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 -_-