{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 11-solutions

# HW 11-solutions - suleimenov(bs26835 HW 11 rusin(55565 This...

This preview shows pages 1–3. Sign up to view the full content.

suleimenov (bs26835) – HW 11 – rusin – (55565) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Sorry for the delay – I got wrapped up grading the tests... This HW will be due Monday evening. 001 10.0points Use the Chain Rule to find the partial derivative ∂w ∂s for w = x 2 + y 2 + z 2 , x = st, y = s cos t, z = s sin t when s = 3, t = 0. 1. ∂w ∂s = 10 2. ∂w ∂s = 3 3. ∂w ∂s = 5 4. ∂w ∂s = 6 correct 5. ∂w ∂s = 8 Explanation: By the Chain Rule for Partial Differentia- tion ∂w ∂s = ∂w ∂x ∂x ∂s + ∂w ∂y ∂y ∂s + ∂w ∂z ∂z ∂s . Here, we have ∂w ∂x = 2 x, ∂x ∂s = t while ∂w ∂y = 2 y, ∂y ∂s = cos t and ∂w ∂z = 2 z, ∂z ∂s = sin t . Thus ∂w ∂s = 2 xt + 2 y cos t + 2 z sin t . Note that when s = 3 and t = 0, it follows that x = 0, y = 3, z = 0. Consequently, for these values, ∂w ∂s = 6 . keywords: 002 10.0points Use partial differentiation and the Chain Rule applied to F ( x, y ) = 0 to determine dy/dx when F ( x, y ) = cos( x 5 y ) xe 3 y = 0 . 1. dy dx = sin( x 5 y ) + e 3 y 5 xe 3 y 3 sin( x 5 y ) 2. dy dx = sin( x 5 y ) + 3 xe 3 y 5 sin( x 5 y ) e 3 y 3. dy dx = sin( x 5 y ) 3 e 3 y 3 sin( x 5 y ) 5 xe 3 y 4. dy dx = sin( x 5 y ) 3 xe 3 y 3 sin( x 5 y ) 5 e 3 y 5. dy dx = sin( x 5 y ) + e 3 y 3 xe 3 y 5 sin( x 5 y ) 6. dy dx = sin( x 5 y ) + e 3 y 5 sin( x 5 y ) 3 xe 3 y correct 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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
suleimenov (bs26835) – HW 11 – rusin – (55565) 2 When F ( x, y ) = cos( x 5 y ) xe 3 y = 0 , therefore, dy dx = sin( x 5 y ) e 3 y 5 sin( x 5 y ) 3 xe 3 y . Consequently, dy dx = sin( x 5 y ) + e 3 y 5 sin( x 5 y ) 3 xe 3 y . 003 10.0points Use the equation ∂z ∂y = ∂F ∂y ∂F ∂z to find ∂z ∂y for xe 2 y + 5 yz + ze 5 x = 0 . 1. ∂z ∂y = 2 e 2 y + 5 y 5 z + e 5 x 2. ∂z ∂y = 2 xe 2 y + 5 z 5 y + e 5 x correct 3. ∂z ∂y = 2 e 2 y + 5 y 5 z + 5 e 5 x 4. ∂z ∂y = 2 xe 2 y + 5 z 5 y + 5 e 5 x 5. ∂z ∂y = e 5 x + 5 y 5 z + 2 xe 2 y Explanation: 004 10.0points The temperature at a point ( x, y ) in the plane is T ( x, y ) C. If a bug crawls on the plane so that its position in the plane after t minutes is given by ( x ( t ) , y ( t )) where x = 5 + t , y = 3 + 3 4 t , determine how fast the temperature is rising on the bug’s path at t = 4 when T x (3 , 6) = 18 , T y (3 , 6) = 8 . 1. rate = 10 C / min 2. rate = 11 C / min 3. rate = 12 C / min 4. rate = 9 C / min correct 5. rate = 13 C / min Explanation: By the Chain Rule for partial differentia- tion, the rate of change of temperature on the bug’s path at ( x ( t ) , y ( t )) is given by dT dt = dT ( x ( t ) , y ( t )) dt = ∂T ∂x dx dt + ∂T ∂y dy dt . Now dx dt = 1 2 5 + t , dy dt = 3 4 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}