Polar Coordinates: Find dy/dx Problem Solution

[HeLiXe]

Homework Statement

Find dy/dx
Problem in picture below

Homework Equations

The Attempt at a Solution

[PLAIN]http://img28.imageshack.us/img28/7162/76013837.png

The answer for this is
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.

 

[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.

 

[HeLiXe]

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

 

[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.

 

[HeLiXe]

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

 

[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.

 

[HeLiXe]

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

Thx peng!

 

[HeLiXe]

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.

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