# Homework Help: Wrong answer on Differentiation

1. Feb 7, 2009

### roeb

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

In polar coords: y = rsin($$\theta$$)
y' = dy/dt = r'sin($$\theta$$) + rcos($$\theta$$) $$\theta$$'

If I want $$\partial r' / \partial y'$$ can I just solve for r' and take the derivative of y'?

so

$$\partial r' / \partial y' = 1 / sin \theta$$

I am hoping this is incorrect because I am getting the wrong answer on another part of a problem...

2. Relevant equations

3. The attempt at a solution

2. Feb 7, 2009

### Staff: Mentor

Re: Differentiation

Starting with y = r sin($\theta$),
$\partial y/\partial r = \partial (r sin(\theta)) = sin(\theta)$

So $\partial r/\partial y = 1/sin(\theta)$

Why do you have the primes on your partial derivative? I'm pretty sure that's an oversight on your part.

If you're trying to calculate dy/dt, and y is a function of both r and $\theta$, and r and $\theta$ are functions of t alone, then
dy/dt = $\partial y/\partial r * dr/dt + \partial y/\partial \theta * d \theta/dt$

3. Feb 7, 2009

### roeb

Re: Differentiation

The primes actually are indicating time derivatives, let me expand upon my question

I'm actually trying to do the following

$$\frac{\partial L}{\partial y'} = \frac{\partial L}{\partial r} \frac{\partial r}{\partial y'} + \frac{\partial L}{\partial \theta} \frac{\partial \theta}{ \partial y'}+ \frac{\partial L }{\partial r'} \frac{\partial r'}{\partial y'} + \frac{\partial L}{\partial \theta '} \frac{\partial \theta '}{\partial \y '}$$

I am having a hard time find the various partial derivs with respect to y' (or maybe I'm not).

L is actually L = T-U where T = KE = 1/2 m (x'^2 + y'^2) and U doesn't matter in this case
so in other words, y' and x' are the velocities in y and x.

where of course
x = rcos(theta)
y = rsin(theta)

Last edited: Feb 7, 2009