# A 4 0 1 1 cos 2 d 32 2 r dr d 0 0 1 cos4 d

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Unformatted text preview: θ + cos θ. ∂y ∂r ∂y ∂θ ∂y ∂r r ∂θ 39. (a) 1 = −r sin θ (d) Square and add the results of parts (a) and (b). (e) From Part (c), 1 ∂z ∂ ∂z ∂2z cos θ − sin θ = 2 ∂x ∂r ∂r r ∂θ = 1 ∂z ∂z cos θ − sin θ ∂r r ∂θ ∂θ ∂x 1 ∂2z 1 ∂z ∂2z sin θ − sin θ cos θ cos θ + 2 ∂r2 r ∂θ r ∂r∂θ + = ∂ ∂r + ∂x ∂θ ∂z 1 ∂2z 1 ∂z ∂2z cos θ − sin θ − cos θ sin θ − 2 ∂θ∂r ∂r r ∂θ r ∂θ − sin θ r 2 ∂2z 1 ∂2z 2 ∂z 1 ∂z ∂2z sin θ cos θ − sin θ cos θ + 2 2 sin2 θ + sin2 θ. cos2 θ + 2 ∂r2 r ∂θ r ∂θ∂r r ∂θ r ∂r Exercise Set 15.4 542 Similarly, from Part (c), 2 ∂2z 1 ∂2z ∂2z 2 ∂z 1 ∂z ∂2z sin θ cos θ + sin θ cos θ + 2 2 cos2 θ + cos2 θ. = 2 sin2 θ − 2 2 ∂y ∂r r ∂θ r ∂θ∂r r ∂θ r ∂r Add to get 40. zx = ∂2z ∂2z 1 ∂2z 1 ∂z ∂2z . + 2= 2+ 2 2+ ∂x2 ∂y ∂r r ∂θ r ∂r −2y 4xy 2x 4xy , zxx = 2 , zy = 2 , zyy = − 2 , zxx + zyy = 0; x2 + y 2 (x + y 2 )2 x + y2 (x + y 2 )2 z = tan−1 2r2 cos θ sin θ = tan−1 tan 2θ = 2θ, zr = 0, zθθ = 0 r2 (cos2 θ − sin2 θ) 41. (a) By the chain rule, ∂u ∂u ∂v ∂v ∂v ∂u = cos θ + sin θ and = − r sin θ + r cos θ, use the ∂r ∂x ∂y ∂θ ∂x ∂y Cauchy-Riemann conditions ∂u ∂v ∂u ∂v ∂u = and =− in the equation for to get ∂x ∂y ∂y ∂x ∂r ∂v ∂v ∂v ∂u 1 ∂v ∂v 1 ∂u ∂u = cos θ − sin θ and compare to to see that = . The result =− ∂r ∂y ∂x ∂θ ∂r r ∂θ ∂r r ∂θ can be obtained by considering (b) ux = uy = x2 ∂v ∂u and . ∂r ∂θ 2x 1 2x 1 , vy = 2 =2 = ux ; 2 2 +y x 1 + (y/x) x + y2 2y y 2y 1 , vx = −2 2 =− 2 = −uy ; x2 + y 2 x 1 + (y/x)2 x + y2 u = ln r2 , v = 2θ, ur = 2/r, vθ = 2, so ur = 1 1 vθ , uθ = 0, vr = 0, so vr = − uθ r r 42. z = f (u, v ) where u = x − y and v = y − x, ∂z ∂u ∂z ∂v ∂z ∂z ∂z ∂z ∂u ∂z ∂v ∂z ∂z ∂z ∂z ∂z = + = − and = + =− + so + =0 ∂x ∂u ∂x ∂v ∂x ∂u ∂v ∂y ∂u ∂y ∂v ∂y ∂u ∂v ∂x ∂y 43. (a) ux = f (x + ct), uxx = f (x + ct), ut = cf (x + ct), utt = c2 f (x + ct); utt = c2 uxx (b) Substitute g for f and −c for c in Part (a). (c) Since the sum of derivatives equals the derivative of the sum, the result follows from Parts (a) and (b). 1 (d) sin t sin x = (− cos(x + t) + cos(x − t)) 2 44. fx (x0 , y0 ) = y0 , fy (x0 , y0 ) = x0 , ∆f = (x0 + ∆x)(y0 + ∆y ) − x0 y0 = y0 ∆x + x0 ∆y + ∆x∆y = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + (0)∆x + (∆x)∆y where 1 = 0 and 2 = ∆x. 45. fx (x0 , y0 ) = 2x0 , fy (x0 , y0 ) = 2y0 , 2 ∆f = (x0 + ∆x)2 + (y0 + ∆y )2 − x2 + y0 = 2x0 ∆x + (∆x)2 + 2y0 ∆y + (∆y )2 0 = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + (∆x)∆x + (∆y )∆y where 1 = ∆x and 2 = ∆y. 543 Chapter 15 46. fx (x0 , y0 ) = 2x0 y0 , fy (x0 , y0 ) = x2 , 0 ∆f = (x0 + ∆x)2 (y0 + ∆y ) − x2 y0 = 2x0 y0 ∆x + x2 ∆y + y0 (∆x)2 + (∆x)2 ∆y + 2x0 ∆x∆y 0 0 = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + (y0 ∆x + ∆x∆y )∆x + (2x0 ∆x)∆y where 1 = y0 ∆x + ∆x∆y and 2 = 2x0 ∆x. 47. fx (x0 , y0 ) = 3, fy (x0 , y0 ) = 2y0 , 2 ∆f = 3(x0 + ∆x) + (y0 + ∆y )2 − 3x0 − y0 = 3∆x + 2y0 ∆y + (∆y )2 = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + (0)∆x + (∆y )∆y where 48. (a) (b) lim (x,y )→(0,0) lim ∆x→0 1 = 0 and 2 = ∆y. f (x, y ) = 0 = f (0, 0) f (0 + ∆x, 0) − f (0, 0) f (∆x, 0) |∆x| = lim = lim , which does not exist because ∆x→0 ∆x→0 ∆x ∆x ∆x lim |∆x|/∆x = 1 and ∆x→0+ lim |∆x|/∆x = −1. ∆x→0− f (∆x, 0) − f (0, 0) −3∆x = lim = −3, ∆x→0 ∆x ∆x f (0, ∆y ) − f (0, 0) −2∆y = lim = −2; fy (0, 0) = lim ∆y →0 ∆y →0 ∆y ∆y lim f (x, y ) does not exist because f (x, y ) → 5 if (x, y ) → (0, 0) where x ≥ 0 or y ≥ 0, but 49. fx (0, 0) = lim ∆x→0 (x,y )→(0,0) f (x, y ) → 0 if (x, y ) → (0, 0) where x < 0 and y < 0, so f is not continuous at (0,0). 50. fx (0, 0) = lim ∆x→0 f (∆x, 0) − f (0, 0) 0−0 = lim = lim 0 = 0 ∆x→0 ∆x ∆x→0 ∆x f (0, ∆y ) − f (0, 0) 0−0 0 = lim = 0; along y = 0, lim 2 = lim 0 = 0, x→0 x x→0 ∆y →0 ∆y ∆y 2 x lim f (x, y ) does not exist. along y = x, lim 2 = lim 1/2 = 1/2 so x→0 2x x→0 (x,y )→(0,0) fy (0, 0) = lim ∆y →0 51. (a) fx (0, 0) = lim ∆x→0 f (∆x, 0) − f (0, 0) = 0; similarly, fy (0, 0) = 0 ∆x f (∆x, y ) − f (0, y ) y (∆x2 − y 2 ) = lim = −y ∆x→0 ∆x→0 ∆x2 + y 2 ∆x (b) fx (0, y ) = lim f (x, ∆y ) − f (x, 0) x(x2 − ∆y 2 ) = lim =x ∆y →0 ∆y →0 x2 + ∆y 2 ∆y fy (x, 0) = lim fx (0, ∆y ) − fx (0, 0) −∆y = lim = −1 ∆y →0 ∆y →0 ∆y ∆y fy (∆x, 0) − fy (0, 0) ∆x = lim =1 fyx (0, 0) = lim ∆x→0 ∆x→0 ∆x ∆x (c) fxy (0, 0) = lim (d) No, since fxy and fyx are not continuous. 52. Represent the line segment C that joins A and B by x = x0 + (x1 − x0 )t, y = y0 + (y1 − y0 )t for 0 ≤ t ≤ 1. Let F (t) = f (x0 + (x1 − x0 )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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