This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: romasko (qrr58) – Assignment 5 – luecke – (55035) 1 This printout should have 18 questions. Multiplechoice 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 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 = 0 . keywords: surface, partial derivative, first or der partial derivative, graphical interpreta tion 002 10.0 points 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( 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 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 = 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 ....
View
Full
Document
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.
 Fall '07
 Gilbert

Click to edit the document details