If f x y x 2 y 2 3 xy y 3 then the gradient vector of f x y at the point x 2 y

If f x y x 2 y 2 3 xy y 3 then the gradient vector of

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11. If f(x, y) = x2y2+ 3xy+ y3, then the gradient vector of f(x, y) at the point x= 2, y = −1 is
MAT1322D Solution to Final Examination December 2017 6 Solution. (D) f x= 2xy2+ 3y, f y= 2x2y+ 3x+ 3y2. When x= 2 and y= −1, grad f(2, −1) = (1, 1). 12. The directional derivative of the function z= ln (2x2y2) at the point x= 3 and y= 4 in the direction of the vector u= (3, 4) is Solution. (D) The unit vector in the direction of uis v = 34,55. zx = 2242xxy, zy= 2222yxy. When x= 3 and y= 4, zx(3, 4) = 6, zy(3, 4) = −4. Du(z) = 3426( 4)555 . Part II. Long Answer Questions (26 marks) 1.(5 marks) Use the definition of improper integralsto determine whether improper integral 220(1)xdxxis convergent or divergent. If it is convergent, find its value. Solution. 21222222001000111111limlimlim(1)(1)22112bbbbbxxdxdxduxxub. This improper integral is convergent, and its value is 12. 2.(5 marks) Find function y(t), where y(t) is the solution to the initial-value problem y'= ysin t, y(0) = −1.Solution. 1sindyt dty. ln | y| = −cos t+ C, | y| = K1ecos t, where K1= eC> 0. Then y= Kecost, where K= K10. When t= 0, cos t= 1. y(0) = Ke1= −1.Then K= −e. Hence, y(t) = −e ecos t= −e1 − cos t. 3. (6 marks) Use an appropriate test method to determine whether each of the following series is convergent or divergent. 21nnnn; 1( 1)1nnn; 213nnnn.
MAT1322D Solution to Final Examination December 2017 7 Solution. (a) Since this series is positive, we can use the limit comparison test. Let an= 221nnn, and let bn = 1n. Then 222221/limlimlim2111/nnnnnannnnbnn. Since series 11nndiverges, series 2121nnnndiverges. This question can also be solved by comparison test. 2221122nnnnnn. Since series 1111122nnnndiverges, series 2121nnnndiverges. (b) Since function f(n) =211nis decreasing, and 21lim1nn= 0, By the alternating series test, this series converges. (c) Use the root test. Let an= 13nnn. 11limlim33nnnnnan. This series is convergent. This question may also be solved by the ratio test. Since 111123(1)231111limlimlimlim1lim13(1)131313nnnnnnnnnnnnnnnannannnnnn< 1.

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