Partial Derivative of z w/ Respect to x: Theta Constant

[jesuslovesu]

Whoops got it now, didn't carry out my substitutions far enough.

Homework Statement

<br /> z = x^2 + 2y^2<br />
<br /> x = rcos(\theta)<br />
<br /> y = rsin(\theta)<br />

Homework Equations

The Attempt at a Solution

Find (\partial z/\partial x) (theta is constant)

dz = 2xdx + 4ydy
dx = cos(\theta)dr - rsin(\theta)d\theta
dy = sin(\theta)dr + rcos(\theta)d\theta

Unfortunately I'm not really quite sure where to go from here, I know that
(\frac{ \partial z } { \partial x} ) is 2x when y is constant. But how to factor in theta being constant?
I suppose I could reduce
dx to dx = cos(\theta)dr
and dy = sin(\theta)dr

 

[HallsofIvy]

Whoops got it now, didn't carry out my substitutions far enough.

Homework Statement

<br /> z = x^2 + 2y^2<br />
<br /> x = rcos(\theta)<br />
<br /> y = rsin(\theta)<br />

Homework Equations

The Attempt at a Solution

Find (\partial z/\partial x) (theta is constant)

dz = 2xdx + 4ydy
dx = cos(\theta)dr - rsin(\theta)d\theta
dy = sin(\theta)dr + rcos(\theta)d\theta

Unfortunately I'm not really quite sure where to go from here, I know that
(\frac{ \partial z } { \partial x} ) is 2x when y is constant. But how to factor in theta being constant?

If \theta is a constant, then d\theta= 0

I suppose I could reduce
dx to dx = cos(\theta)dr
and dy = sin(\theta)dr

Yes, that is exactly correct. Then dz= dx+ dy= ?