seemingly simple question about polar coordinates

Question

I was recently looking at a problem that looked like this:

Let $x = rcos(t)$ and $y = rsin(t)$. Assuming that x is held constant, what is $\frac{\partial t}{\partial r}$?

Apparently the correct answer is $\frac{cot(t)}{r}$, but this is not what I am getting. I am just wondering what is wrong with my method.

I thought that an appropriate way to solve the problem would be to use the chain rule:

$\frac{\partial t}{\partial r} = \frac{\partial t}{\partial y}*\frac{\partial y}{\partial r}$.

In this case, we have the equations t = arctan(y/x) and y=rsin(t), so

$\frac{\partial t}{\partial y} = \frac{1}{(1+(y/x)^2)}*2(y/x)*(1/x)$, and $\frac{\partial y}{\partial r} = sin(t)$.

Putting this all together, it can be simplified to cot(r)/r.

$\frac{\partial r}{\partial r} = 2*\frac{(sin(t))^2}{r}$

which is certainly not equivalent to $\frac{cot(t)}{r}$

Could someone help me figure out where I am going wrong?

Thanks,

Paul

Answer 1

Here it is easy to isolate the variables in which you are interested by noting that$$t=\arccos \frac xr$$

Now take the derivative with respect to $r$ noting that $x$ is constant to obtain $$-\frac x{r^2}\cdot -\frac {1}{\sqrt{\left(1-\frac {x^2}{r^2}\right)}}=\frac 1r\cdot\cos t\cdot\frac 1{\sin t}$$

Because you are taking the square root, this method requires checking that the sign comes out correctly.

Answer 2

You ask where you went wrong. How about the step where you asserted that $\frac {\partial y}{\partial r} = \sin t?$ This assertion is true only if $\frac {\partial y}{\partial r}$ denotes how $y$ varies when you hold $t$ constant and allow $r$ to vary. But you have already assumed that $x$ is held constant, and since $x = r \,\cos\, t,$ it follows that $r$ is a function of $t.$ It is then impossible to vary $r$ while holding $t$ constant.