# How do you differentiate the following parametric equation:  (tsec(t/3-pi),t cos(pi-t))?

Aug 9, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\cos}^{2} t \left(\cos t - t \sin t\right)}{\cos t + t \sin t}$

#### Explanation:

$x = t \sec \left(\frac{t}{3} - \pi\right) = t \sec \left(\pi - t\right) = - t \sec t$

$\frac{\mathrm{dx}}{\mathrm{dt}} = - \sec t - t \sec t \tan t = - \sec t \left(1 + t \tan t\right)$

$y = t \cos \left(\pi - t\right) = - t \cos t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = - \cos t + t \sin t$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} / \frac{\mathrm{dx}}{\mathrm{dt}}$

$= \frac{\cos t - t \sin t}{\sec t \left(1 + t \tan t\right)}$

$= \frac{{\cos}^{2} t \left(\cos t - t \sin t\right)}{\cos t + t \sin t}$