Find the equations of the lines tangent to $r=4\sin(3\theta)$ at the pole.
Firstly I used the facts that $x=r\cos \theta$ and $y=r\sin \theta$ and took their derivatives to get $dx/d\theta$ and $dy/d\theta$ so that I could get $dy/dx$. Since everything is in $\theta$, I try to find the $\theta$s that put a point at the pole (so basically, find the thetas that satisfy $4\sin (3\theta) = 0$), which are $0,\pi,2\pi,…$ However, when I do plug that into $dy/dx$, the numerator cancels to $0$ because of the sines, and it seems like I only have a horizontal tangent. Is that right though?
Best Answer
$r=0$ at $\theta=0$, $\theta=\pi/3$, $\theta=2\pi/3$.
The cartesian equations are $$x=4\sin(3\theta)\cos(\theta)$$ $$y=4\sin(3\theta)\sin(\theta)$$ Taking the derivative $$x'=12\cos(3\theta)\cos(\theta) - 4\sin(3\theta)\sin(\theta)$$ $$y'=12\cos(3\theta)\sin(\theta) + 4\sin(3\theta)\cos(\theta)$$ At $\theta=0$: $$x'=12$$ $$y'=0$$ so you have a horizontal line (tangent vector (1,0))
At $\theta=\pi/3$: $$x'=-12\cos(\pi/3)=-6$$ $$y'=-12\sin(\pi/3)=-6\sqrt{3}$$ so you have another line with tangent vector (1,$\sqrt{3}$)
Repeat for $\theta=2\pi/3$...