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

Dec 23, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \frac{\frac{t}{3}}{- t \cos \left(\frac{t}{3}\right) - 3 \sin \left(\frac{t}{3}\right) + 3}$

#### Explanation:

We know that:
$x = t - t \sin \left(\frac{t}{3}\right)$
$y = - \cos \left(\frac{\pi}{2} - \frac{t}{3}\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left[- \cos \left(\frac{\pi}{2} - \frac{t}{3}\right)\right] = \sin \left(\frac{\pi}{2} - \frac{t}{3}\right) \cdot - \frac{1}{3} = - \sin \frac{\frac{\pi}{2} - \frac{t}{3}}{3}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left[t - t \sin \left(\frac{t}{3}\right)\right] = \frac{d}{\mathrm{dt}} \left[t\right] + \frac{d}{\mathrm{dt}} \left[- t \sin \left(\frac{t}{3}\right)\right] = 1 + \frac{d}{\mathrm{dt}} \left[- t\right] \sin \left(\frac{t}{3}\right) - t \frac{d}{\mathrm{dt}} \left[\sin \left(\frac{t}{3}\right)\right] = 1 - \sin \left(\frac{t}{3}\right) - \frac{t \cos \left(\frac{t}{3}\right)}{3}$

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

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

$= - \cos \frac{\frac{t}{3}}{- t \cos \left(\frac{t}{3}\right) - 3 \sin \left(\frac{t}{3}\right) + 3}$