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Unformatted text preview: gutierrez (ig3472) HW10 Neitzke (56585) 1 This printout should have 19 questions. Multiplechoice 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 > 2. f x = 0 , f y < 3. f x < , f y > 4. f x = 0 , f y = 0 5. f x < , f y = 0 6. f x > , f y > 7. f x < , f y < correct 8. f x > , 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 ydirection. Thus the sign of f x indicates whether f is increasing or decreasing in the xdirection, 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 ydirection, and so the sign of f y indicates whether f is increasing or decreasing in the ydirection, 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 < . keywords: surface, partial derivative, first or der partial derivative, graphical interpreta tion 002 10.0 points Determine f x when f ( x, y ) = x cos( x + 2 y ) + sin( x + 2 y ) . 1. f x = 2 x sin( x + 2 y ) 2. f x = x cos( x + 2 y ) 3. f x = 2 cos( x + 2 y ) + x sin( x + 2 y ) 4. f x = x sin( x + 2 y ) 5. f x = 2 x cos( x + 2 y ) 6. f x = 2 cos( x +2 y ) x sin( x +2 y ) correct 7. f x = 2 sin( x + 2 y ) x cos( x + 2 y ) 8. f x = 2 sin( x + 2 y ) + x cos( x + 2 y ) Explanation: From the Product Rule we see that f x = cos( x +2 y ) x sin( x +2 y )+cos( x +2 y ) . Consequently, f x = 2 cos( x + 2 y ) x sin( x + 2 y ) . 003 10.0 points gutierrez (ig3472) HW10 Neitzke (56585) 2 Find the slope in the xdirection at the point P (0 , 2 , f (0 , 2)) on the graph of f when f ( x, y ) = 2(2 x + y ) e xy . 1. slope = 4 correct 2. slope = 8 3. slope = 10 4. slope = 2 5. slope = 6 Explanation: The graph of f is a surface in 3space and the slope in the xdirection 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 f xy when f ( x, y ) = 2 x tan 1 parenleftBig y x parenrightBig . 1. f xy = 4 x 2 y ( x 2 + y 2 ) 2 2. f xy = xy 2 ( x 2 + y 2 ) 2 3. f xy = x 2 y x 2 + y 2 4. f xy = 4 xy 2 ( x 2 + y 2 ) 2 correct 5. f xy = 4 xy x 2 + y 2 6. f xy = xy x 2 + y 2 Explanation: Differentiating f partially with respect to x , using also the Chain Rule and d dx tan 1 x = 1 1 + x 2 , we see that f x = 2 tan 1 parenleftBig y x parenrightBig 2 x parenleftBig y x 2 parenrightBigparenleftBig 1 1 + ( y/x ) 2 parenrightBig ....
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This note was uploaded on 04/13/2010 for the course MATH 408L taught by Professor Gogolev during the Spring '09 term at University of Texas at Austin.
 Spring '09
 GOGOLEV
 Calculus, Slope

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