# Given x = cost y=sin2t, how do you find the dy/dx terms parameter t and find the values parameter t points dy/dx = 0?

Jul 3, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ at $t = \left(2 m + 1\right) \frac{\pi}{4}$

#### Explanation:

In parametric equations $x = x \left(t\right)$ and $y = y \left(t\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $y = \sin 2 t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = \cos 2 t \times 2 = 2 \cos 2 t$

and as $x = \cos t$, $\frac{\mathrm{dx}}{\mathrm{dt}} = - \sin t$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 \cos 2 t}{\sin} t$

As $\sin t \ne 0$, when $t = n \pi$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$, when $\cos 2 t = 0$ but $t \ne n \pi$ i.e. $2 t = \left(2 m + 1\right) \frac{\pi}{2}$

or $t = \left(2 m + 1\right) \frac{\pi}{4}$, where $m$ is an integer

But note that $x$ and $y$ both are sinusoidal functions and hence their domain is limited to $\left[- 1 , 1\right]$ and hence as $x = \cos t$, $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ at $x = \pm \frac{1}{\sqrt{2}}$

graph{2xsqrt(1-x^2) [-2.527, 2.473, -1.11, 1.39]}