# To find the slope of a curve of given polar co-ordinates.

I successfully finished the first part of the question that i.e sketching the graph of the same. I sketched the cartesian curves and sketched the locus using that. The second part of the question is what I'm unsure of. Will finding the derivate of dr/dtheta of the given polar equation and then substituting in theta=pie/3 be the right way to go about it? Or should I take x= rcos(theta) and y=rsin(theta) , substitute r into the equations and find the dy/dx

$$m=\frac{dy}{dx}=\frac{d(r\sin\theta)/d\theta}{d(r\cos\theta)/d\theta}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}=\frac{\frac{dr}{d\theta}\tan\theta+r}{\frac{dr}{d\theta}-r\tan\theta}$$
Another way to look at this is in the complex plane. If you say $z=re^{i\theta}$, then the first derivative, $\dot z=dz/d\theta,\$ gives the slope. Specifically, $m=\frac{\mathfrak{Im}\{\dot z\}}{\mathfrak{Re}\{\dot z\}}$, or the tangent angle is given by $\arg(z)$. I always find that working in the complex plane is the easiest.