# 1 3 0 is on the line so d 21 23 20 3 4

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Unformatted text preview: ) horizontal tangent line when dy/dx = 0, or dy/dt = 0, so cos t = 0, t = π/2, 3π/2; 1 1 vertical tangent line when dx/dt = 0, so − csc2 t − 4 sin t = 0, t = π +sin−1 √ , 2π − sin−1 √ , 3 3 4 4 t = 3.823, 5.602 Chapter 12 Supplementary Exercises 442 (d) r2 = x2 + y 2 = (cot t + 4 cos t)2 + (1 + 4 sin t)2 = (4 + csc t)2 , r = 4 + csc t; with t = θ, f (θ) = 4 + csc θ; m = dy/dx = (f (θ) cos θ + f (θ) sin θ)/(−f (θ) sin θ + f (θ) cos θ); when √ √ θ = π + sin−1 (1/4), m = 15/15, when θ = 2π − sin−1 (1/4), m = − 15/15, so the tangent √ lines to the conchoid at the pole have polar equations θ = ± 15/15. 15. A = 2 π /6 0 π /4 1 (2 sin θ)2 dθ + 2 π/6 1 12 1 , dθ = 2 θ − sin 2θ 2 4 π /6 0 √ 5π π 3 − + ,A = 24 12 2 16. The circle has radius a/2 and lies entirely inside the cardioid, so 2π A= 0 a2 5a2 3a2 12 a (1 + sin θ)2 dθ − πa2 /4 = π− π= π 2 2 4 4 π /2 17. (a) r = 1/θ, dr/dθ = −1/θ2 , r2 + (dr/dθ)2 = 1/θ2 + 1/θ4 , L = π/4 1 θ2 1 + θ2 dθ ≈ 0.9457 by Endpaper Table Formula 93. +∞ 1 θ2 1 arc length is inﬁnite. (b) The integral 1 + θ2 dθ diverges by the comparison test (with 1/θ), and thus the 18. (a) When the point of departure of the thread from the circle has traversed an angle θ, the amount of thread that has been unwound is equal to the arc length traversed by the point of departure, namely aθ. The point of departure is then located at (a cos θ, a sin θ), and the tip of the string, located at (x, y ), satisﬁes the equations x − a cos θ = aθ sin θ, y − a sin θ = −aθ cos θ; hence x = a(cos θ + θ sin θ), y = a(sin θ − θ cos θ). (b) Assume for simplicity that a = 1. Then dx/dθ = θ cos θ, dy/dθ = θ sin θ; dx/dθ = 0 has solutions θ = 0, π/2, 3π/2; and dy/dθ = 0 has solutions θ = 0, π, 2π . At θ = π/2, dy/dθ > 0, so the direction is North; at θ = π, dx/dθ < 0, so West; at θ = 3π/2, dy/dθ < 0, so South; at θ = 2π, dx/dθ > 0, so East. Finally, lim+ dy/dx = lim+ tan θ = 0, so East. θ →0 (c) u x y 0 1 0 p/ 2 p/ 2 1 p –1 p 3p/ 2 –3p/ 2 –1 θ →0 2p 1 –2p Note that the parameter θ in these equations does not satisfy equations (1) and (2) of Section 12.1, since it measures the angle of the point of departure and not the angle of the tip of the thread. 4 -5 2 -8 443 Chapter 12 √ a2 +b2 π b2 x2 /a2 − b2 dx 19. (a) V = y a = 2 πb2 2 (b − 2a2 ) a2 + b2 + ab2 π 3a2 3 x √ a2 +b2 x b2 x2 /a2 − b2 dx = (2b4 /3a)π (b) V = 2π y a x p/ 2 20. (a) (b) θ = π/2, 3π/2, r = 1 5 -5 5 0 -5 r cos θ + (dr/dθ) sin θ ; at θ = π/2, m1 = (−1)/(−1) = 1, m2 = 1/(−1) = −1, −r sin θ + (dr/dθ) cos θ m1 m2 = −1; and at θ = 3π/2, m1 = −1, m2 = 1, m1 m2 = −1 (c) dy/dx = 22. The tips are located at r = 1, θ = π/6, 5π/6, 3π/2 and, for example, √ d = 1 + 1 − 2 cos(5π/6 − π/6) = 2(1 − cos(2π/3)) = 3 23. (a) x = r cos θ = cos θ + cos2 θ, dx/dθ = − sin θ − 2 sin θ cos θ = − sin θ(1 + 2 cos θ) = 0 if sin θ = 0 or cos θ = −1/2, so θ = 0, π, 2π/3, 4π/3; maximum x = 2 at θ = 0, minimum x = −1/4 at θ=π (b) y = r sin θ = sin θ + sin θ cos θ, dy/dθ = 2 cos2 θ + cos θ − 1 = 0 at cos θ = 1/2, −1, so √ √ θ = π/3, 5π/3, π ; maximum y = 3 3/4 at θ = π/3, minimum y = −3 3/4 at θ = 5π/3 √ 2θ cos θ − sin θ = 0 if 2θ cos θ = sin θ, tan θ = 2θ which 24. (a) y = r sin θ = (sin θ)/ θ, dy/dθ = 2θ3/2 only happens once on (0, π ]. Since lim y = 0 and y = 0 at θ = π , y has a maximum when tan θ = 2θ. (b) θ ≈ 1.16556 √ (c) ymax = (sin θ)/ θ ≈ 0.85124 θ →0 + Chapter 12 Supplementary Exercises 444 25. The width is twice the maximum value of y for 0 ≤ θ ≤ π/4: = y = r sin θ = sin θ cos 2θ = sin θ − 2 √ 3 θ, dy/dθ = cos θ − 6 sin2 θ cos θ √ 0 when cos θ = 0 or sin √ √ √ sin θ = 1/ 6, y = 1/ 6 − 2/(6 6) = 6/9, so the width of the petal is 2 6/9. 26. (a) h/2 y2 x2 − = 1, so V = 2 225 1521 y2 1521 225π 1 + 0 h/2 25 πh3 + 225πh ft3 . 2028 h/2 2πx 1 + (dx/dy )2 dy = 4π (b) S = 2 dy = 225 + 1521 225 + y 2 0 0 5 πh = 26 h2 6084 + h+ + 1170π ln √ 6084 + h2 78 225 1521 2 dy ft2 27. (a) The end of the inner arm traces out the circle x1 = cos t, y1 = sin t. Relative to the end of the inner arm, the outer arm traces out the circle x2 = cos 2t, y2 = − sin 2t. Add to get the motion of the center of the rider cage relative to the center of the inner arm: x = cos t + cos 2t, y = sin t − sin 2t. (b) Same as part (a), except x2 = cos 2t, y2 = sin 2t, so x = cos t + cos 2t, y = sin t + sin 2t 2π dx dt (c) L1 = 0 2π L2 = √ 2 + 2 1/2 dy dt 2π dt = √ 5 − 4 cos 3t dt ≈ 13.36489321, 0 5 + 4 cos t dt ≈ 13.36489322; L1 and L2 appear to be equal, and indeed, with the 0 substitution u = 3t − π and the periodicity of cos u, L1 = 1 3 π /2 5π −π dx dt 29. C = 4 0 2π 5 − 4 cos(u + π ) du = √ 5 + 4 cos u du = L2 . 0 2 dy dt + 2 1/2 π /2 (a2 sin2 t + b2 cos2 t)1/2 dt dt = 4 0 π /2 π /2 (a2 sin2 t...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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