{k=1}^{∞} \frac{k^{2} + k + 1}{k^{2} − k + 1} = \frac{3}{1} + \frac{7}{3} + \frac{13}{7} + ⋯, a_{n} → 1 as n → ∞

(e) \begin{matrix}∑_{j=1}^{∞} cos (jπ) & = cos π + cos 2π + cos 3π + ⋯ \\ & = −1 + 1 − 1 + ⋯, a_{n} has no limit ‡ \end{matrix}

Even if a_{n} → 0 as n → ∞, the series ∑_{n=1}^{∞} a_{n} may diverge. There, a_{n} = ln (1 + 1/n) → 0

as n → ∞. Yet the series diverges. Such examples are easy to construct. Consider, for example, the series

∑_{n=1}^{∞} \frac{1}{\sqrt{n}} = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + · · · .

The nth term a_{n} = \frac{1}{\sqrt{n}} certainly tends to 0 as n → ∞. Nonetheless, the series diverges because

s_{n} = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ⋯ + \frac{1}{\sqrt{n}} > \frac{n}{\sqrt{n}} = \sqrt{n} → ∞ as n → ∞.

The identical phenomenon is present in the harmonic series

(8)  ∑_{n=1}^{∞} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + ⋯

whose partial sums grow ever so slowly; but they do grow. One must add 12,367 terms to obtain a partial sum that exceeds 10, and 33,617 terms are needed to exceed 11. At this latter stage, where each additional term adds less than 3 × 10^{−5}, it seems far more likely that the partial sums will level off and approach a limit. That they continue to grow may be seen by clever grouping:

\begin{aligned}s_{2} & = 1 + \frac{1}{2} \\ s_{4} & = (1 + \frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) > 1 + \frac{1}{2} + \frac{1}{2} \\ s_{8} & = (1 + \frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \\ s_{2^{m}} & = (1 + \frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) + ⋯ + (\frac{1}{2^{m−1}+1} + ⋯ + \frac{1}{2^{m}} ) > 1 + \frac{m}{2}. \end{aligned}

This means that the partial sums s_{2}, s_{4}, s_{8}, s_{16}, s_{32}, s_{64}, ... diverge to infinity. However, s_{1} < s_{2} < s_{3} < s_{4} < ⋯, so we conclude that the entire sequence of partial sums diverges to infinity and the series diverges. This property of the harmonic series merits a special statement.

Theorem 8. The harmonic series diverges.

The simple rules for adding two convergent series and multiplying a series by a constant are set forth in the following theorem. The more complicated rules for multiplying and dividing infinite series will be touched on later.

Theorem 9. Suppose ∑_{n=1}^{∞} (a_{n} + b_{n}) is a convergent series and

(9)  ∑_{n=1}^{∞} (a_{n} + b_{n}) = ∑_{n=1}^{∞} a_{n} + ∑_{n=1}^{∞} b_{n} = A + B.

Also ∑_{n=1}^{∞} ca_{n} is convergent for every constant c and

160 20. INFINITE SERIES

(10)  ∑_{n=1}^{∞} ca_{n} = c ∑_{n=1}^{∞} a_{n} = cA.

Proof. By assumption s_{n} = ∑_{n=1}^{∞} a_{k} → A and t_{n} = ∑_{k=1}^{∞} b_{k} → B as n → ∞, and s_{n} + t_{n} =

∑_{k=1}^{∞} (a_{k} + b_{k}) is the nth partial sum for ∑_{k=1}^{∞} (a_{k} + b_{k}). The limit of these partial sums is lim_{n→∞} (s_{n} + t_{n}) = lim_{n→∞} s_{n} + lim_{n→∞} t_{n} = A + B, by Theorem 3. This proves (9). For (10), write p_{n} = ∑_{k=1}^{∞} ca_{k} = c ∑_{k=1}^{∞} a_{k} = cs_{n} so that lim_{n→∞} cs_{n} = c lim_{n→∞} s_{n} = cA, again by Theorem 3. This proves (10). ‡

Notice that rules (9) and (10) resemble the two basic rules of integration

∫_{a}^{b} (f(x) + g(x) dx = ∫_{a}^{b} f(x) dx + ∫_{a}^{b} g(x) dx

and

∫_{a}^{b} cf(x) dx = c ∫_{a}^{b} f(x) dx.

In practice, they are applied in an identical manner. For example, consider the series

∑_{n=1}^{∞} \frac{1}{n(n + 1)} = 1 and ∑_{n=1}^{∞} \frac{1}{n(n + 2)} = \frac{3}{4}.

By (9),

\begin{aligned}\frac{7}{4} = 1 + \frac{3}{4} & = ∑_{n=1}^{∞} \frac{1}{n(n+1)} + ∑_{n=1}^{∞} \frac{1}{n(n+2)} \\ & = ∑_{n=1}^{∞} [\frac{1}{(n+1)} + \frac{1}{(n+2)}] = ∑_{n=1}^{∞} \frac{2n+3}{n(n+1)(n+2)} \end{aligned}

Also,

\begin{aligned}\frac{1}{4} = 1 − \frac{3}{4} & = ∑_{n=1}^{∞} − \frac{1}{n(n+1)} − ∑_{n=1}^{∞} \frac{1}{n(n+2)} \\ & = ∑_{n=1}^{∞} [\frac{1}{n(n+1)} − ∑_{n=1}^{∞} \frac{1}{n(n+2)}] = ∑_{n=1}^{∞} \frac{1}{n(n+1)(n+2)}. \end{aligned}

This obvious rule for subtracting two convergent series results from a combination of (9) and (10), using the choice c = −1:

\begin{aligned}∑_{n=1}^{∞} (a_{n} − b_{n}) = ∑_{n=1}^{∞} (a_{n} + (−b_{n})) & = ∑_{n=1}^{∞} a_{n} + ∑_{n=1}^{∞} (−b_{n}) \\ & = ∑_{n=1}^{∞} a_{n} − ∑_{n=1}^{∞} b_{n} = A − B. \end{aligned}

Example 6. Find the sum of the series

(\frac{1}{2})^{3} + (\frac{1}{2})^{4} + (\frac{1}{2})^{5} + · · · .

Solution. This is a geometric series a+ax+ax^{2} +⋯ in which a = (\frac{1}{2})^{3}. We simply factor a out:

\begin{aligned}(\frac{1}{2}) + (\frac{1}{2})^{4} + (\frac{1}{2})^{5} + ⋯ + & = (\frac{1}{2})^{3} [1 + \frac{1}{2} + (\frac{1}{2})^{2} + ⋯] \\ & = (\frac{1}{2})^{3} \frac{1}{1−1/2} = \frac{1}{4}. ‡ \end{aligned}

Rule (10) applies only to convergent series. But there is a related idea which is often useful in deciding a series diverges. It is this:

(11) if ∑_{n=1}^{∞} a_{n} diverges and c ≠ 0, then ∑_{n=1}^{∞} ca_{n} diverges.

Thus a known divergent series can’t be turned into a convergent series by multiplying each term by a constant, unless that constant is zero. For each partial sum of the second series is the product of c and the corresponding partial sum of the first series; therefore if one diverges so must the other.

Example 7.

(a)  ∑_{n=1}^{∞} \frac{5}{n} diverges because ∑_{n=1}^{∞} \frac{1}{n} diverges.

(b)  ∑_{n=1}^{∞} \frac{3n}{n+1} diverges because ∑_{n=1}^{∞} \frac{n}{n+1} diverges. ‡

To this point we have used only series starting with a_{0} or a_{1}, but it is also sometimes necessary to use series that begin with a_{k}:

∑_{n=k}^{∞} a_{n} = a_{k} + a_{k+1} + a_{k+2} + · · · .

Indeed, any convergent series can be written as the sum of an initial block of terms plus another convergent series. Thus,

∑_{n=1}^{∞} a_{n} = a_{1} + ∑_{n=2}^{∞} a_{n},

∑_{n=1}^{∞} a_{n} = a_{1} + a_{2} + ∑_{n=3}^{∞} a_{n},

and, in general,

(12)  ∑_{n=1}^{∞} a_{n} = a_{1} + a_{2} + ⋯ + a_{k} + ∑_{n=k+1}^{∞} a_{n}

for each positive integer k. This formula, an easy consequence of comparing partial sums of the two series, makes no sense for divergent series.

162 20. INFINITE SERIES

Example 8.

(a)  ∑_{n=1}^{∞} x^{n} = 1 + ∑_{n=1}^{∞} x^{n} if |x| < 1.
(b)  ∑_{n=1}^{∞} \frac{n}{n(n+1)} = \frac{1}{2} + \frac{1}{6} + ∑_{n=3}^{∞} \frac{1}{n(n+1)}.

Example 9. Find the sum of the series ∑_{k=3}^{∞} 1/2^{k}.

By (12),

∑_{k=0}^{∞} \frac{1}{2^{k}} = 1 + \frac{1}{2} + \frac{1}{2^{2}} + ∑_{k=3}^{∞} \frac{1}{2^{k}};

also

∑_{k=0}^{∞} \frac{1}{2^{k}} = \frac{1}{1 − 1/2} = 2.

Therefore,

\begin{aligned}∑_{k=3}^{∞} \frac{1}{2^{n}} & = ∑_{k=0}^{∞} \frac{1}{2^{k}} − 1\frac{1}{2} − \frac{1}{2^{2}} \\ & = 2 − \frac{7}{4} = \frac{1}{4}. \end{aligned}

‡ Whether a series converges or diverges has nothing to do with where the indexing starts. This simple observation saves considerable fussing in practice. Specifically, if the series ∑_{n=k}^{∞} a_{n} converges for some starting index k, it will converge for any other starting index;

and if ∑_{n=k}^{∞} a_{n} diverges for some starting index, it will diverge for any other starting index. To illustrate, since

∑_{n=1}^{∞} \frac{1}{2^{n}} = 1 + \frac{1}{2} + \frac{1}{2^{2}} + ⋯

is a convergent series, so is the series

∑_{n=100}^{∞} \frac{1}{2^{n}} = \frac{1}{2^{100}} + \frac{1}{2^{101}} + \frac{1}{2^{102}} + ⋯;

and since the harmonic series

∑_{n=1}^{∞} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + ⋯

diverges, the series

∑_{n=1}^{∞} \frac{1}{n + 5} = \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + ⋯

is also divergent.

4. Series of Positive Terms

More often than not the question of convergence for an infinite series cannot be settled with the ease suggested by the preceding section, the difficulty being that it is not usually possible to find a simple formula for the nth partial sums s_{n} = a_{1} + a_{2} + a_{3} + ⋯ + a_{n} of the series ∑ a_{n}. Consequently, the technique of establishing convergence by directly finding the limit of the sequence s_{n} must be abandoned in favor of various indirect approaches. The most basic of these methods of testing a series for convergence apply to series in which all the terms are positive. For brevity we shall refer to a series ∑ a_{n} in which a_{n} > 0 as a series of positive terms. For such a series we always have

s_{n} = s_{n−1} + a_{n} > s_{n−1}

so that the partial sums form a strictly increasing sequence. If the partial sums are also bounded, that is, if there is a constant M such that s_{n} ≤ M for all n, then the series must converge because, by Theorem 5, every increasing and bounded sequence has a limit. But if the partial sums are not bounded, then the series diverges. These remarks establish a basic principle about series of positive terms.

Theorem 10. A series ∑ a_{n} of positive terms converges if its partial sums are bounded, and diverges if they are not.

In many series the nth term a_{n} is the value of a simple function f at n,

a_{n} = f(n)

and it becomes fruitful to analyze the partial sums of

a_{1} + a_{2} + a_{3} + ⋯ = f(1) + f(2) + f(3) + ⋯

by comparing areas. The comparison is particularly simple if the associated function f is continuous, positive, and decreasing.

Theorem 11. The Integral Test Let f be continuous, positive, and decreasing on the interval x ≥ 1. Then the series

(1) f(1) + f(2) + f(3) + ⋯

and the improper integral

(2) ∫_{1}^{∞} f(x) dx

either both converge or both diverge.

Proof. Since f is decreasing on each of the unit intervals n ≤ x ≤ n + 1, the area under the graph from n to n + 1 is less than that of a rectangle of area f(n) (Figure 14.6) and greater than that of a rectangle of area f(n + 1) (Figure 14.5). In particular,

f(2) < ∫_{1}^{2} f(x) dx < f(1),

f(3) < ∫_{2}^{3} f(x) dx < f(2),

f(4) < ∫_{3}^{4} f(x) dx < f(3),

f(n + 1) < ∫_{n}^{n+1} f(x) dx < f(n),

Addition yields

f(2) + f(3) + ⋯ + f(n + 1) < ∫_{1}^{n+1} f(x) dx < f(1) + f(2) + ⋯ + f(n).

In terms of the partial sums s_{n} = ∑_{k=1}^{n} f(k), these two inequalities state that

(3) s_{n+1} − f(1) < ∫_{1}^{n+1} f(x) dx and ∫_{1}^{n+1} f(x) dx < s_{n}

These comparisons form the basis of the integral test.

The second inequality shows that if the improper integral (2) diverges, that is, if

∫_{1}^{n+1} f(x) dx → ∞ as n → ∞, then s_{n} → ∞ as n → ∞ and the series diverges. But

if the improper integral converges, then ∫_{1}^{n+1} f(x) dx increases to a limit L as n → ∞ and

s_{n+1} < f(1) + ∫_{1}^{n+1} f(x) dx < f(1) + L

for all n. Hence {s_{n}}, being increasing and bounded, has a limit by Theorem 10. ‡

As a first application of the integral test, consider the harmonic series ∑ 1/n, a known divergent series. Clearly, the test function is f(x) = 1/x because f(n) = 1/n. And since 1/x is continuous, positive and decreasing on the range x > 0, we conclude from

∫_{1}^{∞} \frac{1}{x} dx = lim_{b→∞} ∫_{1}^{b} \frac{1}{x} dx = lim_{b→∞} ln x]_{1}^{b} = lim_{b→∞} ln b = ∞

that the harmonic series diverges. Moreover, the second inequality in (3) provides a rough estimate of the rate at which its partial sums grow:

ln (n + 1) = ∫_{1}^{n+1} \frac{1}{x} dx < 1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{n} = s_{n}.

It is interesting to compare the estimates given by this inequality with some computed values of s_{n} for selected n:

Estimates  Computed Values
s_{83} > 4.4308  s_{83} = 5.002068...
s_{227} > 5.4293  s_{227} = 6.004366...
s_{616} > 6.4248  s_{616} = 7.001274...
s_{1,674} > 7.4235 s_{1,674} = 8.000485...
s_{4,550} > 8.4231 s_{4,550} = 9.000208...
s_{12,367} > 9.4228 s_{12,367} = 10.000043...

The harmonic series is a special case of the more general p-series

(4)  ∑_{n=1}^{∞} \frac{1}{n^{p}} = 1 + \frac{1}{2^{p}} + \frac{1}{3^{p}} + · · · .

Example 1. Show that the p-series converges if p > 1 and diverges if p ≤ 1.

Solution. Here the test function is f(x) = 1/x^{p} because f(n) = 1/n^{p}. It is continuous, positive, and since f′(x) = −p/x^{p+1} < 0, it is decreasing on the interval x > 0. If p = 1, the series (4) is the divergent harmonic series. If p ≠ 1, the p-series converges or diverges with the improper integral

∫_{1}^{∞} \frac{1}{x^{p}} dx = lim_{b→∞} ∫_{1} b x^{−p} dx = lim_{b→∞} \frac{x^{−p+1}}{−p + 1} |_{1}^{b} = lim_{b→∞} (\frac{b^{−p+1}}{1 − p} − \frac{1}{1 − p}) .

Whether this limit is finite or infinite depends on the exponent −p + 1. If −p + 1 > 0 or p < 1, b^{−p+1} will increase without bound as b increases; but if −p + 1 < 0, the exponent on b is negative and b^{−p+1} → 0 as b → ∞. Hence

∫_{1}^{∞} \frac{1}{x^{p}} dx = \frac{1}{p − 1} if p > 1 and ∫_{1}^{∞} \frac{1}{x^{p}} dx diverges if p < 1.

According to the integral test this means that the p-series (4) converges if p > 1 and diverges if 0 < p < 1. ‡

In particular, this means that

∑_{n=1}^{∞} \frac{1}{n^{2}} and ∑_{n=1}^{∞} \frac{1}{n^{4}}

are convergent series (p = 2 and p = 4) while

∑_{n=1}^{∞} \frac{1}{\sqrt{n}} and ∑_{n=1}^{∞} \frac{1}{n^{1/3}}

are divergent series (p = 1/2 and p = 1/3). the conclusions appear reasonable because, for instance, 1/n^{2} certainly goes to 0 much faster than 1/\sqrt{n}. But, intuitively, no meaningful distinction is apparent between the series

∑_{n=1}^{∞} \frac{1}{n^{1.001}} and ∑_{n=1}^{∞} \frac{1}{n^{.999}} .

The first converges, the second diverges; yet, for large n their nth terms would seem to be going to zero at about the same rate. But there is a difference, and that difference is enough to force radically different behavior on the respective partial sums.

If a series is shown to converge by the integral test, the sum of the series is never exactly the value of the integral. In fact, if a_{n} = f(n) and f satisfies the conditions of the integral test, then, by (3), the best we can say is that

(5) ∫_{1}^{∞} f(x) dx < ∑_{n=1}^{∞} a_{n} < a_{1} + ∫_{1}^{∞} f(x) dx.

Hence

1 = ∫_{1}^{∞} \frac{1}{x^{2}} < ∑_{n=1}^{∞} \frac{1}{n^{2}} < 1 + ∫_{1}^{∞} \frac{1}{x^{2}} dx = 1 + 1 = 2

and

\frac{1}{p − 1} = ∫_{1}^{∞} \frac{1}{x^{p}} dx < ∑_{n=1}^{∞} \frac{1}{n^{p}} < 1 + ∫_{1}^{∞} \frac{1}{x^{p}} dx = 1 + \frac{1}{p − 1} = \frac{p}{p − 1} if p > 1.

Used in this fashion the integral test provides an upper and lower bound for the sum of a convergent series, but the estimates it provides aren’t very accurate, as you can tell by examining finite 14.5. A more useful approach consists of calculating the sum of the first few terms and using an integral to estimate the tail-end or remainder. Specifically, we write

S = ∑_{n=1}^{∞} a_{n} = a_{1} + a_{2} + a_{3} + ⋯ + a_{n} + ∑_{k=n+1}^{∞} a_{k}

and call

(6) R_{n} = ∑_{k=n+1}^{∞} a_{k}

the remainder after n-terms. In the setting of the integral test, a comparison of areas shows that

(7) R_{n} < ∫_{n}^{∞} f(x) dx.

Example 2. Estimate the sum of ∑_{n=1}^{∞} 1/n^{2} with an error less than or equal to .005.

Solution. Write

S = ∑_{n=1}^{∞} \frac{1}{n^{2}} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + ⋯ + \frac{1}{n^{2}} + R_{n},

where R_{n} = ∑_{n=1}^{∞} 1/k^{2}. Using f(x) = 1/x^{2} we find that

R_{n} < ∫_{n}^{∞} \frac{1}{x^{2}} dx = \frac{1}{n}.

Now the idea is to choose n so large that R_{n} < .005 and this certainly occurs when n = 200. Since the partial sums increase to S we have

\frac{1}{2^{2}} + ⋯ + \frac{1}{200^{2}} < S < 1 + \frac{1}{2^{2}} + ⋯ + \frac{1}{200^{2}} + .005

and, with the aid of a calculator, find that

1.639946... < S < 1.644946...

Actually it can be shown by advanced methods that

S = ∑_{n=1}^{∞} \frac{1}{n^{2}} = \frac{π^{2}}{6} = 1.64493406...

Example 3. Use the integral test to show that the series ∑_{n=2}^{∞} \frac{1}{n ln n} diverges.

Solution. The test function f(x) = \frac{1}{x ln x} is continuous, positive, and decreasing on the range x ≥ 2, and

∫_{2}^{∞} \frac{dx}{x ln x} = lim_{b→∞} ∫_{2}^{b} \frac{dx}{x ln x} = lim_{b→∞} ln (ln x) |_{2}^{b}
= lim_{b→∞} [ln (ln b) − ln (ln 2)] = ∞.

Hence the series diverges, even though its terms tend to zero faster than the terms of the divergent series ∑ 1/n. ‡

In applying the integral test in the preceding example we deviated from the integral test recipe as given in Theorem 10 in that everything was shifted to the right by one unit. A quick review of the idea behind the integral test should convince you that nothing is amiss. Indeed, the integral test applies equally well if the beginning index in (1) and the lower limit in (2) are changed from 1 to any positive integer k.

COMPARISON TESTS

The integral test is of no use in showing that the series

(8) 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ⋯ + \frac{1}{n!} + ⋯

converges, even though the terms are positive and decreasing. The difficulty, of course, is that there is no apparent candidate for f(x). Nonetheless, this series can be analyzed by the idea upon which the integral test is based; namely, that of comparing the terms of one series with those of another. One comparison arises from observing that 2! ≥ 2, 3! > 2^{2}, 4! > 2^{3} and, in general, if k > 1 and one knows that if

k! > 2^{k−1}, then (k + 1)! = k!(k + 1) > 2^{k−1}(k + 1) > 2^{k}.

This inductive argument shows that n! ≥ 2^{n−1} for each positive integer (in fact, equality holds only at n = 1). Hence 1/n! ≤ 1/2^{n−1} and

\begin{aligned}s_{n} & = 1 + \frac{1}{2} + \frac{1}{2^{2}} + ⋯ + \frac{1}{2^{n−1}} = \frac{1 − 1/2^{n}}{1 − 1/2} \\ & = 2 − \frac{1}{2^{n−1}} < 2 \end{aligned}

for each positive integer n. This last inequality shows that the partial sums of (8) are bounded above by 2 and that, according to Theorem 10, it converges. Later on we will show that the sum in (8) is e − 1 = 1.718281828...

The preceding argument shows that the series ∑_{n=1}^{∞} 1/n! converges because its terms are

less than the corresponding terms of the convergent series ∑_{n=1}^{∞} 1/2^{n−1}. In general, given a

series ∑ a_{n} of positive terms and inequalities

0 < a_{n} ≤ b_{n},

for all n, we say that the series ∑ b_{n} dominates the series ∑ a_{n}.

Theorem 12. Comparison Test Let ∑ a_{n} and ∑ b_{n} be series of positive terms.

Convergence Test. If 0 ≤ a_{n} ≤ b_{n} for all n and ∑ b_{n} converges, then ∑ a_{n} is also convergent.

Divergence Test. If 0 ≤ b_{n} ≤ a_{n} for all n and ∑ b_{n} diverges, then ∑ a_{n} is also divergent.

Proof. Let s_{n} = ∑_{k=1}^{n} a_{k} and t_{n} = s_{n} = ∑_{k=1}^{n} b_{k}. For the Convergence Test, s_{n} ≤ t_{n} for all n

and the partial sums t_{n} increase to the sum T = ∑ b_{n}. Hence s_{n} ≤ t_{n} ≤ T and the partial sums {s_{n}} form an increasing and bounded sequence, that is,

s_{1} ≤ s_{2} ≤ s_{3} ≤ ⋯ ≤ s_{n} ≤ ⋯ ≤ T.

The convergence of ∑ a_{n} is guaranteed by Theorem 8.

In the Divergence Test, the relation between s_{n} and t_{n} is reversed. Now, t_{n} ≤ s_{n} because b_{n} ≤ a_{n} for all n and it is clear that s_{n} must grow large if t_{n} does. But t_{n} tends to infinity because ∑ b_{n} is divergent; hence s_{n} → ∞ as n → ∞ and ∑ a_{n} diverges. ‡

Example 4. Use the comparison test to determine whether the series ∑_{n=1}^{n} \frac{\sqrt{n}}{n^{2} + 1} converges.

Solution. The series is convergent because

0 ≤ \frac{\sqrt{n}}{n^{2} + 1} = \frac{1}{n^{3/2} + 1/\sqrt{n}} < \frac{1}{n^{3/2}}

and ∑_{n=1}^{n} 1/n^{3/2} is a known convergent series – the p-series for p = 3/2. ‡ Example 5. Apply the comparison test to the series ∑_{n=1}^{n} \frac{n^{3/2}}{3n^{2} − n}.

Solution. The key observation is that the nth term of this series behaves like 1/3\sqrt{n} for large n and that ∑ \frac{1}{3\sqrt{n}} = \frac{1}{3} ∑ \frac{1}{\sqrt{n}} is a divergent series. This suggests we use the comparison test in its divergence form. Specifically, the inequality

\frac{n^{3/2}}{3n^{2} − n} = \frac{1}{3\sqrt{n} − \frac{1}{n}} > \frac{1}{3\sqrt{n}} > 0

shows that the given series is divergent. ‡

WARNING. In applying the comparison test 0 < a_{n} ≤ b_{n}, you can conclude nothing about ∑ a_{n} if the series ∑ b_{n} is divergent. Thus, 0 < 1/n^{2} < 1/n and 0 < 1/2n < 1/n; yet ∑ 1/n^{2} converges and ∑ 1/2n diverges. Similarly, if 0 < a_{n} ≤ b_{n} and ∑ b_{n} and ∑ a_{n} converges, you can conclude nothing about the series ∑ b_{n}.

5. Alternating Series

In this section we turn our attention to series that have positive as well as negative terms. Of these, the simplest are those whose terms are alternating positive and negative. An alternating series may be written conveniently in the form

∑_{n=1}^{∞} (−1)^{n−1}a_{n} = a_{1} − a_{2} + a_{3} − a_{4} + ⋯

where each a_{n} > 0. The series

\frac{1}{\sqrt{3}} − \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{7}} − \frac{1}{\sqrt{9}} + ⋯,

1 − \frac{1 · 2}{1 · 3} + \frac{1 · 2 · 3}{1 · 3 · 5} − \frac{1 · 2 · 3 · 4}{1 · 3 · 5 · 7} + ⋯,

and

1 − 1 + 1 − 1 + ⋯

are examples of alternating series. Also, for x in the range −1 < x < 0, the geometric series

1 + x + x^{2} + ⋯ = \frac{1}{1 − x} (|x| < 1)

becomes a convergent alternating series; at x = −1/2, it reduces to

1 − \frac{1}{2} + \frac{1}{2^{2}} − \frac{1}{2^{3}} + ⋯ = \frac{2}{3}.

The remarkable feature of an alternating series is that it will converge on the sole assumption that the a_{n} decrease to zero.

Theorem 14. Alternating Series Test If (1) a_{n} ≥ a_{n+1} > 0 and a_{n} → 0 as n → ∞, then the alternating series ∑_{n=1}^{∞}(−1)^{n−1}a_{n} converges. Moreover, if S = ∑_{n=1}^{∞} (−1)^{n−1}a_{n} and s_{n} is the nth partial sum, then

(2) |S − s_{n}| < a_{n+1}

Proof. the idea: partial sums with an odd index decrease toward S while partial sums with an even index increase toward S. Let us first look at the even partial sums – those with even indices. We have

s_{4} = s_{2} + (a_{3} − a_{4}) ≥ s_{2} because a_{3} ≥ a_{4},
s_{6} = s_{4} + (a_{5} − a_{6}) ≥ s_{4} because a_{5} ≥ a_{6},

and, in general,

s_{2n+2} = s_{2n} + (a_{2n+1} − a_{2n+2}) ≥ s_{2n} because a_{2n+1} ≥ a_{2n+2}

For the odd partial sums,

s_{3} = s_{1} − a_{2} + a_{3} = s_{1} − (a_{2} − a_{3}) ≤ s_{1} because a_{2} ≥ a_{3},
a_{5} = s_{3} − a_{4} + a_{5} = s_{3} − (a_{4} − a_{5}) ≤ s_{3} because a_{4} ≥ a_{5},

and, in general,

s_{2n+1} = s_{2n−1} = a_{2n} + a_{2n+1} = s_{2n−1} − (a_{2n} − a_{2n+1}) ≤ s_{2n−1} because a_{2n} ≥ a_{2n+1}

Further, s_{2} + a_{3} = s_{3}, s_{4} + a_{4} = s_{5}, and

(3) s_{2n} + a_{2n+1} = s_{2n+1}

for all n, so that s_{2n} < s_{2n+1}. Hence the situation is exactly as shown in the figure:

s_{2} ≤ s_{4} ≤ s_{6} ≤ ⋯ ≤ s_{2n−2} ≤ s_{2n} ≤ s_{2n−2} ≤ s_{2n} < s_{2n+1} ≤ s_{2n−1} ≤ ⋯ ≤ s_{5} ≤ s_{3} ≤ s_{1}.

The information we wish to extract from this string of inequalities is that the even partial sums {s_{2n}} form an increasing sequence bounded above by s_{1} and that the odd partial sums {s_{2n+1}} form a decreasing sequence bounded below by s_{2}. We now call on Theorem 5; it tells us that both sequences converge, one to a number \underline{S}, say, the other to \overline{S}:

(4)  lim_{n→∞} s_{2n} = \underline{S}, lim_{n→∞} = \overline{S}.

To this point, everything we have said depends on the assumptions a_{n} ≥ a_{n+1} > 0. The next step uses the assumption a_{n} → 0 and, in particular, that a_{2n+1} → 0 as n → ∞. From the identities (3) and the limits in (4) it follows that

\overline{S} = lim_{n→∞} s_{2n+1} = lim_{n→∞} (s_{2n} + a_{2n+1}) = \underline{S} + 0 = \underline{S},

that is, \overline{S} = \underline{S}. This common value, call it S, is clearly the limit of the entire sequence {s_{n}}. We have proved that the partial sums have a limit and hence that the alternating series is convergent.

Proof of inequality (2) rests on the observation that S, the sum of the series, always lies between any two consecutive partial sums; precisely

(5) s_{n+1} < S < s_{n} if n is odd. and

(6) s_{n} < S < s_{n+1} if n is even.

Since |s_{n+1} − s_{n}| = a_{n+1}, we see that the inequality |S − s_{n}| < a_{n+1} is merely the statement that the distance from a point inside an interval to one of its endpoints is less than the length of the interval. ‡

The first step in applying the Alternating Series Test is visual: you have to check to be the terms actually alternate in sign. Clearly, the series

(7) 1 − \frac{1}{2} − \frac{1}{3} + \frac{1}{4} + \frac{1}{5} − \frac{1}{6} − \frac{1}{7} − \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16} − \frac{1}{17} − ⋯

has positive and negative terms; yet, it is not an alternating series and the Alternating Series Test cannot be applied. Actually, the series (7) diverges. Nor does the alternating series

e − e^{1/2} + e^{1/3} − e^{1/4} + ⋯

meet the requirements of the Alternating Series Test, even though a_{n} = e^{1/n} is positive and decrease with n. The difficulty is that lim_{n→∞} a_{n} ≠ 0 (see Theorem 1). Thus the series diverges by Theorem 7. But in the series

(8) 1 − \frac{1}{2} + \frac{1}{3} − \frac{1}{4} + ⋯,

a_{n} = 1/n is positive, a_{n} > a_{n+1} and lim_{n→∞} a_{n} = 0. The series is therefore convergent because all conditions in (1) are satisfied; accordingly, we say that the series (8) converges by the Alternating Series Test.

Further, according to (2), if S denotes the sum of the series in (8) and we truncate the series after the 199th term, then the partial sum s_{199} approximates S with an error less than a_{200} = 1/200 = .005. More accurately, by (5), s_{199} is an overapproximation and s_{200} is an underapproximation to S. Aided by a calculator, we find that

.690653... = s_{200} < S < s_{199} = .695653...

Naturally, more accurate estimates require more terms; it takes more than a million terms to estimate S to six place accuracy. Later we shall show that

S = ln 2 = .693147...

Example 1. Show that the series

∑_{n=1}^{∞} (−1)^{n−1} \frac{n}{n^{2} + 1} = \frac{1}{2} − \frac{2}{5} + \frac{3}{10} − \frac{4}{17} + ⋯

is convergent.

Solution. In this alternating series a_{n} = \frac{n}{n^{2}+1} and the conditions in (1) are most easily verified by examining the continuous function f(x) = \frac{x}{1+x^{2}} . It is clearly positive in the interval x ≥ 1 and decreasing there since

f′(x) = \frac{1 − x^{2}}{(x^{2} + 1)^{2}} < 0

if 1 < x. Because a_{n} = f(n), it follows that a_{n} > a_{n+1} > 0. And it is clear that lim_{n→∞} a_{n} = lim_{n→∞} \frac{n}{n^{2}+1} = 0. Therefore the series is convergent by the Alternating Series Test. ‡

Example 2. Show that the series

1 − \frac{1}{2!} + \frac{1}{3!} − \frac{1}{4!} + ⋯

is convergent and estimate the error in approximating its sum S by the first nine terms.

Solution. Here a_{n} = 1/n! and a_{n} > a_{n+1} > 0 because n! increases with n. Also a_{n} → 0 because n! increases without bound. [Recall that n! ≥ 2^{n−1}.] Hence the series converges by the Alternating Series Test.

The error in approximating the sum S by

s_{9} = 1 − \frac{1}{2!} − \frac{1}{3!} − \frac{1}{4!} + \frac{1}{5!} − \frac{1}{6!} + \frac{1}{7!} − \frac{1}{8!} + \frac{1}{9!} = .63212081...

is less than

a_{10} = \frac{1}{10!} = .000000275...;

therefore

S ≈ .632121,

rounded to six decimal places. ‡

Example 3. Show that the series

∑_{n=1}^{∞} (−1)^{n−1} \frac{2n + 1}{n(n + 1)} = \frac{3}{2} − \frac{5}{2.3} + \frac{7}{3.4} − \frac{9}{4.5} + ⋯

is convergent.

Solution. In this alternating series a_{n} = \frac{2n+1}{n(n+1)}. Clearly, a_{n} > 0 and lim_{n→∞} a_{n} = 0. Rather than proceed as in Example 1, we can establish the inequality a_{n} ≥ a_{n+1} directly by computing the ratio

\frac{a_{n+1}}{a_{n}} = \frac{2n + 3}{(n + 1)(n + 2)} ˙ \frac{n(n + 1)}{2n + 1} = \frac{2n^{2} + 3n}{2n^{2} + 5n + 2}

and then noting the denominator is larger than the numerator if n ≥ 1. Hence a_{n+1}/a_{n} < 1 or a_{n} > a_{n+1}, and by Theorem 14 the series converges. ‡

PROBLEM SET 20.1

1. Compute the first 5 terms of each of the sequences \{a_{n}\}. (a) a_{n} = sin \frac{nπ}{2} (b) a_{n} = \frac{1}{2}[(1 + \sqrt{5})^{n} + (1 − \sqrt{5}^{n}] (c) a_{n} = 2^{cos n π/3}

Problems 3–15 give the nth term of a sequence. Find the limit if it exists. If the sequence diverges, say so. Use Theorem 1, 2, and 3, the techniques introduced in the Examples, and known properties of functions.

3. a_{n} = \frac{n^{2} + 1}{n^{2}}  4. a_{n} = \frac{3n^{4} − 5n^{3} + 7n − 1}{5n^{4} + 4n^{3} − 8n + 15}
5. a_{n} = e^{1/n}  6. a_{n} = (0.7)^{n}

7. a_{n} = (1.01)^{n}  8. a_{n} = \frac{ln n)^{2}}{n}

9. a_{n} = ∫_{−1/n}^{1/n} \sqrt{1 + t^{4}} dt  10. a_{n} = ∫_{−n}^{n} \frac{dt}{1 + t^{2}}
11. a_{n} = (1 + \frac{3}{n})^{n}  12. a_{n} = (n + 1)^{1/n}

13. a_{n} = (1 − \frac{1}{n^{2}} )^{1/n}

14. a_{n} = [sin 1/n]^{1/n}  [Hint: \frac{2x}{π} < sin x < x for 0 < x < π/2.]

15. a_{n} = \frac{4^{n}}{n!}

16. Does the convergence of {|a_{n}|} always imply the convergence of {a_{n}}?

17. If a_{n} → 3 and b_{n} → 4 as n → ∞, what about the sequence a_{1}, b_{1}, a_{2}, b_{2}, a_{3}, b_{3}, ...?

174 20. INFINITE SERIES

CALCULATOR PROBLEM

18. (a) The sequence a_{n} = 50^{1/n} has limit 1. Compute a table comparing a_{n} − 1 to \frac{1}{n} ln 50 for the values n = 50, 100, 1000, 2000 and observe that, of the two, a_{n} − 1 is lightly larger. (b) Apply the Mean Value Theorem for derivatives to 50^{x} = e^{x ln 50} to conclude that 50^{1/n} − 1 = e^{θ ln 50} · \frac{1}{n} ln 50 for some θ between 0 and 1/n. (c) Use the estimates e < 3 and ln 50 < 4 to conclude that 50^{1/n} − 1 < 12/n if n > 4.

5. ALTERNATING SERIES 175

PROBLEM SET 20.2

In Problems 1–5, decide whether the sequence is eventually increasing, eventually decreasing, or neither.

1. a_{n} = \frac{1}{1 + n^{2}}  2. a_{n} = \frac{2^{n}}{n!}
3. a_{n} = \frac{10^{n}}{n!}  4. a_{n} = ∫_{n}^{n+1} \frac{dx}{1 + x^{4}}

5. a_{n} = \frac{1}{n} − ln \frac{n + 1}{n}

Use Theorem 5 to show that the next four sequences are convergent.

\begin{matrix}6. a_{n} = \frac{2 · 4 · 6 ⋯ 2n}{1 · 3 · 5 ⋯ (2n + 1)} & 7. a_{n} = \frac{(n!)^{2}}{(2n)!} \\ 8. a_{n} = ∫_{0}^{n} e^{−x^{2}} dx & 9. a_{n} = \frac{1}{(n + 1)^{2}} + \frac{1}{(n + 2)^{2}} + ⋯ + \frac{1}{(2n)^{2}} \end{matrix}

CALCULATOR PROBLEMS

10. Choose x_{1} positive and define the sequence \{x_{n}\} by x_{n+1} = \frac{5}{1 + x_{n}} . (a) Construct tables for various choices of x_{1} to show that the terms seem to lie alternately to the left and to the right of L = 1.7912878... (b) Argue geometrically that the terms a_{n} oscillate about the limit L with the even terms forming an increasing sequence and the odd terms forming a decreasing sequence. [Look at the graphs y = x and y = \frac{5}{1+x}.] (c) Show that L is the positive root of the quadratic x^{2} + x − 5 = 0.

11. Compute the nth term of the sequence a_{n} = \frac{1}{n + 1} + \frac{1}{n + 2} + ⋯ + \frac{1}{2n}. for n = 25, 50, 100, 200 and show that ln \frac{2n + 1}{n + 1} < a_{n} < ln 2.

176 20. INFINITE SERIES

PROBLEM SET 20.3

In Problems 1–7, decide whether the series converges or diverges. If it diverges, state why. If it converges, find its sum.

1. 1 + \frac{1}{2} + \frac{1}{2^{2}} − \frac{1}{2^{3}} + ⋯  2. 1 + e^{−1} + e^{−2} + e^{−3} + ⋯
3. 1 + 0 + \frac{1}{2} + 0 + \frac{1}{3} + 0 + \frac{1}{4} + ⋯ 4. ∑_{n=0}^{∞} (−1)^{n+2}(.9)^{n}
5. ∑_{n=0}^{∞} \frac{1}{R} (\frac{r}{R})^{n} (0 < r < R)  6. ∑_{n=1}^{∞} ln (\frac{3n+5}{2n+1})

7. ∑_{n=0}^{∞} \frac{e^{−ks}}{s^{k}} (s = positive constant)

8. Use the geometric series formula and (10) to find the rational representation for each of the following infinite decimals. (a) .494949...  (b) .123123... (e) .ababab...  (a and b one of 0, 1, 2, ..., 9)

9. Replace x by x^{2} in the geometric series (6) to find a series representation for \frac{1}{1−x^{2}} . Then obtain the same series another way: add the series 1 + x + x^{2} + ⋯ and 1 − x + x^{2} − · · ·.

10. Use the geometric series and (10) to find the sum of each of the following series. Assume |x| < 1. (a) x^{2} + x^{3} + x^{4} + ⋯ (b) 1 + sin \frac{πx}{2} + sin^{2} \frac{πx}{2} + ⋯

11. Choose a > 0 and define the function f by f(x) = \frac{1}{a} − \frac{1}{a} (\frac{(x − a)}{a}) + \frac{1}{a} (\frac{(x − a)}{a})^{2} − \frac{1}{a} (\frac{(x − a)}{a})^{3} + ⋯ Find the domain of f(x) and a simple formula for f(x) in that domain.

12. Find the terms of the infinite series having partial sums s_{n} = \frac{1}{2} + \frac{1}{2} \frac{sin (n + \frac{1}{2}) θ}{sin \frac{θ}{2}} , n = 0, 1, 2, ... .

PROBLEM SET 20.4

In Problems 1–6, show that the Integral Test can be applied and then use it to determine whether the series converges or diverges.

\begin{matrix}1. ∑_{n=1}^{∞} \frac{1}{5n^{2}+1} & 2. ∑_{n=1}^{∞} \frac{1}{(3n+1)^{2}} \\ 3. ∑_{n=1}^{∞} \frac{2n+1}{n^{2}+n+1} & 4. ∑_{n=1}^{∞} ne^{−n} \\ 5. ∑_{n=1}^{∞} \frac{tan^{−1} n}{1+n^{2}} & 6. ∑_{n=1}^{∞} \frac{1}{\sqrt{n}+5} \end{matrix}

In the next seven problems establish the convergence or divergence of each series by application of one of the comparison tests.

\begin{matrix}7. ∑_{n=1}^{∞} \frac{1}{10n^{2}+17} & 8. ∑_{n=1}^{∞} \frac{3^{n}+2}{4^{n}+3} \\ 9. ∑_{n=1}^{∞} \frac{1}{n(n^{2}+n+1)} & 10. ∑_{n=1}^{∞} \frac{1}{n(n^{2}+n+1)} \\ 11. ∑_{n=1}^{∞} \frac{1}{\sqrt{n^{2}+n+1}} & 12. ∑_{n=1}^{∞} \frac{\sqrt{n}}{n^{2}+e^{−k}} \end{matrix}

13. Use the Mean Value Theorem to show that 0 < x − ln (1 + x) < x^{2} if x > 0, and hence that the series ∑_{n=1}^{∞} (\frac{1}{n} − ln (1 + \frac{1}{n})) converges while both ∑_{n=1}^{∞} \frac{1}{n} and ∑_{n=1}^{∞} ln (1 + \frac{1}{n}) are divergent series.

In the next two problems, first use the Mean Value Theorem to estimate the nth term; then determine convergence or divergence by one of the comparison tests.

14. ∑_{n=1}^{∞} sin^{2} \frac{1}{n}  15. ∑_{n=1}^{∞} (e^{1/n} − 1)

16. From the inequality ∫_{1}^{1+1/n} \frac{dx}{n} > \frac{1}{1+n} and by comparing areas, show that s_{n} = 1 = \frac{1}{2} + \frac{1}{3} + ⋯ + +\frac{1}{n} − ln n is a decreasing sequence of positive numbers.

CALCULATOR PROBLEMS In Problems 17 and 18, find an index N for which the partial sum s_{N} approximates the sum S of the given convergent series with an error less than the indicated value ϵ. Then compute s_{N} and express the results in the form s_{N} < S < s_{N} + ϵ. Follow the procedure outlined in Example 2.

17. ∑_{n=1}^{∞} \frac{1}{n^{4}} , ϵ = .00005  18. ∑_{n=2}^{∞} \frac{1}{n(ln n)^{2}} , ϵ = .1

PROBLEM SET 20.5

In Problems 1–9, determine whether the given series is convergent or divergent.

1. 1 − \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} − \frac{1}{\sqrt{4}} + ⋯  2. 1 − e^{−2} + e^{−3} − e^{−4j} + ⋯
3. ∑_{n=1}^{∞} (−1)^{n−1} \frac{1}{4n^{2}}  4. ∑_{n=1}^{∞} (−1)^{n−1}(ln n)^{−2}
5. ∑_{n=1}^{∞} (−1)^{n−1} (\frac{1}{2} − \frac{1}{n})^{−n}  6. ∑_{n=1}^{∞} (−1)^{n−1} \frac{n^{2}}{n^{3} + n}
7. ∑_{n=0}^{∞} \frac{(−2)^{n}}{n!}  8. ∑_{n=1}^{∞} (−1)^{n−1} \frac{n^{2}}{n^{2} + n}

9. ∑_{n=1}^{∞} (−1)^{n−1}ne^{−n}

In Problems 10–12 indicate the values of x for which the given series alternates in sign.

\begin{matrix}10. 1 − \frac{x^{2}}{2!} + \frac{x^{4}}{4!} − \frac{x^{6}}{6!} + ⋯ & 11. x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − \frac{x^{7}}{7!} + ⋯ \\ 12. (x − 1)^{2} − (x − 1)^{3} + (x − 1)^{4} − ⋯ \end{matrix}

13. ∑_{n=1}^{∞} (−1)^{n−1} \frac{1}{n!} , error < .0001

14. ∑_{n=1}^{∞} (−1)^{n−1} \frac{1}{n(n + 1)} , error < .01

6. Absolute Convergence

As we have seen, the alternating series

(1) 1 − \frac{1}{2} + \frac{1}{3} − \frac{1}{4} + ⋯ = 1 + (\frac{1}{2}) + (\frac{1}{3}) + (−\frac{1}{4}) + ⋯

is convergent, but the harmonic series

|1| + |−\frac{1}{2}| + |\frac{1}{3}| + |−\frac{1}{4}| + ⋯ = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ⋯

obtained from it by taking the absolute value of each term, is not. Thus tampering with its signs may destroy the convergence of a series – but not necessarily as we see by examining the series

(2) 1 − \frac{1}{2} + \frac{1}{2^{2}} − \frac{1}{2^{3}} + · · · .

This is a convergent geometric series ∑_{n=0}^{∞}x^{n} with x = −1/2 and the series of absolute values

|1| + |−\frac{1}{2}| + |\frac{1}{2^{2}} | + |−\frac{1}{2^{3}} | + ⋯ = 1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ⋯

is also a convergent geometric series. The point is that some convergent series having both positive and negative terms persist in their convergence after each term is replaced by its absolute value – others do not. The underlying concept is called absolute convergence.

Definition. Absolute Convergence The series

∑_{n=1}^{∞} a_{n} = a_{1} + a_{2} + a_{3} + ⋯ + a_{n} + ⋯

is said to be absolutely convergent if the series

(3)  ∑_{n=1}^{∞} |a_{n}| = |a_{1}| + |a_{2}| + |a_{3}| + ⋯ + |a_{n}| + ⋯

is convergent.

According to his definition and the preceding discussion, the series (2) is absolutely convergent; the series (1) is not. Of course, if a_{n} ≥ 0 for all n, then

∑_{n=1}^{∞} |a_{n}| = |a_{1}| + |a_{2}| + |a_{3}| + ⋯ = a_{1} + a_{2} + a_{3} + · · · .

Hence a convergent series of positive terms is always absolutely convergent.

Example 1. The series

(a) 1 − \frac{1}{3^{2}} + \frac{1}{5^{2}} − \frac{1}{7^{2}} + ⋯ + \frac{(−1)^{n+1}}{(2n − 1)^{2}} + ⋯

is absolutely convergent because

1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + ⋯ + \frac{1}{(2n − 1)^{2}} + ⋯

is convergent. For instance, apply the integral test to f(x) = \frac{1}{(2x − 1)^{2}} . But the series

(b) 1 − \frac{1}{3} + \frac{1}{5} − \frac{1}{7} + ⋯ + \frac{(−1)^{n+1}}{(2n − 1)} + ⋯

is not absolute convergent because

1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + ⋯ + \frac{1}{(2n − 1)} + ⋯

is a divergent series, as may be seen by applying the integral test to f(x) = \frac{1}{2x − 1}.

It is traditional to give a name to a series which converges but does not converge absolutely. Definition The series ∑ a_{n} is called conditionally convergent if it is convergent but ∑ |a_{n}| diverges.

This definition simply divides all convergent series into two groups – those that converge absolutely and those that do not. The series

ln 2 − ln 3/2 + ln 4/3 − ln 5/4 + ⋯

and

1 − 1/\sqrt{2} − 1/\sqrt{3} + 1/\sqrt{4} + 1/\sqrt{5} − 1/\sqrt{6} − 1/\sqrt{7} + ⋯

are conditionally convergent: each series converges but the corresponding series, with all connecting signs positive, diverge. In such series you should think of convergence as being the result of delicately balanced internal cancellation, a balance that is shattered by making each connecting sign positive. On the other hand, in an absolutely convergent series like

1 − \frac{1}{2!} + \frac{1}{3!} − \frac{1}{4!} + ⋯,

you may change the signs any way you please and the resulting series remains convergent. We close this section with an important test for absolute convergence.

Theorem 15. The Ratio Test In the series ∑ a_{n} suppose that lim_{n→∞} |\frac{a_{n+1}}{a_{n}} | = L. Then

(a) if 0 ≤ L < 1, the series converges absolutely;

(b) if L > 1, the series diverges; and

(c) if L = 1, the test gives no information concerning convergence.

Proof. We begin with (a). The assertion is that the rations |\frac{a_{n+1}}{a_{n}} | tend to some number

L in the interval 0 ≤ L < 1. Since the number r = \frac{L + 1}{2} satisfies the inequality L < r < 1, it is possible to choose N large enough so that

|\frac{a_{n+1}}{a_{n}} | < r if n ≥ N.

In this inequality we give n, successively, the values N, N +1, N +2, ... to obtain interlocking inequalities which, in tandem, imply that

|a_{N+1}| < |a_{N}|r,

|a_{N+2}| < |a_{N+1}|r < |a_{N}|r^{2},
|a_{N+3}| < |a_{N+2}| < |a_{N}|r^{3},

⋮

At the kth level,

|a_{N+k}| < |a_{N}|r^{k} (k ≥ 1).

These inequalities are the key because, recalling that r < 1, they show that the series

|a_{N+1}| + |a_{N+2}| + |a_{N+3}| + ⋯

is dominated by the convergent geometric series

|a_{N}|r + |a_{N}|r^{2} + |a_{N}|r^{3} + ⋯ = |a_{N}| ∑_{n=1}^{∞} r^{n}.

Hence ∑_{n=1}^{∞} |a_{N+k}| converges by the Comparison Test. Now add on the first N terms, and

we see that

∑_{n=1}^{∞} |a_{n}| = {|a_{1}| + |a_{2}| + ⋯ + |a_{N}|} + ∑_{n=1}^{∞} |a_{N+k}|

converges.

Turning to (b) we need only observe that if L > 1 or L = ∞ then the ratios ultimately become and remain greater than 1 – meaning that, for some N,

|\frac{a_{n+1}}{a_{n}} | ≥ 1 if n ≥ N.

Again, by giving n the values N, N + 1, N + 2, ..., there results a string of inequalities

|a_{N}| < |a_{N+1}| < |a_{N+2}| < ⋯

from which it is clear that the positive numbers |a_{n}| increase for n ≥ N and, in particular, lim_{n→∞} a_{n} ≠ 0. Hence the series ∑ a_{n} must diverge.

The proof of part (c) is left as a problem. ‡

In a series of positive terms, the Ratio Test is applied without absolute values, in the form

lim_{n→∞} \frac{a_{n+1}}{a_{n}} = L.

Then ∑ a_{n} converges if L < 1, diverges if L > 1, and the test is inconclusive if L = 1.

Example 2. Use the Ratio Test to determine the convergence or divergence of the following series.

(a) ∑_{n=1}^{∞} \frac{n}{2^{n}} = \frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \frac{4}{2^{4}} + ⋯

(b) ∑_{n=1}^{∞} \frac{n^{n}}{n!} = 1 + \frac{2^{2}}{2!} + \frac{3^{3}}{3!} + ⋯

(c) ∑_{n=1}^{∞} \frac{1}{3n + 1} = \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + ⋯

Solution. (a) Here a_{n} = \frac{n}{2^{n}} , a_{n+1} = \frac{n + 1}{2^{n+1}} and

lim_{n→∞} \frac{a_{n+1}}{a_{n}} = lim_{n→∞} \frac{n + 1}{2^{n+1}} \frac{2^{n}}{n} = lim_{n→∞} \frac{1}{2} \frac{n + 1}{n} = \frac{1}{2}.

Because \frac{1}{2} < 1, the series converges. ‡

(b) We have

\frac{a_{n+1}}{a_{n}} = \frac{(n + 1)^{n+1}}{(n + 1)!} \frac{n!}{n^{n}} = \frac{(n + 1)(n + 1)^{n}}{(n + 1)n!}

and

lim_{n→∞} \frac{a_{n+1}}{a_{n}} = lim_{n→∞} (1 + \frac{1}{n})^{n} = e.

Because e > 1, the series diverges. ‡

(c) In this series the Ratio Test is useless inasmuch as

lim_{n→∞} \frac{a_{n+1}}{a_{n}} = lim_{n→∞} \frac{3n + 1}{3(n + 1) + 1} = 1.

That the series is actually divergent may be seen by applying the Integral Test with f(x) =

\frac{1}{3x + 1}. ‡

Example 6. Use the Ratio Test to show that series

∑_{n=1}^{∞} (−1)^{n+1} \frac{1 · 3 · 5 ⋯ (2n − 1)}{(2n)!} = \frac{1}{2} − \frac{1 · 3}{4!} + \frac{1 · 3 · 5}{6!} − \frac{1 · 3 · 5 · 7}{8!} + ⋯

is absolutely convergent. Solution. In this more complicated example

a_{n} = (−1)^{n+1} \frac{1 · 3 · 5 ⋯ (2n − 1)}{(2n)!}, a_{n+1} = (−1)^{n+2} \frac{1 · 3 · 5 ⋯ (2n − 1)(2n + 1)}{(2n + 2)!}

and

\frac{a_{n+1}}{a_{n}} = −\frac{1 · 3 · 5 ⋯ (2n − 1)(2n + 1)}{(2n + 2)!} \frac{(2n)!}{1 · 3 · 5 ⋯ (2n − 1)}
= \frac{2n + 1}{(2n + 1)(2n + 2)}. Note that [(2n + 1)! = (2n + 2)(2n + 1)(2n)!]

Hence

lim_{n→∞} |\frac{a_{n+1}}{a_{n}} | = lim_{n→∞} \frac{1}{2n + 2} = 0

and the series converges absolutely. ‡

184 20. INFINITE SERIES

At the elementary level, the Ratio Test is our best convergence test, especially for series in which a_{n} contains n!; (2n)! or some other factorial expression.

CHAPTER 21 Power Series

The ideas of the foregoing sections culminate in the study of infinite series in which each term is a function. Chief among these are power series, series that are important primarily because they provide a natural and powerful method of defining and studying functions – the ordinary as well as the exotic.

1. Power Series

By a power series in x we mean a series of the form

(1)  ∑_{n=0}^{∞} a_{n}x^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ⋯ + a_{n}x^{n} + ⋯

in which the a_{n} denote real numbers called coefficients and x is considered a variable.

The first question in discussing a power series is this: For which values of x does the series converge? It is the nature of a power series in x to converge absolutely in some symmetric open interval of the form |x| < r and to diverge if |x| > r or else to converge absolutely for −∞ < x < ∞. This open interval of absolute convergence is most conveniently found by applying the Ratio Test.

Example 1. Find the values of x at which the power series

∑_{n=1}^{∞} (−1)^{n+1} \frac{x^{n}}{n · 3^{n}} = \frac{1}{3}x − \frac{1}{2 · 3^{2}} x^{2} + \frac{1}{3 · 3^{3}} x^{3} + ⋯

converges.

Solution. In this power series a_{0} = 0, a_{1} = \frac{1}{3}, a_{2} = −\frac{1}{2·3^{2}} , etc. To apply the Ratio Test, we put

u_{n} = (−1)^{n+1} \frac{x^{n}}{n 3^{n}} , u_{n+1} = (−1)^{n+2} \frac{x^{n+1}}{(n + 1)3^{n+1}}

and calculate

\frac{u_{n+1}}{u_{n}} = (−1)^{n+2} \frac{x^{n+1}}{(n + 1)3^{n+1}} \frac{n 3^{n}}{(−1)^{n}3^{n+1}x^{n}} = −\frac{n}{3(n + 1)}x.

For fixed x,

lim_{n→∞} |\frac{u_{n+1}}{u_{n}} | = lim_{n→∞} |−\frac{n}{3(n + 1)}x| = lim_{n→∞} \frac{n}{3(n + 1)} |x| = \frac{|x|}{3}.

According to the Ratio Test, the series converges absolutely if \frac{|x|}{3} < 1 and diverges of

\frac{|x|}{3} > 1. Thus the series converges absolutely for x in the open interval

|x| < 3, −3 < x < 3

and diverges if

x < −3 or x > 3.

At x = 3, the power series reduces to the conditionally convergent alternating series

∑_{n=1}^{∞} (−1)^{n+1} \frac{3^{n}}{n 3^{n}} = ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n} = 1 − \frac{1}{2} + \frac{1}{3} − ⋯

and at x = −3, it reduces to the divergent series

∑_{n=1}^{∞} (−1)^{n+1} \frac{(−3)^{n}}{n 3^{n}} = ∑_{n=1}^{∞} \frac{(−1)^{2n+1}}{n} = −1 − \frac{1}{2} − \frac{1}{3} − · · · .

To sum up: This power series converges for every x in the interval −3 < x ≤ 3 and diverge otherwise. In the open interval −3 < x < 3, the convergence is absolute, at the endpoint x = 3 the convergence is conditional. ‡

Example 2. Find the values of x at which the power series

∑_{n=0}^{∞} \frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ⋯

converges.

Solution. To properly interpret the summation, it is necessary to remember that 0! = 1, 1! = 1 and x^{0} = 1.

Applying the Ratio Test with u_{n} = x^{n}/n! gives

lim_{n→∞} |\frac{u_{n+1}}{u_{n}} | = lim_{n→∞} |\frac{x^{n+1}}{(n + 1)!} \frac{n!}{x^{n}} | = lim_{n→∞} \frac{|x|}{(n + 1)} = 0,

no matter what the values of x. Conclusion: the series converges absolutely for all x, for −∞ < x < ∞. Later we shall show that this series has sum e^{x}. ‡

The following theorem provides a precise description of the possible situations than can occur in analyzing the convergence of a power series in x. A proof is outlined in the problems.

Theorem 18. Given a power series in x, exactly one of the following hold:

(a) The series converges only at x = 0.

(b) For some positive number r, the series converges absolutely in the open interval |x| < r and diverges if |x| > r.

(c) The series converges absolutely for all x.

Several remarks are in order. First, the situation in (a) is extreme and uninteresting because to be useful a power series must converge throughout some interval. The series

1 + x + 2!x^{2} + 3!x^{3} + 4!x^{4} + ⋯ + n!x^{n} + ⋯

illustrates the possibility. Like every power series in x, it converges at x = 0, but if x ≠ 0 the Ratio Test gives

lim_{n→∞} |\frac{(n + 1)!x^{n+1}}{n!x^{n}} | = lim_{n→∞} (n + 1) |x| = ∞

and the series diverges. The problem is caused by the rapid rate at which the coefficients a_{n} = n! increase.

Second, the endpoint r and −r of the open interval −r < x < r of absolute convergence set forth in part (b) must be analyzed separately. At these two points the series may diverge or converge, either absolutely or conditionally.

Thus the convergence set may be a single point, any of the intervals (−r, r), (−r, r], [−r, r), [−r, r], or (−∞, ∞). In all cases we shall refer to the set of x where a power series converges as its interval of convergence and to r as its radius of convergence. The possibilities corresponding to parts (a), (b), and (c) respectively are that r is 0, r is finite, and r is infinite.

Example 3. Find the interval of convergence and the radius of convergence of the power series

∑_{n=1}^{∞} \frac{2^{n}}{n^{2}} x^{n} = 2x + x^{2} + \frac{8}{9}x^{3} + · · · .

Solution. The Ratio Test with

u_{n} = \frac{2^{n}}{n^{2}} x^{n} and u_{n+1} = \frac{2^{n+1}}{(n + 1)^{2}} x^{n+1}

gives

lim_{n→∞} |\frac{u_{n+1}}{u_{n}} | = lim_{n→∞} |\frac{2^{n+1}}{(n + 1)^{2}} x^{n+1} · \frac{n^{2}}{2^{n}x^{n}} | = lim_{n→∞} \frac{2n^{2}}{(n + 1)^{2}} |x| = 2 |x|.

Hence the power converges absolutely in the open interval 2 |x| < 1 or |x| < 1/2, that is, for

−1/2 < x < 1/2;

and diverges if

x < −1/2 or x > 1/2.

Hence the radius of convergence is r = 1/2.

The series is also absolutely convergent at x = 1/2, where it reduces to the convergent p-series

∑_{n=1}^{∞} \frac{2^{n}}{n^{2}} (\frac{1}{2})^{n} = ∑_{n=1}^{∞} \frac{1}{n^{2}} = 1 + \frac{1}{2^{2}} + \frac{1}{2^{n}3} + ⋯,

and at x = −1/2 where it reduces to the absolutely convergent alternating series

∑_{n=1}^{∞} \frac{2^{n}}{n^{2}} (−\frac{1}{2})^{n} == ∑_{n=1}^{∞} \frac{(−1)^{n}}{2^{n}} = −\frac{1}{2} + \frac{1}{2^{2}} − −\frac{1}{3^{2}} + · · · .

Consequently, the interval of convergence is −1/2 ≤ x ≤ 1/2 and the convergence is absolute throughout. ‡

The series

(2)  ∑_{n=1}^{∞} a_{n}(x − a)^{n} = a_{0} + a_{1}(x − a) + a_{2}(x − a)^{2} + ⋯ + a_{n}(x − a)^{n} + ⋯,

obtained from (1) by replacing x by x − a, is called a power series in x − a. Its interval of absolute convergence is symmetric about a.

Example 4. Find the radius of convergence and the interval of convergence of the power series

∑_{n=0}^{∞} \frac{1 · 3 · 5 · 7 ⋯ (2n + 1)}{2^{2n}n!}(x − 1)^{n} = 1 + \frac{1 · 3}{2^{2}}(x − 1) + \frac{1 · 3 · 5}{2^{4} · 2!}(x − 1)^{2} + \frac{1 · 3 · 5 · 7}{2^{6} · 3!}(x − 1)^{3} + · · · .

Solution. We apply the Ratio Test with

u_{n} = \frac{1 · 3 · 5 ⋯ (2n + 1)}{2^{2^{n}}n!}(x − 1)^{n} and u_{n+1} = \frac{1 · 3 · 5 ⋯ (2n + 1)(2n + 3)}{2^{2n+2}(n + 1)!}(x − 1)^{n+1}.

Note: The next odd integer after 2n + 1 is 2(n + 1) + 1 = 2n + 3 and (n + 1)! = (n + 1)n!. Thus

\frac{u_{n+1}}{u_{n}} = \frac{1 · 3 · 5 ⋯ (2n + 1)(2n + 3)}{2^{2n+2}(n + 1)!}(x − 1)^{n+1} \frac{2^{2n}n!}{1 · 3 · 5 ⋯ (2n + 1)} \frac{1}{(x − 1)^{n}} = \frac{2n + 3}{2^{2}(n + 1)}(x − 1)

and

lim_{n→∞} |\frac{u_{n+1}}{u_{n}} | = lim_{n→∞} \frac{2n + 3}{4(n + 1)} |x − 1| = \frac{1}{2} |x − 1|.

Consequently, there is absolute convergence where \frac{1}{2} |x − 1| < 1 or |x − 1| < 2. This means that the radius of convergence is r = 2 and that the interval −1 < x < 3, with center 1 and radius 2, is an interval of absolute convergence.

Of course, the series diverges where |x − 1| > 2. It also diverges at the endpoints x = 3 and x = −1 because the terms of the corresponding series do not tend to zero as n increases. For instance, at x = 3 the original power series reduces to

∑_{n=0}^{∞} \frac{1 · 3 · 5 ⋯ (2n + 1)}{2^{2n}n!}(2)^{n} \frac{1 · 3 · 5 ⋯ (2n + 1)}{2^{n}n!}

and

∑_{n=0}^{∞} \frac{1 · 3 · 5 ⋯ (2n + 1)}{2^{n}n!} = \frac{1 · 3 · 5 ⋯ (2n + 1)}{2 · 4 · 6 ⋯ 2n} = \frac{3}{2} · \frac{5}{4} · \frac{7}{6} ⋯ \frac{2n + 1}{2n} > 1

for all n. Therefore −1 < x < 3 is the interval of convergence. ‡

2. Power Series Calculus

On its interval of convergence a power series ∑_{n=0}^{∞} a_{n}x^{n} defines a function

(1) f(x) = ∑_{n=0}^{∞} a_{n}x^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ⋯

and inside that interval of convergence f can be differentiated and integrated according to the following theorem whose proof we omit.

Theorem 19. Suppose the series (1) has a nonzero radius of convergence r. Then

(2) f′(x) = ∑_{n=1}^{∞} na_{n}x^{n−1} = a_{1} + 2a_{2}x^{2} + 3a_{3}x^{3} + ⋯ if |x| < r

and

(3) \begin{matrix}∫_{0}^{x} f(t) dt & = ∫_{0}^{x} (∑_{n=0}^{∞} a_{n}t^{n}) dt = ∑_{n=0}^{∞} a_{n}∫_{0}^{x} t^{n} dt = ∑_{n=0}^{∞} \frac{a_{n}}{n + 1}x^{n+1} \\ & = a_{0}x + \frac{a_{1}}{2} x^{2} + \frac{a_{2}}{3} x^{3} + ⋯ if |x| < r. \end{matrix}

Further, the three series (1), (2), and (3) have the same radius of convergence.

The point is that inside its radius of convergence a power series is a continuous and differentiable function whose derivatives and integrals are found by operating on the series t erm-by-term. It is important to understand that the operations in (2) and (3) can be applied repeatedly. Thus the derivative of f′(x) in (2) is

f′′(x) = ∑_{n=2}^{∞} n(n − 1)x^{(n−2)} if |x| < r.

In other words, insofar as differentiation and integration are concerned, we operate on a power series exactly as if it were a polynomial. Of course, like any other convergent series, power series can be multiplied by a constant and added together term by term.

In this section we shall apply (2) and (3) to the problem of expressing some of the elementary functions of calculus as power series. That is, given a function f(x) we will attempt to write it as a convergent power series

f(x) = a_{0} + a_{1}x + a_{2}x^{2} + ⋯ + a_{n}x^{n} + ⋯

in some interval |x| < r. The problem is to find the coefficient a_{n}, and the process is called expanding f in a power series or representing f by a power series. Our initial efforts revolve around the geometric series

(4) 1 + x + x^{2} + ⋯ + x^{n} + ⋯ = \frac{1}{1 − x}(|x| < 1).

Example 1. Find a power series representation for \frac{1}{3 − 2x}.

Solution. First we write the function in the form \frac{1}{3} \frac{1}{1 − \frac{2}{3}x}, then apply the geometric series

representation (4) with \frac{2}{3}x in place of x. The result is

\begin{aligned}\frac{1}{3 − 2x} = \frac{1}{3} \frac{1}{1 − \frac{2}{3}x} & = \frac{1}{3} [1 + \frac{2}{3}x + \frac{2^{2}}{3^{2}} x^{2} + \frac{2^{3}}{3^{3}} x^{3} + ⋯] \\ & = \frac{1}{3} + \frac{2}{3^{2}} x + \frac{2^{2}}{3^{3}} x^{3} + ⋯ (|x| < \frac{3}{2}) . ‡ \end{aligned}

Example 2. Find a power series representation for \frac{1}{(1 − x)^{2}} .

Solution. Let f(x) = \frac{1}{1 − x}. Then, from (4),

f(x) = 1 + x + x^{2} + x^{3} + ⋯ + x^{n} + ⋯ (|x| < 1).

According to (2), the derivative f′(x) = \frac{1}{(1 − x)^{2}} is the term-by-term derivative of the

series. Thus

\frac{1}{(1 − x)^{2}} = 1 + 2x + 3x^{2} + ⋯ + nx^{n−1} + ⋯ (|x| < 1). ‡

Example 3. Find a power series expansion for ln (1 + x).

Solution. Because 1/(1 + x) is the derivative of ln (1 + x), we start with the series

\frac{1}{1 + t} = 1 − t + t^{2} − t^{3} + ⋯ + (−1)^{n}t^{n} + ⋯ (|t| < 1)

and integrate both sides. First,

∫_{0}^{x} \frac{1}{1 + t} dt = ln (1 + t)]_{t=0}^{x} = ln (1 + x).

Hence, following (3),

ln (1 + x) = ∫_{0}^{x} \frac{1}{1 + t} dt = ∫_{0}^{x} (∑_{0}^{∞} (−1)^{n}t^{n}) dt = ∑_{n=0}^{∞} (−1)^{n} ∫_{0}^{x} t^{n} dt = ∑_{n=0}^{∞} (−1)^{n} \frac{x^{n+}}{n + 1},

that is,

(5) ln (1 + x) = x − \frac{x^{2}}{2} + \frac{x^{3}}{3} − \frac{x^{4}}{4} + ⋯ + (−1)^{n} \frac{x^{n+1}}{n+} + ⋯

if |x| < 1. ‡

Remark 1. It is important to understand that the equality in (5) is good only so long as |x| < 1. This is because the starting point was the geometric series which itself converges only where |x| < 1. In contrast, the function ln (1 + x) is defined and has derivatives of all orders where x > −1.

Remark 2. The series expansion in (5) provides a reasonable procedure for computing ln (1 + x) if x is near zero. For example, being an alternating series we know that

ln \frac{25}{24} = ln (1 + \frac{1}{24}) ≈ \frac{1}{2}4 − \frac{1}{2} − \frac{1}{2} (\frac{1}{24})^{2} + \frac{1}{3} (\frac{1}{24})^{3}

≈ 0.04166667 − 0.00086806 + 0.00002411

≈ 0.04082272

with an error in absolute value less than \frac{1}{4} (\frac{1}{24})^{4} < 0.0000008. Hence ln \frac{25}{24} = 0.04082,

accurate to five decimal digits.

Example 5. Find a power series representation for ln \frac{1 + x}{1 − x}.

Solution. In representation (5)

ln (1 + x) = x − \frac{x^{2}}{2} + \frac{x^{3}}{3} − \frac{x^{4}}{4} + ⋯,

2. POWER SERIES CALCULUS 191

we replace x by −x to obtain

ln (1 − x) = −x − \frac{x^{2}}{2} − \frac{x^{3}}{3} − \frac{x^{4}}{4} + ⋯

and subtract the second series from the first. The resulting series representation is

(6) ln \frac{(1 + x)}{(1 − x)} = 2 (x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + ⋯ + \frac{x^{2n−1}}{2n − 1} + ⋯) ,

valid if |x| < 1. Example 6. Find a power series representation for tan^{−1} x. Solution. Again the geometric series is involved because 1/(1 + x^{2}) is the derivative of tan^{−1} x. Integration of both sides of

\frac{1}{1 + t^{2}} = 1 − t^{2} + t^{4} − t^{6} + ⋯

from 0 to x gives

\begin{aligned}∫_{0}^{x} \frac{1}{1 + t^{2}} dt & = ∫_{0}^{x} (1 − t^{2} + t^{4} − t^{6} + ⋯) dt \\ & = x − \frac{x^{3}}{3} + \frac{x^{5}}{5} − \frac{x^{7}}{7} + · · · . \end{aligned}

Hence

(7) tan^{−1} x = x − \frac{x^{3}}{3} + \frac{x^{5}}{5} − \frac{x^{7}}{7} + ⋯ + (−1)^{n} \frac{x^{2n+1}}{2n + 1} + ⋯

if |x| < 1. ‡

An analysis of the remainder in (7) shows that the representation is valid at the endpoints x = ±1. Since tan^{−1} 1 = π/4, this means that

(8) \frac{π}{4} = 1 − \frac{1}{3} + \frac{1}{5} − \frac{1}{7} + \frac{1}{9} − \frac{1}{11} + ⋯,

but its agonizingly slow convergence makes it a poor vehicle for computing π. It is better to observe that tan^{−1} (1/\sqrt{3}) = π/6 and hence that

(9) \frac{π}{6} = \frac{1}{\sqrt{3}} − \frac{1}{3} (\frac{1}{\sqrt{3}})^{3} + \frac{1}{5} (\frac{1}{\sqrt{3}})^{5} − · · · .

Another approach to computing π exploits the trigonometric identity

\frac{π}{4} = tan^{−1} \frac{1}{2} + tan^{−1} \frac{1}{3}

Using the first six terms in (7), we find that

tan^{−1} \frac{1}{2} ≈ 0.45363989 with error < 0.0000094,
tan^{−1} \frac{1}{3} ≈ 0.32175051 with error < 0.00000005

and hence that π ≈ 3.141565 with error < 0.00004. If we note further that the seventh term in (7) is positive if x > 0, we see that our estimate is too small and that π ≈ 3.1416 accurate to four decimal places.

3. Taylor Series

As we have just seen, special purpose techniques can be used to find power series representations for certain of the elementary calculus functions. For other such functions it is preferable to adopt a more theoretical approach based on Taylor’s formula About some point x = a that formula represents f in the form

(1) f(x) = f(x)+f′(x)(x−a)+ \frac{f′′(a)}{2!}(x−a)^{2}+⋯+f^{(n)}(a)\frac{(x − a)^{n}}{n!} +R_{n}(x)

where the remainder R_{n} is given by

(2) R_{n}(x) = f^{(n+1)}(c)\frac{(x − a)^{n+1}}{(n + 1)!}

for some c between a and x. Recall that we have frequently used the polynomial part to approximate f(x) and it is natural to examine that approximation as n → ∞. Specifically, suppose that f is a function that has derivatives of all orders (infinitely differentiable) in some interval about a and that

(3)  lim_{n→∞} R_{n}(x) = 0

for every x in that interval. Denoting the polynomial part in (1) by P_{n}(x), we have

f(x) = P_{n}(x) + R_{n}(x)

and

f(x) = P_{n}(x) + R_{n}(x)

= lim_{n→∞} [P_{n}(x) + R_{n}(x)]

= lim_{n→∞} P_{n}(x).

This mean that the polynomials P_{n}(x) form the partial sums of a series whose limit is f(x).

(4) f(x) = f(a)+f′(a)(x−a)+f′′(a)\frac{(x − a)^{2}}{2!} +⋯+f^{(n)}(a)\frac{(x − a)^{n}}{n!} +· · · .

Put differently, the series converges throughout some interval about a and its sum is f(x). Note that the expansion (4) is a power series in x − a; it is called the Taylor series for f at x = a. If a = 0− in (4) the expansion reduces to

(5) f(x) = f(0) + f′(0)x + f′′(0)\frac{x^{2}}{2!} + ⋯ + f^{(n)}(0)\frac{x^{n}}{n!} + · · · .

This representation of f as a power series in x is called the Maclaurin series for f.

The crucial point is that the validity of (4) or (5) depends directly on the limit calculation in (3). We begin by rederiving the power series expansion for e^{x}.

Example 1. Find the Maclaurin series for e^{x}.

Solution. Since f(x) = e^{x} is its own derivative, f^{(n)}(x) = e^{x} for all n and f^{(n)}(0) = e^{0} = 1. Consequently, the Maclaurin series (5) becomes

e^{x} = 1 + x + \frac{x^{2}}{2!} + ⋯ + \frac{x^{n}}{n!} + · · · .

To justify the claim that this expansion is valid for all x, we must show that

R_{n}(x) = f^{(n+1)}(c)\frac{x^{n+1}}{(n + 1)!} = e^{c} \frac{x^{n+1}}{(n + 1)!}

tends to zero as n → ∞. Since 0 < c < x, we see that e^{c} < e^{x} if x > 0, and e^{c} < 0 if x < 0. In either case e^{c} < e^{|x|}; hence

|R_{n}(x)| ≤ e^{|x|} \frac{|x|^{n+1}}{(n + 1)!}.

At this point, we need to recall that the ratio test shows the series ∑ \frac{x^{n}}{n!} to be a convergent series for all x. Hence

(6)  lim_{n→∞} \frac{x^{n+1}}{(n + 1)!} = 0

because the terms of a convergent series must tend to zero. Therefore,

0 ≤ lim_{n→∞} |R_{n}(x)| ≤ lim_{n→∞} e^{|x|} \frac{|x|^{n+1}}{(n + 1)!} = 0

or lim_{n→∞} |R_{n}(x)| = 0.

It should be emphasized that there are two phases to finding the Taylor or Maclaurin series for a function. The first is formal – just compute the derivatives f^{(n)}(a) or f^{(n)}(0) and substitute into the appropriate formula; then, to prove the validity of the representation, you must wrestle with the problem of showing that lim_{n→∞} R_{n}(x) = 0.

Example 2. Find the Maclaurin series for sin x.

Solution. The even derivatives of sin x at x = 0 are zero and its odd derivatives of order 2n + 1 are (−1)^{n}. Substituted in (5), this information gives the Maclaurin series

(7) sin x = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − \frac{x^{7}}{7!} + ⋯ + (−1)^{n} \frac{x^{2n+1}}{(2n + 1)!} + ⋯ (−∞ < x < ∞)

The validity of this expansion rests on the observation that if f denotes sin x, then |f^{(n)}(c)| ≤ 1 for any number c and any positive integer n and hence that

|R_{n}(x)| = |f^{(n+1)}(c)| |\frac{x^{n+1}}{(n + 1)!}| = |f^{(n+1)}(c)| \frac{|x|^{n+1}}{(n + 1)!} ≤ \frac{|x|^{n+1}}{(n + 1)!}.

Therefore lim_{n→∞} |R_{n}(x)| = 0, by (6). ‡

Example 3. Find the Taylor series expansion for sin x about a = π/6.

Solution. Let f(x) = sin x. The derivative pattern is

f(x) = sin x, f′(x) = cos x

194 21. POWER SERIES

f′′(x) = − sin x, f^{(3)}(x) = − cos x

and, in general,

f^{(2n)}(x) = (−1)^{n} sin x, f^{(2n+1)}(x) = (−1)^{n} cos x.

And at a = π/6,

f^{(2n)}(π/6) = (−1)^{n} \frac{1}{2}, f^{(2n+1)}(π/6) = (−1)^{n}\sqrt{3}/2.

Accordingly, the Taylor series is

sin x = \frac{1}{2} + \frac{\sqrt{3}}{2}(x − π/6) − \frac{1}{2 · 2!}(x − π/6)^{2} − \frac{\sqrt{3}}{2 · 3!}(x − π/6)^{3} + · · · .

The proof that this expansion is valid for all x is exactly as in the previous example because |f^{(n)}(c) ≤ 1 for all x. Example 4. Find the Maclaurin series for (1 + x)e^{x}. Solution. There are two ways to attack this problem. (A) Put f(x) = (1 + x)e^{x} and calculate

f′(x) = e^{x} + (1 + x)e^{x} = (2 + x)e^{x},

f′′(x) = e^{x} + (2 + x)e^{x} = (3 + x)e^{x},

f′′′(x) = e^{x} + (3 + x)e^{x} = (4 + x)e^{x}.

The pattern is clear: for each n

f^{(n−1)}(x) = (n + 1 + x)e^{x}.

Hence f^{(n−1)}(0) = n + 1 for n ≥ 0 and the Maclaurin series is

(1 + x)e^{x} = 1 + 2x + \frac{3x^{2}}{2} + \frac{4x^{3}}{3!} + \frac{5x^{4}}{4!} + ⋯ + \frac{(n + 1)x^{n}}{n!} + · · · .

This expansion is valid for all x because

|R_{n}(x)| = |(n + 2 + c)e^{c} \frac{x^{n+1}}{(n + 1)!}| ≤ (n + 2) \frac{|x|^{n+1}}{(n + 1)!} + |x|e^{|x|} \frac{|x|^{n+1}}{(n + 1)!}

and both terms on the right tend to zero as n → ∞. (B) Start with

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ⋯ + \frac{x^{n}}{n!} + ⋯,

multiply by x to get

xe^{x} = x + x^{2} + \frac{x^{3}}{2!} + \frac{x^{4}}{3!} + ⋯ + \frac{x^{n}}{(n + 1)!} + · · · .

Then add:

(1 + x)e^{x} = (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ⋯ + \frac{x^{n}}{n!} + ⋯) + (x + x^{2} + \frac{x^{3}}{2!} + \frac{x^{4}}{3!} + ⋯ + \frac{x^{n}}{(n + 1)!} + ⋯)

= 1 + 2x + (\frac{1}{2!} + 1) x^{2} + (\frac{1}{3!} + \frac{1}{2}) x^{3} + ⋯ + (\frac{1}{n!} + \frac{1}{(n + 1)!}) x^{n} + ⋯

= 1 + 2x + 3\frac{x^{2}}{2!} + 4\frac{x^{3}}{3!} + ⋯ + (n + 1) \frac{x^{n}}{n!} + · · · .

In the preceding example, it is no accident that the two methods yield identical results.

Theorem 21. If a function f can be represented by a power series

(9) \begin{matrix}f(x) & = ∑_{n=0}^{∞} b_{n}(x − a)^{n} \\ & = b_{0} + b_{1}(x − a) + b_{2}(x − a)^{2} + ⋯ + b_{n}(x − a)^{n} + ⋯ \end{matrix}

in some interval about x = a, then the series is the Taylor series expansion of f at x = a and the coefficients b_{n} are given by

(10) b_{n} = \frac{f^{(n−1)}(a)}{n!} for all n.

Proof. Inside its interval of convergence the power series on the right in (9) represents an infinitely differentiable function that can be differentiated term-by-term. Clearly, f(a) = b_{0} and from

f′(x) = b_{1} + 2b_{2}(x − a) + 3b_{3}(x − a)^{2} + ⋯ + nb_{n}(x − a)^{n−1} + ⋯

it follows that f′(a) = b_{1}. Letting x = a in the second derivative

f′′(x) = 2b_{2} + 2 · 3b_{3}(x − a) + ⋯ + n(n − 1)b_{n}(x − a)^{n−2} + ⋯

gives

f′′(a) = 2b_{2} or b_{2} = \frac{1}{2}f′′(a).

From the third derivative,

f^{(3)}(x) = 2 · 3b_{3} + 2 · 3 · 4b_{4}(x − a) + ⋯ + n(n − 1)(n − 2)b_{n}(x − a)^{n−3} + ⋯,

we conclude that

f^{(3)}(a) = 2 · 3b_{3} or b_{3} = \frac{f^{(n−1)}(a)}{3!}.

In general,

f^{(n−1)}(x) = n(n − 1)(n − 2) ⋯ (2)b_{n} + (n + 1)(n)(n − 1) ⋯ (2)b_{n+1}(x − a) + ⋯

and

f^{(n)}(a) = n(n − 1)(n − 2) ⋯ (2)b_{n} = \frac{f^{(n)}(a)}{n!}. ‡

The point of this theorem is that the Taylor series representation for a function is unique, and that no matter how you arrive at a power series representation for a function, the series in hand is the Taylor series. In particular, this means that the series representations of the last section are Maclaurin expansions.

Example 5. Find the Maclaurin series expansion for e^{−x^{2}}.

Solution. In the series

e^{x} = 1 + x + \frac{x^{2}}{2!} + ⋯ + \frac{x^{n}}{n!} + ⋯

replace x by −x^{2} to get

e^{−x^{2}} = 1 − x^{2} + \frac{x^{4}}{2!} − \frac{x^{6}}{3!} + ⋯ + (−1)^{n} \frac{x^{2n}}{n!} + ⋯

The power series converges for all x and for all x it converges to e^{−x^{2}}. Conclusion: It must be the Maclaurin series for e^{−x^{2}}. ‡

Knowledge of coefficients in a Maclaurin expansion often provides an efficient procedure for computing derivatives at the origin. For example, suppose you want to know the sixth derivative of f(x) = e^{−x^{2}} at x = 0. Straightforward differentiation is tedious; however, according to (10),

f^{(6)}(0) = 6!b_{6} = 6! (−\frac{1}{3}) = −120

because −\frac{1}{3!} is the coefficient of x^{6} in the Maclaurin expansion of e^{−x^{2}}.

3. TAYLOR SERIES 197

PROBLEM SET 20.6

In Problems 1–12 test the series for absolute and for conditional convergence.

1. 1 − 1/3 + 1/9 − 1/27 + ⋯  2. 1 − 1/\sqrt[2]{2} + 1/\sqrt[3]{3} − 1/\sqrt[4]{4} + ⋯
3. ∑_{n=1}^{∞} \frac{cos n^{2}}{n^{3}}  4. ∑_{n=1}^{∞} (−1)^{n} \frac{(ln n)^{2}}{n^{3}}
5. ∑_{n=1}^{∞} \frac{100 − n}{n^{3}}  6. ∑_{n=1}^{∞} \frac{10^{n}}{n!}
7. ∑_{n=1}^{∞} \frac{(n + 2)!}{n^{2}n!2^{n}}  8. ∑_{n=1}^{∞} (−1)^{n} \frac{1 · 3 · 9 ⋯ (3n)}{(2n)!}
9. ∑_{n=1}^{∞} (−1)^{n} \frac{(ln 2n)^{n}}{(2n)^{n}}  10. ∑_{n=1}^{∞} (−1)^{n} \frac{2^{3n+1}}{3^{2n+3}}
11. ∑_{n=2}^{∞} (−1)^{n} \frac{ln n^{n}}{n^{2}}  12. ∑_{n=2}^{∞} (−1)^{n} \frac{ln n}{(1.02)^{n}}

13. For which values of a does the series ∑_{n=1}^{∞} (a^{−1} − \frac{1}{n})^{n} converge absolutely?

14. Find the values of x for which the series ∑_{n=1}^{∞} (x + 1)^{n}(x + 2)^{n} is absolutely convergent.

CALCULATOR PROBLEMS

In Problems 15 and 16, let S denote the sum of the series and estimate the error in the approximation S ≅ s_{N} for the indicated value of N.

15. ∑_{n=1}^{∞} \frac{sin n^{2}}{n^{3}} ; N = 15  16. ∑_{n=1}^{∞} (−1)^{n} \frac{1}{(3n + 7)^{5}} ; N = 100

PROBLEM SET 20.7

For the power series in Problems 1–6, find the interval of convergence; in Problems 7 and 8, find only the radius of convergence.

\begin{matrix}1. ∑_{n=0}^{∞} 3^{n}x^{n} & 2. ∑_{n=1}^{∞} \frac{x^{n}}{n} \\ 3. ∑_{n=0}^{∞} \frac{x^{2n+1}}{2n + 1} & 4. ∑_{n=1}^{∞} e^{−n}x^{n} \\ 5. ∑_{n=1}^{∞} \frac{x^{n}}{n ln (n + 1)} & 6. ∑_{n=0}^{∞} (−1)^{n} \frac{n^{3}}{5^{n}} (x − 1)^{n} \\ 7. ∑_{n=0}^{∞} (−1)^{n} \frac{1 · 3 · 6 ⋯ (3n)}{(n + 1)^{3}}x^{n} & 8. ∑_{n=0}^{∞} \frac{2 · 5 · 8 ⋯ (3n − 4)}{n!}x^{n} \end{matrix}

In Problems 9–11, determine the values of x for which the series converges.

\begin{matrix}9. ∑_{n=1}^{∞} \frac{1}{x^{n}} & 10. ∑_{n=0}^{∞} \frac{2^{n}}{(x + 1)^{n}} \\ 11. ∑_{n=1}^{∞} (−1)^{n} (\frac{2x − 1}{x + 2})^{2} \end{matrix}

12. Construct a power series whose interval of convergence is (−2, 2).

PROBLEM SET 20.8

With one of the series (4), (10), or (11), as a starting point derive a power series expansion for the functions in Problems 1–5. State the interval of convergence.

1. \frac{1}{(1 + x)^{2}}  2. \frac{1}{(1 − x)^{5}}
3. \frac{3x^{2}}{(2 − 5x)^{2}}  4. ∫_{0}^{x} ln (1 − t) dt

5. (2 − x) − (1 − x) ln (1 − x) In Problems 6–8, verify the power series expansion.

6. \frac{1 + x}{1 − x} = 1 + 2x + 2x^{2} + 2x^{3} + ⋯ if |x| < 1.

7. \frac{x^{2} + 6x + 1}{(1 − x)^{3}} = 1 + 3^{2}x + 5^{2}x^{2} + 7^{2}x^{3} + ⋯ if |x| < 1.

8. sin^{−1} x = ∫_{0}^{x} \frac{dt}{\sqrt{1 − t^{2}}} = x + \frac{1}{2} \frac{x^{3}}{3} + \frac{1 + 3}{2 · 4} \frac{x^{5}}{5} + \frac{1 + 3 · 5}{2 · 4 · 6} \frac{x^{7}}{7} + ⋯ if |x| < 1.

CALCULATOR PROBLEMS In Problems 9–12, use (5), (7), (10), and (11) to estimate the following with an error less than 0.001.

\begin{matrix}9. ln 1.2 & 10. e^{−0.3} \\ 11. \sqrt[3]{1.2} & 12. (0.95)^{−1/4} \end{matrix}

13. Because π/6 = sin^{−1} 0.5, the series in Problem 18 can be used to approximate π. Construct a table of its first fifteen partial sums at x = 1/2. Compare the resulting approximations to the calculator value for π.

14. Repeated applications of the identity tan^{−1} \frac{1}{p} = tan^{−1} \frac{1}{p + q} + tan^{−1} \frac{1}{p + r} (qr = 1 + p^{2}) of C.L. Dodgson (Lewis Carroll) leads to the formula \frac{π}{4} = 5 tan^{−1} \frac{1}{7} + 2 tan^{−1} \frac{3}{79}. Use this and the Maclaurin expansion for tan^{−1} x to find the first eight digits in the decimal expansion of π.

PROBLEM SET 20.9

In Problems 1–4, find the Maclaurin series expansion for f(x) and use (3) to establish its validity for all x. The symbols a and b denote constants.

1. f(x) = e^{ax}  2. f(x) = sinh (x)

3. f(x) = \frac{x^{3}}{6} − x + sin x  4. f(x) = sin^{2} x = \frac{1 − cos 2x}{2}

In Problems 5–8 derive a formula for the general nth derivative f^{(n)} and use it to write the Taylor series (4) for f(x) about x = a You may omit the verification (3).

\begin{matrix}5. f(x) = \frac{1}{1 + x}; a = 0 & 6. f(x) = \frac{1}{2x + 3}; a = −1 \\ 7. f(x) = x^{5}; a = 1 & 8. f(x) = x^{2}; a = 100 \end{matrix}

Use known series expansions from this and the preceding section together with algebraic manipulation and substitution to obtain the following Maclaurin expansions.

200 21. POWER SERIES

9. x^{2}e^{−x} = x^{2} − x^{3} + \frac{x^{4}}{2!} − \frac{x^{5}}{3!} + ⋯

10. (1 + x)^{−1/3} = 1 − \frac{1}{3}x + \frac{1 · 4}{3^{2} · 2}x^{2} − \frac{1 · 4 · 7}{3^{3} · 3!}x^{3} + ⋯

11. sin^{−1} x − x = \frac{1}{2} \frac{x^{3}}{3} + \frac{1 · 3}{2 · 4} \frac{x^{5}}{5} + \frac{1 · 3 · 5}{2 · 4 · 6} \frac{x^{7}}{7} + ⋯

12. (1 − x)^{2}) ln (1 − x^{2}) = −x^{2} + \frac{1}{2x^{4}} + \frac{1}{6}x^{6} + \frac{1}{12}x^{8} + \frac{1}{20}x^{10} + ⋯

13. Integrate both sides of \frac{1}{1 + t} = 1 − t + t^{2} − t^{3} + ⋯ (−1)^{n−1} t^{n−1} + \frac{t^{n}}{1 + t} from 0 to 1 to obtain ln 2 = 1 − \frac{1}{2} + \frac{1}{3} − \frac{1}{4} + ⋯ + (−1)^{n−1} \frac{1}{n} + (−1)^{n} ∫_{0}^{1} \frac{t^{n}}{1 + t} dt. Then show that R_{n} = (−1)^{n} ∫_{0}^{1} \frac{t^{n}}{1 + t} dt → 0 as n → ∞, thereby verifying that ln 2 = 1 − 1/2 + 1/3 − 1/4 + · · ·.

14. Use integration by parts in the formula tan^{−1} x = ∫_{0}^{x} \frac{1}{1 + t^{2}} dt to obtain the series expansion tan^{−1} x = \frac{x}{1 + x^{2}} + \frac{2}{3} \frac{x^{3}}{(1 + x^{2})^{2}} + \frac{2}{3} · \frac{4}{5} \frac{x^{5}}{(1 + x^{2})^{3}} + ⋯

Lab Projects for Ma114

Lab projects are generally stated as a topic of investigation rather then as a specific exercise. You are required to perform a detailed analysis of the problem. This will include: setting up the problem, making a labeled diagram of the problem with information acquired, performing the necessary calculations and visually displaying the results (usually as graphs of functions). You must then organize this information into a full written report.

Lab projects are done by collaborative groups of three or four students who are expected to organize themselves, do the work, and prepare the report as a group. Each person in the group must sign the final report and be aware that in doing so he or she is guaranteeing that everyone who signed has done his or her fair share of the work. This does not mean that everyone did exactly the same number of tasks, spent the same amount of time, etc. It may well be the case that one member of the group provides the majority of the ideas, another provides the organization and direction, and another actually prepares the ”write-up”. It does mean that each person who signs the final report is personally satisfied that every author contributed in fair measure and fully understands the report.

The reports will be graded with each student in the group receiving the group grade. The grade may be determined either by the written report alone or in equal measure by an oral presentation by a randomly selected member of the group. Report grading will be on correctness, completeness, correct English, demonstration of understanding of the problem and solution, and style. Bonus points may be given for originality.

The following are general guidelines and expectations:

The Lab Session:

• Get started immediately. You will not be able to complete the project all in the last minute. Do not waste this valuable time.

• Read over the laboratory material carefully before you begin discussing or completing any part of it. This should be done before the lab sessions designated for this activity. Give some advanced thought to what you are going to do and the anticipated form of the report you are going to prepare.

• Initially you may have some trouble getting started. Don’t panic, a discussion with your group partners will usually generate some ideas. Your TA can also give you a ‘push’ in the right direction. There will likely be a general discussion on the problem in lecture. Your TA may also be able to make suggestions as to how to organize your efforts.

• The generality of projects like these tend to make them quite different from the standard ”exercises” of the homework. You will need to make assumptions to simplify the problem. Be sure to discuss such assumptions in detail in your final write-up.

• Plan on spending some time outside of lab on the report. In particular, plan on writing the final report outside of class.

• Your TA will be happy to go over complete drafts of your final report with you if they are ready at least two days before the submission deadline. He or she will then be able to point out areas of concern and make suggestions on improving the final submission.

The Lab Report: The report should be readable by anyone who can understand the mathematics with no reference to any other source It must at least have the following identifiable features:

1. A Cover Page signed by each member of the team which produced the report. A signature on a report means that each person signing attests to the fact that each contributed approximately equally to the work. Each member of the team is responsible for any errors; each member of the team may be asked to explain any detail of a report.

2. A complete statement of the problem

3. An analysis of the problem, complete with carefully drawn sketch which contains a complete explanation of the solution, derives and explains all equations and sets up all calculations that need to be done. This may be done as part of a Maple worksheet and should contain carefully annotated graphs of the pertinent functions. This is the most important part of the report.

4. Details of the calculations (this can be an annotated Maple worksheet. It must be clear to the reader what each calculation is intended to do. This can be done with comments.

The report should be free of any extraneous calculations. Do not include any intermediate calculations unless you need to refer to them or you feel that their presence makes the report easier to read.

Most of the graphs will be in this section. It should be clear to the reader why one doing a certain graph before it is actually done. This can be done with a comment. Features of importance on a graph should be labeled and interpreted either on the graph itself or immediately following it. For instance the axes of all graphs should be labeled and a caption provided for each graph which explains its purpose. This labeling can be done by hand on computer-generated drawings. The function(s) being graphed should be in the caption or penciled on the graph.

5. Conclusions and interpretations . At the very least this should briefly explain how what has been done answers the original question or solves the original problem. If appropriate, it should refer back to graphs and relate features of the graphs to numerical calculations. Finally, there should be a summary paragraph or sentence describing what has been accomplished and (possibly) suggesting directions for further study. problem(s).

A good report will not be very long - probably three to five typed pages, excluding graphs.

The report should be free of any extraneous calculations (e.g. material that should have been cut out of a worksheet.)

LAB PROJECTS FOR MA114 203

Finally, there should be a summary paragraph or sentence describing what has been accomplished and (possibly) suggesting directions for further study. problem(s).

The Tilted Bottle

Date:

Team Members:

Investigate the following question:

A cylindrical container 6 inches high and 2.5 inches in diameter is 80% full of water and sits on a level table. The cylinder is tilted so that the plane of its base makes an acute angle of θ radians with the table.

At which value of theta will water just begin to flow out of the container?

NOTE: This problem was originally suggested by Professor Peter Russell of McGill University. During a drive from the Montreal airport he was behind a beer delivery truck on whose sides cases of bottled beer sat in racks at an angle which made it easy for the delivery man to remove them. Dr. Russell noted that the fluid in the necks of the bottles didn’t touch the lids even though the angle of the cases was actually rather steep and wondered exactly what angle would cause that to happen.

The Hot Water Tank Problem

Date:

Team Members:

Newpark Shipyards manufactures and repairs large barges. Many of the barges brought in for repair are tankers with substantial amounts of fuel vapor in the tanks. In the past this explosive gas was vented to the atmosphere but clean air standards now require that it be burned cleanly. The shipyard had to choose between a $250,000 flare system or a $500,000 hot water system which would use the fuel vapor to produce hot water which would then be used to clean the barge. The decision on which alternative to use boiled down to whether the hot water produced would on one day would be hot enough to use the next. This was because the vapor had to be removed but the tanks would not be safe for workers to enter for up to eight hours more.

Here is a modified version of one the problems that was solved to help with the decision. At six a.m. a tank contains 20,000 gallons of water at 90 degrees Fahrenheit. One hour earlier the water in the tank had been at At six a.m. fuel vapor from a barge is burned and the heat transferred to the water in the tank via a heat exchanger at a rate of 80,000 BTU per minute for four hours. Assuming that the ambient air temperature remains at a constant 80 degrees and that only Newton’s Law of Cooling applies (ignore radiant heat loss) what will be the temperature at the end of the four hours and how long (if at all) will the water in the tank remain above 170 degrees Fahrenheit.

Investigate the problem above, answering the question posed. Then investigate the sensitivity of the answer to the ”constant ambient temperature” assumption by assuming that the temperature is a constant during each clock hour, starting with 40 degrees from 6:00 to 7:00 and that it increases two degrees per hour until 6:00 p.m and then decreases two degrees per hour until it gets back to 40 degrees at 6:00 a.m. at which time the cycle repeats.

208 LAB PROJECTS FOR MA114

Review Problem Sets: First Midterm

Review Problem Set #1: First Midterm. (50 points in all, time = 1 hour)

1. (13 points) A particle is constrained to move only in the x-direction. A rotating force of constant magnitude causes it to accelerate in the x-direction with a(t) = sin(t/3). The particle’s initial velocity is 1.5 units/sec to the left. Find the TOTAL distance the particle travels between t = 0 and t = 5 sec.

2. (12 points) Write each of the following in terms of definite integrals. Do NOT attempt to evaluate the integrals.

(a) The volume of the solid of revolution obtained by rotating around the x-axis the shape in the first quadrant bounded by the line 2x + y = 1 and the curve y = cos(πx).

(b) The center of mass of a uniform plate in the 1st and 4th quadrants that is bounded by the line from the origin to (4, 3), the line from the origin to (4, −3), and the circumference of the circle of radius 5 centered at the origin.

3. (12 points) An ellipse is widest in the x-direction, where it measures 10 units across, and narrowest in the y-direction, where it measures 6 units across. The bottom half of this ellipse is rotated around its minor axis to form a container, into which water is poured at the rate of 5 cubic units/sec.

(a) Find the length of time T before the container is filled.

(b) At time T (i.e., the instant when the container is about to overflow) find the rate at which the water level is rising. Compare this with the average rate at which the water level rose between time t = 0 and t = T.

4. (13 points) A pebble is caught half-way up the spoke of a wheel of radius r. The wheel is rolling along a flat surface without slipping.

(a) Set up a definite integral for the distance along its path that the pebble travels as the wheel goes through half a revolution, during which time the pebble goes from a position directly below the center of the wheel to a position directly above the center.

(b) Find this distance using the trapezoid rule with n = 6 subdivisions. Express your answer as a multiple of r. To receive credit you must show clearly the steps in applying the trapezoid rule, indicating what numbers you are substituting into the rule.

Review Problem Set #2: First Midterm. (50 points in all, time = 1 hour)

1. (12 points) In this problem neglect air resistance, choose upward to be the positive direction, and take g = 9.8 m/sec^{2}. You know that a ball was thrown down from a window at 10 m/sec, but you do not know the angle α below the horizontal at which it was thrown. You also do not know the height s_{0} of the window. Suppose that the ball hits the ground after 5 seconds at a point on the ground that is 25 meters from the side of the building from which the ball was thrown. Find α and s_{0}.

2. (12 points) Write each of the following in terms of definite integrals. Do NOT attempt to evaluate the integrals.

(a) The volume of the solid of revolution obtained by rotating around the y-axis the shape bounded by the x-axis, the line y = 2, and the curve y = 3 − (1/x^{2}).

(b) The distance along its path traveled in the first tenth of a second by a pebble which at time t = 0 is caught in the tread of a wheel of radius 40 cm that rolls along the pavement without slipping at 5 m/sec.

3. (14 points) A bug is on the circumference of the ellipse pictured at the right. It crawls in such a way that its x-coordinate increases at the rate of 0.20 units/sec. The sun is shining from the left from an angle of 45^{∘} above the horizontal, casting the bug’s shadow on the x-axis. At the instant when the x-coordinate of the bug is 3, find:

(a) the rate at which its y-coordinate is changing,

(b) the rate at which its shadow is moving along the x-axis,

(c) the speed of the bug along the ellipse.

−5  4  5

4. (12 points) The graph to the right shows the function f(x) = (sin x)/x between x = 0 and x = π. Actually, when x = 0 the ratio (sin x)/x is 0/0, which is undefined; however, the expression (sin x)/x has the limit 1 as x approaches 0, so we take f(0) = 1. A uniform plate is bounded by f(x) and the x- and yaxes.

(a) Write the x-coordinate \overline{x} of its center of mass in terms of definite integrals. One of the two definite integrals is easy to evaluate by the anti-derivative method.

1 π

(b) Use the trapezoid rule with n = 6 subdivisions to evaluate the second of the two definite integrals in part (a). To receive credit you must show clearly the steps in applying the trapezoid rule, indicating what numbers you are substituting in the rule.

(c) Find \overline{x}.

Review Problem Set #3: First Midterm. (50 points in all, time = 1 hour)

1. (10 points) Give the equation of the sinusoidal function below, and find the area of the region between the portion of the curve shown and the line y = 1.

2

2. (20 points) Set up each of the following in terms of definite integrals. Do NOT evaluate the integrals.

(a) The distance along its trajectory traveled in the first second by an object thrown at 15 m/sec directed 30^{∘} below the horizontal from a window 20 meters high. (Neglect air resistance, take upward to be the positive direction, and take g = 9.8 m/sec^{2}.)

(b) The entire circumference of an ellipse with semimajor axis a = 10 and eccentricity e = 0.6. (First find b, and write the ellipse in parametric form.)

(c) The net distance traveled between time t = 0 and t = 2 if the acceleration is a(t) = 1/(0.1t + 1)^{2} and the initial velocity is 0.

3. (20 points) A wheel of radius 1 rotates clockwise at the rate of 1 revolution every four seconds as the center of the wheel moves along the straight line joining the points A = (0, 2) and B = (3, 0) (see the drawing at the right). The center moves at constant velocity, starting at A at time t = 0 and ending up at B at time t = 1. P is the point on the rim of the wheel which is directly above the center

at time 0. P t = 0 P^{θ} A B

(a) Find parametric formulas in terms of t for the motion of the center of the wheel.

(b) After the wheel has rotated through an angle of θ, how far is the point P from the center in the x-direction and in the y-direction?

(c) Express θ in terms of t.

(d) Find parametric formulas in terms of t for the motion of P.

(e) Write an integral for the distance traveled by the point P between time 0 and 1.

(f) Using the trapezoid rule with n = 4 subdivisions, evaluate the integral in part (e). Please show your work carefully.

Review Problem Set #4: First Midterm. (50 points in all, time = 1 hour)

1. (8 points) Find the area in the first quadrant between the graphs of y = cos(πx/2) and x + y = 1.

2. (16 points) Set up each of the following in terms of definite integrals. Do NOT evaluate the integrals.

(a) The distance along its trajectory until reaching the peak of the trajectory that is traveled by an object thrown at 9.7 m/sec directed 45^{∘} above the horizontal. (Neglect air resistance, take upward to be the positive direction, and take g = 9.8 m/sec^{2}.)

(b) The volume of the solid obtained by rotating around the y-axis the region in the first quadrant bounded by the axes and the curve which is given parametrically by the formulas x = 3 cos θ, y = 2 sin θ.

3. (12 points) Find the center of mass of a uniform flat plate bounded by y = sin x, y = − sin x, and x = π/2. Use the trapezoid rule with n = 6 subdivisions to find the value of the top integral. Please show your work clearly.

4. (14 points) In this problem you will need the parametric formulas for a cycloid: x = r(θ − sin(θ)), y = r(1 − cos(θ)). A car accelerates at the constant rate of 6 m/sec^{2}. When it reaches a speed of 20 m/sec, it picks up a pebble in the tread of a wheel. The wheel has radius 20 cm. Let t denote the time measured from the instant the wheel picks up the pebble.

(a) Express in terms of t the angle θ through which the pebble has rotated around the center of the wheel. (θ is measured in radians.)

(b) Draw a circle with an arrow pointing to the position of the pebble on the wheel when t = 0.5 sec.

(c) Find the speed of the pebble along its trajectory when t = 0.5 sec.

Review Problem Set #5: First Midterm. (50 points in all, time = 1 hour)

1. (12 points) Find the coordinates of the center of mass of a uniform flat plate in the shape shown below to the left.

(2, 2)  −25  25 7 −7 −4

2. (12 points) The ellipse pictured above to the right is centered at the origin of the xy-plane. Its semi-major axis is 25 and its semi-minor axis is 7. The units are meters.

(a) Find the coordinates of its foci.

(b) Find the eccentricity.

(c) Suppose that the bottom half of the ellipse is rotated around the y-axis to form a container. Water is entering the container at the rate of 0.7 m^{3}/sec. How fast is the water level rising when the water is 3 meters deep at its deepest point (i.e., when the water level is 4 meters below the top of the container)?

3. (12 points) A bicycle has a wheel of radius 0.4 meter. It is traveling at a constant speed of 6 m/sec. A pebble is lodged in a spoke of the wheel at a distance of 0.3 m from the center. At time t = 0 the pebble is at its lowest point (0.1 m above the pavement). Suppose that the wheel is moving from left to right along the x-axis, and the origin is the point where the wheel meets the pavement at time t = 0. In this problem, use the parametric formulas x = rθ − a sin θ, y = r − a cos θ for the coordinates of the pebble after the wheel has rotated through an angle of θ radians.

(a) Find formulas for the coordinates of the pebble at time t.

(b) When t = 0.4 sec, find the coordinates of the pebble, and draw a diagram showing the relative positions of the center of the wheel and the pebble at that instant.

(c) When t = 0.4 sec, find the speed of the pebble along its trajectory, and draw a picture of the velocity vector showing the pebble’s direction of motion.

4. (14 points) At time t = 0 a particle is at rest (zero velocity) at the origin of the xy-plane. In the horizontal direction its acceleration at time t is given by the formula a_{horiz}(t) = t, and in the vertical direction its acceleration at time t is given by the formula a_{vert}(t) = cos(πt/6).

(a) Find formulas for the horizontal velocity and the vertical velocity at time t.

(b) Express the arclength of the particle’s path during the first 3 seconds (the distance it has traveled) as a definite integral.

(c) Use the trapezoid rule with n = 6 to find the value of the integral in part (b). You do NOT have to do the final numerical computations, but be sure you show clearly the values that you are substituting into the trapezoid rule formula.

214 REVIEW PROBLEM SETS: FIRST MIDTERM

Review Problem Sets : Second Midterm

Review Problem Set #1: Second Midterm. (50 points in all, time = 1 hour)

1. (15 points) A shape is bounded by the curve y = x^{2}/\sqrt{4 − x^{2}}, the x-axis, and the line x = 1. Find the x-coordinate of the center of mass of a uniform plate in that shape. Please show all your work carefully (including the triangle you use for the inverse trig substitution, and the trig identities you use to integrate powers of sine or cosine).

2. (10 points) The shape in Problem 1 is rotated around the x-axis. Find the volume of the resulting solid of revolution. Again please show all of your work.

3. (10 points) Find the maximum possible volume of the cylinder obtained by rotating the rectangle shown at the right around the x-axis. (The upperright corner of the rectangle is at the point (x, e^{−x}) on the curve y = e^{−x}. Volume of cylinder = πr^{2}h.)

4. (15 points) At time t = 0 a ball is thrown horizontally at 8 m/sec from a point 40 meters directly above the observer’s eye O. (See the drawing at the right, neglect air resistance, and take g = 9.8 m/sec^{2}.) At the same instant a 20-meter span of a bridge extends from O at an angle of 30^{∘} above the horizontal. The bridge is being lowered by a rope (the vertical dotted line) which is pulled downward at the steady rate of 2 m/sec.

(a) Express the x- and y-coordinates of the ball as functions of time.

(b) Express the height of the point B (the tip of the bridge span) in terms of t.

(c) Express the angle AOB (the angle between the observer’s line of vision to the ball and her line of vision along the bridge) in terms of t.

(d) Find a formula for the rate in rad/sec at which the angle AOB is decreasing at time t.

x y = e^{−x}_{40 θ AB O}

Review Problem Set #2: Second Midterm. (50 points in all, time = 1 hour)

1. (15 points) Let y = f(x) = \sqrt{x} (ln x) for x > 0.

(a) What are the values of f(x) at x = 1 and as x ⟶ 0?

(b) Find the x- and y-coordinates of the minimum point of f(x). Express your answer in terms of e (not as a decimal).

(c) Find the x-coordinate of the center of mass of a uniform flat plate bounded by y = f(x) and the x-axis between x = 0 and x = 1.

2. (12 points) Find the integral of the following two functions from 0 to ∞:

(a) \frac{1}{x^{2} + 5x + 4}, (b) \frac{1}{(x + 1)\sqrt{x^{2} + 2x}}.

3. (8 points) The first region in the 1st quadrant bounded by the curve \sqrt{x} cos x and the x-axis is revolved around the x-axis to form a solid of revolution. Find its volume.

4. (15 points) At time t = 0 an experimenter counts 5 million of bacteria A and 15 million of bacteria B. For each type of bacteria the rate of change of its population at any time t is proportional to its population at time t; the constant of proportionality is the instantaneous birthrate minus the instantaneous deathrate. For bacteria A, the instantaneous birthrate exceeds its instantaneous death rate by 3% per hour; but for bacteria B, the instantaneous birthrate is less than its instantaneous deathrate by 5% per hour. Find the time t when the total population of bacteria (A plus B) is minimal.

Review Problem Set #3: Second Midterm. (50 points in all, time = 1 hour)

1. (24 points) Evaluate the following integrals. In parts (a) and (c) express your answers in terms of π, \sqrt{2}, ln 2, etc. (not as decimals).

(a) ∫_{0}^{1}(1 + 2y)Arctan(y) dy;  (b) ∫ \frac{x dx}{\sqrt{x − x^{2}}} ;  (c) ∫_{π}^{4π} cos^{3}(t/4)dt.

2. (8 points) Find the derivative of (ln x)^{e^{x}}.

3. (18 points) A wheel 20 cm in radius rolls in a straight line without slipping. Its velocity at time 0 is 5 m/sec, but friction decelerates the wheel at a rate proportional to the velocity. Let k denote the constant of proportionality. The only force on the wheel is friction. That is, the instantaneous rate of change of the wheel’s velocity at time t equals minus k times the velocity at time t.

(a) Write a differential equation for the wheel’s velocity v(t).

(b) Find a formula for v(t). The constant k should appear in your formula.

(c) Find a formula for the distance s(t) traveled in the first t sec. The constant k should appear in your formula.

(d) If k = 0.1, find all of the following at time t = 8: (1) v(8); (2) s(8); (3) the angle θ in radians through which the wheel rotates in the first 8 sec; (4) the angular velocity in rad/sec at time t = 8.

(e) Suppose that the wheel picked up a pebble at time t = 0. Use the parametric formulas for a cycloid (x = r(θ − sin θ), y = r(1 − cos θ)) to determine the speed of the pebble at time t = 8.

Review Problem Set #4: Second Midterm. (50 points in all, time = 1 hour)

1. (24 points) Evaluate the following integrals. In part (a) express your answer in terms of π (not as a decimal), and in part (c) express your answer in terms of e (not as a decimal).

(a) ∫_{10\sqrt{3}}^{∞} \frac{dx}{100 + x^{2}} ;  (b) ∫ \frac{x + 2}{x^{2} − 3x + 2} dx;

(c) ∫_{−1/2}^{1/2} x e^{2(x+1)} dx;  (d) ∫_{0}^{π} sin^{5} (\frac{x}{3}) dx.

2. (10 points) Point A moves at constant velocity from the point (3,7) to the point (10,2) in 5 seconds, starting at t = 0. Meanwhile, point B rotates counterclockwise at 1 revolution per second around the circle of radius 2 centered at the origin. At time t = 0 the point B is at (2,0).

(a) Find a formula in terms of t for the angle above the horizontal of the line of vision from B to A.

(b) Find a formula in terms of t for the rate in degrees per second that the angle in part (a) is decreasing.

B A

3. (16 points) Let f(x) = \sqrt{x} − ln x, x > 0.

(a) Find f′(x) and all points where f′(x) = 0.

(b) Find f′′(x), and use the second derivative test to determine whether the point(s) in part (a) is (are) maximum or minimum point(s).

(c) Find all points of inflection, and regions of upward and downward concavity.

(d) What happens to f(x) as x ⟶ 0 and as x ⟶ ∞?

(e) Sketch the graph of f(x).

218 REVIEW PROBLEM SETS : SECOND MIDTERM

Review Problem Sets: Final Exam

Review Problem Set #1: Final Exam.

(100 points in all; time = 3 hours, checking your work carefully)

1. (20 points) Let f(x) = ln x − 2.5 Arctan(0.5x), x > 0.

(a) Find all critical points (give both x- and y-coordinates), and use the 2nd derivative test to determine whether they are maximum or minimum points.

(b) When x approaches 0, what simple function does f(x) become very close to?

(c) When x approaches ∞, what simple function does f(x) become very close to?

(d) Use parts (a)–(c) to give a rough sketch of the graph of f(x).

2. (15 points) A container is obtained by rotating around the y-axis a function x = g(y) whose values at regular intervals are given by the following table (the units are cm):

\begin{matrix}y & 0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 \\ g(y) & 2.34 & 2.63 & 2.90 & 3.16 & 3.40 & 3.63 & 3.85 & 4.06 \end{matrix}

At a certain instant the water level in the container is 12 cm, and it is increasing at the rate of 0.05 cm/sec. At this instant find:

(a) the volume of water in the container (use the trapezoid rule);

(b) the rate in cm^{3}/sec at which water is being poured into the container.

3. (25 points) A wire extending infinitely far in one direction (say, along the positive x-axis from the origin) has c coulombs/meter of charge per unit length. A particle with q coulombs of charge is at a distance r from the end of the wire in a direction perpendicular to the wire (i.e., the particle is located at −r on the y-axis). Use the physics principle that the force of repulsion between two point charges is equal to k(product of charges)/(dist)^{2}, where k is a universal constant.

(a) Find the magnitude of the force of repulsion caused by the small segment of wire between x and x+Δx. For simplicity of notation, set K = kqc. Then find (i) the component of this force in the direction perpendicular to the wire, and (ii) the component of this force in the direction parallel to the wire. Use parts (i) and (ii) to write an improper integral for the total force perpendicular to the wire and another improper integral for the total force parallel to the wire.

(b) Find both components of the total force from the wire on the particle.

(c) Find the direction of the force on the particle.

4. (25 points) You drop a rock (with zero initial velocity) from a bridge. Take g = 9.8 m/sec^{2}. Because of air resistance (which we assume to be proportional to the velocity v(t)), the rock would eventually approach a downward terminal velocity of 140 m/sec if it fell indefinitely through the air. However, it does not fall indefinitely: after exactly 5 seconds it hits the water.

(a) Write out the differential equation satisfied by v(t) while the rock falls through the air.

(b) Solve the differential equation in part (a).

(c) How fast is the rock falling when it hits the water?

(d) After the rock hits the water, the water resistance is great enough to slow it down. If it fell indefinitely through the water, its downward velocity would decrease to the terminal velocity of 4.9 m/sec. Find a formula for the rock’s velocity while it falls through the water.

(e) How many seconds after the rock hits the water does it slow to a downward velocity of 6 m/sec?

5. (15 points) At time t = 0 a container holds 10 liters of water with 1 gram of impurities. Water containing 0.28 gm/liter of impurities enters the container at a rate r(t) that is 2 liters/min when t = 0 but decreases steadily by 0.2 liters/min each minute. Water leaves the container at the constant rate of 2 liters/min. Assume that the impurities are at all times uniformly mixed in the water in the container.

(a) Write a formula for the rate of the incoming water at time t.

(b) Find a formula for the volume of water in the container at time t. When will the container be empty?

(c) Set up a differential equation for the amount y(t) of impurity in the water at time t.

Review Problem Set #2: Final Exam. (100 points; time = 3 hours, checking your work carefully)

1. (20 points) Let f(x) = x e^{−(1.5x^{2})}.

(a) Find the x- and y-coordinates of all critical points, and use the second derivative test to determine if they are minimum or maximum points.

(b) Find the x- and y-coordinates of all points of inflection.

(c) What happens to f(x) when x ⟶ +∞ and when x ⟶ −∞?

(d) Use parts (a)–(c) to sketch the curve. Your sketch should show the concavity (upward or downward) correctly, and also the behavior as x ⟶ ±∞.

2. (15 points) As a bug crawls up the y-axis, the sun shines from the left from an angle θ above the horizontal, casting the bug’s shadow on the x-axis. At a certain instant you measure the following information: the bug’s position a on the y-axis, its velocity ȧ (in units/min), the position b of its shadow on the x-axis, and the shadow’s velocity ḃ (in units/min). Express the following in terms of these measured values: (a) the angle θ; (b) the rate in DEGREES per minute at which the sun is rising.

3. (15 points) The following table shows the horizontal and vertical velocities (in m/sec) of a particle at intervals of 2 sec from t = 0 to t = 8. Using the trapezoid rule, find the distance the particle traveled along its trajectory during the 8-second period.

\begin{matrix}t & 0 & 2 & 4 & 6 & 8 \\ ẋ & 10.0 & 9.8 & 9.4 & 9.0 & 8.4 \\ ẏ & 0.0 & −8.0 & −15.2 & −22.0 & −28.2 \end{matrix}

4. (15 points) A container contains a volume V of liquid. At time t = 0, the liquid contains y_{0} grams of impurity. Fresh water is pumped into the container at rate r, and the well-mixed liquid is drained from the container at the same rate.

(a) Find a differential equation for the amount y(t) of impurity at time t.

(b) Find a formula for the length of time it takes for the amount of impurity to be reduced to 1% of its original amount.

5. (25 points) At time t = 0 a rhinoceros farm has 8000 rhinos. The instantaneous annual population growth rate (birth rate minus death rate) is 12%. The farm sells rhinos steadily at the rate of 1500 each year.

(a) Write a differential equation for the number y(t) of rhinos the farm has at time t.

(b) Solve the differential equation in part (a). Please show all of your steps clearly.

(c) In how many years will the farm have no more rhinos left?

(d) How many rhinos will the farm have sold during this time?

(e) Make up an installment buying problem which leads to the identical mathematics as this rhino population problem (i.e., the exact same differential equation and the same initial value of y(t)). Write out your problem clearly (use complete sentences).

6. (10 points) At time t = 0 a rocket-car traveling at 100 m/sec starts to decelerate. Its deceleration is caused by two things: a constant braking force that produces a negative acceleration of −8 m/sec^{2}, and a small parachute which causes an air resistance equal to 0.05 sec^{−1} times the velocity with which the air is rushing against the car. If there were no wind, the latter velocity would be simply the velocity of the car. At first there is no wind, but at time t = 0 a wind starts blowing in the opposite direction to the car’s motion. The wind increases steadily by 5 m/sec each second.

(a) Write a differential equation for the velocity v(t) of the car.

(b) Arrange the differential equation in part (a) in the form dv/dt = k(t)y + l(t).

Review Problem Set #3: Final Exam.

(100 points in all, time = 3 hours, checking your work carefully)

1. (20 points) At time t = 0 an object is located 50 meters above your eye level (directly overhead). At that instant it is thrown at an initial velocity of 20 m/sec directed 30^{∘} below the horizontal. Neglect air resistance, and take g = 9.8 m/sec^{2}.

(a) Find parametric equations for the object’s motion.

(b) Express the angle θ at which you observe the object (i.e., the angle above the horizontal of your line of sight to the object) as a function of t.

(c) Find the rate of change of θ in rad/sec when t = 2.

2. (20 points) At time t = 0 you start driving at the speed limit of 55 miles/hr. As your confidence builds up, noticing no cop for miles around, you gradually start accelerating. Your acceleration is

a(t) = \frac{24}{\sqrt{1 − t^{2}}} miles/hr^{2}.

(a) Find the velocity v(t).

(b) After a half-hour you are arrested. What is your velocity at that time?

(c) Find the distance you traveled in the half-hour before you’re arrested.

3. (25 points) Let f(x) = \frac{1}{x} + ln x, x > 0.

(a) Find the x- and y-coordinates of any max/min points and any points of inflection.

(b) For x very close to 0, what does f(x) do?

(c) As x approaches ∞, what does f(x) do?

(d) Sketch the graph of f(x).

(e) Find the arclength of the curve between x = 1/2 and x = 1, using the trapezoid rule with n = 4 subdivisions.

4. (25 points) A vat holds 5 m^{3} of liquid. At time t = 0 the liquid contains y_{0} grams of impurity. One pipe carries water with k gm/m^{3} of impurity into the vat at rate r m^{3}/min, and a second pipe carries the uniformly mixed impurities from the vat at the same rate r. In this problem r is not constant, but rather varies sinusoidally according to the following formula: r(t) = 3 + 2 sin(πt/10).

(a) Write a differential equation for the amount y(t) of impurity in the vat at time t. The constant k should appear in your differential equation.

(b) Solve the differential equation. Show all your work clearly. The initial value y_{0} (as well as k) should appear in your final formula for y(t).

(c) Does the amount of impurity in the vat increase or decrease with time?

5. (10 points) A country with a population of 100 million has instantaneous growth rate (birth rate minus death rate) of 2.5% and a net immigration that equals 1.0 million people/year at time t = 0 and increases steadily by 100000 people/year each year. Set up a differential equation for the population y(t) (measured in units of millions of people).

Review Problem Set #4: Final Exam.

(100 points in all, time = 3 hours, checking your work carefully)

1. (15 points) Suppose you are 5 feet tall, and the sun casts a shadow y feet long when it is at an angle θ above the horizon.

(a) Express the angle θ in terms of the length y of your shadow.

(b) Shortly before sunset, your shadow is 25 feet long, and it is lengthening at the rate of 0.5 feet/min. At what angular speed, expressed in deg/min, is the sun setting?

2. (15 points) Find any critical points of the function f(x) = x^{x}, x > 0; then use the second derivative test to determine whether the point(s) is (are) maxima or minima.

3. (10 points) You measure the slope of the tangent line to the curve y = f(x) at intervals of 0.5 from x = 0 to x = 3, and find the following values:

\begin{matrix}x & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\ slope & 1.00 & 0.91 & 0.74 & 0.56 & 0.41 & 0.21 & 0.15 \end{matrix}

Use the trapezoid rule to find the arclength of the graph of y = f(x) between x = 0 and x = 3.

4. (20 points) At time t = 0 a little stick gets lodged in a spoke of a wheel at a point half way between the center of the wheel and the point where the wheel is touching the ground. The wheel has radius 2 meters and is traveling from left to right along the x-axis (see the diagram at the right) at 5 m/sec. The sun is shining from behind the wheel at an angle of 45^{∘}, casting the stick’s shadow on the x-axis. Find the speed with which the shadow is moving along the xaxis when the stick has gone one third the way around the wheel (as in the diagram). (To find the coordinates of the stick, use: x = rθ − a sin θ, y = r − a cos θ.)

45^{∘}

5. (20 points) An item is bought using a loan of y_{0} dollars, which is paid back at the steady rate of a dollars/year. The instantaneous interest rate is k (assume continuous compounding), and the term of the loan is T.

(a) Give the differential equation for the amount y(t) still owed after t years.

(b) Solve the differential equation for y(t).

(c) Derive a formula for a in terms of k, y_{0}, and T.

(d) Now suppose that the interest rate is 10%, and the term is 20 years. At what instant t is exactly half of the money going for interest (i.e., half of the a dollars per year being paid at that instant) and half for principal?

6. (10 points) In a certain population the instantaneous deathrate always exceeds the instantaneous birthrate by k. Meanwhile, individuals are immigrating to the region. The initial rate of immigration is I_{0} individuals per year, but this rate is increasing exponentially in such a way so as to double every D years. Write a differential equation for the population y(t).

7. (10 points) At time t = 0 you borrow y_{0} dollars to make a purchase. The instantaneous annual interest rate is 12%, compounded continuously. You will pay back the loan initially at the rate of α dollars per month, but the monthly amount you pay will decrease steadily by 0.01α each month, so that after 100 months your payments will decrease to zero. Set up a differential equation for the amount y(t) that you owe after t months.

(b) Find y(t). Your answer should involve the constants y_{0} and α.

Answers

4. Answers to Selected Homework Problems

Homework 1

1. (a) 8\sqrt{2}/15 = 0.754, (b) 1/12, (c) (n − 1)/(n + 1), (d) 4.5 (the curves intersect at (−3, 2) and (0, −1)), (e) 4/3, (f) (2/3) − (2/π) = 0.030, (g) (3/π)(1 − 0.5\sqrt{3}) − (1/8) = 0.00294.

2. (a) \sqrt{2} − 1 = 0.414, (b) 46/3.

3. (a) net=1/π, total=5/π, (b) net=0, total=245, (c) net=20, total=28, (d) net=(1.5/π) − (1/2) = −0.0225, total=0.25 + (4.5 − 3\sqrt{3})/π = 0.0284.

4. (a) a_{x}(t) = (F/m) cos(ωt), a_{y}(t) = (F/m) sin(ωt), (b) v_{x}(t) = (F/mω) sin(ωt), v_{y}(t) = (F/mω)(1− cos(ωt)), (c) x(t) = (F/mω^{2}) (1 − cos(ωt)), y(t) = (F/mω^{2}) (ωt − sin(ωt)), (d) a cycloid running up the y-axis (rather than the x-axis), (e) a circle if v_{0} = −F/mω.

Homework 2

1. hits ground at (a) 3 sec, (b) 4.5 sec, (c) 2 sec. 2. 44.3 m/sec, 0.232 sec.

3. v_{0} = −38 ft/sec, s_{0} = 590 ft. 4. 1032 ft.

5. (a) 8π/3, (b) π/30, (c) 0.5π^{2} − π = 1.793. 6. (a) 22.8π, (b) 14.8π, (c) 20π, (d) 4π.

7. (a) \frac{4}{3}πab^{2}, (b) \frac{4}{3}πa^{2}b. 8. πh^{2} (r − \frac{1}{3}h). 9. ∫_{0}^{H} \frac{\sqrt{3}}{4}(\frac{x}{H} s)^{2}dx = \frac{\sqrt{3}}{12}s^{2}H.

Homework 3

1. (a) 6 min 17 sec, (b) 0.03183 units/min. 2. (a) 20 min 57 sec, (b) 0.03183 units/min. 3. 0.6283 ft^{3}/min. 4. dV/dt = 0.04π (4 − (−2 + 0.04t)^{2}). 5. dV/dt = 0.09π (4 − (−2 + 0.04t)^{2}). 6. 1.21 cm/sec. 7. cross-sectional area = \frac{1}{4} \sqrt{3}(\frac{h}{H} s)^{2}.

Homework 4

1. (a) x^{2}/4 + 4y^{2} = 1, (b) x^{2}/100 + y^{2}/64 = 1, (c) x^{2}/625 + y^{2}/576 = 1, (d) x^{2} + y^{2}/9 = 1, (e) x^{2}/16 + y^{2}/25 = 1.

2. (a) (±3, 0), e = 0.6; (b) (0, ±24), e = 0.96; (c) (±0.3727, 0), e = 0.7454. 3. 10.

4. (a) ∫_{0}^{b} \sqrt{1 + n^{2}x^{2(n−1)}} dx; (b) \frac{8}{27} ((1 + 9b/4)^{3/2} − 1). 5. (a) ∫_{0}^{4} \sqrt{98 + (−9.8t + 7\sqrt{2})^{2}} dt; (b)

∫_{0}^{π/2} \sqrt{a^{2} sin^{2} θ + b^{2} cos^{2} θ} dθ; (c) r∫_{0}^{2π} \sqrt{1.25 − cos θ} dθ. 6. (a) 1/\sqrt{7} = 0.3780; (b) \frac{d}{dt}(x + 2.5y) = 0.3780(1 + 2.5\sqrt{3}/2) = 1.1964; (c) u = x + y cot α, so \frac{du}{dt} = \frac{dx}{dt} + \frac{dy}{dt} cot α + y(− csc^{2} α)\frac{dα}{dt} = 1.1964 − y(csc^{2} α) (−15 · \frac{π}{180}) = 1.1964 + 0.5 · 7.2509 · π/12 = 2.1455 units/hour.

Homework 5

1. 55.4 m/sec 2. (a) 0.790766, (b) 1.183465, (c) 0.987116, (d) 0.997943. 3. 8.878.

4. (a) 0.58726, (b) 0.58622 (limits of integration are π/3 and π/2).

5. (b) 0.2402 m, (c) 0.4541 m.

6. (a) x(8)−x(0) = ∫_{0}^{8} ẋ(t)dt ≈ 34.45, y(8)−y(0) = ∫_{0}^{8} ẏ(t)dt ≈ 46.35, and finally \sqrt{34.45^{2} + 46.35^{2}} = 57.75; (b) ∫_{0}^{8} \sqrt{(dx/dt)^{2} + (dy/dt)^{2}}dt ≈ 59.29.

Homework 6

1. (a) (2,1), (b) (1.5,1.2), (c) (0.4244r, 0), (d) (0, −0.8377).

2. at 39.37^{∘} northern latitude & 116.6^{∘} western longitude (right between Summit Mountain and Hickison Summit on the border between Lander and Eureka counties). 3. 0.5466.

Homework 7

1. (a) 2x^{2}/(4 + x^{2}) + 2x arctan(x/2).

(b) (arcsin(x^{2}) − 2x(x + 1)(1 − x^{4})^{−1/2}) / (arcsin(x^{2}))^{2}.

(c) −1/2\sqrt{x − x^{2}}, (d) arcsin x. 2. (a) 0.25 arcsin(0.4x) + C, (b) 0.5\sqrt{2} arctan(\sqrt{2} x) + C. 3. 31.32^{∘}.

4. π^{2}/4. 5. (0.2559,0).

6. (a) max: (1, 0.2854), min: (−1, −0.2854), (b) max: (−\sqrt{3}, 0.685), min: (\sqrt{3}, −0.685).

7. (a) π − (2/3), (b) 2\sqrt{3} − 2π/3. 8. 4.584 deg/sec.

9. (a) arccos ((a^{2} + b^{2} − x^{2})/2ab), (b) 2c / \sqrt{(2ab)^{2} − (a^{2} + b^{2} − c^{2})^{2}}.

Homework 8

1. (a) 1/x, (b) (4 ln^{3} x)/x, (c) (2 ln(ln x))/(x ln x), (d) 1/(sin x cos x), (e) − arctan(x/4).

2. (a) both derivatives are 0; (b) use the fact that only a constant function has derivative zero, find the constant by setting x = 1, and then replace x by b in the first function and by a in the second function.

3. (a) ln 3, (b) \frac{3}{2} ln 2, (c) \frac{2}{3} ln 3, (d) \frac{3}{2} ln 3 − ln 2, (e) ln π, (f) \frac{1}{2} ln(r + x).

4. (a) − ln 2, (b) ln sin x + C, (c) ln(ln x) + C, (d) (1/ ln 2) − (1/ ln 3) = 0.5325, (e) 2 ln 3.

5. (a) 1.4427, (b) 1.3863, (c) 0.4413. 6. (a) 0.4449, (b) 0.09436. 7. 4.3552.

8. 11.18, 14.14, 22.36.

9. (a) 0, (b) 0, (c) 0, (d) 0. 10. (a) (1 + ln x)x^{x}, (b) (\sqrt{x} cot x + (ln sin x)/2\sqrt{x})(sin x)^{\sqrt{x}},

(c) (\frac{1}{ln x} + ln(ln x))(ln x)^{x}, (d) (\frac{2}{x} ln x) x^{ln x}, (e) (−(arctan x) tan x + \frac{ln cos x}{1+x^{2}}) (cos x)^{arctan x}.

11. Let a be the number of hours before noon when it started snowing, and let s(t) be the distance traveled after t hours of plowing. Show that ds/dt is inversely proportional to t+a, and then use the method in section 2 of Chapter 1 to find a formula for s(t); you should find that s(t) = C ln((t+a)/a). (The value of the constant C makes no difference.) You’re given that s(2) = \frac{3}{2}s(1). Use the properties of ln and some algebra to get a quadratic equation for a, from which you find that a = (−1 + \sqrt{5})/2 (this number is called the “golden ratio”). Snow started falling at 11:23 a.m.

Homework 9

1. (a) 2, (b) ∞, (c) ∞, (d) 2, (e) π/2, (f) π, (g) π/3, (h) π/2, (i) π, (j) 1, (k) ∞, (l) π/4.

2. πn/(2 − n), infinite volume if n ≥ 2. 3. (n − 1)/(n − 2).

Homework 10

1. (a) 1, (b) 0.9917. 2. (a) 3x, (b) −x^{2}, (c) x/y^{2}, (d) x, (e) \frac{1}{2}x^{2}. 3. (a) 20 ln x, (b) x^{2} + x,

(c) arcsin e^{x}, (d) ±\sqrt{2 ln x}, (e) 1/(1 + e^{0.01x}). 4. (a) e^{2x} (\frac{1}{1+x^{2}} + 2 arctan x),

(b) e^{− sin(πx)}(1 − πx cos(πx)), (c) e^{x+e^{x}}, (d) e^{− cos x}(2x + x^{2} sin x), (e) e^{arcsin x}/\sqrt{1 − x^{2}}. 5. (a) \frac{d}{dx} cosh x = sinh x, \frac{d}{dx} sinh x = cosh x, (b) 1.

6. (a) 3.6, (b) \frac{1}{2} − \frac{1}{2e}, (c) e − \frac{1}{e}, (d) 1/ ln 10. 7. \frac{15π}{16 ln 4} ≈ 2.1245.

8. (a) max at (π/6, 1.9477), inflection points at (π/18, 1.3956) and (5π/18, 1.3956).

(b) max at (0.6065, 0.1839), inflection at (0.2231, 0.0747).

Homework 11

1. $4932. 2. (a) 11.55 yrs, (b) 6.93 yrs, (c) 3.85 yrs; (a) 11.90 years, (b) 7.27 yrs, (c) 4.19 yrs. 3. 20480.

4. 1020201, 1491825, 34.66 yrs. 5. 250e^{0.007675(t−1650)}, A.D. 2250. 6. 11.33%, 6.12 yrs. 7. 6.05 times

8. (a) $4034300, (b) 100e^{0.13t}, (c) ∫_{0}^{10} 52 · 100e^{0.13t} = 40000e^{0.13t}|_{0}^{10} = $106772, (d) $44241000, (e) about 2/3 day. 9. 8033 ln(y_{0}/y_{1}). 10. 6.64 hrs. 11. 100e^{−0.0112t}, 62.1 yrs. 12. $11477. 13. (a) v(t) = v_{0}e^{−kt}; (b) s(t) = \frac{v_{0}}{k} (1−e^{−kt}); (c)arctan(ky_{0}/v_{0}). Homework 12

1. 2.5587. 2. (a) −\frac{1}{π} \sqrt{2 + cos(2πt)} + C, (b) −1/ (7a sin^{7}(ax + b)) + C. 3. 2.0191.

4. (a) 4/(3π) = 0.4244, (b) 3/8 = 0.375, (c) 16/(15π) = 0.3395, (d) 5/16 = 0.3125, (e) 32/(35π) = 0.2910.

Homework 13

1. (a) −\frac{x}{a} cos(ax + b) + \frac{1}{a^{2}} sin(ax + b) + C, (b) x^{2} sin x + 2x cos x − 2 sin x + C,

(c) x arctan x − \frac{1}{2} ln (1 + x^{2}) + C, (d) (x^{3} − 3x^{2} + 6x − 6) e^{x} + C,

(e) \frac{1}{2}x^{2} arctan x − \frac{x}{2} + \frac{1}{2} arctan x + C.

2. (a) (−\frac{1}{k} , \frac{1}{4}), (b) (9/16, 0), (c) ((\frac{n+1}{n+2})^{2} , 0), (d) 0.18484. 3. 0.60672.

Homework 14

1. (a) 0.2747, (b) 0.09861, (c) 0.05371, (d) 0.4621, (e) 0.06112, (f) 0.04251, (g) 0.2118, (h) 0.2, (i) 0.1903, (j) −\frac{1}{5} ln(6) = −0.35835.

2. (a) (0.7095,0), (b) (0.7912,0). 3. \frac{1}{2} ln \frac{1+sin x}{1−sin x} + C.

Homework 15

1. arcsin(x/r)+C. 2. (a) ln |x+\sqrt{x^{2} + r^{2}}|+C, (b) ln |x+\sqrt{x^{2} − r^{2}}|+C. 3. (0.37616, 0). 4. 0.93313.

5. 0.48121.

Homework 16

1. $27.98, $1007.30. 2. (a) $379.15, $27749, (b) 7.7 years, (c) $2613.91 (1st year), $1574.15 (5th year). 3. $48633.80, $9726.76. 4. (a) $5613.14, (b) only $17.01 more, (c) 22.57 years.

5. T − (ln 2)/k = 13 years and 8.22 months.

6. (a) $44754, (b) $328.12, (c) 64.9 years, (d) $298.36. 8. 9:43 p.m. 9. 7.27 hours

10. 4:58 a.m. 11. 8:57 a.m. 12. (a) 0.2, (c) 40.3 m/sec upward, (d) 1.71 sec.

13. s(t) = \begin{cases}98t + 980 (e^{−0.1t} − 1) , & for 0 ≤ t ≤ 10; \\ 7t − 39.25e^{−1.4(t−10)} + 329.77, & for t > 10, \end{cases} 24.3 sec.

14. v(t) = \begin{cases}9.8t, & for 0 ≤ t ≤ 7; \\ 19.6 + 49.0e^{−0.5(t−7)}, & for 7 ≤ t ≤ 15. \end{cases}

15. (a) y = 10 − 5e^{−0.03t}, (b) y = 10 + 10e^{−0.03t}, (c) y = 10. 16. 7.42 lb.

17. (a) y(t) = 48 (1 − e^{−t/12000}), (b) 36 min.

18. (c) V k + V P/r, (e) Superior 431 yr, Michigan 71.4 yr, Erie 6.05 yr, Ontario 17.6 yr.

19. (a) dy/dt = −ky + I, (b) 339 million, 400 million. 20. R = 80565 per year; if R = 300000, then y(t) = 3147600 − 147600e^{0.0953t}, and the salmon will die out after 32.1 years.

Homework 17

1. (a) y = \sqrt{3x/(6 + x)}, (b) y = 1/(10 − \frac{1}{3}x^{3}), (c) y = \sqrt{\frac{2}{π} sin(πx) + 1}.

2. (a) y(t) = M/(1 + Ce^{−kt}), (b) 100/(1 + 19e^{−0.9474t}), 6.216 years.

3. y_{0}(1 − 0.02t)^{2.5}, domain is 0 ≤ t ≤ 50. 4. (a) v_{∞} = 49 m/sec, (b) dv/(100 + v)(49 − v) = 0.002dt, then get (100 + v)/(49 − v) = e^{C}e^{0.298t}, where e^{C} = 100/49 (since v(0) = 0), finally end up with v(t) = 49 − (73.01/(e^{0.298t} + 0.49)), (c) ∫_{0}^{3} 49 − (73.01/(e^{0.298t} + 0.49)).

(d) ds/dv = v/(9.8 − 0.102v − 0.002v^{2}), so ∫_{0}^{48} v dv/(9.8 − 0.102v − 0.002v^{2}) = 508 m.

5. 205 m. 6. (a) \frac{dy}{dt} = − \frac{r(t)}{V} y, (b) y_{0}e^{−0.02t−0.0025t^{2}}, (c) y_{0}e^{−0.01t+ \frac{0.01}{π} cos(π(t−0.5))}.

7. dy/dt = −2y/(100 + 0.25t^{2}), y(t) = y_{0}e^{−0.4 arctan(0.05t)}, y(∞) = y_{0}e^{−0.2π} = 0.5335y_{0}.

8. (a) \frac{dy}{dt} = −\frac{r(t)}{V}(y − kV), where V = 100. (b) y(t) = 100k + (y_{0} − 100k)e^{−0.01 ∫_{0}^{t} r(u)du}.

(c) ∫_{0}^{10} r(u)du = 26.73; e^{−0.01·26.74} = 0.765, i.e., 23.5% decrease.

(d) y(10) = 40 + (30 − 40)e^{−0.01·26.74} = 32.35 gm.

9. (a) \frac{dv}{dt} = −8 − 0.4v − 0.01v^{2}, (b) v(t) = 20(tan(1.19 − 0.2t) − 1), (c) 2.024 sec.

10. (a) (m − αt)\frac{dv}{dt} = F − Kv, (b) v(t) = \frac{F}{K} + (v_{0} − \frac{F}{K} ) (1 − \frac{αt}{m})^{K/α}.

11. y(t) = y_{0} sin(ωt + \frac{π}{2} ) = y_{0} cos(ωt). 12. y(t) = \frac{v_{0}}{ω} sin(ωt) + y_{0} cos(ωt).

Homework 18

2. (a) y_{0} = D − Akα/(k^{2} + α^{2}), (b) for k small the time delay is nearly 6 hours and the amplitude is small, while for k large the time delay is small and the amplitude is only a little less than A, (c) for k near zero, time delay= \frac{1}{α}(π/2 − arctan (\frac{k}{α})) ≈ 6 − k/α^{2} = 6 − 144k/π^{2} (since arctan(x) ≈ x

if x is near 0), while for k large, time delay= \frac{12}{π} arctan (\frac{π}{12k} ) ≈ 1/k, (d) θ = 60^{∘}, so time delay is 4 hrs, hottest: 7 p.m.

3. dy/dt = −0.01y + (2 + 0.25t). 4. \frac{dy}{dt} = −k(y − (b + at)).

5. dy/dt = −2y/(100 + 0.25t^{2}) + (4 + t). 6. dy/dt = −\frac{r}{V} y + r(k + αt).

7. (a) dy/dt = ky + (18.5 sin (\frac{2πt}{365}) − 20), (b) $29200, (c) you can use the solution to Example 18.2 by replacing k by −k throughout Example 18.2, where k = 0.00025 day^{−1}, and choosing α = −2π/365, θ = arctan(2π/365k) = 1.5563 rad, D = 80000, A = 7400; then y(t) = 80000 + (y_{0} − 78925.5)e^{0.00025t} − 1074.58 sin (\frac{2π}{365}t + 1.5563); (d) $24136.

8. (a) r(t) = 12 − at, V (t) = 100 − 12t + \frac{1}{2}at^{2}; (b) \frac{dy}{dt} = −\frac{r(t)}{V (t)}y + l.

9. (a) x_{0}e^{−k_{1}t}, (b) dy/dt = −k_{2}y + k_{1}x_{0}e^{−k_{1}t}.

10. (a) x(t) = kV + (x_{0} − kV)e^{−rt/V}, (b) dy/dt = −\frac{r}{W} y + (rk + (\frac{x_{0}}{V} − k) re^{−rt/V} ).

5. Answers to Review Problem Sets

Answers to Review Problem Set #1: First Midterm

1. v(t) = −1.5 + ∫_{0}^{t} sin(u/3)du = 1.5 − 3 cos(t/3); velocity is at first negative, then becomes positive after v(t) crosses the t-axis, i.e., when: 1.5 − 3 cos(t/3) = 0; cos(t/3) = 0.5; t/3 = π/3; t = π. So total distance = − ∫_{0}^{π}(1.5 − 3 cos(t/3))dt + ∫_{π}^{5}(1.5 − 3 cos(t/3))dt = = −(1.5t − 9 sin(t/3))|_{0}^{π} + (1.5t − 9 sin(t/3))|_{π}^{5} = −1.5π+9(\sqrt{3}/2)+7.5−9 sin(5/3)−(1.5π−9(\sqrt{3}/2)) = −3π+9\sqrt{3}+7.5−9·0.9954 = 4.705. 2.

(a) π ∫_{0}^{1/2} cos^{2}(πx) − (1 − 2x)^{2}dx; (b) \overline{x} = 0, \overline{y} = \frac{∫_{0}^{4} 1.5x^{2}dx + ∫_{4}^{5} 2x\sqrt{25 − x^{2}}dx}{∫_{0}^{4} 1.5x dx + ∫_{4}^{5} 2\sqrt{25 − x^{2}}dx}.

3. (a) \frac{1}{5} · π ∫_{−3}^{0} x^{2}dy = \frac{π}{5} ∫_{−3}^{0} 25(1 − y^{2}/9)dy = 5π(y − y^{3}/27)|_{−3}^{0} = 10π sec.

(b) 5/(cross-sectional area)= 5/(π5^{2}) = 0.2/π units/sec; the average rate is total rise divided by T, i.e., 3/10π = 0.3/π, which is 1 1/2 times as great.

4. (a) r ∫_{0}^{π} \sqrt{(1 − 0.5 cos θ)^{2} + (0.5 sin θ)^{2}} dθ = r ∫_{0}^{π} \sqrt{1.25 − cos θ} dθ.

(b) r \frac{π}{6} (\frac{1}{2} · 0.5 + \sqrt{1.25 − 0.5\sqrt{3}} + \sqrt{1.25 − 0.5} + \sqrt{1.25} + \sqrt{1.25 + 0.5} + \sqrt{1.25 + 0.5\sqrt{3}}

+\frac{1}{2} · 1.5) = r \frac{π}{6} (0.25 + 0.6197 + 0.8660 + 1.1180 + 1.3229 + 1.4547 + 0.75) = 3.3412r.

Answers to Review Problem Set #2: First Midterm

1. x(t) = (10 cos α)t, y(t) = −4.9t^{2} − (10 sin α)t + s_{0}. Since y(5) = 0 and x(5) = 25, from the second equation we get cos α = 0.5, i.e., α = 60^{∘}, and from the first equation we then get 0 = −4.9 · 5^{2} − (10 · 0.5\sqrt{3})5 + s_{0}, so that s_{0} = 165.8 m. 2.

(a) π ∫_{0}^{2} \frac{1}{3 − y} dy (b) ∫_{0}^{0.1} 5\sqrt{(1 − cos(12.5t))^{2} + sin^{2}(12.5t)} dt.

3. (a) Taking d/dt of \frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 gives y ẏ = −0.64x ẋ = −0.384. When x = 3, we have y = 3.2, so that ẏ = −0.384/3.2 = −0.12. (b) The position of the shadow is x + y, so that its rate is ẋ + ẏ = 0.20 − 0.12 = 0.08. (c) \sqrt{ẋ^{2} + ẏ^{2}} = \sqrt{0.20^{2} + 0.12^{2}} = 0.2332.

4. (a) ∫_{0}^{π} sin(x)dx/ ∫_{0}^{π} \frac{sin x}{x}dx. Top integral = 2.

(b) Bottom integral = \frac{π}{6} (\frac{1}{2} · 1 + \frac{0.5}{π/6} + \frac{0.5\sqrt{3}}{π/3} + \frac{1}{π/2} + \frac{0.5\sqrt{3}}{2π/3} + \frac{0.5}{5π/6} + 0) = 1.84465.

(c) \overline{x} = 1.0842.

Answers to Review Problem Set #3: First Midterm 1. y = 2 sin (\frac{π}{6} (x − 1)), ∫_{2}^{6}(2 sin (\frac{π}{6} (x − 1)) − 1)dx = 12\sqrt{3}/π − 4 = 2.616.

2. (a) ∫_{0}^{1} \sqrt{13.0^{2} + (−9.8t − 7.5)^{2}}dt.

(b) c = ea = 6, so b = \sqrt{a^{2} − c^{2}} = 8, 4 ∫_{0}^{π/2} \sqrt{100 sin^{2} θ + 64 cos^{2} θ} dθ.

(c) ∫_{0}^{2} (10 − \frac{10}{0.1t+1}) dt.

3. ∫_{0}^{3} \sqrt{ẋ(t)^{2} + ẏ^{2}}dt ≈ 65.98 meters (in practical terms 66 meters, since your data are given to only two places of accuracy).

4. (a) x = 3t, y = 2 − 2t. (b) sin θ in x-direction, cos θ in y-direction.

(c) θ = πt/2. (d) x = 3t + sin(πt/2), y = 2 − 2t + cos(πt/2).

Answers to Review Problem Set #4: First Midterm

1. ∫_{0}^{1} cos(πx/2)dx − ∫_{0}^{1}(1 − x)dx = (2/π) − (1/2) = 0.13662.

2. (a) ∫_{0}^{0.7} \sqrt{6.86^{2} + (−9.8t + 6.86)^{2}}dt. (b) π ∫_{0}^{2} 9(1 − y^{2}/4)dy.

3. \overline{x} = ∫_{0}^{π/2} x sin x dx / ∫_{0}^{π/2} sin x dx, bottom= 1,

top≈ \frac{π}{12}(\frac{π}{12} sin \frac{π}{12} + \frac{π}{6} sin \frac{π}{6} + \frac{π}{4} sin \frac{π}{4} + \frac{π}{3} sin \frac{π}{3} + \frac{5π}{12} sin \frac{5π}{12} + \frac{1}{2} · \frac{π}{2} ) = 1.00573, so \overline{x} = 1.00573. (Note: Later, when we come to integration by parts, we will be able to compute the top integral exactly, and find that it is actually equal to 1.)

4. (a) linear distance car traveled = 3t^{2}+20t, angular distance θ = 15t^{2}+100t radians. (b) About 20^{∘} beyond the top point (clockwise). (c) dx/dt = (6t + 20)(1 − cos(15t^{2} + 100t)) = 44.661 when t = 0.5, dy/dt = (6t + 20) sin(15t^{2} + 100t) = −7.734 when t = 0.5, so speed= \sqrt{44.661^{2} + 7.734^{2}} = 45.33 m/sec.

Answers to Review Problem Set #5: First Midterm

1. Note that \overline{y} = \overline{x} because of symmetry; the bottom integral in \overline{x} is ∫_{0}^{1}(0.5x + 1)dx + ∫_{1}^{2}(0.5x + 1 − (2x−2))dx = 5/4 + 3/4 = 2; the top integral in \overline{x} is ∫_{0}^{1} x(0.5x+1)dx+∫_{1}^{2} x(0.5x+1−(2x−2))dx = ∫_{0}^{1}(0.5x^{2} + x)dx + ∫_{1}^{2}(−1.5x^{2} + 3x)dx = 5/3, so that \overline{x} = (5/3)/2 = 5/6.

2. (a) c = \sqrt{a^{2} − b^{2}} = 24, foci at (−24, 0) and (24, 0). (b) e = c/a = 24/25 = 0.96. (c) x^{2}/25^{2} + y^{2}/7^{2} = 1, so x^{2} = 625(1 − y^{2}/49), which is 625 · 33/49 when y = −4; \frac{dh}{dt} = \frac{dV/dt}{dV/dh} = \frac{0.7}{π·625·33/49} = 0.00052936 m/sec.

3. (a) x = 6t − 0.3 sin(15t), y = 0.4 − 0.3 cos(15t). (b) When t = 0.4, x = 2.4838 and y = 0.1119, which is below and slightly to the right of the center, whose coordinates at that time are (2.4, 0.4).

(c) dx/dt = 1.6792, dy/dt = −1.2573, speed= 2.098 m/sec directed 36.8^{∘} below the horizontal.

4. (a) v_{horiz}(t) = \frac{1}{2}t^{2}, v_{vert}(t) = \frac{6}{π} sin(πt/6). (b) ∫_{0}^{3} \sqrt{(0.5t^{2})^{2} + (\frac{6}{π} sin(πt/6))^{2}}dt.

(c) \frac{1}{2}(\frac{1}{2}f(0) + f(0.5) + f(1) + f(1.5) + f(2) + f(2.5) + \frac{1}{2}f(3)), where

f(t) = \sqrt{0.25t^{4} + (36/π^{2}) sin^{2}(πt/6)} (final answer = 6.007).

Answers to Review Problem Set #1: Second Midterm

1. If you take 2 as the hypotenuse of the triangle and x as the leg opposite θ, then x = 2 sin θ, dx = 2 cos θ dθ, \sqrt{4 − x^{2}} = 2 cos θ, and the limits of integration become 0 to π/6. The top integral in \overline{x} becomes: ∫_{0}^{π/6} 8 sin^{3} θ dθ = 8 ∫_{0}^{π/6} (1 − cos^{2} θ) sin θ dθ = 8 ∫_{1}^{\sqrt{3}/2}(1 − u^{2})(−du) (where u = cos θ). This leads to the value \frac{16}{3} − 3\sqrt{3} = 0.13718 for the top integral. The bottom integral

is: ∫_{0}^{π/6} 4 sin^{2} θ dθ = 4 ∫_{0}^{π/6} (\frac{1−cos(2θ)}{2}) dθ = (2θ − sin(2θ)) |_{0}^{π/6}= \frac{π}{3} − \frac{\sqrt{3}}{2} = 0.18117. So \overline{x} = 0.13718/0.18117 = 0.75718.

2. Since x^{4}/(4 − x^{2}) = −x^{2} − 4 + 16/(4 − x^{2}) = −x^{2} − 4 − \frac{16}{4} (\frac{1}{x−2} − \frac{1}{x+2}), we have:

Vol = π ∫_{0}^{1} (−x^{2} − 4 − \frac{16}{4} (\frac{1}{x−2} − \frac{1}{x+2})) dx = π (−\frac{x^{3}}{3} − 4x − 4 ln |\frac{x−2}{x+2} |) |_{0}^{1} = π (−\frac{13}{3} + 4 ln 3) =

0.1920.

3. Vol= f(x) = π(e^{−x})^{2}x = πxe^{−2x}. Set 0 = f′(x) = π (−2xe^{−2x} + e^{−2x}) = πe^{−2x}(−2x + 1), so −2x + 1 = 0, i.e., x = 1/2. So the maximum volume is π · \frac{1}{2} · e^{−1} = π/2e = 0.57786.

4. (a) (x_{A}, y_{A}) = (8t, 40 − 4.9t^{2}), (b) 10 − 2t, (c) arctan((40 − 4.9t^{2})/8t) − arcsin((10 − 2t)/20), (d)

\frac{−8t(9.8t) − 8(40 − 4.9t^{2})}{(8t)^{2} + (40 − 4.9t^{2})^{2}} + \frac{0.1}{\sqrt{1 − (0.5 − 0.1t)^{2}}} .

Answers to Review Problem Set #2: Second Midterm

1. (a) f(1) = 0 and lim_{x→0} f(x) = 0. (b) f′(x) = \frac{\sqrt{x}}{x} + \frac{ln x}{2\sqrt{x}} = 0 gives ln x = −2, so x = 1/e^{2} and the point is (1/e^{2}, −2/e). (c) The top integral is

∫_{0}^{1} x^{3/2} ln x dx = \frac{2}{5}x^{5/2} ln x |_{0}^{1} − ∫_{0}^{1} \frac{2}{5}x^{5/2} · \frac{1}{x} dx = −\frac{2}{5} ∫_{0}^{1} x^{3/2}dx = −\frac{4}{25}.

Similarly, the bottom integral is ∫_{0}^{1} x^{1/2} ln x dx = −\frac{4}{9}. So \overline{x} = 9/25 = 0.36.

2. (a) Write 1/(x^{2} + 5x + 4) = \frac{1}{−1−(−4)} (\frac{1}{x+1} − \frac{1}{x+4}), so we get \frac{1}{3} ln (\frac{x+1}{x+4}) |_{0}^{∞}= \frac{1}{3} ln 4 = 0.4621. (b)

Write x^{2} + 2x = (x + 1)^{2} − 1^{2}; if you take x + 1 for the hypotenuse of the triangle and 1 for the leg adjacent to θ, then you have x + 1 = sec θ, dx = sec θ tan θ dθ, \sqrt{x^{2} + 2x} = tan θ, and the limits of integration become 0 to π/2 (because the triangle has θ = 0 when x = 0 and has angle θ approaching π/2 as x ⟶ ∞). In the integral everything cancels, and you get simply ∫_{0}^{π/2} dθ = π/2 = 1.5708.

3. π ∫_{0}^{π/2} x cos^{2} x dx = π ∫_{0}^{π/2} x\frac{1+cos(2x)}{2}dx = π (\frac{π^{2}}{16} + \frac{1}{2} ∫_{0}^{π/2} x cos(2x)dx) =

= \frac{π^{3}}{16} + \frac{π}{2} (\frac{1}{2}x sin(2x) |_{0}^{π/2} −\frac{1}{2} ∫_{0}^{π/2} sin(2x)dx) = \frac{π^{3}}{16} − \frac{π}{4} = 1.1525.

4. y_{A}(t) = 5e^{0.03t}, y_{B}(t) = 15e^{−0.05t}, so minimize f(t) = 5e^{0.03t} + 15e^{−0.05t}. Set 0 = f′(t) = 0.15e^{0.03t} − 0.75e^{−0.05t}, so 0.15e^{0.03t} = 0.75e^{−0.05t}, and e^{0.03t}/e^{−0.05t} = 0.75/0.15, i.e., e^{0.08t} = 5. Take ln of both sides and divide by 0.08 to get t = (ln 5)/0.08 = 20.118 hours.

Answers to Review Problem Set #3: Second Midterm

1. (a) ∫_{0}^{1}(1 + 2y) arctan(y)dy = (y + y^{2}) arctan(y) |_{0}^{1} − ∫_{0}^{1} \frac{y+y^{2}}{1+y^{2}} dy = \frac{π}{2} − ∫_{0}^{1} (1 + \frac{y}{1+y^{2}} − \frac{1}{1+y^{2}} ) dy

= \frac{π}{2} − (y + \frac{1}{2} ln(1 + y^{2}) − arctan y) |_{0}^{1} = \frac{3}{4}π − 1 − \frac{1}{2} ln 2.

(b) Write x − x^{2} = 0.5^{2} − (x − 0.5)^{2}, take 0.5 as the hypotenuse and x − 0.5 as the leg opposite θ; then ∫ \frac{x dx}{\sqrt{x−x^{2}}} = ∫ \frac{0.5(sin θ+1)0.5 cos θ dθ}{0.5 cos θ} = −0.5 cos θ + 0.5θ + C = −\sqrt{x − x^{2}} + \frac{1}{2} arcsin(2x − 1) + C.

(c) ∫_{π}^{4π} cos^{3}(t/4)dt = ∫_{π}^{4π} (1 − sin^{2}(t/4)) cos(t/4)dt = ∫_{\sqrt{2}/2}^{0}(1 − u^{2})(4 du) (where u = sin(t/4)); this is computed to be −5\sqrt{2}/3.

2. ln y = ln ((ln x)^{e^{x}}) = e^{x} ln(ln x), which leads to \frac{1}{y} \frac{dy}{dx} = e^{x}(\frac{1}{x ln x} + ln(ln x)), and then \frac{dy}{dx} =

(ln x)^{e^{x}} (\frac{1}{x ln x} + ln(ln x)).

3. (a) dv/dt = −kv; (b) v(t) = 5e^{−kt}; (c) s(t) = ∫_{0}^{t} 5e^{−ku}du = \frac{5}{k} (1 − e^{−kt}); (d) (1) 5e^{−0.8} = 2.247 m/sec, (2) 50(1 − e^{−0.8}) = 27.53 m, (3) s/r = 137.67 rad, (4) v/r = 11.23 rad/sec; (e)

\frac{dL}{dt} = \frac{dL}{dθ} \frac{dθ}{dt} = (\sqrt{(\frac{dx}{dθ})^{2} + (\frac{dy}{dθ} )^{2}}) · 11.23 = 11.23\sqrt{r^{2}(1 − cos θ)^{2} + r^{2} sin^{2} θ} = 1.247 \frac{m}{sec}

when r = 0.2, θ = 137.67 rad.

Answers to Review Problem Set #4: Second Midterm 1. (a) ∫_{10\sqrt{3}}^{∞} \frac{1}{100} \frac{dx}{1+(x/10)^{2}} = \frac{1}{10} ∫_{√\overline{3}}^{∞} \frac{du}{1+u^{2}} = \frac{1}{10} arctan u |_{√\overline{3}}^{∞}= \frac{1}{10} (\frac{π}{2} − \frac{π}{3} ) = π/60.

(b) ∫ (x + 2) (\frac{1}{x−2} − \frac{1}{x−1}) dx = ∫ \frac{x+2}{x−2}dx − ∫ \frac{x+2}{x−1}dx = ∫ (1 + \frac{4}{x−2}) dx − ∫ (1 + \frac{3}{x−1}) dx = 4 ln |x − 2| − 3 ln |x − 1| + C. (c) Set u = 2(x + 1), so that: ∫_{−1/2}^{1/2} x e^{2(x+1)}dx = \frac{1}{2} ∫_{1}^{3}(\frac{1}{2}u − 1)e^{u}du = \frac{1}{4} ∫_{1}^{3} ue^{u}du − \frac{1}{2} ∫_{1}^{3} e^{u}du = \frac{1}{4} (ue^{u}|_{1}^{3} − e^{u}|_{1}^{3}) − \frac{1}{2}e^{u}|_{1}^{3} = \frac{1}{2}e.

(d) ∫_{0}^{π} sin^{5}(x/3)dx = ∫_{0}^{π} (1 − cos^{2}(\frac{x}{3}))^{2} sin(\frac{x}{3})dx = ∫_{1}^{1/2}(1 − u^{2})^{2}(−3 du) (where u = cos(\frac{x}{3})). This gives 3(−\frac{1}{5}u^{5} + \frac{2}{3}u^{3} − u)|_{1}^{1/2} = 0.331.

2. (a) arctan((7 − t − 2 sin(2πt)) / (3 + 1.4t − 2 cos(2πt))). (b) Let y(t) be the top function and

x(t) the bottom function in the Arctan. Then \frac{d}{dt} arctan (\frac{y}{x}) = \frac{1}{1+(y/x)^{2}} · \frac{xẏ−yẋ}{x^{2}} = \frac{xẏ−yẋ}{x^{2}+y^{2}}; multiply by 180/π to convert to degrees; the result is:

\frac{180}{π} \frac{(3 + 1.4t − 2 cos(2πt))(−1 − 4π cos(2πt)) − (7 − t − 2 sin(2πt))(1.4 + 4π sin(2πt))}{(3 + 1.4t − 2 cos(2πt))^{2} + (7 − t − 2 sin(2πt))^{2}}.

3. (a) 0 = f′(x) = (1/2\sqrt{x})−(1/x) gives x = 4, i.e., the point (4, 0.6137). (b) f′′(x) = −\frac{1}{4}x^{−3/2}+x^{−2}, which is positive if x = 4, so that the point in part (a) is a minimum. (c) f′′(x) = 0 when x = 16, and the point (16, 1.2274) is an inflection; the curve is concave up for x < 16 and concave down for x > 16. (d) f(x) approaches +∞ as x ⟶ 0 and as x ⟶ ∞. (e)

Answers to Review Problem Set #1: Final Exam

1. (a) 0 = f′(x) = \frac{1}{x} − \frac{2.5·0.5}{1+(0.5x)^{2}} , so that 1/x = 1.25/(1 + 0.25x^{2}), i.e., 1 + 0.25x^{2} − 1.25x = 0, or x^{2} − 5x + 4 = 0, which has roots x = 1 and x = 4; hence, the critical points are (1, −1.159) and (4, −1.382); by the 2nd derivative test, the first is a max and the second is a min; (b) ln x; (c) ln x − (5π/4) = ln x − 3.927.

2. (a) π ∫_{0}^{12} g(y)^{2}dy ≈ 2π(\frac{1}{2} · 5.4756 + 6.9169 + 8.4100 + 9.9856 + 11.5600 + 13.1769 + \frac{1}{2} · 14.8225) = 378 cm^{3}; (b) \frac{dV}{dt} = \frac{dV}{dh} · \frac{dh}{dt} = π(3.85)^{2} · 0.05 = 2.33 cm^{3}/sec.

3. (a) KΔx/(r^{2}+x^{2}), (i) KrΔx/(r^{2}+x^{2})^{3/2}, (ii) KxΔx/(r^{2}+x^{2})^{3/2}; (b) in the direction perpendicular to wire: ∫_{0}^{∞} \frac{Kr dx}{(r^{2}+x^{2})^{3/2}} , use inverse trig substitution with x = r tan θ, dx = r sec^{2} θ dθ, \sqrt{r^{2} + x^{2}} =

r sec θ, to transform the integral to ∫_{0}^{π/2} \frac{Kr^{2} sec^{2} θ dθ}{r^{3} sec^{3} θ} = \frac{K}{r} ∫_{0}^{π/2}(1/ sec θ)dθ = \frac{K}{r} ∫_{0}^{π/2} cos θ dθ = \frac{K}{r}; in the direction parallel to wire: ∫_{0}^{∞} \frac{Kx dx}{(r^{2}+x^{2})^{3/2}} , use u-substitution u(x) = r^{2} + x^{2} to get:

∫_{r^{2}}^{∞} \frac{1}{2}Ku^{−3/2}du = −Ku^{−1/2} |_{r^{2}}^{∞}= \frac{K}{r}; (c) the horizontal and vertical components of the force are the same (K/r in both cases), so the force points at a 45^{∘} angle.

4. (a) \frac{dv}{dt} = g − 0.07v; (b) v(t) = −140 (1 − e^{−0.07t}); (c) 41.34 m/sec downward;

(d) \frac{dv}{dt} = g − 2.0v, v(t) = −4.9 − 36.44e^{−2t}; (e) 1.75 sec.

5. (a) 2 − 0.2t; (b) 10 − 0.1t^{2}, empty when t = 10 min; (c) \frac{dy}{dt} = −\frac{2}{10−0.1t^{2}} y + 0.56(1 − 0.1t).

Answers to Review Problem Set #2: Final Exam

1. (a) 0 = f′(x) = (−3x^{2}+1)e^{−1.5x^{2}}, so x = ±1/\sqrt{3}, and the critical points are (0.57735, 0.35018) and (−0.57735, −0.35018); since f′′(x) = (−3x(−3x^{2} + 1) − 6x)e^{−1.5x^{2}} = 9x(x^{2} − 1)e^{−1.5x^{2}}, the second derivative test shows that the first of these points is a max, the second is a min; (b) inflections at (0, 0), (1, 0.22313) and (−1, −0.22313); (c) f(x) rapidly approaches the x-axis; (d) (see the figure below).

2. (a) θ = arctan(a/b); (b) \frac{dθ}{dt} = \frac{180}{π}(bȧ − aḃ)/(a^{2} + b^{2}).

3. ∫_{0}^{8} \sqrt{ẋ^{2} + ẏ^{2}}dt ≈ 2(\frac{1}{2} \sqrt{10.0^{2} + 0^{2}}+\sqrt{9.8^{2} + 8.0^{2}}+\sqrt{9.4^{2} + 15.2^{2}}+\sqrt{9.0^{2} + 22.0^{2}}+\frac{1}{2} \sqrt{8.4^{2} + 28.2^{2}}) = 148 meters.

4. (a) \frac{dy}{dt} = −\frac{r}{V} y; (b) 0.01y_{0} = y_{0}e^{−(r/V)T}, so T = −(V/r) ln(0.01) = V ln 100/r = 4.605V/r.

5. (a) dy/dt = 0.12y − 1500; (b) y = 12500 − 4500e^{0.12t}; (c) 8.51 years; (d) 12770 rhinos; (e) if you buy an $8000 car, no money down, 12% interest on the loan, making monthly payments of $125, how long does it take you to pay off the car loan, and how much do you end up having paid for the car in all?

6. (a) \frac{dv}{dt} = −8 − 0.05(v + 5t); (b) \frac{dv}{dt} = −0.05v − (8 + 0.25t).

Answers to Review Problem Set #3: Final Exam

1. (a) x = 17.32t, y = −4.9t^{2} − 10t + 50; (b) θ = arctan((−4.9t^{2} − 10t + 50) /17.32t);

(c) \frac{dθ}{dt} = (17.32t(−9.8t − 10) − 17.32(−4.9t^{2} − 10t + 50))/(300t^{2} + (−4.9t^{2} − 10t + 50)^{2}); when t = 2 this gives −0.9215 rad/sec.

2. (a) v(t) = 55 + ∫_{0}^{t} (24/\sqrt{1 − u^{2}}) du = 24 arcsin(t) + 55; (b) v(\frac{1}{2}) = 67.57 miles/hr;

(c) s(\frac{1}{2}) − s(0) = ∫_{0}^{1/2} v(t)dt = ∫_{0}^{1/2} 24 arcsin(t) + 55 dt =

= 24 (t arcsin(t) |_{0}^{1/2} − ∫_{0}^{1/2} t(1 − t^{2})^{−1/2}dt) + 55t |_{0}^{1/2}=

= 24 · \frac{1}{2} · \frac{π}{6} − 24 (−\sqrt{1 − t^{2}} |_{0}^{1/2}) + \frac{1}{2} · 55 = 2π + 12\sqrt{3} + 3.5 = 30.57 miles.

3. (a) y = (1/x) + ln x; y′ = (−1/x^{2}) + (1/x) = 0 gives the point x = 1, y = 1 (a min pt); y′′ = (2/x^{3}) − (1/x^{2}) = 0 gives the point x = 2, y = 0.5 + ln 2 = 1.193;

(b) approaches +∞; (c) approaches +∞; (e) ∫_{1/2}^{1} \sqrt{1 + (−(1/x^{2}) + (1/x))^{2}} dx ≈ ≈ \frac{1}{8} (\frac{1}{2}f(\frac{1}{2}) + f(\frac{5}{8}) + f(\frac{3}{4}) + f(\frac{7}{8}) + \frac{1}{2}f(1)) = \frac{1}{8}(1.1180 + 1.3862 + 1.0943 + 1.0132 + 0.5) = 0.6390.

4. (a) \frac{dy}{dt} = −\frac{r}{V} (y − kV) = −(0.6 + 0.4 sin(\frac{πt}{10}))(y − 5k); (b) separation of variables leads to y − 5k = Ce^{−0.6t+(4/π) cos(πt/10)}, where C = (y_{0} −5k)e^{−4/π}, i.e., y(t) = 5k +(y_{0} −5k)e^{−0.6t−(4/π)(1−cos(πt/10))};

(c) increases if y_{0} < 5k, decreases if y_{0} > 5k.

5. dy/dt = 0.025y + (1.0 + 0.1t).

Answers to Review Problem Set #4: Final Exam

1. (a) θ = arctan(\frac{5}{y}); (b) \frac{dθ}{dt} = \frac{dθ}{dy} · \frac{dy}{dt} = 1/ (1 + \frac{25}{y^{2}} ) · \frac{−5}{y^{2}} · 0.5 = −\frac{2.5}{y^{2}+25} = −\frac{1}{260} rad/min = −0.22 deg/min.

2. f′(x) = x^{x}(ln x + 1); critical point at (1/e, 0.692); f′′(x) = x^{x} (\frac{1}{x} + (ln x + 1)^{2}), which equals 0.692e at the critical point — so the critical point is a minimum.

3. Δx(\frac{1}{2} \sqrt{1 + f′(0)^{2}} + \sqrt{1 + f′(0.5)^{2}} + \sqrt{1 + f′(1)^{2}} + \sqrt{1 + f′(1.5)^{2}} + \sqrt{1 + f′(2)^{2}}+

+\sqrt{1 + f′(2.5)^{2}} + \frac{1}{2} \sqrt{1 + f′(3)^{2}}) = \frac{1}{2}(0.7071 + 1.3521 + 1.2440 + 1.1461 + 1.0808+ +1.0218 + 0.5056) = 3.529.

4. position of shadow= u = x+y = 2θ +2−sin θ −cos θ; \frac{dθ}{dt} = \frac{1}{r} ·speed = 2.5 rad/sec; \frac{du}{dt} = \frac{du}{dθ} · \frac{dθ}{dt} = (2 − cos θ + sin θ) · 2.5; for θ = 2π/3 this gives 8.415 m/sec.

5. (a) \frac{dy}{dt} = ky−a; (b) y(t) = \frac{a}{k} −(\frac{a}{k} −y_{0})e^{kt}; (c) a = ky_{0}/(1−e^{−kT}). (d) Note that ky_{0} = a(1−e^{−kT}). To find when ky(t) = \frac{1}{2}a, we set \frac{1}{2}a = ky = a − (a − ky_{0})e^{kt}. This gives (a − ky_{0})e^{kt} = a − \frac{1}{2}a = \frac{1}{2}a. Replacing ky_{0} by a(1 − e^{−kT}) and dividing through by a, we get (1 − (1 − e^{−kT}))e^{kt} = 1/2, i.e., e^{k(t−T)} = 1/2. Taking ln of both sides and dividing through by k, we finally get t = T − \frac{ln 2}{k}. In this problem, T = 20 years and k = 10% = 0.1, so that t = 20 − 10 ln 2 = 13.07 years.

6. \frac{dy}{dt} = −k_{1}y + I_{0}e^{k_{2}t}. 7. time units are months, so convert 12% annual interest to 1% monthly interest; \frac{dy}{dt} = 0.01y − α(1 − 0.01t).

234 ANSWERS

Antiderivative Formulas

We will use the notation ∫ f(x)dx to denote the antiderivative of f(x). In the following formulas, f(x) and g(x) are functions and c is a constant:

∫ f + gdx = ∫ f + ∫ g
∫ f − gdx = ∫ f − ∫ g
∫ cfdx = c ∫ fdx
∫ f′(g(x))g′(x) = f(g(x)) + C

Formulas for common functions:

∫ x^{a} dx = \frac{x^{a+1}}{a+1} + C if a ≠ −1  ∫ x^{−1} dx = ln x + C
∫ a^{x} dx = \frac{a^{x}}{ln a} + C  ∫ ln x dx = x ln x − x + C
∫ sin x dx = − cos x + C  ∫ cos x dx = sin x + C
∫ tan x dx = − ln | cos x| + C  ∫ cot x dx = ln | sin x| + C
∫ sec x dx = ln | sec x + tan x| + C ∫ csc x dx = ln | csc x − cot x| + C
∫ sec^{2} x dx = tan x + C  ∫ csc^{2} x dx = − cot x + C
∫ sec x csc x dx = sec x + C  ∫ csc x cot x dx = − csc x + C
∫ \frac{1}{\sqrt{1−x^{2}}} dx = arcsin x + C  ∫ \frac{1}{x^{2}+1} dx = arctan x + C

∫ \frac{1}{x\sqrt{x^{2}−1}} dx = arcsec x + C

∫ sinh x dx = cosh x + C  ∫ cosh x dx = sinh x + C

Index

ds/dt, 20 p-series, 165 u-substitution technique of integration, 79 eccentricity of an ellipse, 19 square cross-sections, 12 Taylor series for f at x = a, 192 power series in x − a, 187 remainder after n-terms, 166 net distance traveled, 5 qualitative information, 105 second order differential equation, 114 total distance traveled, 5

caption, 202 collaborative groups, 201 Conclusions, 202 Cover Page, 202 English, 201 guidelines, 201 interpretations, 202 Lab projects, 201 Maple worksheet, 202

a solid with finite volume but infinite surface area,

59 alternating series, 169 angular velocity, 18 antiderivative method, 21, 22 antiderivative of x^{−1}, 50 applying Taylor’s formula, 129 approximation, 125 approximation method, 21 arclength, 20 arctangent, 42 area enclosed by two curves, 1 area out to infinity, 57 ATE, iii average value of a function, 76 axis of symmetry, 29

bad notation, 4 bad notation, tan^{−1} x, 42 balancing point, 33 bank interest problem, 63, 69 best view problem, 44 bounded above, 153 bounded sequence, 153 Brook Taylor, 126 bug problem, 22 bug’s shadow, 23

córdobas, 71 cardiac output, 29 center of mass, 33 centroid, 34 centroid of Nicaragua, 37 circle, 17 circular cross-section, 28 Combined circular motion, 17 common ratio, 157 companion antiderivative formula, 45 compounded continuously, 98 concentration, 103 continuous compounding, 64 converge absolutely, 185 convergent sequence, 147 convergent series, 155 converges to the limit, 147 cross sectional area, 10 cycloid, 18

daily compounding, 64 damped harmonic oscillation, 116 decreasing, 152 degree/radian warning, 42 differential equation, 69 differential equation y′ = ±ky, 97 Divergence Test, 168 divergent, 147 divergent series, 155 dominates, 168 doubling time, 71 draw a picture, 14

effective annual interest rate, 70 ellipse, 18, 19 ellipse parameterized from a focus, 20 erm-by-term, 189 expanding f in a power series, 189 exponential function, 65 exponential term, 102

finite sequences, 145 first term, 145 focus of an ellipse, 19 force of repulsion, 94 Formulas, v freeway problem, 59, 94 fundamental constant of mathematics, 63 Fundamental Theorem of Calculus, 1 fundamental theorem of calculus, 49

goes backwards, 41 greatest lower bound, 154

half-life, 72 harmonic oscillation, 115 harmonic series, 159 hemisphere, 8 home mortgages, 99 horizontal asymptote, 102 horizontal velocity, 17 hypotenuse, 93

implicitly differentiate, 43 improper integral, 57 increasing, 152 Indeterminate Form, 134 indeterminate form 0/0, 135 infinite decimal, 157 infinite series, 145, 155 initial velocity, 7, 17 instantaneous birth rate, 71 instantaneous interest rate, 70 instantaneous rate of interest, 97 integration by parts, 79 integration of even powers of sin and cos, 76 integration of odd powers of sin and cos, 75 integration of rational functions, 85 interpolating, 29 interpolation, 28 interval of convergence, 187 inverse function, 41 inverse sine function, 45 inverse tangent function, 42 inverse trig substitution, 91, 94

Koblitz, iii

L’Hospital’s Rule, 134

law of conservation of energy, 115

least upper bound, 154

left endpoint Riemann sum, 27

limit, 147

limit problems, 139 log to the base a = 2, 53 logarithmic differentiation, 53 logarithms to the base 10, 53

Maclaurin series for f, 192

main uses of integration by parts, 79 major axis, 18 method of partial fractions, 112 method of substitution, 37 minor axis, 18 mirror images, 41, 43 model for the spread of an epidemic, 112 moment, 33 moment integral, 37 monotonic, 152

natural logarithm, 49

Newton’s law of cooling, 101

Newton’s method, 125

nonlinear mixing problems, 113

nonlinear separation of variables, 111 NSF, iii numerical integration, 22

parabola, 2

parameterized by time, 17 parameterized curves, 17 parametric equations, 20 partial fractions, 85 Pinching Principle for Sequences, 151 Plutonium 239, 72 point-masses, 33 power series, 185 principal value, 42 properties of ln, 49 properties of the exponential function, 65 Pythagorean theorem, 60

radius of convergence, 187

rate of inflation, 71 rate of payment, 98 recursively, 146 Related Rate Volume Problems, 13 representing f by a power series, 189 Riemann sum, 25, 59 Riemann sum rectangles, 9 right circular cone, 8 right endpoint Riemann sum, 26 second derivative method, 44 second term, 145 seesaw principle, 33 separation of variables, 69 sequence, 145, 146 sequence of partial sums, 155 shortcuts, 3 significant figures, 28 solid of revolution, 7 sphere, 8, 10 stay close, 147 strictly decreasing, 152 strictly increasing, 152 substitution u = cos θ, 93 sum of the infinite series, 155 summation notation, 9

tangent line approximation, 44 Taylor Polynomials, 126 Taylor’s Theorem, 125, 127 technique of integration: substitution, 75 terms, 145 terms of the series, 156 the exponential function, 63 third degree Taylor polynomial, 126 total moment, 35 trapezoid rule, 21, 27

variant on the mixing problem, 114 velocity v(t) of a falling body, 102 vertical components, 22