# Calc 3 partial derivative review for PDE's class

1. Sep 17, 2009

### Nick Bruno

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

I am suppose to use polar coordinate data to find derivatives, ie

x = r cos(theta)
y = r sin(theta)

r^2 = x^2 + y^2

2. Relevant equations

show dtheta/dy = cos(theta)/r
show dtheta/dx = -sin(theta)/r

in other words since i dont have the math script
find the equation for theta and take the derivatives
These are partial derivatives by the way (as you can tell by inspection)

3. The attempt at a solution

d theta / dy = cos(theta)/r

I separate and integrate

dtheta/cos(theta) = dy/sqrt(x^2+y^2)

ln(cos(theta)) = ln(sqrt(x^2+y^2))/(0.5(x^2+y^2)^-.5*2y) => per chain rule

ln(cos(theta))=ln(r)*r/ r*sin(theta)

ln(cos(theta)) = ln(r)/ sin(theta)

now what?

any help is very much appreciated. Thanks for looking. have a good one.

2. Sep 17, 2009

### lanedance

how about trying implictly differentiating both sides of all your equations w.r.t x & y, then a little subtitution to get the form you want?

you'll need to consider
r = r(x,y)
theta = theta(x,y)

3. Sep 17, 2009

### Nick Bruno

do you mind refreshing me on implicit differentiation?

4. Sep 17, 2009

### Office_Shredder

Staff Emeritus
Implicit differentiation is where you differentiate every term with respect to one of them without solving one term as a function of another.

For example

x2 + y2 = 1

I want to find dy/dx. Rather than calculate y as a function of x, I just differentiate both sides with respect to x. I get a dy/dx because of the chain rule

2x + 2y dy/dx = 0

dy/dx = -x/y

Similarly, you could differentiate your equations implicitly with respect to e.g. y in an attempt to find dtheta/dy

5. Sep 20, 2009

### Nick Bruno

Thanks, this worked and was actually quite a clever way to solve the problem.