The maximum volume is v 131313 127 563 chapter

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Unformatted text preview: 2 (cos ωti +sin ωtj) = − r R 2 (c) F = ma = 70 2π . ω 13 6.5 = , 9600 19,200 cos 13t 19,200 2 kg · km/s2 ≈ 10.77 N 13 11. (a) Let r = xi + y j + z k, then x2 + z 2 = t2 (sin2 πt + cos2 πt) = t2 = y 2 z y x (b) Let x = t, then y = t2 , z = ± 4 − t2 /3 − t4 /6 z x y i + sin 13t 19,200 j Chapter 14 Supplementary Exercises 12. 522 y t= 2 3 t=0 x t=1 13. (a) (b) t= 1 3 er (t) 2 = cos2 θ + sin2 θ = 1, so er (t) is a unit vector; r(t) = r(t)e(t), so they have the same direction if r(t) > 0, opposite if r(t) < 0. eθ (t) is perpendicular to er (t) since er (t) · eθ (t) = 0, and it will result from a counterclockwise rotation of er (t) provided e(t) × eθ (t) = k, which is true. dθ dθ d dθ d dθ er (t) = (− sin θi + cos θj) = eθ (t) and eθ (t) = − (cos θi + sin θj) = − er (t), so dt dt dt dt dt dt d dθ d v(t) = r(t) = (r(t)er (t)) = r (t)er (t) + r(t) eθ (t) dt dt dt (c) From Part (b), a = d v(t) dt = r (t)er (t) + r (t) = d2 r −r dt2 dθ dt dθ dθ d2 θ eθ (t) + r (t) eθ (t) + r(t) 2 eθ (t) − r(t) dt dt dt 2 er (t) + r −→ 2 er (t) d2 θ dr dθ eθ (t) +2 dt2 dt dt 14. The height y (t) of the rocket satisfies tan θ = y/b, y = b tan θ, v = 15. r = r0 + t P Q= (t − 1)i + (4 − 2t)j + (3 + 2t)k; dθ dt dy dθ dθ dy = = b sec2 θ . dt dθ dt dt s−3 12 − 2s 9 + 2s dr = 3, r(s) = i+ j+ k dt 3 3 3 16. By equation (26) of Section 14.6, r(t) = (60 cos α)ti + ((60 sin α)t − 16t2 + 4)j, and the maximum 15 sin α, so the ball clears the height of the baseball occurs when y (t) = 0, 60 sin α = 32t, t = 8 2 152 15 152 sin α 28 ceiling if ymax = (60 sin α) sin α − 16 2 sin2 α + 4 ≤ 25, ≤ 21, sin2 α ≤ . The ball 8 8 4 75 2 hits the wall when x = 60, t = sec α, and y (sec α) = 60 sin α sec α − 16 sec α + 4. Maximize the 28 . Then height h(α) = y (sec α) = 60 tan α − 16 sec2 α + 4, subject to the constraint sin2 α ≤ 75 15 15 60 15 = , so sin α = √ , but for = h (α) = 60 sec2 α − 32 sec2 α tan α = 0, tan α = 2 + 152 32 8 17 8 this value of α the constraint is not satisfied (the ball hits the ceiling). Hence the maximum value of h occurs at one of the endpoints of the α-interval on which the ball clears the ceiling, i.e. 0, sin−1 28/75 . Since h (0) = 60, it follows that h is increasing throughout the interval, since 28 , hmax = 60 tan α − 16 sec2 α + 4 = h > 0 inside the interval. Thus hmax occurs when sin2 α = 75 √ √ 120 329 − 1012 75 28 ≈ 24.78 ft. Note: the possibility that the baseball keeps 60 √ − 16 + 4 = 47 47 47 climbing until it hits the wall can be rejected as follows: if so, then y (t) = 0 after the ball hits 15 15 sin α occurs after t = sec α, hence sin α ≥ sec α, 15 sin α cos α ≥ 8, the wall, i.e. t = 8 8 15 sin 2α ≥ 16, impossible. 523 Chapter 14 17. r (1) = 3i + 10j + 10k, so if r (t) = 3t2 i + 10j + 10tk is perpendicular to r (1), then 9t2 + 100 + 100t = 0, t = −10, −10/9, so r = −1000i − 100j + 500k, −(1000/729)i − (100/9)j + (500/81)k. dy dx = x(t), = y (t), x(0) = x0 , y (0) = y0 , so dt dt t t t x(t) = x0 e , y (t) = y0 e , r(t) = e r0 18. Let r(t) = x(t)i + y (t)j, then dv 2 1 = 2t2 i + j + cos 2tk, v0 = i + 2j − k, so x (t) = t3 + 1, y (t) = t + 2, z (t) = sin 2t − 1, dt 3 2 1 1 1 1 x(t) = t4 + t, y (t) = t2 + 2t, z (t) = − cos 2t − t + , since r(0) = 0. Hence 6 2 4 4 1 12 1 14 t +t i+ t + 2t j − cos 2t + t − k r(t) = 6 2 4 4 19. (a) ds dt (b) 20. v 2 = r (t) t=1 = v(t) · v(t), 2 v t=1 (5/3)2 + 9 + (1 − (sin 2)/2)2 ≈ 3.475 d 1 d v = 2v · a, ( v )= (v · a) dt dt v CHAPTER 15 Partial Derivatives EXERCISE SET 15.1 1. (a) f (2, 1) = (2)2 (1) + 1 = 5 (c) f (0, 0) = (0)2 (0) + 1 = 1 (e) f (3a, a) = (3a)2 (a) + 1 = 9a3 + 1 2. (a) 2t (b) f (1, 2) = (1)2 (2) + 1 = 3 (d) f (1, −3) = (1)2 (−3) + 1 = −2 ( f ) f (ab, a − b) = (ab)2 (a − b) + 1 = a3 b2 − a2 b3 + 1 (c) 2y 2 + 2y (b) 2x 3. (a) f (x + y, x − y ) = (x + y )(x − y ) + 3 = x2 − y 2 + 3 (b) f xy, 3x2 y 3 = (xy ) 3x2 y 3 + 3 = 3x3 y 4 + 3 4. (a) (x/y ) sin(x/y ) 3 5. F (g (x), h(y )) = F x3 , 3y + 1 = x3 ex (3y +1) 6. g (u(x, y ), v (x, y )) = g x2 y 3 , πxy = πxy sin 7. (a) t2 + 3t10 8. √ te x2 y 3 2 (πxy ) = πxy sin π x5 y 7 (b) 0 √ −3 ln(t2 +1) (c) (x − y ) sin(x − y ) (b) xy sin(xy ) = (t2 (c) 3076 t 3 + 1) (b) −9 6 (c) 3 (e) −t + 3 9. (a) 19 (f ) (a + b)(a − b)2 b3 + 3 8 (d) a + 3 10. (a) x2 (x + y )(x − y ) + (x + y ) = x2 x2 − y 2 + (x + y ) = x4 − x2 y 2 + x + y (b) (xz )(xy )(y/x) + xy = xy 2 z + xy 2 11. F x2 , y + 1, z 2 = (y + 1)ex (y +1)z 2 12. g x2 z 3 , πxyz, xy/z = (xy/z ) sin π x3 yz 4 √ n √ 13. (a) f ( 5, 2, π, −3π ) = 80 π (b) f (1, 1, . . . , 1) = k = n(n + 1)/2 k=1 14. (a) f (−2, 2, 0, π/4) = 1 (b) f (1, 2, . . . , n) = n(n + 1)(2n + 1)/6, see (Theorem 2(b), Section 7.4) 15. y 16. y x x 1 2 524 525 Chapter 15 y 17. y 18. x x 19. (a) all points above or on the line y = −2 (b) all points on or within the sphere x2 + y 2 + z 2 = 25 (c) all points in 3-space 20. (a) all points on or between the vertical lines x = ±2. (b) all points above the line y = 2x (c...
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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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