{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ca2-2 - Version PREVIEW FinalReview02 Van Ligten(56650 This...

This preview shows pages 1–3. Sign up to view the full content.

Version PREVIEW – FinalReview02 – Van Ligten – (56650) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. CalC15c10a 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 ) = 4(2 x + y ) e xy . 1. slope = - 10 2. slope = - 14 3. slope = - 6 4. slope = - 12 5. slope = - 8 correct 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 = 8 e xy - 4(2 xy + y 2 ) e xy . Consequently, at P (0 , 2 , f (0 , 2)) slope = - 8 . CalC15c49a 002 10.0 points Determine the second partial f xy of f when f ( x, y ) = 7 x 2 y + y 2 10 x . 1. f xy = 14 x y 2 + y 5 x 2 2. f xy = 14 x - y 3. f xy = - 14 x y 2 - y 5 x 2 correct 4. f xy = 14 x + y 5. f xy = 14 x y 2 - y 5 x 2 Explanation: Differentiating with respect to x , we obtain f x = 14 x y - y 2 10 x 2 , and so after differentiation with respect to y we see that f xy = - 14 x y 2 - y 5 x 2 . CalC16b16s 003 10.0 points Determine the value of the double integral I = integraldisplay integraldisplay A 3 xy 2 16 + x 2 dA over the rectangle A = braceleftBig ( x, y ) : 0 x 3 , - 2 y 2 bracerightBig , integrating first with respect to y . 1. I = 4 ln parenleftBig 16 25 parenrightBig 2. I = 8 ln parenleftBig 25 32 parenrightBig 3. I = 8 ln parenleftBig 25 16 parenrightBig correct 4. I = 8 ln parenleftBig 16 25 parenrightBig 5. I = 4 ln parenleftBig 25 16 parenrightBig 6. I = 4 ln parenleftBig 25 32 parenrightBig Explanation:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Version PREVIEW – FinalReview02 – Van Ligten – (56650) 2 The double integral over the rectangle A can be represented as the iterated integral I = integraldisplay 3 0 parenleftbigg integraldisplay 2 2 3 xy 2 16 + x 2 dy parenrightbigg dx , integrating first with respect to y . Now after integration with respect to y with x fixed, we see that integraldisplay 2 2 3 xy 2 16 + x 2 dy = bracketleftBig xy 3 16 + x 2 bracketrightBig 2 2 = 16 x 16 + x 2 . But integraldisplay 3 0 16 x 16 + x 2 dx = bracketleftBig 8 ln(16 + x 2 ) bracketrightBig 3 0 . Consequently, I = 8 ln parenleftBig 25 16 parenrightBig . CalC16c07b 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, 0 y 1 bracerightBig . 1. I = 9 10 2. I = 4 5 3. I = 7 10 4. I = 3 5 correct 5. I = 1 Explanation: The integral can be written as the repeated integral I = integraldisplay 1 0 bracketleftBigg integraldisplay y y (8 x - y ) dx bracketrightBigg dy. Now integraldisplay y y (8 x - y ) dx = bracketleftBig 4 x 2 - xy bracketrightBig y y = 4 y - y 3 / 2 - 3 y 2 . But then I = integraldisplay 1 0 (4 y - y 3 / 2 - 3 y 2 ) dx = bracketleftBig 2 y 2 - 2 5 y 5 / 2 - y 3 bracketrightBig 1 0 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

ca2-2 - Version PREVIEW FinalReview02 Van Ligten(56650 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online