So there's this curve in polar coordinates
$r=\theta$ for $0\le\theta\le2\pi$
and I need to find the $\theta$ values for all of its vertical and horizontal tangents. so I've got
$x=rcos\theta=\theta cos\theta$ and
$y=\theta sin\theta$,
which gives
$\displaystyle\frac{dx}{d\theta}=cos\theta-\theta sin\theta$ and
$\displaystyle\frac{dy}{d\theta}=sin\theta+\theta cos\theta$.
So the slope should be $\displaystyle\frac{dy}{dx}=\displaystyle\frac{sin\theta+\theta cos\theta}{cos\theta-\theta sin\theta}$.
So for the tangent to be horizontal, I need ${sin\theta+\theta cos\theta}=0$, but I can't see how to solve this, except just by seeing that it works when $\theta=0.$ As for the vertical tangent, I'm not sure how to find $\theta$ so that $cos\theta-\theta sin\theta=0$.