HW9-solutions - wei (jw35975) HW9 milburn (54685) 1 This...

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Unformatted text preview: wei (jw35975) HW9 milburn (54685) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find dy dx when x 3 y 3- y = x . 1. dy dx = 1- 3 x 2 y 3 3 x 3 y 2- 1 correct 2. dy dx = 1- 2 x 3 y 3 2 x 3 y 2- 1 3. dy dx = 1- 3 x 3 y 3 3 x 3 y 2- 2 4. dy dx = 1 + 3 x 2 y 3 3 x 3 y 2- 2 5. dy dx = 1- 3 x 3 y 2 3 x 2 y 2- 1 Explanation: Differentiating x 3 y 3- y = x implicitly with respect to x we see that 3 x 3 y 2 dy dx + 3 y 3 x 2- dy dx = 1 . Thus 3 x 3 y 2 dy dx- dy dx = 1- 3 x 2 y 3 and so dy dx = 1- 3 x 2 y 3 3 x 3 y 2- 1 . 002 10.0 points The points P and Q on the graph of y 2- xy + 6 = 0 have the same x-coordinate x = 5. Find the point of intersection of the tangents to the graph at P and Q . 1. intersect at = parenleftBig 12 5 , 12 5 parenrightBig 2. intersect at = parenleftBig 24 5 , 24 5 parenrightBig 3. intersect at = parenleftBig 8 5 , 6 5 parenrightBig 4. intersect at = parenleftBig 24 5 , 12 5 parenrightBig correct 5. intersect at = parenleftBig 12 5 , 24 5 parenrightBig Explanation: The y-coordinate at P, Q will be the solu- tions of y 2- xy + 6 = 0 at x = 5, i.e. , the solutions of y 2- 5 y + 6 = ( y- 2)( y- 3) = 0 . Thus P = (5 , 2) , Q = (5 , 3) . To determine the tangent lines we need also the value of the derivative at P and Q . But by implicit differentiation, 2 y dy dx- x dy dx- y = 0 . so dy dx = y 2 y- x . Thus dy dx vextendsingle vextendsingle vextendsingle P =- 2 , dy dx vextendsingle vextendsingle vextendsingle Q = 3 . By the point-slope formula, therefore, the equation of the tangent line at P is y- 2 =- 2( x- 5) , while that at Q is y- 3 = 3( x- 5) . wei (jw35975) HW9 milburn (54685) 2 Consequently, the tangent lines at P and Q are- y- 2 x =- 12 and- y + 3 x = 12 respectively. These two tangent lines intersect at = parenleftBig 24 5 , 12 5 parenrightBig . 003 10.0 points Determine d 2 y/dx 2 when x 2 + 2 y 2 = 5 . 1. d 2 y dx 2 =- 5 4 y 3 correct 2. d 2 y dx 2 = 5 4 y 3 3. d 2 y dx 2 = 5 4 y 2 4. d 2 y dx 2 =- 5 4 y 2 5. d 2 y dx 2 =- 1 4 y 3 Explanation: Differentiating implicitly with respect to x we see that 2 x + 4 y dy dx = 0 , which after simplification becomes dy dx =- 1 2 x y . But then d 2 y dx 2 =- d dx parenleftBig 1 2 x y parenrightBig =- parenleftbigg 2 y- 2 x dy dx 4 y 2 parenrightbigg =- 1 4 y 2 parenleftBig 2 y + x 2 y parenrightBig . Consequently, d 2 y dx 2 =- 1 4 y 3 parenleftBig x 2 + 2 y 2 parenrightBig =- 5 4 y 3 . 004 10.0 points Find the rate of change of q with respect to p when p = 20 q 2 + 5 . 1. dq dp =- 10 qp 2. dq dp =- 10 qp 2 correct 3. None of these 4. dq dp =- 5 q 2 p 5....
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This note was uploaded on 10/25/2011 for the course MATHEMATIC 408K taught by Professor Milburn during the Fall '11 term at University of Texas at Austin.

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HW9-solutions - wei (jw35975) HW9 milburn (54685) 1 This...

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