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2002FinalExamSolutions

# 2002FinalExamSolutions - FINAL EXAM SOLUTIONS Math 51...

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FINAL EXAM SOLUTIONS Math 51, Spring 2002. You have 3 hours. No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck! Name ID number 1. (/40 points) 2. (/40 points) 3. (/40 points) 4. (/30 points) 5. (/30 points) 6. (/20 points) Bonus (/20 points) Total (/200 points) “On my honor, I have neither given nor received any aid on this examination. I have furthermore abided by all other aspects of the honor code with respect to this examination.” Signature: Circle your TA’s name: Tarn Adams (2 and 6) Mariel Saez (3 and 7) Yevgeniy Kovchegov (4 and 8) Heaseung Kwon (A02) Alex Meadows (A03) Circle your section meeting time: 11:00am 1:15pm 7pm 1

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1. Suppose that f, g : R 2 R 2 are given by f x y ‚¶ = x 2 y y 2 - x 2 = u v g u v ‚¶ = sin u cos u = s t (a) Evaluate J f, -→ a (where -→ a = x y ) and J g,f ( -→ a ) , in terms of x and y . Solution: The Jacobian matrices are just made up of the partial derivatives of the corresponding functions. J f, -→ a = 2 xy x 2 - 2 x 2 y J g,f ( -→ a ) = cos u 0 - sin u 0 = cos x 2 y 0 - sin x 2 y 0 (b) Without computing the composition function g f , evaluate J g f, -→ a , in terms of x and y . Solution: The Jacobian of the composition is, by the Chain Rule, just the product of the individual Jacobian matrices: J g f, -→ a = J g,f ( -→ a ) J f, -→ a = cos x 2 y 0 - sin x 2 y 0 2 xy x 2 - 2 x 2 y = 2 xy cos x 2 y x 2 cos x 2 y - 2 xy sin x 2 y - x 2 sin x 2 y 2
(c) Using the result from part (b), determine ∂s ∂y without explicitly computing s as a function of x and y . Explain your reasoning. Solution: J g f, -→ a = ∂s ∂x ∂s ∂y ∂t ∂x ∂t ∂y So, ∂s ∂y is the top right component of the matrix J g f, -→ a computed in part (b). So ∂s ∂y = x 2 cos x 2 y 3

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2. (a) Write out the single variable limit that defines the directional derivative D -→ v f ( -→ a ). Solution: D -→ v f ( -→ a ) = lim h 0 f ( -→ a + h -→ v ) - f ( -→ a ) h (b) Let f : R 2 R be given by f x y ‚¶ = x 2 - 3 xy 2 Compute the directional derivative of f at the point 2 1 in the direction 3 2 directly from the definition above. Solution: D -→ v f ( -→ a ) = lim h 0 f ( -→ a + h -→ v ) - f ( -→ a ) h = lim h 0 f 2 + 3 h 1 + 2 h ‚¶ - f 2 1 ‚¶ h = lim h 0 ((2 + 3 h ) 2 - 3(2 + 3 h )(1 + 2 h ) 2 ) - (2 2 - 3(2)(1) 2 ) h = lim h 0 - 36 h 3 - 51 h 2 - 21 h h = lim h 0 ( - 36 h 2 - 51 h - 21) = - 21 4
(c) Write out the definition (involving a multivariable limit) of the derivative transfor- mation D f, -→ a of a differentiable function f. Solution: The derivative transformation D f, -→ a is the linear transformation for which lim -→ x -→ a k f ( -→ x ) - f ( -→ a ) - D f, -→ a ( -→ x - -→ a ) k k -→ x - -→ a k = 0 (d) Suppose that a function f has D -→ v 1 f ( -→ a ) = -→ w 1 D -→ v 2 f ( -→ a ) = -→ w 2 D -→ v 1 + -→ v 2 f ( -→ a ) = -→ w 1 + 2 -→ w 2 for some nonzero vectors -→ v 1 , -→ v 2 , -→ w 1 , -→ w 2 .

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2002FinalExamSolutions - FINAL EXAM SOLUTIONS Math 51...

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