# Partial Derivatives and Polar Coordinates

1. Mar 2, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
Write the chain rule for the following composition using a tree diagram:
z =g(x,y)
where x=x(r,theta) and y=y(r,theta). Write formulas for the partial derivatives dz/dr and dz/dtheta. Use them to answer: Find first partial derivatives of the function z=e^x+yx^2, in polar coordinates, that is find dz/dr and dz/dtheta as a function of polar coordinates (r, theta).

2. Relevant equations

3. The attempt at a solution
Tree diagram was relatively simple
g extends to x and y. x and y both extend to r and theta.

**all "d's" symbolize partial derivatives**
dz/dr = dg/dx*dx/dr+ dg/dy*dy/dr
dz/dtheta = dg/dx*dx/dtheta+dg/dy*dy/dtheta

Ok.
So
partial x = e^x+2yx
partial y = x^2

Now the problem I'm having trouble with is taking the partial derivative of the polar coordinate functions.
If I am right, then:
x=rcos(theta)
y=rsin(theta)
Taking the partial derivatives dx/dr and dy/dr the answers would be cos(theta) for x and sin(theta) for y, as we treat these as 'constants'.
From here it is merely a plug-n-chug.

Is this correct? Do you have any opinions? Thank you.

Last edited: Mar 3, 2015
2. Mar 3, 2015

### Svein

Why is $\frac{\partial g}{\partial x}\frac{\partial x}{\partial r}$ missing?

3. Mar 3, 2015

### RJLiberator

Ah, typo, it should be dx/dr instead of dx/dtheta. My apologies.

4. Mar 3, 2015

### Svein

OK. So we have $\frac{\partial g}{\partial x}=\frac{\partial g}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial g}{\partial \theta}\frac{\partial \theta}{\partial x}$ and $\frac{\partial g}{\partial y}=\frac{\partial g}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial g}{\partial \theta}\frac{\partial \theta}{\partial y}$...

5. Mar 3, 2015

### RJLiberator

I'm not quite sure why you are using that.

I believe this would be correct, no?:
dz/dr = dg/dx*dx/dr+ dg/dy*dy/dr
dz/dtheta = dg/dx*dx/dtheta+dg/dy*dy/dtheta

So if x=x(r,theta) and y=y(r,theta) then x=rcos(theta) and y=rcos(theta) ??

6. Mar 4, 2015

Not quite...
OK.

7. Mar 4, 2015

### RJLiberator

How could I make it correct? I feel like the x=rcos(theta) is a very important part of the problem. Why is it wrong?

8. Mar 4, 2015

### Svein

It isn't wrong. Look at your entire statement.

9. Mar 4, 2015

### RJLiberator

Ugh, another typo. Apologies. y=rsin(theta).
So, if these are right, then their respective derivatives are
cos(theta) with respects to dx/dr
sin(theta) dy/dr
and -rsin(theta) dx/dtheta
rcos(theta) dy/dtheta

And then it's just plugging in to the chain rule equations with the partial derivatives of the g function.