# polar coordinates and derivatives

Using the standard notation $(x,y)$ for cartesian coordinates, and $(r, \theta)$ for polar coordinates, it is true that $$x = r \cos \theta$$ and so we can infer that $$\frac{\partial x}{\partial r} = \cos \theta.$$ This means that if we perturb $r$ to $r+\Delta r$, $x$ gets perturbed to approximately $x+ \cos(\theta) \Delta r$. This is equivalent to saying that if we perturb $x$ to $x+\Delta x$, then we perturb $r$ to approximately $r+ (1/\cos \theta) \Delta x$. Correspondingly, we expect that $$\frac{\partial r}{\partial x} = \frac{1}{\cos \theta}.$$ In fact, $$\frac{ \partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2+y^2}= \frac{2x}{2 \sqrt{x^2+y^2}} = \frac{x}{r} = \cos \theta.$$ What gives?

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If you are in the $(r,\theta)$ plane, a change in $r$ does not affect $\theta$; the change in $x$ is proportional to the change in $r$, with the constant of proportionality given by $\cos\theta$, which is constant because $\theta$ is not changing. But if you are in $(x,y)$-plane, a change in $x$ affects both $r$ and $\theta$, so you cannot consider $\cos\theta$ a fixed quantity; this means that the claim that perturbing $x$ by $\Delta x$ corresponds to perturbing $r$ by $\frac{1}{\cos\theta}\Delta x$ is unjustified and unclear: which $\theta$? $\theta$ changes! –  Arturo Magidin Feb 7 '11 at 22:05
Good question. I had the same questions when I first learnt partial derivatives. My teachers did not emphasize what we were doing and used to simply write out the answer using the appropriate expressions. –  user17762 Feb 7 '11 at 22:05
thanks, everyone –  calc student Feb 7 '11 at 22:33

## 1 Answer

When you take partial derivatives, what you are doing is to hold the other variable fixed. So for instance if $f$ is a function of $n$ variables, $f(x_1,x_2,\ldots,x_n)$, $\frac{\partial f}{\partial x_i}$ means you hold all the other $x_j$'s constant except $x_i$ and vary $x_i$ by $\delta x_i$ and find out what happens to $\delta f$.

We have

$x= r \cos(\theta)$,$y= r \sin(\theta)$, $r^2= x^2 + y^2$ and $\tan(\theta) = \frac{y}{x}$.

We are transforming from the $(x,y)$ space to $(r,\theta)$ space.

In the $(x,y)$ space, $x$ and $y$ are independent variables and In the $(r,\theta)$ space, $r$ and $\theta$ are independent variables.

$\frac{\partial x}{\partial r}$ means you are fixing $\theta$ and finding out how changing $r$ affects $x$.

So $\frac{\partial x}{\partial r} = \cos(\theta)$ since $\theta$ is fixed.

$\frac{\partial r}{\partial x}$ means you are fixing $y$ and finding out how changing $x$ affects $r$.

So $\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}$ since $y$ is fixed.

When you do $r = \frac{x}{\cos(\theta)}$ and argue that $\frac{\partial r}{\partial x} = \frac{1}{\cos (\theta)}$ you are not holding $y$ constant.

If you were to hold $y$ constant, then $\theta$ would change as well.

EDIT:

I am adding this in the hope that this will make it a bit more clear. If you want to use $r = \frac{x}{\cos(\theta)}$ and still derive it, you need to do as follows:

$r = \frac{x}{\cos(\theta)}$, $\delta r = \frac{\delta x}{\cos(\theta)} + \frac{-x}{\cos^2(\theta)} (-\sin(\theta)) \delta \theta$.

$\tan(\theta) = \frac{y}{x} \Rightarrow x \tan(\theta) = y$

$\delta x \tan(\theta) + x \sec^2(\theta) \delta \theta = 0$ (Since $y$ is held constant)

$x \delta \theta = - \frac{\delta x \tan(\theta)}{\sec^2{\theta}} = - \sin(\theta) \cos(\theta) \delta x$.

Plugging the above in the previous expression, we get

$\delta r = \frac{\delta x}{\cos(\theta)} + \frac{\sin(\theta)}{\cos^2(\theta)} (-\sin(\theta) \cos(\theta) \delta x) = \frac{1-\sin^2(\theta)}{\cos(\theta)} \delta x = \cos(\theta) \delta x$ and hence we get

$\frac{\partial r}{\partial x} = \cos(\theta) = \frac{x}{\sqrt{x^2+y^2}}$

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Or, alternatively to "not holding $y$ constant", one is not applying the chain rule properly, since $\theta$ also changes with $x$. –  Arturo Magidin Feb 7 '11 at 22:25
@Arturo: Right. I was adding that when you made this comment. –  user17762 Feb 7 '11 at 22:27
I would only add a reference to the inverse function theorem: en.wikipedia.org/wiki/… –  AppliedSide Feb 8 '11 at 0:08