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Unformatted text preview: wtm369 – Homework 7 – Helleloid – (58250) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the equation of the tangent line to the graph of 3 y 2 − xy − 2 = 0 , at the point P = (5 , 2). 1. 3 y = x + 1 2. 7 y = 2 x + 4 correct 3. 5 y = 2 x 4. 7 y + 2 x = 4 5. 3 y + x = 1 Explanation: Differentiating implicitly with respect to x we see that 6 y dy dx − y − x dy dx = 0 , so dy dx = y 6 y − x . At P = (5 , 2), therefore, dy dx vextendsingle vextendsingle vextendsingle P = 2 7 . Thus by the point slope formula, the equation of the tangent line at P is given by y − 2 = 2 7 ( x − 5) . Consequently, 7 y = 2 x + 4 . 002 10.0 points Find dy dx when √ x + 3 √ y = 5 . 1. dy dx = 1 9 parenleftBig 5 + 1 √ x parenrightBig 2. dy dx = 5 − 1 √ x 3. dy dx = 1 + 5 √ x 4. dy dx = 1 9 parenleftBig 1 − 5 √ x parenrightBig correct 5. dy dx = 1 3 parenleftBig 5 − 1 √ x parenrightBig 6. dy dx = 1 3 parenleftBig 1 − 5 √ x parenrightBig Explanation: Differentiating implicitly with respect to x we see that 1 2 parenleftBig 1 √ x + 3 √ y dy dx parenrightBig = 0 . Thus dy dx = − 1 3 parenleftBig √ y √ x parenrightBig . But √ y = 5 − √ x 3 , so dy dx = 1 9 parenleftBig √ x − 5 √ x parenrightBig . Consequently, dy dx = 1 9 parenleftBig 1 − 5 √ x parenrightBig . 003 10.0 points Find the rate of change of q with respect to p when p = 20 q 2 + 5 . wtm369 – Homework 7 – Helleloid – (58250) 2 1. dq dp = − 10 qp 2. dq dp = − 5 q 2 p 3. dq dp = − 5 qp 2 4. None of these 5. dq dp = − 10 qp 2 correct Explanation: Differentiating implicitly with respect to p we see that 1 = − 2 q parenleftBig 20 ( q 2 + 5) 2 parenrightBig dq dp , and so dq dp = − ( q 2 + 5) 2 40 q . But q 2 + 5 = 20 p . Consequently, dq dp = − 10 qp 2 . 004 10.0 points Find dy dx when tan(2 x + y ) = 2 x . 1. dy dx = 8 x 2 1 + 4 x 2 2. dy dx = − 4 x 2 2 + y 2 3. dy dx = 4 y 2 2 + x 2 4. dy dx = − 8 y 2 2 + x 2 4. dy dx = 8 x 2 1 + 4 y 2 6. dy dx = − 8 x 2 1 + 4 x 2 correct Explanation: Differentiating tan(2 x + y ) = 2 x implicitly with respect to x we see that sec 2 (2 x + y ) parenleftbigg 2 + dy dx parenrightbigg = 2 . Thus sec 2 (2 x + y ) dy dx = 2(1 − sec 2 (2 x + y )) , in which case dy dx = 2(1 − sec 2 (2 x + y )) sec 2 (2 x + y ) . But sec 2 θ = 1 + tan 2 θ , so sec 2 (2 x + y ) = 1 + tan 2 (2 x + y ) = 1 + 4 x 2 , while 1 − sec 2 (2 x + y ) = − tan 2 (2 x + y ) = − 4 x 2 . Consequently, dy dx = − 8 x 2 1 + 4 x 2 . 005 10.0 points The points P and Q on the graph of y 2 − xy + 6 = 0 have the same xcoordinate x = 5. Find the point of intersection of the tangents to the graph at P and Q ....
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This note was uploaded on 12/04/2009 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz
 Calculus

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