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Homework 8 - Garcia Ilse Homework 8 Due 3:00 am Inst Fonken...

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Garcia, Ilse – Homework 8 – Due: Oct 17 2007, 3:00 am – Inst: Fonken 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find dy/dx when 3 x 2 + 2 y 2 = 7 . 1. dy dx = - 3 x y 2. dy dx = x 2 y 3. dy dx = - 3 x 2 y correct 4. dy dx = 3 x 2 y 5. dy dx = - 3 xy 6. dy dx = 2 xy Explanation: Diferentiating 3 x 2 + 2 y 2 = 7 implicitly with respect to x we see that 6 x + 4 y dy dx = 0 . Consequently, dy dx = - 6 x 4 y = - 3 x 2 y . keywords: implicit differentiation 002 (part 1 of 1) 10 points Find dy/dx when y + x = 4 xy . 1. dy dx = 2 q y x - 1 1 + 2 q x y 2. dy dx = q y x + 2 4 - q x y 3. dy dx = 2 q y x + 1 1 - 2 q x y 4. dy dx = q y x - 2 4 + q x y 5. dy dx = q y x + 2 4 + q x y 6. dy dx = 2 q y x - 1 1 - 2 q x y correct Explanation: Differentiating implicitly with respect to x , we see that dy dx + 1 = 2 r y x + r x y dy dx , so dy dx 1 - 2 r x y = 2 r y x - 1 . Consequently, dy dx = 2 q y x - 1 1 - 2 q x y . keywords: implicit differentiation 003 (part 1 of 1) 10 points If y = y ( x ) is defined implicitly by 4 y 2 - xy - 9 = 0 ,
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Garcia, Ilse – Homework 8 – Due: Oct 17 2007, 3:00 am – Inst: Fonken 2 find the value of dy/dx at (9 , 3). 1. dy dx fl fl fl (9 , 3) = - 1 5 2. dy dx fl fl fl (9 , 3) = 3 16 3. dy dx fl fl fl (9 , 3) = 4 15 4. dy dx fl fl fl (9 , 3) = 1 5 correct 5. dy dx fl fl fl (9 , 3) = - 4 15 Explanation: Differentiating implicitly with respect to x we see that 8 y dy dx - y - x dy dx = 0 . Thus dy dx = y 8 y - x . At (9 , 3), therefore, dy dx fl fl fl (9 , 3) = 1 5 . keywords: implicit differentiation 004 (part 1 of 1) 10 points Find the rate of change of q with respect to p when p = 20 q 2 + 5 . 1. dq dp = - 5 qp 2 2. dq dp = - 5 q 2 p 3. None of these 4. dq dp = - 10 qp 5. dq dp = - 10 qp 2 correct Explanation: Differentiating implicitly with respect to p we see that 1 = - 2 q 20 ( q 2 + 5) 2 · dq dp , and so dq dp = - ( q 2 + 5) 2 40 q . But q 2 + 5 = 20 p . Consequently, dq dp = - 10 qp 2 . keywords: implicit differentiation 005 (part 1 of 1) 10 points The graph of the equation of the ( x 2 + y 2 - 4 x ) 2 = 4( x 2 + y 2 ) is shown in the figure -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 2 4 6 8 10 12 2 4 6 - 2 - 4 - 6 P Find the equation of the tangent line to the graph at the point P = (2 , 2 3).
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Garcia, Ilse – Homework 8 – Due: Oct 17 2007, 3:00 am – Inst: Fonken 3 1. y = 1 3 · 3 x - 16 · 2. y = 1 3 · 1 3 x + 4 3 · 3. y = 1 3 · 3 x + 16 · 4. y = 1 3 · 1 3 x + 16 3 · correct 5. y = 1 3 · 1 3 x - 16 3 · Explanation: Differentiating ( x 2 + y 2 - 4 x ) 2 = 4( x 2 + y 2 ) implicitly with respect to x we see that 2( x 2 + y 2 - 4 x ) n 2 x + 2 y dy dx - 4 o = 8 n x + y dy dx o . Thus y dy dx = 4 x + 2(2 - x )( x 2 + y 2 - 4 x ) 2( x 2 + y 2 - 4 x ) - 4 . To simplify the algebra at P = 2 , 2 3 · , we note first that x 2 + y 2 - 4 x = 8 . In this case 4 x + 2(2 - x )( x 2 + y 2 - 4 x ) = 8 , while 2( x 2 + y 2 - 4 x ) - 4 = 12 . Consequently, y dy dx ·fl fl fl P = 2 3 dy dx ·fl fl fl P = 2 3 . The slope of the tangent line at P is thus given by dy dx ·fl fl fl P = 1 3 · 1 3 .
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