10c_04f_2bs

10c_04f_2bs - , 1) = f (0 , 1) u = 2 / 5. 3. The gradient...

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Math 10C Second Midterm Version B Solutions November 17, 2004 1. (a) This can be done in various ways: Best is to take a value on each side of the (50,10) entry and form the Fnite di±erence: w s (50 , 10) 37 21 60 40 = 16 20 and w t (50 , 10) 36 19 15 5 = 17 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 , 10) 37 29 60 50 = 8 10 or w s (50 , 10) 29 21 50 40 = 8 10 and w t (50 , 10) 36 29 15 10 = 7 5 or w t (50 , 10) 29 19 10 5 = 10 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 , 10) + w s (50 , 10) × ( 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) = 1 × 1 + ( 2) × 3 = 5 . (b) Since f xy = f yx , we’ll compute f x Frst. It is 3 x 2 y . Thus f xy = (3 x 2 y ) /∂y = 3 x 2 . (c) |a 1 , 2 A| = 1 2 + 2 2 = 5. Thus u = a 1 / 5 , 2 / 5 A . Since f = a 2 x 2 y, 2 x A , we have f (0 , 1) = a− 2 , 0 A ²inally D u f (0
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Unformatted text preview: , 1) = f (0 , 1) u = 2 / 5. 3. The gradient is a 6 x, 2 y, 4 z A , which equals a 6 , 8 , 4 A at (1 , 4 , 1). Thus the equation of the plane is 0 = a 6 , 8 , 4 A a x 1 , y + 4 , z 1 A = 6 x 8 y + 4 z 42 , which can be rewritten as 3 x 4 y + 2 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 (1 , 2) at (1 , 2) f xx 2 2 2 2 f xy 2 y 2 2 2 2 f yy 2 x 2 2 D 4(1 x y 2 ) 4 8 8 By the second derivative test, (0 , 0) is a local maximum and the other two are saddle points....
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