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

Dec 8, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{24 \sin \left(\frac{t}{3}\right) + 8 t \cos \left(\frac{t}{3}\right)}{12 - \left(6 - 3 \sqrt{3} t\right) \sin \left(\frac{t}{2}\right) - \left(6 \sqrt{3} + 3 t\right) \cos \left(\frac{t}{2}\right)}$

#### Explanation:

The differential of a parametric equation of the type $f \left(x \left(t\right) , y \left(t\right)\right)$ is given by $\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here $y \left(t\right) = 2 t \cos \left(\frac{\pi}{2} - \frac{t}{3}\right)$ and hence

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

= $2 \sin \left(\frac{t}{3}\right) + \frac{2}{3} t \cos \left(\frac{t}{3}\right)$ and

as $x \left(t\right) = t - t \sin \left(\frac{t}{2} + \frac{\pi}{3}\right)$ and hence

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

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

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

= $1 - \frac{1}{2} \sin \left(\frac{t}{2}\right) - \frac{\sqrt{3}}{2} \cos \left(\frac{t}{2}\right) - \frac{t}{2} \left[\frac{1}{2} \cos \left(\frac{t}{2}\right) - \frac{\sqrt{3}}{2} \sin \left(\frac{t}{2}\right)\right]$

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

and hence

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 \sin \left(\frac{t}{3}\right) + \frac{2}{3} t \cos \left(\frac{t}{3}\right)}{1 - \left(\frac{2 - \sqrt{3} t}{4}\right) \sin \left(\frac{t}{2}\right) - \left(\frac{2 \sqrt{3} + t}{4}\right) \cos \left(\frac{t}{2}\right)}$

= $\frac{24 \sin \left(\frac{t}{3}\right) + 8 t \cos \left(\frac{t}{3}\right)}{12 - \left(6 - 3 \sqrt{3} t\right) \sin \left(\frac{t}{2}\right) - \left(6 \sqrt{3} + 3 t\right) \cos \left(\frac{t}{2}\right)}$