10c_04f_2as

10c_04f_2as - A inally D u f (1 , 0) = f (1 , 0) u = 6 / 5....

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Math 10C Second Midterm Version A Solutions November 17, 2004 1. (a) This can be done in various ways: Best is to take a value on each side of the (50,15) entry and form the Fnite di±erence: w s (50 , 15) 47 25 60 40 = 22 20 and w t (50 , 15) 40 29 20 10 = 11 10 . Not as good is to take a value at the point and a point on one side (but this is still acceptable): w s (50 , 15) 47 36 60 50 = 11 10 or w s (50 , 15) 36 25 50 40 = 11 10 and w t (50 , 15) 40 36 20 15 = 4 5 or w t (50 , 15) 36 29 15 10 = 7 5 . (b) The units of w s are feet per knot and those of w t feet per hour. If you have singular or plural where you should not (foot, knots, hours), you will still receive credit. (c) The answer is w (50 , 15) + w s (50 , 15) × 1 + w t (50 , 15) × ( 1). You should plug in the numbers you got in (a). 2. By the chain rule f (1) = g x ( x (1) , y (1)) x (1) + g y ( x (1) , y (1)) y (1) = 2 × 2 + ( 1) × 1 = 3 . (b) Since f xy = f yx , we’ll compute f y Frst. It is 2 xy . Thus f xy = (2 xy ) /∂x = 2 y . (c) |a 2 , 1 A| = 2 2 + 1 2 = 5. Thus u = a 2 / 5 , 1 / 5 A . Since f = a 2 x + 2 y, 2 x A , we have f (1 , 0) = a 2 , 2
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Unformatted text preview: A inally D u f (1 , 0) = f (1 , 0) u = 6 / 5. 3. The gradient is a 4 x, 6 y, 2 z A , which equals a 4 , 6 , 8 A at (1 , 1 , 4). Thus the equation of the plane is 0 = a 4 , 6 , 8 A a x 1 , y + 1 , z 4 A = 4 x 6 y + 8 z 42 , which can be rewritten as 2 x 3 y + 4 z = 21. Any of these forms, including the dot product form, is acceptable. 4. Since we are given the critical points, we only need to compute f xx , f xy , f yy and D = f xx f yy ( f xy ) 2 there. quantity at ( x, y ) at (0 , 0) at ( 2 , 1) at ( 2 , 1) f xx 2 2 y 2 f xy 2 x 2 2 2 2 f yy 2 2 2 2 D 4(1 y x 2 ) 4 8 8 By the second derivative test, (0 , 0) is a local minimum and the other two are saddle points....
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