# How do you differentiate the following parametric equation:  x(t)=tlnt, y(t)= cost-tsin^2t ?

Jul 13, 2016

$\frac{\mathrm{df} \left(t\right)}{\mathrm{dt}} = \left(\ln \left(t\right) + 1 , - \sin \left(t\right) - {\sin}^{2} \left(t\right) - 2 t \sin \left(t\right) \cos \left(t\right)\right)$

#### Explanation:

Differentiating a parametric equation is as easy as differentiating each individual equation for its components.

If $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ then $\frac{\mathrm{df} \left(t\right)}{\mathrm{dt}} = \left(\frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}} , \frac{\mathrm{dy} \left(t\right)}{\mathrm{dt}}\right)$

So we first determine our component derivatives:
$\frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}} = \ln \left(t\right) + \frac{t}{t} = \ln \left(t\right) + 1$
$\frac{\mathrm{dy} \left(t\right)}{\mathrm{dt}} = - \sin \left(t\right) - {\sin}^{2} \left(t\right) - 2 t \sin \left(t\right) \cos \left(t\right)$

Therefore the final parametric curve's derivatives is simply a vector of the derivatives:
$\frac{\mathrm{df} \left(t\right)}{\mathrm{dt}} = \left(\frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}} , \frac{\mathrm{dy} \left(t\right)}{\mathrm{dt}}\right)$
$= \left(\ln \left(t\right) + 1 , - \sin \left(t\right) - {\sin}^{2} \left(t\right) - 2 t \sin \left(t\right) \cos \left(t\right)\right)$