# Polar Coordinates

1. Sep 4, 2011

### HeLiXe

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

Find dy/dx
Problem in picture below

2. Relevant equations

3. The attempt at a solution
[PLAIN]http://img28.imageshack.us/img28/7162/76013837.png [Broken]

dy/dx = $$-cos\theta sin\theta + (1-sin\theta)cos\theta$$/$$-cos^2\theta - (1-sin/theta)sin\theta$$ I cannot figure out why I'm not getting the same or equivalent for dy/dtheta.... and trig makes me woozy.

Last edited by a moderator: May 5, 2017
2. Sep 4, 2011

### Pengwuino

Your $dy/d\theta$ is equivalent. Take a look at the second to last line of your calculation before you used the double angle equation.

3. Sep 5, 2011

### HeLiXe

I did Peng Peng, but I just can't see it -_- Which type of identity did they use?

4. Sep 5, 2011

### Pengwuino

They didn't! $\cos(\theta) - 2\sin(\theta)\cos(\theta) = -\cos(\theta)\sin(\theta) +(1-\sin(\theta))\cos(\theta)$

The tex translator seems to be pooping itself.

5. Sep 5, 2011

### HeLiXe

@_@ I really need to take another trig class...let me try to work it out.

6. Sep 5, 2011

### Pengwuino

Just write out the right hand side. There is no trig involved. They just wrote down the answer in a way that the 'r' shows up again. They could have written it

$-\cos(\theta)\sin(\theta) +r\cos(\theta)$

if they wanted to be more explicit.

7. Sep 5, 2011

### HeLiXe

omg I see it now...maybe I need to take another algebra class -_-

Thx peng!

8. Sep 5, 2011

### HeLiXe

yes thx peng I saw it after I started to think "algebra" and not trig -_-