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M408m-Luecke_HW5

# M408m-Luecke_HW5 - romasko(qrr58 Assignment 5 luecke(55035...

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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 < 0 , f y = 0 2. f x = 0 , f y = 0 3. f x = 0 , f y < 0 4. f x < 0 , f y < 0 5. f x = 0 , f y > 0 6. f x > 0 , f y > 0 7. f x < 0 , f y > 0 8. f x > 0 , 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 > 0 , 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 .

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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 . 1. local minimum at ( - 2 , 0) 2. saddle-point at ( - 2 , 0) correct 3. saddle-point at (2 , 0) 4. local maximum at (2 , 0) 5. local minimum at (2 , 0) 6. local maximum at ( - 2 , 0) Explanation: After differentiation once again we see that f xx = 2(2 x + 2) , f yy = 2 , f xy = 2 , and so at the critical point ( - 2 , 0), A = f xx vextendsingle vextendsingle vextendsingle ( - 2 , 0) = - 4 , and C = f yy vextendsingle vextendsingle vextendsingle ( - 2 , 0) = 2 > 0 , while B = f xy vextendsingle vextendsingle vextendsingle ( - 2 , 0) = 2 .
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