How do you differentiate the following parametric equation:  x(t)=-t-2, y(t)= t^2/(-t+4?

1 Answer
Mar 19, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{t \left(t - 8\right)}{4 - t} ^ 2$

Explanation:

Differential of a parametric of the type $y = y \left(t\right)$ and $x = x \left(t\right)$ is given by

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

Here $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\left(- t + 4\right) \times 2 t - {t}^{2} \left(- 1\right)}{- t + 4} ^ 2$

= $\frac{\left(8 t - 2 {t}^{2}\right) + {t}^{2}}{- t + 4} ^ 2 = \frac{t \left(8 - t\right)}{4 - t} ^ 2$

As $x = - t - 2$, $\frac{\mathrm{dx}}{\mathrm{dt}} = - 1$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{t \left(8 - t\right)}{4 - t} ^ 2 \times \left(- 1\right) = \frac{t \left(t - 8\right)}{4 - t} ^ 2$