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21c_02f_2as - y =-2 x 11 1 4(a We need ∇ f = From Problem...

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Math 21C Second Midterm Version A Nov. 25, 2002 1. Use the chain rule. (a) g s = g x x s + g y y s = g x + g y . (b) Let h = g s . Then g ts = g st = h t = h x x t + h y y t = - h x + 3 h y = - ( g xx + g xy ) + 3( g xy + g yy ) = - g xx + 2 g xy + 3 g yy . 2. Note that f = h 2 x + 2 y, 3 y 2 + 8 y + 2 x i and f (0 , 1) = h 2 , 11 i . (a) u = f/ |∇ f | = 125 - 1 / 2 h 2 , 11 i . (b) The maximum is |∇ f | = 125 = 5 5. (c) We need u · ∇ f = 0. There are two possible answers: 125 - 1 / 2 h 11 , - 2 i and 125 - 1 / 2 h- 11 , 2 i . 3. This could be done it at least two ways. The tangent line is in a direction in which D u = 0. Such a vector was found in 2(c). Then h x, y i = t u + h 0 , 1 i . Since all we need is a vector parallel to u , we can drop the factor of 125 - 1 / 2 if we wish to get the cleaner formula h x, y i = t h 11 , - 2 i + h 0 , 1 i . Since dy/dx = - f x /f y , we have dy/dx = - 2 / 11. Thus the line is y - 1 = ( - 2 / 11)( x -
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Unformatted text preview: y =-2 x/ 11 + 1. 4. (a) We need ∇ f = . From Problem 2, this gives us 2 x +2 y = 0 and 3 y 2 +8 y +2 x = 0. The first equation give x =-y , which turns the second equation into 3 y 2 +6 y = 0. The solutions are y = 0 and y =-2. Since x =-y , the critical points are (0 , 0) and (2 ,-2). (b) We have f xx = 2, f xy = 2 and f yy = 6 y + 8. Thus f xx > 0. At (0 , 0), f yy = 8 and f xx f yy-( f xy ) 2 > 0 so the point is a (local) minimum. At (2 ,-2), f yy =-4 and f xx f yy-( f xy ) 2 < 0 so the point is a saddle....
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