tutorial3_sol

# tutorial3_sol - MATH1211/10-11(2)/Tu3sol/TNK Department of...

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Unformatted text preview: MATH1211/10-11(2)/Tu3sol/TNK Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010-11 2nd semester) Solution to Tutorial 3 Note : Some solutions given below are outline only. You may need to give more details in your solution. 1. Write y (2 , 1) = y , so that when ( s,t ) = (2 , 1), ( x,y ) = (2 ,y ). Using chain rule we obtain ∂z ∂t (2 , 1) = ∂z ∂x ( x,y )=(2 ,y ) · ∂x ∂t ( s.t )=(2 , 1) + ∂z ∂y ( x,y )=(2 ,y ) · ∂y ∂t ( s,t )=(2 , 1) = 2 x ( x,y )=(2 ,y ) · s ( s.t )=(2 , 1) + ∂z ∂y ( x,y )=(2 ,y ) · 0 = (4)(2) + 0 = 8 . 2. As w = f ( u ) and u = x 2- y 2 x 2 + y 2 , we have ∂w ∂x = df du ∂u ∂x = df du 2 x ( x 2 + y 2 )- ( x 2- y 2 )2 x ( x 2 + y 2 ) 2 = df du 4 xy 2 ( x 2 + y 2 ) 2 , ∂w ∂y = df du ∂u ∂y = df du- 2 y ( x 2 + y 2 )- ( x 2- y 2 )2 y ( x 2 + y 2 ) 2 = df du- 4 x 2 y ( x 2 + y 2 ) 2 . Therefore, x ∂w ∂x + y ∂w ∂y = x df du 4 xy 2 ( x 2 + y 2 ) 2 + y df du- 4 x 2 y ( x 2 + y 2 ) 2 =...
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## This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

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