Hw14-3

# Hw14-3 - /r ∂ 2 f ∂x 2 =-2 xy x 2 y 2 2 ∂ 2 f ∂y 2...

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MATH 1920 Homework 4, Chapter 14.3 10. f ( x,y ) = x x 2 + y 2 ∂f ∂x = ( x 2 + y 2 )(1) - x (2 x ) ( x 2 + y 2 ) 2 = y 2 - x 2 ( x 2 + y 2 ) 2 ∂f ∂y = ( x 2 + y 2 )(0) - x (2 y ) ( x 2 + y 2 ) 2 = - 2 xy ( x 2 + y 2 ) 2 18. f ( x,y ) = cos 2 (3 x - y 2 ) ∂f ∂x = 2cos(3 x - y 2 )( - sin(3 x - y 2 ))(3) = - 6cos(3 x - y 2 )sin(3 x - y 2 ) ∂f ∂y = 2cos(3 x - y 2 )( - sin(3 x - y 2 ))( - 2 y ) = 4 y cos(3 x - y 2 )sin(3 x - y 2 ) 32. f ( x,y,z ) = e - xyz f x = - yze - xyz f y = - xze - xyz f z = - xye - xyz 34. f ( x,y,z ) = sinh( xy - z 2 ) f x = y cosh( xy - z 2 ) f y = x cosh( xy - z 2 ) f z = - 2 z cosh( xy - z 2 ) 45. r ( x,y ) = ln( x + y ) r x = 1 x + y r y = 1 x + y r xx = - 1 ( x + y ) 2 r xy = - 1 ( x + y ) 2 r yx = - 1 ( x + y ) 2 r yy = - 1 ( x + y ) 2 1

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54. w = x sin y + y sin x + xy w x = sin y + y cos x + y = w xy = cos y + cos x + 1 w y = x cos y + sin x + x = w yx = cos y + cos x + 1 58. f ( x,y ) = 4 + 2 x - 3 y - xy 2 , ﬁnd ∂f ∂x and ∂f ∂y at ( - 2 , 1) ∂f ∂x ( x,y ) = lim h 0 f ( x + h,y ) - f ( x,y ) h = lim h 0 2 h - hy 2 h = 2 - y 2 so at ( - 2 , 1) we have f x = 1. ∂f ∂x ( x,y ) = lim h 0 f ( x,y + h ) - f ( x,y ) h = lim h 0 - 3 h - x ( y + h ) 2 + xy 2 h = lim h 0 - 3 h - 2 xyh - xh 2 h = - 3 - 2 xy so at ( - 2 , 1) we have f y = 1. 78. f ( x,y ) = tan - 1 ( x/y ), which simply measures the angle that ( x,y ) makes with the positive y -axis. ∂f ∂x = 1 /y 1 + ( x/y ) 2 = y x 2 + y 2 ∂f ∂y = - x/y 2 1 + ( x/y ) 2 = - x x 2 + y 2 Note that the gradient (as a vector ﬁeld) goes around the origin, and the magnitude falls oﬀ as 1
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Unformatted text preview: /r . ∂ 2 f ∂x 2 =-2 xy ( x 2 + y 2 ) 2 ∂ 2 f ∂y 2 = 2 xy ( x 2 + y 2 ) 2 Therefore Δ f = ∂ 2 f ∂x 2 + ∂ 2 f ∂y 2 = 0 so the function f is harmonic (on its domain). Note that we can’t make f deﬁned on the entire plane. 80. f ( x,y,z ) = e 3 x +4 y cos5 z f x = 3 e 3 x +4 y cos5 z = ⇒ f xx = 9 e 3 x +4 y cos5 z f y = 4 e 3 x +4 y cos5 z = ⇒ f yy = 16 e 3 x +4 y cos5 z f z = e 3 x +4 y (-5sin5 z ) = ⇒ f zz = e 3 x +4 y (-25cos5 z ) Therefore Δ f = f xx + f yy + f zz = 0 because 9 + 16 = 25. 2...
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## This note was uploaded on 06/11/2011 for the course MATH 1920 at Cornell.

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Hw14-3 - /r ∂ 2 f ∂x 2 =-2 xy x 2 y 2 2 ∂ 2 f ∂y 2...

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