Calc 3 partial derivative review for PDE's class

[Nick Bruno]

1. Homework Statement

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. Homework Equations

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

in other words since i don't 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.

 

[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)

 

[Nick Bruno]

do you mind refreshing me on implicit differentiation?

 

[Office_Shredder]

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

 

[Nick Bruno]

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