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Homework 9

# Homework 9 - tgo72 Homework 9 Gompf(58370 This print-out...

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tgo72 – Homework 9 – Gompf – (58370) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 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 correct 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 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 . 002 10.0 points Determine f x when f ( x, y ) = cos(4 y - x ) - x sin(4 y - x ) . 1. f x = - x sin(4 y - x ) 2. f x = x cos(4 y - x ) - sin(4 y - x ) 3. f x = x sin(4 y - x ) 4. f x = - x cos(4 y - x ) 5. f x = - cos(4 y - x ) - x sin(4 y - x ) 6. f x = - 2 sin(4 y - x ) - x cos(4 y - x ) 7. f x = 2 sin(4 y - x ) - x cos(4 y - x ) 8. f x = x cos(4 y - x ) correct Explanation: From the Product Rule we see that f x = sin(4 y - x ) - sin(4 y - x )+ x cos(4 y - x ) . Consequently, f x = x cos(4 y - x ) . 003 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 .

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tgo72 – Homework 9 – Gompf – (58370) 2 1. slope = - 10 2. slope = - 12 3. slope = - 6 4. slope = - 4 correct 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 = 4 e - xy - 2(2 xy + y 2 ) e - xy . Consequently, at P (0 , 2 , f (0 , 2)) slope = - 4 . 004 10.0 points Determine h = h ( x, y ) so that ∂f ∂x = h ( x, y ) ( x 2 + 4 y 2 ) 2 when f ( x, y ) = 2 x 2 y x 2 + 4 y 2 . 1. h ( x, y ) = 8 xy 3 2. h ( x, y ) = 16 xy 2 3. h ( x, y ) = 16 x 3 y 4. h ( x, y ) = 16 xy 3 correct 5. h ( x, y ) = 8 xy 2 6. h ( x, y ) = 8 x 3 y Explanation: Differentiating with respect to x using the quotient rule we obtain ∂f ∂x = 4 xy ( x 2 + 4 y 2 ) - 4 x 3 y ( x 2 + 4 y 2 ) 2 . Consequently, h ( x, y ) = 16 xy 3 . 005 10.0 points Find f y when f ( x, y ) = integraldisplay x y cos ( t 4 ) dt . 1. f y = 4 x 3 sin ( y 4 ) 2. f y = - cos ( y 4 ) correct 3. f y = cos ( y 4 ) 4. f y = - sin ( y 4 ) 5. f y = sin ( y 4 ) 6. f y = 4 x 3 cos ( y 4 ) Explanation: Now integraldisplay x y cos ( t 4 ) dt = - integraldisplay y x cos ( t 4 ) dt .
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Homework 9 - tgo72 Homework 9 Gompf(58370 This print-out...

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