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Make up 2

# Make up 2 - Rehman(aar638 – Makeup Exam 2 – sachse...

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Unformatted text preview: Rehman (aar638) – Makeup Exam 2 – sachse – (56620) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the slope in the x-direction at the point P (0 , 2 , f (0 , 2)) on the graph of f when f ( x, y ) = 2(2 x + y ) e − xy . 1. slope =- 10 2. slope =- 4 correct 3. slope =- 8 4. slope =- 12 5. slope =- 6 Explanation: The graph of f is a surface in 3-space and the slope in the x-direction at the point P (0 , 2 , f (0 , 2)) on that surface is the value of the partial derivative f x at (0 , 2). Now f x = 4 e − xy- 2(2 xy + y 2 ) e − xy . Consequently, at P (0 , 2 , f (0 , 2)) slope =- 4 . 002 10.0 points Determine the second partial f xy of f when f ( x, y ) = 6 x 2 y + y 2 6 x . 1. f xy = 12 x- y 2. f xy = 12 x y 2 + y 3 x 2 3. f xy =- 12 x y 2- y 3 x 2 correct 4. f xy = 12 x y 2- y 3 x 2 5. f xy = 12 x + y Explanation: Differentiating with respect to x , we obtain f x = 12 x y- y 2 6 x 2 , and so after differentiation with respect to y we see that f xy =- 12 x y 2- y 3 x 2 . 003 10.0 points Determine the value of the double integral I = integraldisplay integraldisplay A 3 xy 2 9 + x 2 dA over the rectangle A = braceleftBig ( x, y ) : 0 ≤ x ≤ 4 ,- 4 ≤ y ≤ 4 bracerightBig , integrating first with respect to y . 1. I = 32 ln parenleftBig 25 9 parenrightBig 2. I = 32 ln parenleftBig 25 18 parenrightBig 3. I = 64 ln parenleftBig 25 9 parenrightBig correct 4. I = 64 ln parenleftBig 25 18 parenrightBig 5. I = 32 ln parenleftBig 9 25 parenrightBig 6. I = 64 ln parenleftBig 9 25 parenrightBig Explanation: The double integral over the rectangle A can be represented as the iterated integral I = integraldisplay 4 parenleftbiggintegraldisplay 4 − 4 3 xy 2 9 + x 2 dy parenrightbigg dx , Rehman (aar638) – Makeup Exam 2 – sachse – (56620) 2 integrating first with respect to y . Now after integration with respect to y with x fixed, we see that integraldisplay 4 − 4 3 xy 2 9 + x 2 dy = bracketleftBig xy 3 9 + x 2 bracketrightBig 4 − 4 = 128 x 9 + x 2 . But integraldisplay 4 128 x 9 + x 2 dx = bracketleftBig 64 ln(9 + x 2 ) bracketrightBig 4 . Consequently, I = 64 ln parenleftBig 25 9 parenrightBig . 004 10.0 points Find the value of the double integral I = integraldisplay integraldisplay A (8 x- y ) dxdy when A is the region braceleftBig ( x, y ) : y ≤ x ≤ √ y, ≤ y ≤ 1 bracerightBig . 1. I = 7 10 2. I = 1 3. I = 9 10 4. I = 4 5 5. I = 3 5 correct Explanation: The integral can be written as the repeated integral I = integraldisplay 1 bracketleftBigg integraldisplay √ y y (8 x- y ) dx bracketrightBigg dy....
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Make up 2 - Rehman(aar638 – Makeup Exam 2 – sachse...

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