Homework Help: 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.