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Review_2_Answers - Miller Kierste Review 2 Due May 6 2007...

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Miller, Kierste – Review 2 – Due: May 6 2007, 6:00 pm – Inst: Gary Berg 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find f y when f ( x, y ) = x 3 e - x 3 - y 4 . 1. f y = - 3 x 3 y 4 e - x 3 - y 4 2. f y = 4 x 3 y 3 e - x 3 - y 4 3. f y = 3 x 4 y 3 e - x 3 - y 4 4. f y = 3 x 3 y 4 e - x 3 - y 4 5. f y = - 4 x 3 y 3 e - x 3 - y 4 correct 6. f y = - 4 x 4 y 3 e - x 3 - y 4 Explanation: Differentiating with respect to y keeping x fixed, we see that f y = - 4 x 3 y 3 e - x 3 - y 4 . keywords: partial derivative, multi-variable, exponential function, chain rule 002 (part 1 of 1) 10 points Determine f yx when f ( x, y ) = 2 y 2 sin xy . 1. f yx = y 2 (3 sin xy - xy cos xy ) 2. f yx = - x 2 (3 cos xy + xy sin xy ) 3. f yx = - 2 x 2 (3 sin xy + xy cos xy ) 4. f yx = x 2 (3 sin xy - xy cos xy ) 5. f yx = 2 y 2 (3 cos xy - xy sin xy ) correct 6. f yx = 2 x 2 (3 cos xy - xy sin xy ) 7. f yx = - y 2 (3 cos xy + xy sin xy ) 8. f yx = - 2 y 2 (3 sin xy + xy cos xy ) Explanation: By the Product Rule, f y = 4 y sin xy + 2 xy 2 cos xy . But then f yx = 4 y 2 cos xy + 2 y 2 cos xy - 2 xy 3 sin xy . Consequently, f xy = 2 y 2 (3 cos xy - xy sin xy ) . keywords: partial derivative, mixed second order derivative, trig function 003 (part 1 of 1) 10 points Evaluate the double integral I = Z Z A 3 + x 2 1 + y 2 dxdy when A = n ( x, y ) : 0 x 1 , 0 y 1 o . 1. I = 1 2 π 2. I = 5 6 π correct 3. I = 2 3 π 4. I = 1 6 π 5. I = 1 3 π
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Miller, Kierste – Review 2 – Due: May 6 2007, 6:00 pm – Inst: Gary Berg 2 Explanation: Since A = n ( x, y ) : 0 x 1 , 0 y 1 o is a rectangle with sides parallel to the coor- dinate axes, the double integral can be repre- sented as the iterated integral I = Z 1 0 Z 1 0 3 + x 2 1 + y 2 dx dy . Now Z 1 0 3 + x 2 1 + y 2 dx = 1 1 + y 2 h 3 x + 1 3 x 3 i 1 0 . Thus I = 10 3 Z 1 0 1 1 + y 2 dy = 10 3 h tan - 1 y i 1 0 . Consequently, I = 5 6 π . keywords: 004 (part 1 of 1) 10 points Reverse the order of integration in the inte- gral I = Z 4 3 ˆ Z 1 x/ 2 f ( x, y ) dy ! dx, but make no attempt to evaluate either inte- gral. 1. I = Z 3 / 2 0 ˆ Z 4 x 2 3 f ( x, y ) dx ! dy 2. I = Z 1 3 / 2 ˆ Z 2 y 2 3 f ( x, y ) dx ! dy 3. I = Z 1 3 / 2 ˆ Z 4 y 2 3 f ( x, y ) dx !
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