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Homework #7 - jiwani(amj566 Homework07 Fouli(58395 This...

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jiwani (amj566) – Homework07 – Fouli – (58395) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find y when xy + 5 x + 2 x 2 = 4 . 1. y = y + 5 + 2 x x 2. y = 5 + 2 x - y x 3. y = - ( y + 5 + 4 x ) 4. y = - y + 5 + 2 x x 5. y = y + 5 + 4 x x 6. y = - y + 5 + 4 x x correct Explanation: Differentiating implicitly with respect to x we see that d dx ( xy + 5 x + 2 x 2 ) = d dx (4) . Thus ( xy + y ) + 5 + 4 x = 0 , and so xy = - y - 5 - 4 x . Consequently, y = - y + 5 + 4 x x . 002 10.0 points Find dy dx when 4 x + 1 y = 7 . 1. dy dx = 4 parenleftBig y x parenrightBig 3 / 2 2. dy dx = 4( xy ) 1 / 2 3. dy dx = 1 4 ( xy ) 1 / 2 4. dy dx = 1 4 parenleftBig x y parenrightBig 3 / 2 5. dy dx = - 1 4 parenleftBig x y parenrightBig 3 / 2 6. dy dx = - 4 parenleftBig y x parenrightBig 3 / 2 correct Explanation: Differentiating implicitly with respect to x , we see that - 1 2 parenleftBig 4 x x + 1 y y dy dx parenrightBig = 0 . Consequently, dy dx = - 4 parenleftBig y x parenrightBig 3 / 2 . 003 10.0 points Find dy dx when tan(2 x + y ) = 2 x . 1. dy dx = - 4 x 2 2 + y 2 2. dy dx = 8 x 2 1 + 4 x 2 3. dy dx = 4 y 2 2 + x 2 4. dy dx = - 8 x 2 1 + 4 x 2 correct 5. dy dx = - 8 y 2 2 + x 2
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jiwani (amj566) – Homework07 – Fouli – (58395) 2 6. dy dx = 8 x 2 1 + 4 y 2 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 . 004 10.0 points Find the equation of the tangent line to the graph of y 2 - xy - 20 = 0 , at the point P = (1 , 5). 1. y = 5 x 2. 9 y + 5 x = 40 3. 8 y + 5 x = 35 4. 9 y = 5 x + 40 correct 5. 8 y = 5 x + 35 Explanation: Differentiating implicitly with respect to x we see that 2 y dy dx - y - x dy dx = 0 , so dy dx = y 2 y - x . At P = (1 , 5), therefore, dy dx vextendsingle vextendsingle vextendsingle P = 5 9 . Thus by the point slope formula, the equation of the tangent line at P is given by y - 5 = 5 9 ( x - 1) . Consequently, 9 y = 5 x + 40 . 005 10.0 points Find the slope of the tangent line to the graph of 2 x 3 + y 3 + xy = 0 at the point P (1 , - 1). 1. slope = 3 2 2. slope = - 2 3 3. slope = 4 5 4. slope = - 4 5 5. slope = 5 4 6. slope = - 5 4 correct
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jiwani (amj566) – Homework07 – Fouli – (58395) 3 Explanation: Differentiating implicitly with respect to x we see that 6 x 2 + 3 y 2 dy dx + y + x dy dx = 0 . Consequently, dy dx = - 6 x 2 + y 3 y 2 + x . Hence at P (1 , - 1) slope = dy dx vextendsingle vextendsingle vextendsingle P = - 5 4 . 006 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 .
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