implicitly differentiating polar equations

Question

For polar coordinates, we have the following equations. $x^2 + y^2 = r^2 $, $x= r \cos(\theta) $, and $y= r \sin(\theta)$.

When I find $ \frac {\partial r}{\partial x}$, I have the following: $$\frac {\partial}{\partial x} (x^2 + y^2 = r^2) = 2x = 2r \frac {\partial r}{\partial x} \implies \frac {\partial r}{\partial x} = \frac {x}{r} = \cos(\theta)$$

Then, when I find $ \frac {\partial x}{\partial r}$, I have the following: $$ \frac {\partial}{\partial r} (x = r \cos(\theta))= \frac {\partial x}{\partial r} = (1) \cos(\theta) = \cos(\theta) $$

How can $ \frac {\partial r}{\partial x}$ = $ \frac {\partial x}{\partial r} = \cos(\theta)?$ I was under the impression that $$ \frac {\partial r} {\partial x} = \frac {1} {\frac {\partial x} {\partial r}} = \frac {1}{\cos(\theta)}.$$

Where did I go wrong?

Answer 1

So you want to find $\dfrac {\partial r}{\partial x}$. Let's use the chain rule: $dr=\dfrac {\partial r}{\partial x}dx+\dfrac {\partial r}{\partial y}dy$

Holding $y$ constant, we have $dr=\dfrac {\partial r}{\partial x}dx = \dfrac {\partial }{\partial x}(\sqrt{x^2+y^2})dx = \dfrac {x}{r}dx = \cos(\theta)dx$ using the equation $r=\sqrt{x^2+y^2}$.

Or using $r=\dfrac {x}{\cos(\theta)}$, we've got $dr=\dfrac {\partial r}{\partial x}dx$ $= \dfrac {\partial }{\partial x}\left(\dfrac {x}{\cos(\theta)}\right)dx$ $= \dfrac {\partial }{\partial x}\left(\dfrac {x}{\cos(\arctan(\frac yx))}\right)dx$ $= \dfrac {\partial }{\partial x}\left(\dfrac {x\sqrt{x^2+y^2}}{x}\right)dx$ $=\dfrac {\partial }{\partial x}(\sqrt{x^2+y^2})dx$ which reduces to the above.

So $\dfrac {\partial r}{\partial x} = \cos(\theta)$.

EDIT: The reason that $\frac{\partial x}{\partial r}\ne (\frac{\partial r}{\partial x})^{-1}$ is that the conditions of the inverse function theorem do not apply to partial derivatives. The multivariable inverse function theorem says that the total derivative of an inverse function is the inverse of the total derivative of the function. You can verify for yourself that this in fact does hold for the polar transform by calculating the Jacobian of the transform and its inverse.

Answer 2

Keeping $\theta$ fixed, we have $$ r = \left[ \frac{1}{\cos \theta} \right] x $$ so $$ \frac{\partial r} {\partial x} = \left[ \frac{1}{\cos \theta} \right] $$ so your impression was right.

But you did your implicit differentiation wrong. You should have started out as $$ d\left( x^2 + y^2 \right) = 2x\,dx + 2y\,dy \\ d\left( r^2 \right) = 2r\,dr $$ and if $y$ is held constant, you drop the $dy$ derm to get $$ 2x\,dx = 2r\, dr \\ \frac{\partial r} {\partial x} = \frac{x}{r} = \frac{1}{\cos \theta} $$