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Unformatted text preview: FINAL EXAM SOLUTIONS Math 51, Spring 2003. 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. (/40 points) 5. (/40 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: Byoungdu Kim (2 and 6) Ted Hwa (3 and 7) Jacob Shapiro (4 and 8) Ryan Vinroot (A02) Michel Grueneberg (A03) Circle your section meeting time: 11:00am 1:15pm 7pm 1 1. Consider the function f : R 2 → R 2 given by f parenleftbiggbracketleftbigg x y bracketrightbiggparenrightbigg = xy ( x 2 + y 2 ) 2 if bracketleftbigg x y bracketrightbigg negationslash = bracketleftbigg bracketrightbigg if bracketleftbigg x y bracketrightbigg = bracketleftbigg bracketrightbigg (a) Do ∂f ∂x and ∂f ∂y exist at the origin? If yes, compute them; if not, explain why. Solution: We can compute these partial derivatives directly from the definition: ∂f ∂x = lim h → f parenleftbiggbracketleftbigg 0 + h bracketrightbiggparenrightbigg f parenleftbiggbracketleftbigg bracketrightbiggparenrightbigg h = lim h → (0) (0) h = 0 ∂f ∂y = lim h → f parenleftbiggbracketleftbigg 0 + h bracketrightbiggparenrightbigg f parenleftbiggbracketleftbigg bracketrightbiggparenrightbigg h = lim h → (0) (0) h = 0 2 (b) Is the function f continuous at the origin? Explain your reasoning. Solution: We first note that the function takes the value 0 everywhere on both the x and y axes. However, something very different happens if we approach the origin along the line y = x : lim x → xy ( x 2 + y 2 ) 2 = lim x → x 2 (2 x 2 ) 2 = lim x → 1 4 x 2 Of course, this limit diverges. So, the limit of this function can not exist at the origin, and therefore it is not continuous there. (c) Is the function f differentiable at the origin? Explain your reasoning. Solution: The fact that the function is not continuous at the origin immediately rules out the possibility that it is differentiable. We can also observe this as follows. Suppose that this function were differentiable at the origin. Since we have computed already the partial derivatives of f at the origin, and they are both zero, we know that the gradient vector of f must be the zero vector, which implies that all vector derivatives must be zero at the origin. However, D 2 4 1 1 3 5 f (→ a ) = lim h → f parenleftbiggbracketleftbigg h h bracketrightbiggparenrightbigg f parenleftbiggbracketleftbigg bracketrightbiggparenrightbigg h = lim h → (1 / 4 h 2 ) (0) h = lim h → 1 / 4 h 3 which also does not exist, contradicting our previous conclusion. So, f cannot be differentiable at the origin....
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 Spring '06
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 Math, Linear Algebra, Algebra, Derivative, Differential Calculus, Gradient, Jacobian, Jacobian matrix

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