EXAM 2 Review-solutions

EXAM 2 Review-solutions - Van Ligten(hlv63 EXAM 2 Review...

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Van Ligten (hlv63) – EXAM 2 Review – Gilbert – (56650) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Use this Review as part of your preparation for Exam 2. Remember that the Review will not count towards the Course Grade. JEG 001 10.0 points Find f x when f ( x, y ) = ln( radicalbig x 2 + y 2 - x ) . 1. f x = x radicalbig x 2 + y 2 2. f x = - 1 radicalbig x 2 - y 2 3. f x = - 1 radicalbig x 2 + y 2 correct 4. f x = 1 radicalbig x 2 + y 2 5. f x = 1 radicalbig x 2 - y 2 6. f x = x radicalbig x 2 - y 2 Explanation: Differentiating with respect to x keeping y fixed, we see that f x = x radicalbig x 2 + y 2 - 1 radicalbig x 2 + y 2 - x . But x radicalbig x 2 + y 2 - 1 = x - radicalbig x 2 + y 2 radicalbig x 2 + y 2 . Consequently, f x = - 1 radicalbig x 2 + y 2 . 002 10.0 points Determine f xy when f ( x, y ) = (3 x + 2 y )ln( xy ) . 1. f xy = 3 x + 2 y xy correct 2. f xy = 3 x + 2 y y 3. f xy = 3 x - 2 y xy 4. f xy = 2 x + 3 y xy 5. f xy = 2 x - 3 y x 6. f xy = 2 x - 3 y xy Explanation: Since ln( xy ) = ln x + ln y , we see that f ( x, y ) = (3 x + 2 y )(ln x + ln y ) . Thus f x = 3(ln x + ln y ) + 3 x + 2 y x = 3(ln x + ln y ) + 3 + 2 y x . Consequently, after differentiating with re- spect to y we see that f xy = 3 y + 2 x = 3 x + 2 y xy . 003 10.0 points Evaluate the iterated integral I = integraldisplay 5 1 braceleftBig integraldisplay 5 1 parenleftBig x y + y x parenrightBig dy bracerightBig dx . 1. I = 24 ln 5 correct
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Van Ligten (hlv63) – EXAM 2 Review – Gilbert – (56650) 2 2. I = 5 ln 12 3. I = 24 ln 12 4. I = 12 ln 5 5. I = 12 ln 24 6. I = 5 ln 24 Explanation: Integrating with respect to y keeping x fixed, we see that integraldisplay 5 1 parenleftbigg x y + y x parenrightbigg dy = bracketleftbigg x ln y + y 2 2 x bracketrightbigg 5 1 = (ln 5) x + 12 parenleftbigg 1 x parenrightbigg . Thus I = integraldisplay 5 1 bracketleftbigg (ln 5) x + 12 parenleftbigg 1 x parenrightbiggbracketrightbigg dx = bracketleftbigg parenleftbigg x 2 2 parenrightbigg ln 5 + 12 ln x bracketrightbigg 5 1 . Consequently, I = 24 ln 5 . 004 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A xy 2 x 2 + 3 dxdy when A = braceleftBig ( x, y ) : 0 x 1 , - 3 y 3 bracerightBig . 1. I = 9 2 ln parenleftbigg 4 3 parenrightbigg 2. I = 9 ln (2) 3. I = 9 2 ln (2) 4. I = 27 4 ln parenleftbigg 4 3 parenrightbigg 5. I = 27 4 ln (2) 6. I = 9 ln parenleftbigg 4 3 parenrightbigg correct Explanation: Since A = braceleftBig ( x, y ) : 0 x 1 , - 3 y 3 bracerightBig is a rectangle with sides parallel to the coor- dinate axes, the double integal can be inter- preted as the iterated integral integraldisplay 3 3 parenleftbiggintegraldisplay 1 0 xy 2 x 2 + 3 dx parenrightbigg dy . But to integrate integraldisplay 1 0 xy 2 x 2 + 3 dx with respect to x with y fixed we use the substitution u = x 2 + 3. For then du = 2 x dx while x = 0 = u = 3 , x = 1 = u = 4 . In this case integraldisplay 1 0 xy 2 x 2 + 3 dx = 1 2 integraldisplay 4 3 y 2 u du = y 2 2 bracketleftBig ln u bracketrightBig 4 3 .
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