# How do you differentiate the following parametric equation:  (3t-sin(t/3-pi/4), 2tcos(pi/3-t/4))?

Nov 27, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \left(4 \cos \left(\frac{\pi}{3} - \frac{t}{4}\right) - t \sin \left(\frac{\pi}{3} - \frac{t}{4}\right)\right)}{18 - 2 \cos \left(\frac{t}{3} - \frac{\pi}{4}\right)}$

#### Explanation:

In a parametric equation of the type f(x(t),y((t)), $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$.

Here $x \left(t\right) = 3 t - \sin \left(\frac{t}{3} - \frac{\pi}{4}\right)$ and hence $\frac{\mathrm{dx}}{\mathrm{dt}} = 3 - \cos \left(\frac{t}{3} - \frac{\pi}{4}\right) \times \frac{1}{3}$

and as $y \left(t\right) = 2 t \cos \left(\frac{\pi}{3} - \frac{t}{4}\right)$ $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \left(\cos \left(\frac{\pi}{3} - \frac{t}{4}\right) - t \sin \left(\frac{\pi}{3} - \frac{t}{4}\right) \times \left(- \frac{1}{4}\right)\right)$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(\cos \left(\frac{\pi}{3} - \frac{t}{4}\right) + \frac{t}{4} \sin \left(\frac{\pi}{3} - \frac{t}{4}\right)\right)}{3 - \cos \left(\frac{t}{3} - \frac{\pi}{4}\right) \times \frac{1}{3}}$

= $\frac{\frac{2}{4} \left(4 \cos \left(\frac{\pi}{3} - \frac{t}{4}\right) + t \sin \left(\frac{\pi}{3} - \frac{t}{4}\right)\right)}{\frac{1}{3} \left(9 - \cos \left(\frac{t}{3} - \frac{\pi}{4}\right)\right)}$

= $\frac{3 \left(4 \cos \left(\frac{\pi}{3} - \frac{t}{4}\right) - t \sin \left(\frac{\pi}{3} - \frac{t}{4}\right)\right)}{18 - 2 \cos \left(\frac{t}{3} - \frac{\pi}{4}\right)}$