I was given the following problems as practice, and I've sold all but one. However, I am not sure that my answers are correct.

Calculate $(\partial r/\partial x)_y$ , $(\partial r/\partial y)_x$ , $(\partial θ/\partial x)_y$ ,$(\partial y/\partial x)_r$, $(\partial r/\partial \theta)_x$

$x=r\cos(\theta)$, $y=r\sin(\theta)$

I have that

$(\partial r/\partial x)_y=\cos(\theta)$

$(\partial r/\partial y)_x=\sin(\theta)$

$(\partial y/\partial x)_r=-\cot(\theta)$

$(\partial \theta/\partial x)y=-\sin(\theta)/r$

Would $(\partial r/\partial \theta)_x$ then be equal to $r\tan\theta$?

  • $\begingroup$ Please use math formatting to remove ambiguities.. Are you asking about $$ \left(\frac{\partial r}{\partial x}\right) y = \frac{y}{ \frac{\partial x}{\partial r} }= \frac{r \sin \theta}{ \frac{\partial }{\partial r}( r \cos \theta) } = r \tan \theta$$ for example? $\endgroup$ Feb 28, 2017 at 3:55
  • 1
    $\begingroup$ I think he meant the number right ahead the brakets to mean that this is made constant. $\endgroup$
    – R.W
    Feb 28, 2017 at 4:09
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    $\begingroup$ @SchrodingersCat Why do you say that? All of the answers are correct. $\endgroup$
    – Mark Viola
    Feb 28, 2017 at 4:41

1 Answer 1


Note that $r=\sqrt{x^2+y^2}$. Holding $y$ fixed we have

$$\begin{align} \frac{\partial r}{\partial x}&=\frac{x}{\sqrt{x^2+y^2}}\\\\ &=\frac xr\\\\ &=\cos(\theta) \end{align}$$

Similarly, holding $x$ fixed we have

$$\begin{align} \frac{\partial r}{\partial y}&=\frac{y}{\sqrt{x^2+y^2}}\\\\ &=\frac yr\\\\ &=\sin(\theta) \end{align}$$

Then, holding $r$ fixed we see that $\sqrt{x^2+y^2}$ is constant. Hence,

$$\begin{align} \frac{\partial r}{\partial x}&=0\\\\ &=\frac{x}{\sqrt{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}}\frac{\partial y}{\partial x} \end{align}$$

whereupon solving for $\frac{\partial y}{\partial x}$ we find that

$$\begin{align} \frac{\partial y}{\partial x}&=-\frac xy\\\\ &=-\cot(\theta) \end{align}$$

Finally, holding $y$ fixed, we have $x=y\cot(\theta)$ so that

$$\begin{align} \frac{\partial x}{\partial x}&=1\\\\ &=y\left(-\csc^2(\theta)\frac{\partial \theta}{\partial x}\right) \end{align}$$

whereupon solving for $\frac{\partial \theta}{\partial x}$ we find that

$$\begin{align} \frac{\partial \theta}{\partial x}&=-\sin^2(\theta)/y\\\\ &=-\sin^2(\theta)/(r\sin(\theta))\\\\ &=-\sin(\theta)/r \end{align}$$


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