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A function say $f$ of $x$, $y$ is away from the origin. This function can be written in polar coordinates as a function of $r$ and $\theta$. Now, if we know what $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$, how can we find $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial \theta}$ and vice versa. Additionally, if we know what $\frac{\partial^2 f}{\partial x^2}$, $\frac{\partial^2 f}{\partial x \partial y}$, and $\frac{\partial^2 f}{\partial y^2}$ is, how can we find $\frac{\partial^2 f}{\partial r^2}$, $\frac{\partial^2 f}{\partial r \partial \theta}$, and $\frac{\partial^2 f}{\partial \theta^2}$ and vice versa.

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2 Answers 2

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You can consider it to be a derivative of a composite function. Given $$ f(x,y) $$ And any differentiable transformation $(x,y) \to (u,v)$ $$ x = g(u,v), $$ $$ y = h(u,v) $$ you can rewrite $$ f(x,y) = f(g(u,v), h(u,v)) $$ and so $$ \frac{\partial f(x,y)}{\partial u} = \frac{\partial f(g(u,v),h(u,v))}{\partial u} = \frac{\partial f(x,y)}{\partial x} \frac{\partial g(u,v)}{\partial u} + \frac{\partial f(x,y)}{\partial y} \frac{\partial h(u,v)}{\partial u} $$ In particular, if you change from cartesian to polar coordinates $$ x = r\cos\theta, $$ $$ y = r\sin\theta $$ you get $$ \frac{\partial f(x,y)}{\partial r} = \frac{\partial f(x,y)}{\partial x} \frac{\partial (r\cos\theta)}{\partial r} + \frac{\partial f(x,y)}{\partial y} \frac{\partial (r\sin\theta)}{\partial r} = \frac{\partial f(x,y)}{\partial x} \cos\theta + \frac{\partial f(x,y)}{\partial y} \sin\theta $$ Example: $$ f(x,y) = x^2y = r^3\cos^2\theta\sin\theta $$ direct derivation will give you $$ \frac{\partial f}{\partial r} = 3r^2\cos^2\theta\sin\theta $$ By the formula above you get: $$ \frac{\partial f}{\partial r} = \frac{\partial f(x,y)}{\partial x} \cos\theta + \frac{\partial f(x,y)}{\partial y} \sin\theta = 2xy \cos\theta + x^2 \cos\theta= 2r^2\cos^2\theta\sin\theta + r^2\cos^2\theta\sin\theta=3r^2\cos^2\theta\sin\theta $$

This can be generalised to any number of variables and any differentiable transformation of coordinates.

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You can use the fact that $x = rCos\theta$ and $y = rSin\theta$ and the Jacobian to transform it. Let me know if anything isn't clear.

For example if f is a function of x and y, you can express f in terms of $r$ and $\theta$ and then find those partial derivatives. Perhaps a more specific example might help.

e.g.

If $ f(x, y) = xy $, then $f(r, \theta) = rCos\theta\times rSin\theta = r^2 Sin\theta Cos\theta = \frac{r^2}{2}Sin2\theta $

Once written in terms of $r$ and $\theta$, it's very straight forward to find those partial derivatives.

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  • $\begingroup$ I havent learned Jacobian yet. $\endgroup$
    – Anne
    Jul 8, 2015 at 10:20
  • $\begingroup$ I'll try to explain that further. $\endgroup$
    – John_dydx
    Jul 8, 2015 at 10:21
  • $\begingroup$ Can you post the specific example you're trying to solve? It'll be easier for me to deal with that. $\endgroup$
    – John_dydx
    Jul 8, 2015 at 10:26
  • $\begingroup$ Its not a problem. It is something I am having trouble understanding. $\endgroup$
    – Anne
    Jul 8, 2015 at 10:27
  • $\begingroup$ I've edited my post to explain more. $\endgroup$
    – John_dydx
    Jul 8, 2015 at 10:29

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