# How do you find the points of horizontal tangency of r=asinthetacos^2theta?

Mar 24, 2017

Horizontal tangents occur when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

For polar equations, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / d \theta}{\mathrm{dx} / d \theta}$ where $x = r \cos \theta$ and $r = \sin \theta$. Then, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{d}{d \theta} r \sin \theta}{\frac{d}{d \theta} r \cos \theta}$.

So, horizontal tangents occur when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$, which is the same as when $\frac{\mathrm{dy}}{d \theta} = 0$, or when $\frac{d}{d \theta} r \sin \theta = 0$.

Here $r = a \sin \theta \cos \theta$, so $y = r \sin \theta = a {\sin}^{2} \theta {\cos}^{2} \theta$.

We can simplify this by noting that $y = \frac{1}{4} a \left(4 {\sin}^{2} \theta {\cos}^{2} \theta\right) = \frac{1}{4} a {\left(2 \sin \theta \cos \theta\right)}^{2} = \frac{1}{4} a {\sin}^{2} 2 \theta$.

Differentiating this with the chain rule, we see that $\frac{\mathrm{dy}}{d \theta} = \frac{1}{4} a \left(2 {\sin}^{1} 2 \theta\right) \left(\cos 2 \theta\right) \left(2\right) = a \sin 2 \theta \cos 2 \theta$.

$\frac{\mathrm{dy}}{d \theta} = 0$ then when either $\sin 2 \theta = 0$ or $\cos 2 \theta = 0$, which occur at $\theta = \frac{k \pi}{4} , k \in \mathbb{Z}$.