# Homework 9 answers - ejm753 – Homework 9 – Cepparo...

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Unformatted text preview: ejm753 – Homework 9 – Cepparo – (58400) 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 < , f y < 2. f x > , f y = 0 3. f x = 0 , f y = 0 4. f x < , f y > 5. f x = 0 , f y < correct 6. f x = 0 , f y > 7. f x < , f y = 0 8. f x > , f y > 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 < . 002 10.0 points Determine f x when f ( x, y ) = x sin(4 y- x ) + cos(4 y- x ) . 1. f x =- cos(4 y- x )- x sin(4 y- x ) 2. f x =- x cos(4 y- x ) 3. f x = 2 sin(4 y- x )- x cos(4 y- x ) correct 4. f x = x cos(4 y- x )- sin(4 y- x ) 5. f x = x cos(4 y- x ) 6. f x = x sin(4 y- x ) 7. f x =- 2 sin(4 y- x )- x cos(4 y- x ) 8. f x =- x sin(4 y- x ) Explanation: From the Product Rule we see that f x = sin(4 y- x )- x cos(4 y- x )+sin(4 y- x ) . Consequently, f x = 2 sin(4 y- 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 ) = 3(2 x + y ) e- xy . ejm753 – Homework 9 – Cepparo – (58400) 2 1. slope =- 2 2. slope =- 6 correct 3. slope =- 4 4. slope = 0 5. slope = 2 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 = 6 e- xy- 3(2 xy + y 2 ) e- xy . Consequently, at P (0 , 2 , f (0 , 2)) slope =- 6 . 004 10.0 points Determine h = h ( x, y ) so that ∂f ∂x = h ( x, y ) (4 x 2 + 3 y 2 ) 2 when f ( x, y ) = 5 x 2 y 4 x 2 + 3 y 2 . 1. h ( x, y ) = 15 xy 3 2. h ( x, y ) = 15 xy 2 3. h ( x, y ) = 30 xy 3 correct 4. h ( x, y ) = 15 x 3 y 5. h ( x, y ) = 30 xy 2 6. h ( x, y ) = 30 x 3 y Explanation: Differentiating with respect to x using the quotient rule we obtain ∂f ∂x = 10 xy (4 x 2 + 3 y 2 )- 40 x 3 y (4 x 2 + 3 y 2 ) 2 ....
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Homework 9 answers - ejm753 – Homework 9 – Cepparo...

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