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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?

Przemysław Scherwentke
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shinify
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  • You went wrong at assuming that derivatives behave like fractions. :) – Mark Fantini Dec 11 '14 at 22:53
  • Can you explain this a little more? I was under the impression that this relationship held. Does it not hold in this case since x is not dependent solely on r, but also $\theta$? – shinify Dec 11 '14 at 22:59
  • Am I having deja vu or did you just ask this like $2$ days ago @shinify? –  Dec 11 '14 at 23:21
  • Haha it was a slightly different question @Bye_World – shinify Dec 11 '14 at 23:21
  • @shinify I finally got answered why $\dfrac {\partial r}{\partial x}$ can equal $\dfrac {\partial x}{\partial r}$. It's because the Jacobians are the things that have to be inverses (via the inverse function theorem) -- not the individual partials themselves. See here. –  Dec 13 '14 at 16:07
  • @Bye_World Ahhh, thanks so much! (Have my final exam today so this definitely helps). – shinify Dec 13 '14 at 16:10
  • @shinify Good luck! :) –  Dec 13 '14 at 16:10

2 Answers2

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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.

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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 $$ 2x\,dx = 2r\, dr \\ \frac{\partial r} {\partial x} = \frac{x}{r} = \frac{1}{\cos \theta} $$

Mark Fischler
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  • Wouldn't $\frac {x} {r} = \frac {r cos \theta} {r} = cos \theta $? – shinify Dec 11 '14 at 23:23
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    @MarkFischler You literally did exactly the same computation as OP, but at the end you incorrectly state that $\dfrac xr = \dfrac 1{\cos(\theta)}$. –  Dec 11 '14 at 23:41