how to find slope of this polar curve: $r^2=\sin(2\theta)$.

Question

Given $r^2=\sin(2\theta),\;$ how to find the slope of the tangent line at $x=0$ ?

If the question were $r=\sin(2\theta)$, it would be o.k.

but since it is $r^2=\sin(2\theta)$, I don't know how to handle this.

Answer 1

$$ \frac{ dy}{dx} = \frac{d (r \sin \theta)}{d ( r \cos \theta) } = \frac{ \sin \theta dr + r \cos \theta d \theta }{ \cos \theta dr - r \sin \theta d \theta} = \frac{ \sin\theta (dr/d\theta) + r \cos \theta }{ \cos \theta (dr /d \theta) - r \sin \theta}$$

Then use:

$$r^2 = \sin ( 2 \theta) \iff r = \pm \sqrt{ \sin (2 \theta)} $$