# R distance between 000 and 3 2 4 1 2 x2

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Unformatted text preview: nates (1/2, 2π/3). 67. (a) Let (x1 , y1 ) and (x2 , y2 ) be the rectangular coordinates of the points (r1 , θ1 ) and (r2 , θ2 ) then d= = (x2 − x1 )2 + (y2 − y1 )2 = (r2 cos θ2 − r1 cos θ1 )2 + (r2 sin θ2 − r1 sin θ1 )2 2 2 r1 + r2 − 2r1 r2 (cos θ1 cos θ2 + sin θ1 sin θ2 ) = 2 2 r1 + r2 − 2r1 r2 cos(θ1 − θ2 ). (b) Let P and Q have polar coordinates (r1 , θ1 ), (r2 , θ2 ), respectively, then the perpendicular from OQ to OP has length h = r2 sin(θ2 − θ1 ) and A = 1 hr1 = 1 r1 r2 sin(θ2 − θ1 ). 2 2 413 Chapter 12 √ 13 − 6 3 ≈ 1.615 (c) From Part (a), d = 9 + 4 − 2 · 3 · 2 cos(π/6 − π/3) = 1 (d) A = 2 sin(5π/6 − π/3) = 1 2 68. (a) 0 = (x2 + y 2 + a2 )2 − a4 − 4a2 x2 = x4 + y 4 + a4 + 2x2 y 2 + 2x2 a2 + 2y 2 a2 − a4 − 4a2 x2 = x4 + y 4 + 2x2 y 2 − 2x2 a2 + 2y 2 a2 = (x2 + y 2 )2 + 2a2 (y 2 − x2 ) = r4 + 2a2 r2 (sin2 θ − cos2 θ) = r4 − 2a2 r2 cos 2θ, so r2 = 2a2 cos 2θ (b) (x2 + a2 + y 2 )2 − 4x2 a2 = a4 ; (x + a)2 + y 2 69. 70. (x − a)2 + y 2 (x + a)2 + y 2 = a4 ; (x − a) + y 2 = a2 lim y = lim r sin θ = lim sin θ =1 θ lim y = lim± r sin θ = lim± sin θ sin θ 1 1 lim = 1 · lim , so lim y does not exist. = lim θ2 θ →0 ± θ θ →0 ± θ θ →0 ± θ θ →0 ± θ →0+ θ →0+ θ →0± θ →0+ θ →0 θ →0 71. Note that r → ±∞ as θ approaches odd multiples of π/2; x = r cos θ = 4 tan θ cos θ = 4 sin θ, y = r sin θ = 4 tan θ sin θ so x → ±4 and y → ±∞ as θ approaches odd multiples of π/2. 72. lim θ →(π/2)− x= lim θ →(π/2)− r cos θ = lim θ →(π/2)− r u -4 4 2 sin2 θ = 2, x = 2 is a vertical asymptote. 73. Let r = a sin nθ (the proof for r = a cos nθ is similar). If θ starts at 0, then θ would have to increase by some positive integer multiple of π radians in order to reach the starting point and begin to retrace the curve. Let (r, θ) be the coordinates of a point P on the curve for 0 ≤ θ < 2π . Now a sin n(θ + 2π ) = a sin(nθ + 2πn) = a sin nθ = r so P is reached again with coordinates (r, θ + 2π ) thus the curve is traced out either exactly once or exactly twice for 0 ≤ θ < 2π . If for 0 ≤ θ < π , P (r, θ) is reached again with coordinates (−r, θ + π ) then the curve is traced out exactly once for 0 ≤ θ < π , otherwise exactly once for 0 ≤ θ < 2π . But a sin nθ, −a sin nθ, a sin n(θ + π ) = a sin(nθ + nπ ) = n even n odd so the curve is traced out exactly once for 0 ≤ θ < 2π if n is even, and exactly once for 0 ≤ θ < π if n is odd. EXERCISE SET 12.2 1. (a) dy/dx = 1/2 = 1/(4t); dy/dx 2t t=−1 (b) x = (2y )2 + 1, dx/dy = 8y, dy/dx = −1/4; dy/dx y =±(1/2) t=1 = 1/4 = ±1/4 2. (a) dy/dx = (4 cos t)/(−3 sin t) = −(4/3) cot t; dy/dx t=π/4 = −4/3, dy/dx (b) (x/3)2 + (y/4)2 = 1, 2x/9 + (2y/16)(dy/dx) = 0, dy/dx = −16x/9y, dy/dx √ x=3/ 2 √ y =4/ 2 = −4/3; dy/dx √ x=3/ 2 √ y =−4/ 2 = 4/3 t=7π/4 = 4/3 Exercise Set 12.2 3. 4. 414 d dy d2 y d dy = = 2 dx dx dx dt dx negative when t = 1 d2 y d = 2 dx dt dy dx 1 dt = − 2 (1/2t) = −1/(8t3 ); positive when t = −1, dx 4t dt −(4/3)(− csc2 t) 4 = = − csc3 t; negative at t = π/4, positive at t = 7π/4. dx −3 sin t 9 5. dy/dx = √ √ 2/ t 2 √ = 4 t, d2 y/dx2 = √ = 4, dy/dx 1/(2 t) 1/(2 t) 6. dy/dx = 1 t2 = t, d2 y/dx2 = , dy/dx t t t=2 = 2, d2 y/dx2 t=1 t=2 = 4, d2 y/dx2 t=1 =4 = 1/2 sec2 t − csc t cot t = csc t, d2 y/dx2 = = − cot3 t, sec t tan t sec t tan t √ √ dy/dx t=π/3 = 2/ 3, d2 y/dx2 t=π/3 = −1/(3 3) 7. dy/dx = 8. dy/dx = 9. dy dy/dθ − cos θ d2 y d = = ; = 2 dx dx/dθ 2 − sin θ dx dθ dy dx 10. d2 y sinh t = tanh t, = sech2 t/ cosh t = sech3 t, dy/dx cosh t dx2 = θ =π/3 −1/2 −1 d2 y √ √; = 2 − 3/2 4 − 3 dx2 dy dx / = 0, d2 y/dx2 t=0 =1 dx 2 1 1 = = ; 2 2 − sin θ dθ (2 − sin θ) (2 − sin θ)3 = θ =π/3 t=0 (2 − 1 √ 3/2)3 = (4 − 8 √ 3)3 3 cos φ d2 y dφ dy d = = −3 cot φ; (−3 cot φ) = −3(− csc2 φ)(− csc φ) = −3 csc3 φ; = 2 dx − sin φ dx dφ dx dy dx φ=5π/6 √ d2 y = 3 3; dx2 = −24 φ=5π/6 −e−t = −e−2t ; for t = 1, dy/dx = −e−2 , (x, y ) = (e, e−1 ); y − e−1 = −e−2 (x − e), et y = −e−2 x + 2e−1 11. (a) dy/dx = (b) y = 1/x, dy/dx = −1/x2 , m = −1/e2 , y − e−1 = − 1 1 2 (x − e), y = − 2 x + e2 e e 12. dy/dx = 16t − 2 = 8t − 1; for t = 1, dy/dx = 7, (x, y ) = (6, 10); y − 10 = 7(x − 6), y = 7x − 32 2 13. dy/dx = 4 cos t = −2 cot t −2 sin t (a) dy/dx = 0 if cot t = 0, t = π/2 + nπ for n = 0, ±1, · · · 1 (b) dx/dy = − tan t = 0 if tan t = 0, t = nπ for n = 0, ±1, · · · 2 14. dy/dx = 2t + 1 2t + 1 = 6t2 − 30t + 24 6(t − 1)(t − 4) (a) dy/dx = 0 if t = −1/2 (b) dx/dy = 6(t − 1)(t − 4) = 0 if t = 1, 4 2t + 1 415 Chapter 12 15. x = y = 0 when t = 0, π ; lines are y = −2x, y = 2x. 2 cos 2t dy dy = ; dx cos t dx = 2, t=0 dy dx = −2, the equations of the tangent t=π 16. y (t) = 0 has thre...
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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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