M408m-Luecke_HW5 - romasko(qrr58 – Assignment 5 – luecke –(55035 1 This print-out should have 18 questions Multiple-choice questions may

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Unformatted text preview: romasko (qrr58) – Assignment 5 – luecke – (55035) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Note the change in due date to Thursday, 10/7, at 3 AM. This is because of the class cancellation on Tuesday. 001 10.0 points Determine whether the partial derivatives f x , f y of f are positive, negative or zero at the point P on the graph of f shown in P x z y 1. f x < , f y = 0 2. f x = 0 , f y = 0 3. f x = 0 , f y < 4. f x < , f y < 5. f x = 0 , f y > 6. f x > , f y > 7. f x < , f y > 8. f x > , f y = 0 correct Explanation: The value of f x at P is the slope of the tangent line to graph of f at P in the x- direction, while f y is the slope of the tangent line in the y-direction. Thus the sign of f x indicates whether f is increasing or decreasing in the x-direction, or whether the tangent line in that direction at P is horizontal. Similarly, the value of f y at P is the slope of the tangent line at P in the y-direction, and so the sign of f y indicates whether f is increasing or decreasing in the y-direction, or whether the tangent line in that direction at P is horizontal. From the graph it thus follows that at P f x > , f y = 0 . keywords: surface, partial derivative, first or- der partial derivative, graphical interpreta- tion 002 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( y 2- x 2 ) ln( x + y ) . 1. slope = 4 correct 2. slope = 0 3. slope = 2 4. slope = 6 5. slope = 8 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 = 2 parenleftbigg- 2 x ln( x + y ) + y 2- x 2 x + y parenrightbigg . Consequently, at P (0 , 2 , f (0 , 2)) slope = 2 × 2 = 4 . romasko (qrr58) – Assignment 5 – luecke – (55035) 2 003 (part 1 of 3) 10.0 points If f is defined by f ( x, y ) = 2 3 x 3 + 2 x 2 + 2 xy + y 2 + 4 y , (i) locate the critical points of f . 1. (2 , 0) , (- 1 , 3) 2. (- 2 , 0) , (- 1 , 3) 3. (- 2 , 0) , (1 ,- 3) correct 4. (- 2 , 0) , (1 , 3) 5. (2 , 0) , (1 ,- 3) 6. (2 , 0) , (1 , 3) Explanation: After differentiation f x = 2 x 2 + 4 x + 2 y, while f y = 2 x + 2 y + 4 . Thus the critical points of f are the common solutions of the equations 2 x 2 + 4 x + 2 y = 0 2 x + 2 y + 4 = 0 . Eliminating y we see that 2 x 2 + 2 x- 4 = 2 ( x + 2)( x- 1) = 0 , the solutions of which are x =- 2 and 1. Consequently, the critical points of f occur at (- 2 , 0) , (1 ,- 3) . 004 (part 2 of 3) 10.0 points (ii) Which of the following most accurately de- scribes the behaviour of f ....
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This note was uploaded on 06/15/2011 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.

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M408m-Luecke_HW5 - romasko(qrr58 – Assignment 5 – luecke –(55035 1 This print-out should have 18 questions Multiple-choice questions may

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