wk7_disc_v1_sol.pdf

# wk7_disc_v1_sol.pdf - Fall 2017 Discussion Set(Week 7...

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Fall 2017 Discussion Set (Week 7) Solutions Calculus II 1. Let f ( x, y ) = 1 1 - x - y . (a) Find f x and f y . Answer: f x = f y = 1 (1 - x - y ) 2 (b) Find f xx , f xy , and f yy . Answer: f xx = f xy = f yy = 2 (1 - x - y ) 3 (c) Find f xxx , f xxy , f xyy , and f yyy . Answer: f xxx = f xxy = f xyy = f yyy = 6 (1 - x - y ) 4 (d) Find the quadratic approximation of f at the origin. Answer: f (0 , 0) = f x (0 , 0) = f y (0 , 0) = 1 and f xx (0 , 0) = f xy (0 , 0) = f yy (0 , 0) = 2, so the quadratic approximation is T 2 ( x, y ) = f (0 , 0) + xf x (0 , 0) + yf y (0 , 0) + 1 2 ( x 2 f xx (0 , 0) + 2 xyf xy (0 , 0) + y 2 f yy (0 , 0) ) = 1 + x + y + 1 2 ( 2 x 2 + 4 xy + 2 y 2 ) = 1 + x + y + x 2 + 2 xy + y 2 . (e) Using the Taylor’s formula, estimate the error associated with the quadratic approximation of f ( x, y ) at the origin if | x | ≤ 0 . 2 and | y | ≤ 0 . 2. Answer: According to Taylor’s formula, the error term is E 2 ( x, y ) = 1 3! ( x 3 f xxx ( cx, cy ) + 3 x 2 yf xxy ( cx, cy ) + 3 xy 2 f xyy ( cx, cy ) + y 3 f yyy ( cx, cy ) ) (1) for some c such that 0 c 1. If | x | ≤ 0 . 2 and | y | ≤ 0 . 2, then | cx | ≤ 0 . 2 and | cy | ≤ 0 . 2, so | f xxx ( cx, cy ) | = · · · = | f yyy ( cx, cy ) | = 6 (1 - cx - cy ) 4 6 (1 - 0 . 2 - 0 . 2) 4 = 6 (0 . 6) 4 = 10 4 6 3 . Hence, | E 2 ( x, y ) | ≤ 1 3! ( (0 . 2) 3 + 3(0 . 2) 3 + 3(0 . 2) 3 + (0 . 2) 3 ) · 10 4 6 3 = 8 · 1 5 3 · 5 4 3 4 = 40 81 0 . 49 , i.e., the error is at most 0 . 5. 1

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Fall 2017 Discussion Set (Week 7) Solutions Calculus II 2. Let f ( x, y ) = x 3 + y 3 + 3 x 2 - 3 y 2 - 8. (a) Find f x and f y . Answer: f x = 3 x 2 + 6 x = 3 x ( x + 2) and f y = 3 y 2 - 6 y = 3 y ( y - 2) . (b) Find the critical points of f . What does the First Derivative Test tell you about these points? Answer: f x and f y exist everywhere, so the critical points are those at which both f x and f y vanish, i.e., (0 , 0) , (0 , 2) , ( - 2 , 0) , and ( - 2 , 2) . The First Derivative Test says that if f has a local extremum, then it must occur at one of these points. (c) Find the Hessian H f = det f xx f xy f yx f yy . Answer: Since f xx = 6 x + 6 , f xy = 0, and f yy = 6 y - 6, the Hessian is H f = det f xx f xy f yx f yy = det 6 x + 6 0 0 6 y - 6 = (6 x + 6)(6 y - 6) = 36( x + 1)( y - 1) . (d) By applying the Second Derivative Test , i. find the local minima of f and their locations ii. find the local maxima of f and their locations. iii. find the saddle points of f and their locations. Answer: We first look at the sign of the Hessian at the critical points: H f (0 , 0) < 0 , H f (0 , 2) > 0 , H f ( - 2 , 0) > 0 , H f ( - 2 , 2) < 0 . Next, we look at the sign of f xx at the critical points: f xx (0 , 0) > 0 , f xx (0 , 2) > 0 , f xx ( - 2 , 0) < 0 , f xx ( - 2 , 2) < 0 . By the Second Derivative Test, f has a local minimum at (0 , 2) , a local maximum at ( - 2 , 0) , and saddle points at (0 , 0) and ( - 2 , 2) . The local minimum value is f (0 , 2) = - 12 , the local maximum value is f ( - 2 , 0) = - 4 , and the saddle points are (0 , 0 , f (0 , 0 , 0)) = (0 , 0 , - 8) and ( - 2 , 2 , f ( - 2 , 2)) = ( - 2 , 2 , - 8) . 3. For each of the following functions defined on R 2 , find the critical points, local extrema, and saddle points, if any. How useful are the First Derivative Test and the Second Derivative Test in finding these points? (a) f ( x, y ) = p x 2 + y 2 Answer: f ( x, 0) = | x | is not differentiable at x = 0, so f x does not exist at (0 , 0). (Neither does f y .) Hence, (0 , 0) is a critical point of f . On the other hand, if ( x, y ) 6 = (0 , 0), then f x = x x 2 + y 2 and f y = y x 2 + y 2 , so f x and f y exist on R 2 - { (0 , 0) } , and they cannot be both zero. It follows that (0 , 0) is the only critical point of f . It is clear that f (0 , 0) = 0 is an absolute minimum, and the First Derivative Test implies that it is the only local extremum.
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