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Homework 7-solutions - wtm369 Homework 7 Helleloid(58250...

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wtm369 – Homework 7 – Helleloid – (58250) 1 This print-out should have 24 questions. Multiple-choice 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 .
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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 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
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wtm369 – Homework 7 – Helleloid – (58250) 3 2. intersect at = parenleftBig 12 5 , 4 5 parenrightBig 3. intersect at = parenleftBig 24 5 , 12 5 parenrightBig correct 4. intersect at = parenleftBig 24 5 , 24 5 parenrightBig 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 3)( y 2) = 0 .
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