# Keywords 016 100 points use the chain rule to find z

• Notes
• 13
• 94% (53) 50 out of 53 people found this document helpful

This preview shows page 7 - 10 out of 13 pages.

keywords: 016 10.0 points Use the Chain Rule to find ∂z ∂u when z = e r cos θ and r = uv , θ = radicalbig u 2 + v 2 . 1. ∂z ∂u = e r parenleftBig v cos θ sin θ 2 u 2 + v 2 parenrightBig 2. ∂z ∂u = ue r parenleftBig v cos θ + sin θ u 2 + v 2 parenrightBig 3. ∂z ∂u = e r parenleftBig v cos θ u sin θ u 2 + v 2 parenrightBig correct 4. ∂z ∂u = e r parenleftBig v cos θ + u sin θ 2 u 2 + v 2 parenrightBig 5. ∂z ∂u = e r parenleftBig v cos θ + sin θ u 2 + v 2 parenrightBig Explanation: By the Chain Rule for Partial Differentia- tion, ∂z ∂u = ∂z ∂r ∂r ∂u + ∂z ∂θ ∂θ ∂u . Thus, since ∂z ∂r = e r cos θ, ∂r ∂u = v and ∂z ∂θ = e r sin θ, ∂θ ∂u = u u 2 + v 2 it follows that ∂z ∂u = e r parenleftBig v cos θ u sin θ u 2 + v 2 parenrightBig . keywords: partial differentiation, Chain Rule, 017 10.0 points 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 = 4, t = 0. 1. ∂w ∂s = 8 correct 2. ∂w ∂s = 12 3. ∂w ∂s = 6 4. ∂w ∂s = 9 5. ∂w ∂s = 11 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
moseley (cmm3869) – HW13 – Gilbert – (56380) 8 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 = 4 and t = 0, it follows that x = 0, y = 4, z = 0. Consequently, for these values, ∂w ∂s = 8 . keywords: 018 10.0 points 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 sin( x 5 y ) 3 xe 3 y correct 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 ) + e 3 y 3 xe 3 y 5 sin( x 5 y ) 4. dy dx = sin( x 5 y ) 3 e 3 y 3 sin( x 5 y ) 5 xe 3 y 5. dy dx = sin( x 5 y ) + e 3 y 5 xe 3 y 3 sin( x 5 y ) 6. dy dx = sin( x 5 y ) 3 xe 3 y 3 sin( x 5 y ) 5 e 3 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 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 . 019 10.0 points The temperature at a point ( x, y ) in the plane is T ( x, y ) measured in degrees Celsius. If a bug crawls so that its position in the plane after t minutes is given by x = 2 + t , y = 5 + 1 2 t , determine how fast is the temperature rising on the bug’s path after 2 minutes when T x (2 , 6) = 20 , T y (2 , 6) = 6 . 1. rate = 7 C / min 2. rate = 4 C / min 3. rate = 8 C / min correct 4. rate = 6 C / min 5. rate = 5 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 2 2 + t T x + 1 2 T y .
moseley (cmm3869) – HW13 – Gilbert – (56380) 9 But when t = 2, the bug is at the point (2 , 6), while 1 2 2 + t vextendsingle vextendsingle vextendsingle t =2 = 1 4 .