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21c_02f_2bs

# 21c_02f_2bs - y =-4 x 5 1 4(a We need ∇ f = By Problem 2...

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Math 21C Second Midterm Version B Nov. 25, 2002 1. Use the chain rule. (a) g s = g x x s + g y y s = 2 g x + g y . (b) Let h = g s . Then g st = g ts = h t = h x x t + h y y t = h x - h y = (2 g xx + g xy ) - (2 g xy + g yy ) = 2 g xx - g xy - g yy . 2. Note that f = h 2 x + 4 y, 3 y 2 + 2 y + 4 x i and f (0 , 1) = h 4 , 5 i . (a) u = f/ |∇ f | = 41 - 1 / 2 h 4 , 5 i . (b) The maximum is |∇ f | = 41. (c) We need u · ∇ f = 0. There are two possible answers: 41 - 1 / 2 h 5 , - 4 i and 41 - 1 / 2 h- 5 , 4 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 41 - 1 / 2 if we wish to get the cleaner formula h x, y i = t h 5 , - 4 i + h 0 , 1 i . Since dy/dx = - f x /f y , we have dy/dx = - 4 / 5. Thus the line is y - 1 = ( - 4 / 5)( x - 0), which can be written
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Unformatted text preview: y =-4 x/ 5 + 1. 4. (a) We need ∇ f = . By Problem 2, we have 2 x +4 y = 0 and 3 y 2 +2 y +4 x = 0. The ﬁrst equation give x =-2 y , which turns the second equation into 3 y 2-6 y = 0. The solutions are y = 0 and y = 2. Since x =-2 y , the critical points are (0 , 0) and (-4 , 2). (b) We have f xx = 2, f xy = 4 and f yy = 6 y + 2. Thus f xx > 0. At (0 , 0), f yy = 2 and f xx f yy-( f xy ) 2 < 0 so the point is a saddle. At (-4 , 2), f yy = 14 and f xx f yy-( f xy ) 2 > 0 so the point is a (local) minimum....
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