# hw9s - sakerwalla(hs9229 – HW9 – milburn –(54685 1...

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Unformatted text preview: sakerwalla (hs9229) – 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 3 y 2 3 x 2 y 2- 1 2. dy dx = 1- 3 x 2 y 3 3 x 3 y 2- 1 correct 3. dy dx = 1 + 3 x 2 y 3 3 x 3 y 2- 2 4. dy dx = 1- 3 x 3 y 3 3 x 3 y 2- 2 5. dy dx = 1- 2 x 3 y 3 2 x 3 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 + 8 = 0 have the same x-coordinate x = 6. Find the point of intersection of the tangents to the graph at P and Q . 1. intersect at = parenleftBig 8 3 , 2 3 parenrightBig 2. intersect at = parenleftBig 8 3 , 8 3 parenrightBig 3. intersect at = parenleftBig 16 3 , 16 3 parenrightBig 4. intersect at = parenleftBig 16 3 , 8 3 parenrightBig correct 5. intersect at = parenleftBig 8 3 , 16 3 parenrightBig Explanation: The y-coordinate at P, Q will be the solu- tions of y 2- xy + 8 = 0 at x = 6, i.e. , the solutions of y 2- 6 y + 8 = ( y- 4)( y- 2) = 0 . Thus P = (6 , 4) , Q = (6 , 2) . 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 =- 1 . By the point-slope formula, therefore, the equation of the tangent line at P is y- 4 = 2( x- 6) , while that at Q is y- 2 =- 1( x- 6) . sakerwalla (hs9229) – HW9 – milburn – (54685) 2 Consequently, the tangent lines at P and Q are 2 y- 4 x =- 16 and 2 y + 2 x = 16 respectively. These two tangent lines intersect at = parenleftBig 16 3 , 8 3 parenrightBig . 003 10.0 points Determine d 2 y/dx 2 when 4 x 2 + 3 y 2 = 5 . 1. d 2 y dx 2 =- 20 9 y 2 2. d 2 y dx 2 = 20 9 y 2 3. d 2 y dx 2 =- 20 9 y 3 correct 4. d 2 y dx 2 =- 4 9 y 3 5. d 2 y dx 2 = 20 9 y 3 Explanation: Differentiating implicitly with respect to x we see that 8 x + 6 y dy dx = 0 , which after simplification becomes dy dx =- 4 3 x y . But then d 2 y dx 2 =- d dx parenleftBig 4 3 x y parenrightBig =- parenleftbigg 12 y- 12 x dy dx 9 y 2 parenrightbigg =- 4 9 y 2 parenleftBig 3 y + 4 x 2 y parenrightBig . Consequently, d 2 y dx 2 =- 4 9 y 3 parenleftBig 4 x 2 + 3 y 2 parenrightBig =- 20 9 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 =- 5 q 2 p 2. None of these 3. dq dp =- 5 qp 2 4. dq dp =- 10 qp 5....
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hw9s - sakerwalla(hs9229 – HW9 – milburn –(54685 1...

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