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MIT18_024S11_soln-pset6

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

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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 = ∂f 1 e x ∂x e x +2 y = 2 +2 y ∂y ∂f 2 = 2 ∂f 2 cos( y + 2 x ) = cos( y + 2 x ) ∂x ∂y So the matrix for the total derivative is: e x +2 y 2 e x +2 y Df ( x, y ) = 2 cos( y + 2 x ) cos( y + 2 x ) 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 + 2 v 2 + 3 w 3 and g 2 ( u, v, w ) = 2 v u 2 we have: ∂g 1 ∂g 1 = 1 = 4 v ∂g 1 = 9 w 2 ∂u ∂v ∂w ∂g 2 = 0 ∂u = 2 u ∂g 2 = 2 ∂g 2 ∂v ∂w And the total derivative is: 1 4 v 9 w 2 Dg ( u, v, w ) = 2 u 2 0 (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 ) = Df ( g ( u, v, w )) Dg ( 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 + 2 g 1 ) cos( g 2 + 2 g 1 ) 2 u 2 0 Now we evaluate at ( u, v, w ) = (1 , 1 , 1) and thus g 1 = 6 , g 2 = 3. As a result g 1 + 2 g 2 = 0 and g 2 + 2 g 1 = 9 and e 0 2 e 0 1 4 9 Dh (1 , 1 , 1) = = 2 cos 9 cos 9 2 2 0 3 0 9 0 6 cos 9 18 cos 9 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
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MIT18_024S11_soln-pset6 - Solutions for PSet 6 1(8.22:14(a...

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