How do you differentiate the following parametric equation:  x(t)=te^t-2t, y(t)=-3t^3-2t^2+3t ?

Aug 11, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{9 {t}^{2} + 4 t - 3}{2 - \left(t + 1\right) {e}^{t}}$.

Explanation:

Let $x = x \left(t\right) , \mathmr{and} , y = y \left(t\right)$ be the given parametric eqns.

Then, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} , \frac{\mathrm{dx}}{\mathrm{dt}} \ne 0. \ldots . . \left(\star\right)$.

Now, $y \left(t\right) = - 3 {t}^{3} - 2 {t}^{2} + 3 t \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = - 9 {t}^{2} - 4 t + 3. \ldots . \left(1\right)$.

$x \left(t\right) = t {e}^{t} - 2 t = t \left({e}^{t} - 2\right)$.

$\Rightarrow x ' \left(t\right) = t \left({e}^{t} - 0\right) + \left({e}^{t} - 2\right) 1 = \left(t + 1\right) {e}^{t} - 2. \ldots \ldots . \left(2\right)$.

Using (1) & (2) in $\left(\star\right)$, we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 9 {t}^{2} - 4 t + 3}{\left(t + 1\right) {e}^{t} - 2} = \frac{9 {t}^{2} + 4 t - 3}{2 - \left(t + 1\right) {e}^{t}}$.