1. Polar Curve

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

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

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

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

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

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

2. Originally Posted by slevvio
A polar curve is defined by $r=1+cos\theta$.

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

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

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

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

I know that $tan(\alpha + \theta) = \frac{dy}{dx}$ but I don't know how to manipulate this. Thanks a lot.
why is this in pre-calculus?

note that we have $r = f( \theta )$

now, $x = r \cos \theta = f ( \theta ) \cos \theta$ and $y = r \sin \theta = f ( \theta ) \sin \theta$

now, $\frac {dy}{dx} = \frac {\frac {dy}{d \theta}}{\frac {dx}{d \theta}} = \frac {\frac {dr}{d \theta} \sin \theta + r \cos \theta}{\frac {dr}{d \theta} \cos \theta - r \sin \theta}$