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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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,...
View Full Document

Ask a homework question - tutors are online