# Partial derivatives using polar coordinates

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 $$r\tan\theta$$?

• 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? Feb 28, 2017 at 3:55
• I think he meant the number right ahead the brakets to mean that this is made constant.
– R.W
Feb 28, 2017 at 4:09
• @SchrodingersCat Why do you say that? All of the answers are correct. Feb 28, 2017 at 4:41

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}