# How do you differentiate the following parametric equation:  x(t)=-t^2+tcost, y(t)=t^2sint ?

Mar 1, 2017

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

$\frac{\mathrm{dy}}{\mathrm{dt}} = {t}^{2} \cos t + 2 t \sin t$

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

#### Explanation:

We have two parametric equations:

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

We can differentiate both equation wrt $t$ (and apply product rule);

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

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

And presumably you also want $\frac{\mathrm{dy}}{\mathrm{dx}}$ which we get from the chain rule:

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

$\setminus \setminus \setminus \setminus \setminus = \frac{{t}^{2} \cos t + 2 t \sin t}{\cos t - t \sin t - 2 t}$