MIT18_024S11_soln-pset6

MIT18_024S11_soln-pset6 - Solutions for PSet 6 1. (8.22:14)...

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Solutions for PSet 6 1. (8.22:14) (a) f ( x, y )= f 1 ( x, y ) i + f 2 ( x, y ) j , where f 1 ( x, y e x +2 y , and f 2 ( x, y sin( y +2 x ). Computing all the partial derivatives ∂f 1 = 1 e x ∂x e x +2 y =2 +2 y ∂y 2 = 2 2 cos( y x ) = cos( y x ) So the matrix for the total derivative is: ± e x +2 y 2 e x +2 y Df ( x, y 2 cos( y x ) cos( y x ) ² Similarly for g ( u, v, w g 1 ( u, v, w ) i + g 2 ( u, v, w ) j , where g 1 ( u, v, w u v 2 +3 w 3 and g 2 ( u, v, w )=2 v u 2 we have: ∂g 1 1 =1 =4 v 1 =9 w 2 ∂u ∂v ∂w 2 =0 = 2 u 2 2 And the total derivative is: 1 4 v 9 w 2 Dg ( u, v, w ± 2 u 2 0 ² (b) The composition h ( u, v, w f ( g ( u, v, w )) = exp ( u +2 v 2 +3 w 3 +4 v 2 u 2 ) i +sin(2 v u 2 +2 u +4 v 2 +6 w 3 ) j (c) The total derivative at a point ( u, v, w ) can be computed using the chain rule: Dh ( u, v, w ( g ( u, v, w )) ( u, v, w ) ± e g 1 +2 g 2 2 e g 1 +2 g 2 1 4 v 9 w 2 = 2 cos( g 2 g 1 ) cos( g 2 g 1 ) ²± 2 u 2 0 ² Now we evaluate at ( u, v, w )=(1 , 1 , 1) and thus g 1 =6 ,g 2 = 3. As a result g 1 g 2 = 0 and g 2 g 1 = 9 and ± e 0 2 e 0 1 49 (1 , 1 , 1) = = 2 cos 9 cos 9 2 2 0 ² ± 3 0 9 0 6 cos 9 18 cos 9 ² 1
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2. (8.24:12) (a) We can compute ± ( 1 ) using the chain rule for the functions r : 3 R R r defined by r ( r )= r · r and g : R 1 R defined by g ( t . With these t functions 1 = g r thus r 1 1 2 r A A ± ( A · g ± ( r ) r · r A · ± ( ( r r ( r 2 ) · 2 r · r r 3 · (b) To evaluate the left hand side in question, we need to first evaluate ± ( A ± ( 1 ) ) . Using part (a), this is equivalent to r A · r f ( r ) ± ± ² = ± r 3 ± h ( r ) ² where f ( r A
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MIT18_024S11_soln-pset6 - Solutions for PSet 6 1. (8.22:14)...

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