# Partial Differentiation of an expression.

1. Oct 11, 2008

1. The problem statement, all variables and given/known data

Find the partial of z with respect to x keeping r constant.

2. Relevant equations

z=x2+y2

x= rcos(t)
y= rsin(t)

3. The attempt at a solution

= r^2(cos(t))^2 + r^2(sin(t))^2

use product rule on "x" and hold r and y constant
= [0(cos(t))^2 + r^2(2cos(t))(-sin(t)))] + 0

simplify terms

= -2r^2(cos(t)sin(t))

I am confused because to hold r constant, does that mean i hold both r and y constant?

thanks

Last edited: Oct 11, 2008
2. Oct 11, 2008

### Staff: Mentor

You're given z as a function of x and y, and asked for $$\frac{\partial z}{\partial x}$$. The bit about holding r constant seems to be a red herring. To help you out a little, $$\frac{\partial z}{\partial y}$$ = 2y.

BTW, I corrected the definitions of x and y per your later submission.

3. Oct 11, 2008

In this case we are holding x constant, but what is happening to r? How can i hold r constant if it is connected to theta? Do I hold theta constant as well?

4. Oct 12, 2008

### Staff: Mentor

It doesn't matter what happens with r and t (your original post gives t, not theta). You have three functions: one with z as a function of x and y, one with x as a function of r and t, and one with y as a function of r and t.

The question you asked was, what is $$\frac{\partial z}{\partial x}$$. As a hint, I showed you what $$\frac{\partial z}{\partial y}$$ was.

You didn't ask, but there are other partial derivatives that could be gotten, namely $$\frac{\partial z}{\partial r}$$ and $$\frac{\partial z}{\partial t}$$. As it turns out, both of these partials are zero. For each of these partials, you do have take r and t into consideration, which you do by using a form of the chain rule for partial derivatives.

For the functions you have in this problem,
$$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} * \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}*\frac{\partial y}{\partial t}$$

The chain rule form for $$\frac{\partial z}{\partial r}$$ is similar, but involves $$\frac{\partial x}{\partial r}$$ and $$\frac{\partial y}{\partial r}$$.

Mark

5. Oct 12, 2008

### Staff: Mentor

It doesn't matter what happens with r and t (your original post gives t, not theta). You have three functions: one with z as a function of x and y, one with x as a function of r and t, and one with y as a function of r and t.

The question you asked was, what is $$\frac{\partial z}{\partial x}$$. As a hint, I showed you what $$\frac{\partial z}{\partial y}$$ was.

You didn't ask, but there are other partial derivatives that could be gotten, namely $$\frac{\partial z}{\partial r}$$ and $$\frac{\partial z}{\partial t}$$. As it turns out, both of these partials are zero. For each of these partials, you do have to take r and t into consideration, which you do by using a form of the chain rule for partial derivatives.

For the functions you have in this problem,
$$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} * \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}*\frac{\partial y}{\partial t}$$

The chain rule form for $$\frac{\partial z}{\partial r}$$ is similar, but involves partials of x and y with respect to r, instead of with respect to t.

Mark