vi

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 95 1. Maxima/Minima 95 2. The Second Derivative Test 97 3. The First Derivative Test 97 4. Concavity and Inflection 97 5. Asymptotes and Other Things to Look For 98

Chapter 14. Maximum/Minimum Problems 105

Chapter 15. Newton’s Method 117

Chapter 16. Indefinite Integrals 123 1. From Acceleration to Velocity to Distance; Falling Bodies 124 2. antiderivative of sinx and cosx 126 3. Substitution 126

Chapter 17. Definite Integrals: Area 131 1. Riemann Sums 131 2. Computing the Area Under a Curve 132 3. Substitution 135 4. Some Properties of Definite Integrals 136 5. Application to Degree-Days 136 6. Velocity and Distance Traveled As Area 137

Chapter 18. The Fundamental Theorem 139 1. The Fundamental Theorem together with the Chain Rule 140

Lab Projects for Ma113 143

Practice First Midterms 151

Practice Second Midterms 155

Practice Final Exams 159

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

Index 186

168 CONTENTS

Answers

1. Answers to Homework Problems

Answers to Selected Problems in Chapter 1. 1: m, b and distance are: (a) 1.5, −3, \sqrt{13}; (b) −3, 2, \sqrt{10}; (c) 1, 0, 2\sqrt{2}; (d) 0, 3, 6; (e) 1, 1, 5\sqrt{2}; (f) −3, 0.02, 0.02\sqrt{10}. 2: y- and x-intercepts are:

(a) 2, −1; (b) 6, 6; (c) 1/2, −1; (d) 3/2, none; (e) −2, −3. 3: (a) x^{2}+y^{2} = 9, (b) (x−5)^{2}+(y−6)^{2} = 9,

(c) (x + 5)^{2} + (y + 6)^{2} = 9, (d) x^{2} + (y − 3)^{2} = 9 (or x^{2} + y^{2} − 6y = 0), (e) x^{2} + y^{2} + 6y = 0, (f) x^{2} − 6x + y^{2} = 0. 4: radius and center are: (a) 5, (0, −5); (b) 7, (5, 0); (c) 5, (3, 4). 5: y = 1.8x+32. 6: y = 0.15x+10. 7: y = 0.08x−3.2. 8(b); y = 3x+120. 9: (a) y = 0 for 0 ≤ x ≤ 100, y = 0.1x − 10 for 100 ≤ x ≤ 1000, y = x − 910 for x ≥ 1000; (b) he’s introducing a “negative income tax” as a rather cheap way of getting good PR — so that, for example, a penniless person receives 10 gold coins from the King (since the y-intercept is −10). 10:(a) P = −0.0001x + 2; (b) x = −10000P + 20000. 11: y = 0.15x for 0 ≤ x ≤ 19450, y = 0.28x − 2528.5 for 19450 ≤ x ≤ 47050, y = 0.33x − 4881 for 47050 ≤ x ≤ 97620.

Answers to Selected Problems in Chapter 2. 1: [1.5, ∞). 2: x ≠ −1. 3: x ≠ ±1. 4: (−∞, 0). 5: all x. 6: [0, ∞). 7: [h−r, h+r]. 8: x negative or in [1, ∞). 9: (−\frac{1}{3}, \frac{1}{3}). 10: x nonnegative but x ≠ 1. 11: x nonnegative but x ≠ 1.

Answers to Selected Problems in Chapter 3. 1: (a) −5, (b) −2.47106, (c) −2.40679, (d) −2.40068, (e) −2.4. 2: (a) −4/3, (b) −24/7, (c) 7/24, (d) 3/4. 3: (a) 5, (b) 4.1, (c) 4.01, (d) 4.001; limit=4. 4: (a) −10.29, (b) −9.849, (c) −9.8049; −9.8 − 4.9Δt; −9.8 m/sec. 5: (a) −0.107527, (b) −0.110742, (c) −0.111074; −1/(3(3 + Δx)); f′(3) = −1/9. 6: (2 + Δx) + 1/(1 + Δx); f′(1) = 3. 7:

(a) 3.31, (b) 3.003001, (c) 3.0000300001; 3 + 3Δx + (Δx)^{2}; f′(1) = 3.

Answers to Selected Problems in Chapter 4. (first two problems) 1: f′(x) = −x/\sqrt{169 − x^{2}}. 2: (a) f′(t) = −9.8t, (b) f′(x) = 2x + (1/x^{2}), (c) f′(x) = 2ax + b, (d) f′(x) = 3x^{2}.

Answers to Selected Problems in Chapter 5. 1:(a) 4.4, (b) 0.9, (c) 1.01, (d) 0.0099, (e) 2.0025,

(f) 5.1, (g) 4.9, (h) 2.001. 2: 3.94 3: 0.9433. 4: 15 ft 2 in. 5: 208.9 ft. 6: 25π ± 4π(Δx). 7: ±30, ±1% and ±3%. 8: (a) 5 sec, (b) ±0.025 sec, ±0.5%, (c) ±\frac{1}{2}p%. 9: (a) r = C A^{1/2}, where C = 1/\sqrt{π}, (b) ±0.5%, (c) ±\frac{1}{2}p%. 10: 8 ± 0.15. 11: ±3p%. 12: (a)±\frac{1}{2}p%, (b) ±\frac{1}{3}p%. 13: (a) ±p%, (b) ±3p%, (c) ±2p%. 14: ±\frac{2}{3}p%. 15: (a) ±2p%, (b) ±p%, (c) ±p%. 16: 1 83/84. 17: Δx = −1/12 cm.

Answers to Selected Problems in Chapter 6. 1: (a) 4x^{3} −9x^{2} +x+7, (b) 3x^{2} −4x+2/\sqrt{x}, (c)- (d) 6x(x^{4} + 2x^{2} + 1),

(e) \sqrt{169 − x^{2}} − x^{2}/\sqrt{169 − x^{2}}, (f) (2x − 4)\sqrt{25 − x^{2}} − x(x^{2} − 4x + 5)/\sqrt{25 − x^{2}},

(g) (−x^{2} + 2x + 2)/(1 − x)^{2}, (h) −25/ (x^{2}\sqrt{25 − x^{2}}), (i)-(j) −\sqrt{x}/\sqrt{169 − x^{2}} − \frac{1}{2} \sqrt{169 − x^{2}}/x^{3/2},

(k) \frac{1}{2} (3x^{2} − 2x + (1/x^{2})) /\sqrt{x^{3} − x^{2} − (1/x)}, (l) 300x/(100 − x^{2})^{5/2},

(m) (\frac{1}{3} +x^{2})(x+x^{3})^{−2/3}. (n) ((x^{2} + 1)^{2} + \sqrt{1 + (x^{2} + 1)^{2}})^{−1/2} x(x^{2} +1)(2+(1 + (x^{2} + 1)^{2})^{−1/2}).

2: (a) −9.8t+20, −9.8; (b) a, 0; (c) \frac{1}{2}a(at+b)^{−1/2},−\frac{1}{4}a^{2}(at+b)^{−3/2}; (d) \frac{1}{2}(2at+b)(at^{2} +bt+c)^{−1/2}, a(at^{2} + bt + c)^{−1/2} − \frac{1}{4}(2at + b)^{2}(at^{2} + bt + c)^{−3/2}, which simplifies to \frac{1}{4}(4ac − b^{2})(at^{2} + bt + c)^{−3/2}. 3: v(t) = −5 + 2t, a(t) = 2; the bicycle is moving leftward initially at 5 m/sec, but its speed is decreasing; after 2.5 sec it has slowed to 0 m/sec, at which point it reverses direction and starts moving to the right. 4: 2 + \frac{n}{4}Δx. 5: ±0.96Δx. 6: \frac{500}{3}π ± 40π(Δx). 7: (a) f(t) = \sqrt{100 + (78 − 4.9t^{2})^{2}}; (b) −9.8t(78 − 4.9t^{2})/\sqrt{100 + (78 − 4.9t^{2})^{2}}.

Answers to Selected Problems in Chapter 7. 1: (a) 1.992, (b) 5.0129, (c) 4.00133. 2: (a) y′ = x/y, (b) y′ = −(2x + y)/(x + 2y), (c) y′ = (3x^{2} + y^{2} − 2xy)/(3y^{2} + x^{2} − 2xy). 3: y′ = 1. 4: y = −\frac{1}{13}x+ \frac{14}{13}, 0.9923. 5: B is 4 times as strong as A. 6: y′ = 3/19. 7: (a) dT/dh = −38.64^{∘}F/hour,

(b) h = −0.0259T +5.493. 8: (a) \sqrt{100 + (10 + 0.3t)^{2}}+\sqrt{(x − 10)^{2} + (10 − 0.3t)^{2}} =const, (b) 12.492 sec.

Answers to Selected Problems in Chapter 8. 1: (a) π/12, (b) 2π/3, (c) π/2, (d) 20π. 2: (a) 25 rad/sec (b) 6.944 rad/sec. 3: (a) 30 sin θ, (b) tan θ, (c) 60(1 − cos θ). 4: sin(α + β) = \overline{BC} = \overline{DF} = \overline{DE} + \overline{EF} = \overline{AE} sin α + \overline{BE} cos α = cos β sin α + sin β cos α. 6: warmest: 62^{∘} on July 12; coolest: −12^{∘} on January 10. 7: F = 50 sin (\frac{π}{2} t) + 150, R = 2500 sin (\frac{π}{2} (t − 3)) + 7500. 8: (a) 20 sin (\frac{2π}{360}(t − 109)) + 40; (b) 2 sin (\frac{2π}{360}(t − 81)) + 7; (c) 5 sin (\frac{2π}{24} (t − 11)) + 20; (d) 10 sin(5πt); (e) 0.4 sin (25(t − \frac{π}{50})) + 0.4 (or, equivalently, 0.4(1 − cos(25t))).

Answers to Selected Problems in Chapter 9. 1: 0.96509 (exact value=0.96569). 2: (a) −csc^{2}x, (b) −cot x csc x. 3: (a) 0.46977, (b) 0.484885, (c) 0.93018. 4: (a) \frac{π}{9} cos (\frac{2π}{360}(t − 109)); (b) \frac{π}{90} cos (\frac{2π}{360}(t − 81)); (c) \frac{5π}{12} cos (\frac{2π}{24} (t − 11)); (d) 50π cos(5πt); (e) 10 sin(25t). 5: (a) cos^{2} x − sin^{2} x, (b) − sin(x) cos(cos x), (c) (x sec^{2} x + tan x) / (2\sqrt{x tan x}),

(d) ((1 + sin x) sec^{2} x − sin x)/(1 + sin x)^{2}. 6: 0.7632. 7: 1.241. 8: (b) 0.5 = 2 sin(πt) sin(πx/3); (c)

x = 0.5; (d) dx/dt = −3; (e) x = 0.5 + (0.2 − \frac{1}{6})(−3) = 0.4.

Answers to Selected Problems in Chapter 10. 1: (a) x = 2 + 3t, y = 3 − 2t; (b) x = 1 + 6t, y = 2 + 2.5t; (c) x = 5\sqrt{3} t, y = 5t. 2: (a) 25 rad/sec; (b) \frac{25}{2π} rev/sec; (c) 4π m/sec; (d) 21.6 km/hr. 3: (a) 34.3 mph; (b) 55 rad/sec, 8.75 rev/sec. 4: (a) 20t; (b) 15π t; (c) 7.5t^{2}. 5: (a) x = 3 cos 0.5t, y = 3 sin 0.5t; (b) x = 3 sin 4πt, y = 3 cos 4πt; (c) x = −3 sin 0.4πt, y = −3 cos 0.4πt;

(d) x = 3 sin 20πt, y = −3 cos 20πt. 6: (a) x = 10t, y = −4.9t^{2} + 24.5t + 1; (b) 2.5 sec; (c) 5.04 sec. 7: (a) x = 49.5t, y = −16t^{2} − 49.5t + 163; (b) (−32t − 49.5)/49.5; (c) 52.9^{∘}, 58.7^{∘}; (d) 124 ft/sec at 66.4^{∘} below horizontal (hits ground at t = 2). 8: (a) x = (v cos α)t, y = \frac{1}{2}gt^{2} + (v sin α)t; (b) 2v sin α/(−g); (c) 2v^{2} sin α cos α/(−g).

Answers to Selected Problems in Chapter 11. 1: (b) sin θ/(1 − cos θ). 2: (a) x = 2t; (b) y = 3 sin(4π(t − \frac{3}{8})); (c) y = 3 sin(2π(x − \frac{3}{4})). 3: y = 3 sin(4π(\sqrt{x} − \frac{3}{8})). 4: x = 2 sin(πt), y = 100 − 16t^{2} + 2 cos(πt). 5: x = 6t − 0.5 sin(8t), y = 0.75 − 0.5 cos(8t). 6: x = 10t + 0.25 sin(40t), y = 0.25 − 4.9t^{2} + 0.25 cos(40t).

Answers to Selected Problems in Chapter 12. 1: 0.0199 cm/sec. 2: 0.25 m/sec. 3: 80π mi/min (=4.2 mi/sec). 4: 6.71 ft/sec. 5: 2.12 cm/sec. 6: 0.65 ft/sec 7: 7.91 m/sec 8: 1.172 m/min 9: 0.5454 m/min 10: 6 ft/sec, 2.5 ft/sec 11: 69.4 mph 12: 0.1234 ft^{3}/sec. 13: (a) (aȧ + bḃ + 0.5aḃ + 0.5bȧ)/\sqrt{a^{2} + b^{2} + ab}; (b) (aȧ + bḃ − aḃ cos θ − bȧ cos θ + abθ̇ sin θ)/\sqrt{a^{2} + b^{2} − 2ab cos θ}. 14: 81.6 m/sec 15: (a) y = a cos(θ) − a sin(θ) cot(β + θ) (or else, using the law of sines, an equivalent answer is: y = a sin β/ sin(θ + β)); (b) dy/dt = (−a sin(θ)(1 − csc^{2}(β + θ)) − a cos(θ) cot(β + θ))dθ/dt (or

else: dy/dt = −a(sin β cos(θ + β)/ sin^{2}(θ + β))dθ/dt); (c) 3.79 cm/sec. 16: (a) a(dθ/dt); (b) closing at a little over 15 rad/sec; (c) no violation of relativity, because no part of the scissors is moving faster than 300 km/sec. 17: (a) -50 m/sec; (b) 68 m/sec; (c)... 18: (a) (aȧ + bḃ)/\sqrt{a^{2} + b^{2} + h^{2}}; (b) (aȧ + bḃ + hḣ)/\sqrt{a^{2} + b^{2} + h^{2}}.

Answers to Selected Problems in Chapter 13. 1: Both f′(x) = 0 and f′′(x) = 0 when x = 0. If n is even, the origin is a minimum, and the curve is always concave up. If n is odd, the origin is a point of inflection, and the curve is concave up for x > 0 and concave down for x < 0. 2: (a) min at (0.5, −0.25), always concave up, no inflection;

(b) max at (1,4), min at (−1, 0), inflection at (0, 2), conc up for x < 0, conc down for x > 0;

(c) max at (2,20), min at (4,16), inflection at (3,18), conc up for x > 3, conc down for x < 3;

(d) max at (0,3), min at (±1, 2), inflections at (±0.577, 2.444), conc down for −0.577 < x < 0.577, otherwise conc up, symmetric with respect to y-axis;

(e) min at (1, −1), inflections at (0,0) and (2/3, −0.5926), conc down for 0 < x < 2/3, otherwise conc up;

(f) no max, min or inflections, y-axis is asymptote, conc up for x < 0, conc down for x > 0, the line y = x is a slanted asymptote;

(g) no max or min, inflections at (±1, 2), y-axis is asymptote, conc up for x < −1 and x > 1, otherwise conc down, symmetric with respect to y-axis;

(h) min when x = 7π/12 (and x = 7π/12±any multiple of π), max when x = 11π/12 (again can add or subtract any multiple of π), inflections when x = π/4, 3π/4 (± any multiple of π), conc up for π/4 < x < 3π/4, general appearance is oscillation around the slanted line y = −x;

(i) max at all points (\frac{π}{4} ± 2kπ, \sqrt{2}) (where k is any integer), min at (\frac{5π}{4} ± 2kπ, −\sqrt{2}), inflections at (\frac{3π}{4} ± kπ, 0), you get a sinusoidal function;

(j) asymptotes at odd multiples of π, max when x = −\frac{π}{2} ± 2kπ, min when x = \frac{π}{2} ± 2kπ, inflections at even multiples of π, in each section between asymptotes the left half is conc down, the right half is conc up. 5: (a) max at (0,0), min at (±\sqrt{10/3}, −500/27); (b) x > \sqrt{2} and x < −\sqrt{2}; (c) (±\sqrt{2}, −12); (d) use 2nd derivative test.

Answers to Selected Problems in Chapter 14. 1: (a) max value f(2) = 5; (b) min value f(0) = 1. 2: P/4 by P/4. 3: \sqrt[3]{2V} by \sqrt[3]{2V} by \frac{1}{2} \sqrt[3]{2V}; ratio=1/2. 4: l^{2}/8. 5: x = 2000, P = $6, profit=$5000. 6: r\sqrt{2} by r\sqrt{2}/2, area=r^{2}. 7: h/r = 2. 8: h/r = 8/π. 9: 4/27. 10: Take x = a, where f(x) has a corner (if w > v, then f′(x) is never zero). 11: h/r = 2; h/r = 3.5. 12: (a) a/6 on a side;

(b) (a + b − \sqrt{a^{2} + b^{2} − ab})/6 on a side. 13: 0.5 − \frac{π}{4} k provided that k < 2/π; otherwise, the optimal

window consists of only the semicircular piece. 14: θ = 30^{∘}, max area= l^{2}\sqrt{3}/12 ≈ 0.144 l^{2}, bigger than in Problem 3. 15: 0.00139 hours (≈ 5 sec), misses by 0.5726 mi. 16: ratio=a/b. 17: \frac{2}{3}r\sqrt{3} by \frac{2}{3}r\sqrt{6}. 18: 1/\sqrt{3} (≈58%). 19: 18 in × 18 in × 36 in. 20: r = 2.71, h = 10.84. 21: h/r = \sqrt{2}. 22: h = 1.1798R, r = 0.9837R, 28.54% 23: c/(1+\sqrt[3]{b/a}) from A. 24: 1/2. 25: x = 2500, P = $5, profit=$7000. 26: Snell’s law: \frac{sin θ_{1}}{v_{1}} = \frac{sin θ_{2}}{v_{2}} .

Answers to Selected Problems in Chapter 15. 1: x_{1} = 1.4765432, x_{2} = 1.4757740, x_{3} = 1.4757732. 2: x_{1} = 2.25 (exactly), x_{3} = x_{2} = 2.19 to three significant figures (x_{2} = 2.19369 …, x_{3} = 2.19192 …). 3: θ_{1} = 1.9359512, θ_{2} = 1.9345639, θ_{3} = 1.9345632. 4: Hint: Set the slope of the line joining (−π/2, −1) to (x, sin x) equal to the derivative of sin x, then multiply through by the denominator to get the f(x) in Newton’s method: f(x) = (\frac{π}{2} + x) cos x − sin x − 1; then x_{0} = 0.7, x_{1} = 0.7633, x_{3} = x_{2} = 0.7603. 5: t = 0.5062 sec, height=6.9938 ft. 6: take t_{0} = 0.7, then t_{1} = 0.68674, t_{2} = 0.68726. 7: x = 40 sin(πt/10)−10 sin(πt/3), y = 50−40 cos(πt/10)−10 cos(πt/3), 9.2655 sec, 98.56 ft. 8: l = (9/ sin(\frac{π}{3} − θ)) + (7/ sin θ), 0.49254 rad (= 28.2^{∘}), 31.89 ft

Answers to Selected Problems in Chapter 16. 1: (a) −\frac{1}{3x^{3}} + \frac{2}{5}x^{5/2}+5x+C, (b) −0.1(1−t)^{10}+ C, (c) \frac{1}{5}x^{5} + \frac{2}{3}x^{3} + x + C, (d) −0.3(1 − 5t)^{2/3} + C,

(e) \frac{1}{4} sin^{4} x + C, (f) −\frac{1}{3}(100 − x^{2})^{3/2} + C, (g) −\frac{2}{3} \sqrt{1 − x^{3}} + C, (h) −\frac{B}{2π} cos (\frac{2π}{B} (t − C)) + Dt + C′, (i)

−\frac{20}{3} (1 − sin \frac{t}{10})^{3/2} +C, (j) \frac{1}{π} sin(sin(πt))+C. 2: −\frac{1}{π} cot(πt)+C. 3: −\frac{1}{2}g(t−t_{0})^{2} +v_{0}(t−t_{0})+s_{0}. 4: \frac{1}{3}t−\frac{1}{2}+1/(6t^{2}). 5: (a) \frac{1}{6}kt^{3}+v_{0}t+s_{0}, (b) \frac{1}{15}(2t+1)^{5/2}+(v_{0}−\frac{1}{3})t+s_{0}−\frac{1}{15}, (c) −\frac{1}{25} sin(5t)+(v_{0}+\frac{1}{5})t+s_{0},

(d) 50 tan(0.1t) + (v_{0} − 5)t + s_{0}.

Answers to Selected Problems in Chapter 17. 1: (a) 6.25 vs. 16/3=5.333, (b) 5.8125 vs. 5.333,

(c) 4.146 vs. 5.333, (d) 1.239 vs. 1, (e) 1.125 vs. 1. 2: (a) 55, (b) 25/12, (c) \frac{3}{2} + \frac{\sqrt{3}}{2}. 3: (a) 28.5,

(b) 14/9, (c) 2/3, (d) 0.24, (e) 2/π. 4: (a) 2/3, (b) 2 − \sqrt{2}, (c) \frac{4}{π} + \frac{4}{3}, (d) π − 2. 5: (a) 13/3, (b) 56/15, (c) 0, (d) 1/4π, (e) 3/4, (f) 9. 6: (a) 14600 degree-days, (b) 4523 degree-days, (c) 277 degree-

days. 7: (a) a(t) = \begin{cases}2t, & 0 ≤ t ≤ 2 \\ 4, & 2 ≤ t ≤ 4 \\ 8 − t, & 4 ≤ t ≤ 8 \end{cases}; v(t) = \begin{cases}t^{2}, & 0 ≤ t ≤ 2 \\ 4t − 4, & 2 ≤ t ≤ 4 \\ 8t − \frac{1}{2}t^{2} − 12, & 4 ≤ t ≤ 8 \end{cases}; s(8) − s(0) = 88. (b)

a(t) = \begin{cases}sin t, & 0 ≤ t ≤ 0.5110 \\ 1 − t, & 0.5110 ≤ t ≤ 1 \end{cases}; v(t) = \begin{cases}1 − cos t, & 0 ≤ t ≤ 0.5110 \\ t − \frac{1}{2}t^{2} − 0.2527, & 0.5110 ≤ t ≤ 1 \end{cases}; s(1) − s(0) = 0.1234.

Answers to Selected Problems in Chapter 18. 1: (a) sin^{2} u, (b) \sqrt{1 + u + u^{2}}, (c) −\sqrt{1 + sin^{2} u},

(d) 2 sin^{2} u; (e) 0. 2: (a) b/(a + bt), (b) (−gt + v_{0}) sin^{2} (−\frac{1}{2}gt^{2} + v_{0}t + s_{0}), (c) 3t^{2} cos(t^{6}) − 2t cos(t^{4}),

(d) Aα cos(α(t−C))\sqrt{1 + (A sin(α(t − C)))^{2}} −b\sqrt{1 + (a + bt)^{2}}. 3: (a) ∫_{0}^{2t} (20 + 10 cos^{2}(0.1πx)) dx,

(b) 2 (20 + 10 cos^{2}(0.2πt)). 4: (a) ∫_{0}^{2t} \frac{40 dx}{1+(0.01x−1)^{2}} , (b) \frac{80}{1+(0.02t−1)^{2}} . 5: (a) 2 ∫_{48−16t^{2}}^{48} 48 cos \frac{πy}{96} dy, (b)

3072t cos (\frac{π}{6} (3 − t^{2})). 6: (a) 2 ∫_{−1}^{u} \sqrt{1 − x^{2}} dx, (b) u(t) = sin(π(t−\frac{1}{2})), (c) 2π cos(π(t−\frac{1}{2}))\sqrt{1 − sin^{2}(π(t − \frac{1}{2}))}.

2. Answers to Practice Midterms

Answers to Practice First Midterm #1

1. (a) 0 ≤ x ≤ 2; (b) −2 ≤ x ≤ 0; (c) all x except 2 and 4; (d) −1 < x < 1.

2. (a) ABIJ; (b) DEFG; (c) GHI; (d) FJ; (e) J; (f) F.

3. t = f(s) = \frac{130}{4s} − \frac{1}{2}s, f(5\frac{1}{12}) ≈ f(5) + \frac{1}{12}f′(5) = 4 + \frac{1}{12}(−1.8) = 3.85 ft. 4. 9\frac{7}{81} = 9.0864 cm. Answers to Practice First Midterm #2

1. (a) like the x^{2}-parabola, but shifted to have its vertex at (−2, −4); (b) bottom half of ellipse, passes through ±1 on the x-axis and −5 on the y-axis. 2. −π/2 < x < π/2.

3. (a) CDJ; (b) AFGH; (c) ABHI; (d) CGJ; (e) C; (f) G. 4. (a) v_{0} = 9.5 m/sec; (b) v_{0} = f(t) = 4.9t − \frac{75}{t}, f(5 ± 0.1) ≈ 9.5 ± 0.1 · 7.9 = 9.5 ± 0.79 m/sec.

5. y = f(x) = 1000/x, f(10.77) ≈ f(10) + (0.77)f′(10) = 100 + (0.77)(−10) = 92.3 m^{2}.

Answers to Practice First Midterm #3

1. 0.5 < x ≤ 2. 2. (a) AFG; (b) CDI; (c) DE; (d) CFI; (e) C.

3. g = f(t) = \frac{22.5}{t^{2}}; (a) f(3) = 2.50; (b) f(3.1) ≈ f(3) + 0.1 · f′(3) = 2.33, so g is between 2.33 and 2.50 m/sec^{2}. 4. y = f(x) = \sqrt{36^{2} − x^{2}}, f(21.6 ± 0.2) ≈ 28.8 in ± 0.15 in.

Answers to Practice First Midterm #4

1. (a) −35 < x < 15; (b) −7 ≤ x ≤ 7. 2. (b) when t = 12 (parachute opens); when t = 0 (drop from plane); right after parachute opens. 3. 12.8. 4. (a) A = \frac{π}{4} r^{2} + rs; (b) 100π + 600;

(c) s = f(r) = \frac{100π+600}{r} − \frac{π}{4} r, f(21) ≈ f(20) + 1 · f′(20) = 26.93 cm.

Answers to Practice Second Midterm #1

1. 19.861. 2. (a) (−1, −1.5) is max; (b) x < −1 and x > 0; (c) y-axis; (d) none; (e) see bottom of page. 3. (b) y = 20 sin (\frac{2π}{12} (t − 9.5)) + 55; (c) mid-March to mid-November. 4. b, C, cos C are all constants, a and c are variables, take d/dt of everything, you know da/dt = −5.70, you get dc/dt = −5.55 m/sec.

Answers to Practice Second Midterm #2

1. (a) (0, 0), (\sqrt{3}, 1.5), (−\sqrt{3}, 1.5); (b) −∞ < x < −\sqrt{3}, and 0 < x < \sqrt{3}; (c) (1, 5/6), (−1, 5/6); (d) −1 < x < 1; (e) see below.

2. (a) 5π rad/sec; (b) ẋ = 5(1 − cos θ), ẏ = 5 sin θ; (c) ẋ = 2.50 m/sec, ẏ = −4.33 m/sec, dy/dx = −1.73. 3. (a) x = \sqrt{100 − 4y^{2}}; (b) A = π(100 − 4y^{2}); (c) 0.63 m^{2}/sec.

4. (a) \sqrt{(x + 5)^{2} + y^{2}} + \sqrt{x^{2} + (y + 5)^{2}} = 20; (b) y′ = 1/5.

Answers to Practice Second Midterm #3 1. (a) (0, 0); (b) (1, 2.75) and (3, 6.75); (c) see bottom of page; (d) 1 < x < 3.

2. C = f(c) (a, b are const), ±0.6175^{∘}. 3. y = 5 cot θ, 0.0436 ft/min.

4. (a) y = 19.6 − 4.9t^{2}; (b) x = 2 sin(π(t − \frac{1}{2})) + 2; (c) ẋ = −4.443 m/sec, ẏ = −12.25 m/sec, dy/dx = 2.757.

Answers to Practice Second Midterm #4

1. (a) max: (0, 0), min: (±\sqrt{6}, −4.5); (b) −\sqrt{6} < x < 0 and \sqrt{6} < x < ∞; (c) (±\sqrt{3}, −2.925);

(d) −\sqrt{3} < x < \sqrt{3}; (e) see bottom of page. 2. πR^{2}T = 1 m^{3}; take d/dt of both sides; 0.804 m/hr.

3. B = f(A) (a, b are const), 61.732^{∘}. 4. (a) V = \frac{1000π}{3} tan^{2}(\frac{1}{2}θ); (b) 349 ft^{3} ± 14 ft^{3}.

3. Answers to Practice Final Exams

Answers to Practice Final Exam #1

1: (a) -15/4; (b) \frac{5}{π} sin^{2} (\frac{π}{10}(t − 1)) + C, or else −\frac{5}{π} cos^{2} (\frac{π}{10}(t − 1)) + C. 2: -78.125 m/sec. 3: (b) v = \frac{1}{2}t^{2}, 0 ≤ t ≤ 1; v = t − \frac{1}{2}, 1 ≤ t ≤ 2; v = 3/2, 2 ≤ t ≤ 3; (c) 8/3 m. 4: area of hexagon = 2\sqrt{3}r^{2}, optimal ratio h/r is 4\sqrt{3}/π = 2.2053 · · ·.

5: (a) 4.60555 cm; (b) x = 2 cos(7πt) + \sqrt{16 − 4 sin^{2}(7πt)};

(c) dx/dt = −14π sin(7πt) − 28π sin(7πt) cos(7πt)(16 − 4 sin^{2}(7πt))^{−1/2};

(d) f′(1/21) = −48.654, f(1/20) ≈ 4.4897 cm. 6: (a) x = 5t − 10 sin(t/4); (b) 8.18361 cm; (c) t_{1} = 3.57285 sec, t_{2} = 3.55148 sec, x(t_{2}) = 10.00011 cm.

Answers to Practice Final Exam #2

1: (a) \frac{1}{4}(1 − 2t)^{−2} + C; (b) 2/9. 2: (a) 20 + \frac{10}{π} rev; (b) 40π + 20 rad;

(c) 109.25 ft. 3. 16 ft/sec. 4: 2.033 ft. 5: s = 1.59 ft, l = 4.76 ft. 6: (b) 2x + 2 cos(2x) = 0; (c) (π/12, π^{2}/144 + 1/2) and (5π/12, 25π^{2}/144 + 1/2); (d) −π/4; (e) x_{1} = −0.5236, x_{2} = −0.5150; (f) no other max/min; (g) graph at right: 7: (a) π/6 rad/sec; (b) x = 20 cos(πt/6), y = 20 + 20 sin(πt/6);

(c) v_{0,horiz} = −30 ft/sec, v_{0,vert} = 52 ft/sec;

(d) x = 75 − 30t, y = −16t^{2} + 52t; (e) 5.6737 ft;

(f) dist= \sqrt{(75 − 30t − 20 cos(πt/6))^{2} + (−16t^{2} + 52t − 20 − 20 sin(πt/6))^{2}}

(g) let f(t) = \frac{d}{dt}(dist^{2}) = 2(75 − 30t − 20 cos(πt/6))(−30 + (20π/6) sin(πt/6))+ +2(−16t^{2} + 52t − 20 − 20 sin(πt/6))(−32t + 52 − (20π/6) cos(πt/6)).

[figure: min; inflec. pts.]

Answers to Practice Final Exam #3 1: (a) −\frac{2}{7} \sqrt{1 − 7z} + C; (b) (4 − \sqrt{2})/3 = 0.8619. 2: (b) 2 − t for 0 ≤ t ≤ 1, 0.5 + 0.5(2 − t)^{2} for 1 ≤ t ≤ 2, 0.5 for 2 ≤ t ≤ 3, 2 − 0.5t for 3 ≤ t ≤ 4;

(c) 2.92 m. 3: −9.823 ft/sec. 4: differentiate x + y + \sqrt{x^{2} + y^{2}} =const, get y′ = −32/49, answer is 24.013. 5: (a) vol= \frac{1}{3}πr^{3}x, area= πr^{2}\sqrt{1 + x^{2}}; (b) r = \sqrt{C/π}(1 + x^{2})^{−1/4}. (c) x = \sqrt{2}.

6: Choose f(x) = x^{1/2} + x^{1/3} − x in Newton’s method, so that f′(x) = \frac{1}{2}x^{−1/2} + \frac{1}{3}x^{−2/3} − 1; then x_{0} = 3, x_{1} = 3.3163, x_{2} = 3.3094. 7: (a) x = \frac{1}{2} sin(2π(t − \frac{1}{4})) + \frac{1}{2}; (b) (4 − 4.9t^{2})^{2} + (\frac{1}{2} sin(2π(t − \frac{1}{4})) + \frac{1}{2})^{2}; (c) f(t) = −19.6t(4 − 4.9t^{2}) + (sin(2π(t − \frac{1}{4})) + 1)π cos(2π(t − \frac{1}{4})).

Answers to Practice Final Exam #4

1: (a) 28/15; (b) 3B/16π. 2: (a) 3 + 3t − \frac{3}{2}t^{2}, 0 ≤ t ≤ 2; 9 − 3t, 2 ≤ t ≤ 3; (b) 9.5 m. 3: (a) Vol= 50 · 4 cos θ(14 + 4 sin θ); (b) d(Vol)/dθ = −400(4X^{2} + 7X − 2) (where X = sin θ); (c) 2904 in^{3}. 4: A good initial guess is s_{0} = 10 (since vol=1000, surface area=600). Choose f(s) = s^{3} +6s^{2} −1500; then s_{1} = 9.762, s_{3} = s_{2} = 9.757. 5: (a) x_{A} = cos(2πt/9), y_{A} = sin(2πt/9), x_{B} = 2 − t, y_{B} = t;

(b) 2(2 − t − cos(2πt/9))(−1 + (2π/9) sin(2πt/9)) + 2(t − sin(2πt/9))(1 − (2π/9) cos(2πt/9)). 6: (a) x = (10 cos α)t, y = −4.9t^{2} + (10 sin α)t + 50; (b) T = 1.0204(sin α + \sqrt{sin^{2} α + 9.8}); (c) X = (10 cos α)T (write T in terms of α using (b)); (d) find \frac{π}{180} · \frac{dX}{dα} and then take α = 60^{∘}: −0.5781 (the minus sign means that a positive error in α leads to a negative error in X).