# How do you differentiate the following parametric equation:  (3sin(t/6-pi/12), 2tcos(pi/3-t/8))?

Sep 18, 2017

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1}{2} \cos \left(\frac{t}{6} - \frac{\pi}{12}\right)$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \left\{\cos \left(\frac{\pi}{3} - \frac{t}{8}\right) + \frac{t}{8} \sin \left(\frac{\pi}{3} - \frac{t}{8}\right)\right\}$

#### Explanation:

$x = 3 \sin \left(\frac{t}{6} - \frac{\pi}{12}\right)$
$\frac{\mathrm{dx}}{\mathrm{dt}} = 3 \left(\frac{1}{6}\right) \cos \left(\frac{t}{6} - \frac{\pi}{12}\right) = \frac{1}{2} \cos \left(\frac{t}{6} - \frac{\pi}{12}\right)$

$y = 2 t \cos \left(\frac{\pi}{3} - \frac{t}{8}\right)$
dy/dt=2{cos(pi/3-t/8)d/dt(t)+td/dt(cos(pi/3-t/8)}
$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \left\{\cos \left(\frac{\pi}{3} - \frac{t}{8}\right) \cdot \left(1\right) - t \left(- \frac{1}{8}\right) \sin \left(\frac{\pi}{3} - \frac{t}{8}\right)\right\}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \left\{\cos \left(\frac{\pi}{3} - \frac{t}{8}\right) + \frac{t}{8} \sin \left(\frac{\pi}{3} - \frac{t}{8}\right)\right\}$