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I have the following question on (i think) using a chain rule...

Let $$\frac{\partial\Theta}{\partial t} = \frac{K}{r^2}\frac{\partial}{\partial r} \left(r^2\frac{\partial\Theta}{\partial r}\right)$$ in which $K$ is a constant. By using the substitution $\Theta(r,t) = W(r,t)/r,$ show that this equation reduces to $$\frac{\partial W}{\partial t} = K\frac{\partial^2 W}{\partial r^2}. $$

I am very, very stuck and any tips would be greatly appreciated. I have tried using various chain rules but it just got all messy!

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  • $\begingroup$ In the future, please take the time to type in your problem instead of posting an image of it. Images are neither searchable nor accessible to people using screen readers. If you expect people to take their time to help you, you might put a little time into the question yourself. $\endgroup$ – amd May 21 '17 at 5:15
  • $\begingroup$ For future reference, so you can see how to format a question like this, I've done it for you. $\endgroup$ – David K May 21 '17 at 13:01
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Let $\Theta(r,t) = \frac{W(r,t)}{r}$. Then we have: $$\frac{1}{r}\frac{\partial W}{\partial t} = \frac{K}{r^2}\frac{\partial}{\partial r} \left(r^2\frac{\partial (\frac{W}{r})}{\partial r}\right) = \\ \frac{K}{r^2}\frac{\partial}{\partial r} \left(r^2 \frac{\frac{\partial W}{\partial r}r- W}{r^2}\right) = \frac{K}{r^2}\frac{\partial}{\partial r} \left( \frac{\partial W}{\partial r} r - W\right) =\frac{K}{r^2}\left[ \left( \frac{\partial^2 W}{\partial r^2}r + \frac{\partial W}{\partial r}- \frac{\partial W}{\partial r}\right)\right] \\ = \frac{K}{r}\frac{\partial^2 W}{\partial r^2} $$ This way we get $$\frac{1}{r}\frac{\partial W}{\partial t} = \frac{K}{r}\frac{\partial^2 W}{\partial r^2} \Rightarrow \frac{\partial W}{\partial t} = K\frac{\partial^2 W}{\partial r^2}$$ As we wanted.

It may be confusing when it comes to doing the derivatives on the right side of the equation but you just have to make sure what is being derived and what isn't. At the end of the day you are only using the Chain Rule.

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  • $\begingroup$ brilliant! so clear, thank you!:) $\endgroup$ – gamma1 May 21 '17 at 2:45

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