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Contents

Preface iii Introduction to the Kentucky Edition: iii Advice for Students iv

Chapter 1. Straight Lines, Polygonal Lines, Slopes, Distance, Circles 1 1. Distance Between Two Points; Circles 4

Chapter 2. Functions 9 1. Shifts and Expansions 13

Chapter 3. Instantaneous Rate Of Change: The Derivative 19

Chapter 4. The Derivative Function 25 1. Corners and Cusps 26 2. Graphical Differentiation 27

Chapter 5. The Tangent Line Approximation 31 1. The Power Rule 31 2. The Tangent Line Approximation 31

Chapter 6. Rules For Finding Derivatives 39 1. Linearity of the Derivative 39 2. The Product Rule 40 3. The Quotient Rule 41 4. The Second Derivative and Acceleration 43 5. The Chain Rule 45

Chapter 7. Implicit Functions and Implicit Differentiation 49 1. Implicit Functions 49 2. Implicit Differentiation 49

Chapter 8. Trig Functions and Sinusoidal Functions 55 1. Graphs of Trig Functions 56 2. Sinusoidal Functions 56

Chapter 9. Derivative of Trig Functions 63 1. The Derivative of sin x 63 2. Other Trig Derivatives 64

Chapter 10. Curves Given Parametrically 67 1. Trajectories 68 2. The Velocity Vector 70 Chapter 11. Combined Parametric Motions 79

Chapter 12. Related Rates 87

Chapter 13. Curve Sketching 97 1. Maxima/Minima 97 2. The Second Derivative Test 99 3. The First Derivative Test 99 4. Concavity and Inflection 100 5. Asymptotes and Other Things to Look For 101

Chapter 14. Maximum/Minimum Problems 107

Chapter 15. Newton’s Method 119

Chapter 16. Indefinite Integrals 127 1. From Acceleration to Velocity to Distance; Falling Bodies 128 2. antiderivative of sinx and cosx 130 3. Substitution 130

Chapter 17. Definite Integrals: Area 135 1. Riemann Sums 135 2. Computing the Area Under a Curve 137 3. Substitution 139 4. Some Properties of Definite Integrals 140 5. Application to Degree-Days 140 6. Velocity and Distance Traveled As Area 141

Chapter 18. The Fundamental Theorem 145 1. The Fundamental Theorem together with the Chain Rule 146

Lab Projects for Ma113 151

Practice First Midterms 161

Practice Second Midterms 167

Practice Final Exams 171

Answers 179 1. Answers to Homework Problems 179 2. Answers to Practice Midterms 182 3. Answers to Practice Final Exams 184

Index 186

62 CONTENTS

CHAPTER 9 Derivative of Trig Functions

1. The Derivative of sin x

To find the slope of the tangent line to the curve y = f(x) = sin x, we must go back to the definition of the derivative as the limit of the difference quotient:

\frac{dy}{dx} = lim_{Δx⟶0} \frac{f(x + Δx) − f(x)}{Δx} = lim_{Δx⟶0} \frac{sin(x + Δx) − sin x}{Δx}.

Recall the procedure used to find the derivative of the much simpler function x^{2}: expand (x + Δx)^{2}, simplify algebraically, and then take the limit as Δx gets closer and closer to 0. We do essentially the same thing for sin x, but we need the following three basic facts in order to do this:

Fact 1. sin(α + β) = sin α cos β + cos α sin β;
Fact 2. lim_{θ⟶0} \frac{sin θ}{θ} = 1;
Fact 3. lim_{θ⟶0} \frac{1 − cos θ}{θ} = 0.

Fact (1) was proved in the last section (Problem 4 of the homework). Facts (2) and (3) can be seen from the diagram below, where θ is a small angle in radians. By the definition of the radian measurement of an angle, θ is the distance along the circumference from A to C. Meanwhile, sin θ = \overline{AB} and 1 − cos θ = \overline{BC}.

θ 1  A B C O  sin(θ) θ

Thus, Fact (2) says that although the line \overline{AB} is smaller than the arc AC, the ratio of the two lengths gets closer and closer to 1 as θ gets smaller. Fact (3) says that the ratio of the line \overline{BC} to the arc AC gets closer and closer to 0 as θ gets smaller. We are now ready to simplify the above difference quotient.

We have:

\frac{sin(x + Δx) − sin x}{Δx} = \frac{sin x cos(Δx) + cos x sin(Δx) − sin x}{Δx}

(by Fact (1) with α = x and β = Δx)

= \frac{cos x sin(Δx) − sin x(1 − cos Δx)}{Δx}

(putting sin x cos(Δx) last on top and factoring out − sin x)

= cos x\frac{sin(Δx)}{Δx} − sin x\frac{1 − cos(Δx)}{Δx}
⟶ (cos x) · 1 − (sin x) · 0 = cos x  as  Δx ⟶ 0.

The last step comes from Facts (2) and (3) with θ replaced by Δx. We therefore conclude the following basic derivative formula:

\frac{d}{dx} sin x = cos x.

Example 9.1. You know the exact value sin(30^{∘}) = 0.5. Find sin(31^{∘}) using the tangent line approximation, and compare with the exact value (found by calculator).

In the tangent line approximation we choose f(x) = sin x, x = 30^{∘} = π/6 and Δx = 1^{∘} = π/180. Here it was important that we changed Δx from degrees to radians. By the tangent line approximation,

sin(31^{∘}) = f(\frac{π}{6} + \frac{π}{180}) ≈ 0.5 + \frac{π}{180}f′(\frac{π}{6} )

= 0.5 + \frac{π}{180} cos(\frac{π}{6} ) = 0.5 + \frac{π}{180} · \frac{\sqrt{3}}{2} = 0.515115;

the exact value is 0.515038 . . ..

When we have a derivative formula for a new function, we can combine it with other rules to find the derivative of a more complicated function that is built up from the basic function.

Example 9.2. Find the derivative of sin^{2}(\sqrt{x}).

The notation sin^{2}(something) means (sin(something))^{2}. To evaluate this derivative, we have to use the chain rule twice:

\frac{d}{dx}(sin \sqrt{x})^{2} = 2 sin \sqrt{x} \frac{d}{dx} sin \sqrt{x} = 2 sin \sqrt{x} cos \sqrt{x} \frac{d}{dx}\sqrt{x}

= 2 sin \sqrt{x} cos \sqrt{x} (1/2\sqrt{x}) = \begin{matrix}sin & \sqrt{x} & cos & \sqrt{x} \\ √ & \overline{x} \end{matrix}.

2. Other Trig Derivatives

All of the other trig functions can be expressed in terms of the sine, and so their derivatives can easily be calculated. The most important ones are:

\frac{d}{dx} cos x = \frac{d}{dx} sin(x + \frac{π}{2} ) = cos(x + \frac{π}{2} ) = sin((x + \frac{π}{2} ) + \frac{π}{2} ) = sin(x + π) = − sin x;

\frac{d}{dx} tan x = \frac{d}{dx} \frac{sin x}{cos x} = \frac{cos x\frac{d}{dx} sin x − sin x\frac{d}{dx} cos x}{cos^{2} x} = \frac{cos^{2} x + sin^{2} x}{cos^{2} x} = \frac{1}{cos^{2} x} = sec^{2} x;
\frac{d}{dx} sec x = \frac{d}{dx}(cos x)^{−1} = (−1)(cos x)^{−2}(− sin x) = \frac{sin x}{cos^{2} x} = tan x sec x.

Homework 9.

1. You know the exact value tan(45^{∘}) = 1. Find tan(44^{∘}) using the tangent line approximation, and compare with the exact value (found by calculator).

2. Find the derivative of (a) y =cot x, (b) y =csc x (csc is the abbreviation of cosecant).

3. Use the tangent line approximation to find (a) cos(62^{∘}), (b) sin(29^{∘}), (c) cot(47^{∘}).

4. Find the derivative of each of the functions in Problem 8(a)–(e) of the last homework set. In each case find the first positive time t for which the derivative is 0 and the first positive t for which the derivative reaches its maximum value; then explain in practical terms what is happening at each of these moments.

5. Find the derivative of (a) y = sin x cos x, (b) y = sin(cos x), (c) y = \sqrt{x tan x}, (d)

y = \frac{tan x}{1 + sin x}.

6. Suppose that the point (π/3, π/4) is on the implicitly defined curve \frac{sin x}{x} + \frac{sin y}{y} = const.

Use the tangent line approximation to determine the y-coordinate of the point on the curve with x-coordinate \frac{π}{3} + \frac{π}{180}.

7. Suppose that the point (1, 1.5) is on the curve x sin(πy/3) + y sin(πx/4) =const. Use the tangent line approximation to determine the y-coordinate of the point on the curve with x-coordinate 1.1.

8. A rope is tied at its left end, and its right end is moved up and down to create a sinusoidal wave of period 6 ft. The equation of the shape of the rope (“standing wave”) is then

y = A sin (\frac{2π}{6} x) = A sin (\frac{π}{3} x) ,

where we are taking the origin to be the point where the rope is fastened, and A is the amplitude. Now actually, the amplitude A varies with time as the rope oscillates. Suppose that the maximum value of A is 2 ft, and it varies sinusoidally with time t with period 2 sec:

A = 2 sin(πt).

Thus, the vertical coordinate y of a point on the rope depends on both its x-coordinate and the time t:

y = 2 sin(πt) · sin (\frac{π}{3} x) .

(a) Sketch the appearance of the rope at quarter-second intervals from t = 0 to t = 2 sec (all on the same graph).

(b) Suppose your vision is blocked by a horizontal ledge at a height of 6 inches above the x-axis, so you see only the tips of the waves. Write an equation that gives the x-value where the wave emerges above the ledge implicitly as a function of t (i.e., write an equation of the form F (x, t) =const).

(c) What is the first x-value where the rope emerges above the ledge when t = 1/6 sec?

(d) How rapidly is that first x-value moving leftward at the instant t = 1/6 sec?

(e) Approximately where will that first point where you see the rope be located when t = 0.2 sec? (Use parts (c) and (d) and the tangent line approximation.)