MATH200 December 2013
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Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Apply the chain rule for multi variable functions

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Solution

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Apply the chain rule at the point $\displaystyle (\rho _{0},r_{0},\theta _{0})=(2,3,\pi /2)$:
${\frac {\partial u}{\partial r}}={\frac {\partial u}{\partial x}}{\frac {\partial x}{\partial r}}+{\frac {\partial u}{\partial y}}{\frac {\partial y}{\partial r}}+{\frac {\partial u}{\partial z}}{\frac {\partial z}{\partial r}}$
Computing all of these elements:
${\frac {\partial u}{\partial x}}=2x=2\rho r\cos(\theta )=2(2)(3)(0)=0$
${\frac {\partial u}{\partial y}}=z=\rho r=(2)(3)=6$
${\frac {\partial u}{\partial z}}=y=\rho r\sin(\theta )=(2)(3)(1)=6$
(Notice that because ${\frac {\partial u}{\partial x}}=0$, we won't need to consider ${\frac {\partial x}{\partial r}}$ any longer)
${\frac {\partial y}{\partial r}}=\rho \sin(\theta )=(2)(1)=2$
${\frac {\partial z}{\partial r}}=\rho =2$
Sewing everything up,
${\frac {\partial u}{\partial r}}=0+{\frac {\partial u}{\partial y}}{\frac {\partial y}{\partial r}}+{\frac {\partial u}{\partial z}}{\frac {\partial z}{\partial r}}=(6)(2)+(6)(2)=\color {blue}24$


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