Originally Posted by

**slevvio** A polar curve is defined by $\displaystyle r=1+cos\theta$.

Show that $\displaystyle \frac{dy}{dx} = -cot\frac{3\theta}{2}$.

I was wondering if anyone could help me with this. This is what I have.

Let $\displaystyle r=f(\theta)$

$\displaystyle tan\alpha=\frac{f(\theta)}{f'(\theta)}=\frac{1+cos \theta}{-sin\theta}=\frac{cos^2\frac{\theta}{2}}{sin\frac{\ theta}{2}cos\frac{\theta}{2}}$

So $\displaystyle tan\alpha = cot\frac{\theta}{2}$

I know that $\displaystyle tan(\alpha + \theta) = \frac{dy}{dx}$ but I don't know how to manipulate this. Thanks a lot.