ps2B-solns-2009

ps2B-solns-2009 - a If a surface is described by z = f x,y...

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18.02A pset 2B, part II solutions, fall 2009 Problem 1 a) ∂z ∂x = - 4 x ∂z ∂x (1 , 1) = - 4 . ∂z ∂y = - 6 y ∂z ∂y (1 , 1) = - 6 . tangent plane: ( z - 995) = - 4( x - 1) - 6( y - 1). Tangent plane approximation: Δ z ≈ - x - y = - 4( - . 05) - 6( - . 05) approx. change in z = . 5 . Exact answer: z ( . 95 ,. 95) - z (1 , 1) = . 487 estimation error = .013. b) First we work in the xy -plane and then transfer our answer to 3 dimensions. The direction in the xy -plane of maximum increase of z is in the direction of z = h- 4 x, - 6 y i ⇒ z | (1 , 1) = h- 4 , - 6 i . in the xy -plane, the fastest increase is in direction of h- 4 , - 6 i . If the projection of her path in the xy -plane has tangent vector a i + b j then the tangent vector of her path on the mountain is a i + b j + ( z · h a,b i ) k . along the mountain the fastest increase is in direction of - 4 i - 6 j + 52 k . x y z = 995 ] ; ; ; ; ; ; ; y = x 3 / 2 Level curves c) We need to show that the tangent at each point along the graph of y = x 3 / 2 is parallel to z . Point on graph: P = ( x, x 3 / 2 ). Tangent vector to y = x 3 / 2 at P is v = h 1 , 3 2 x 1 / 2 i . z | P = h- 4 x, - 6 x 3 / 2 i = - 4 x v . So they are parallel . QED (The picture at right shows the path along the mountain.) x h y = = = z . . . . . . ..... . . . . . . . . . z = f ( x,y ) Problem 2
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Unformatted text preview: a) If a surface is described by z = f ( x,y ) then its normal is N = h f x , f y ,-1 i . ⇒ a normal to the first surface is N 1 = h 2 x, 2 y,-1 i . ⇒ a normal to the second surface is N 2 = h 1 , 1 ,-1 i . At (1 , 1 , 2): N 1 = h 2 , 2 ,-1 i and N 2 = h 1 , 1 ,-1 i . The tangent to the intersection curve is orthogonal to both normals. ⇒ tangent vector = N 1 × N 2 = ± ± ± ± ± ± i j k 2 2-1 1 1-1 ± ± ± ± ± ± =-i + j . b. As in part (a): normal to the surfaces are: N 1 = h f x ( x ,y ) , f y ( x ,y ) ,-1 i and N 1 = h g x ( x ,y ) , g y ( x ,y ) ,-1 i . The tangent to the intersection curve is orthogonal to both normals. ⇒ tangent vector = N 1 × N 2 = ± ± ± ± ± ± i j k f x ( x ,y ) f y ( x ,y )-1 g x ( x ,y ) g y ( x ,y )-1 ± ± ± ± ± ± = ( g y-f y ) i-( g x-f x ) j + ( f x g y-f y g x )) k . 1...
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