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Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$.

I can show using the chain rule that $$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$

What is the method to compute $\dfrac{\partial^2}{\partial x^2}$? I don't know how to do it.

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  • $\begingroup$ $\frac{\partial ^2}{\partial x^2}= \frac{\partial }{\partial x}\bigg( \frac{\partial }{\partial x} \bigg) = \frac{\partial }{\partial x}\bigg(\frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta})$ $\endgroup$
    – user99914
    Nov 2, 2013 at 5:38
  • $\begingroup$ @John Sure, and that's exactly what I don't know how to do. How to continue? $\endgroup$
    – Mika H.
    Nov 2, 2013 at 5:55
  • $\begingroup$ @MikaH. If you got $\frac{\partial }{\partial x}$, just do the same thing to get the second derivative. $\endgroup$ Nov 2, 2013 at 5:59
  • $\begingroup$ @Euler....IS_ALIVE How do I compute $\frac{\partial}{\partial x}(\frac{x}{r}\frac{\partial}{\partial r})$? The product rule? $\endgroup$
    – Mika H.
    Nov 2, 2013 at 6:32

1 Answer 1

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Rewrite your formula as $$\frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r}-\sin\theta \frac{1}{r}\frac{\partial}{\partial\theta}\tag{1}$$ (I like writing in this way because $\frac{1}{r}\frac{\partial}{\partial\theta}$ is more natural than $\frac{\partial}{\partial\theta}$: it means the rate of change in the tangential direction). Now square (1): $$\frac{\partial^2}{\partial x^2} = \left(\cos\theta\frac{\partial}{\partial r}-\sin\theta \frac{1}{r}\frac{\partial}{\partial\theta}\right)\left(\cos\theta\frac{\partial}{\partial r}-\sin\theta \frac{1}{r}\frac{\partial}{\partial\theta}\right)\tag{2}$$ The computation of (2) involves no thinking, just some product rule: $$\begin{split}\cos^2\theta \frac{\partial^2}{\partial r^2}-2\cos\theta\sin\theta \frac{1}{r}\frac{\partial^2}{\partial r\partial\theta}+2\cos\theta\sin\theta \frac{1}{r^2}\frac{\partial }{ \partial\theta} \\ +\sin^2\theta \frac1r\frac{\partial}{\partial r}+\sin^2\theta \frac{1}{r^2}\frac{\partial^2 }{ \partial\theta^2}\end{split}$$

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