HW11S - Gauthier Joseph Homework 11 Due 3:00 am Inst...

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Gauthier, Joseph – Homework 11 – Due: Nov 10 2007, 3:00 am – Inst: JEGilbert 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 Use the Chain Rule to find dw dt when w = xe y/z and x = t 2 , y = 5 - t , z = 5 + 4 t . 1. dw dt = t + x z + 5 xy z 2 · e y/z 2. dw dt = t + x z + 5 xy z · e y/z 3. dw dt = 2 t + x z + 4 xy z 2 · e y/z 4. dw dt = 2 t - x z - 4 xy z · e y/z 5. dw dt = 2 t - x z - 4 xy z 2 · e y/z correct 6. dw dt = t - x z - 5 xy z · e y/z Explanation: By the Chain Rule for Partial Differentia- tion, dw dt = ∂w ∂x dx dt + ∂w ∂y dy dt + ∂w ∂z dz dt . When w = xe y/z when x = t 2 , y = 5 - t , z = 5 + 4 t , therefore, dw dt = 2 te y/z - x z e y/z - 4 xy z 2 e y/z . Consequently, dw dt = 2 t - x z - 4 xy z 2 · e y/z . keywords: 002 (part 1 of 1) 10 points Use partial differentiation and the Chain Rule applied to F ( x, y ) = 0 to determine dy/dx when F ( x, y ) = cos( x - 2 y ) - xe 6 y = 0 . 1. dy dx = sin( x - 2 y ) + e 6 y 2 sin( x - 2 y ) - 6 xe 6 y correct 2. dy dx = sin( x - 2 y ) - 6 e 6 y 6 sin( x - 2 y ) - 2 xe 6 y 3. dy dx = sin( x - 2 y ) + 6 xe 6 y 2 sin( x - 2 y ) - e 6 y 4. dy dx = sin( x - 2 y ) - 6 xe 6 y 6 sin( x - 2 y ) - 2 e 6 y 5. dy dx = sin( x - 2 y ) + e 6 y 2 xe 6 y - 6 sin( x - 2 y ) 6. dy dx = sin( x - 2 y ) + e 6 y 6 xe 6 y - 2 sin( x - 2 y ) Explanation: Applying the Chain Rule to both sides of the equation F ( x, y ) = 0, we see that ∂F ∂x dx dx + ∂F ∂y dy dx = ∂F ∂x + ∂F ∂y dy dx = 0 . Thus dy dx = - ∂F ∂x ∂F ∂y = - F x F y . When F ( x, y ) = cos( x - 2 y ) - xe 6 y = 0 ,
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Gauthier, Joseph – Homework 11 – Due: Nov 10 2007, 3:00 am – Inst: JEGilbert 2 therefore, dy dx = - - sin( x - 2 y ) - e 6 y 2 sin( x - 2 y ) - 6 xe 6 y . Consequently, dy dx = sin( x - 2 y ) + e 6 y 2 sin( x - 2 y ) - 6 xe 6 y . keywords: partial differentiation, Chain Rule, implicit differentiation, Implicit Function Theorem 003 (part 1 of 1) 10 points The temperature at a point ( x, y ) is T ( x, y ) measured in degrees Celsius. If a bug crawls so that its position after t minutes is given by x = 11 + t , y = 3 + 3 5 t , with x, y measured in centimeters, determine how fast the temperature rising on the bug’s path after 5 minutes when T x (4 , 6) = 24 , T y (4 , 6) = 10 . 1. rate = 9 C / min correct 2. rate = 10 C / min 3. rate = 7 C / min 4. rate = 6 C / min 5. rate = 8 C / min Explanation: By the Chain Rule for partial differentia- tion, the rate of change of temperatuure T on the bug’s path is given by dT dt = dT ( x ( t ) , y ( t )) dt = ∂T ∂x dx dt + ∂T ∂y dy dt = 1 11 + t ∂T ∂x + 3 5 ∂T ∂y . But when t = 5, the bug is at the point (4 , 6), while dx dt = 1 4 , dy dt = 3 5 . Consequently, after 5 minutes the tempera- ture on the bug’s path is changing at a rate = 9 C / min . keywords: 004 (part 1 of 1) 10 points The radius of a right circular cylinder is increasing at a rate of 3 inches per minute while the height is decreasing at a rate of 2 inches per minute. Determine the rate of change of the volume when r = 4 and h = 3.
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