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

Mar 26, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 3 t}{\left(1 + 7 t + 3 {t}^{2}\right) {e}^{4 t}} , t \ne \frac{- 7 \pm \sqrt{37}}{2.}$

#### Explanation:

$y \left(t\right) = \frac{t}{e} ^ \left(3 t\right) = t {e}^{- 3 t}$

$\therefore y ' \left(y\right) = t \left\{\frac{d}{\mathrm{dt}} {e}^{- 3 t}\right\} + {e}^{- 3 t} \frac{d}{\mathrm{dt}} \left(t\right)$

=t(e^(-3t))d/dt(-3t)+e^(-3t)(1)...[because," the Chain Rule]."

$\therefore y ' \left(t\right) = \left(1 - 3 t\right) {e}^{- 3 t} \ldots \ldots \ldots . \left(1\right) .$

$x \left(t\right) = \left(t + 3 {t}^{2}\right) {e}^{t}$

:. x"(t)=(t+3t^2)d/dt(e^t)+e^td/dt(t+3t^2)

$= \left(t + 3 {t}^{2}\right) {e}^{t} + {e}^{t} \left(1 + 6 t\right)$

$\therefore x ' \left(t\right) = \left(t + 3 {t}^{2} + 1 + 6 t\right) {e}^{t} = \left(1 + 7 t + 3 {t}^{2}\right) {e}^{t} \ldots \ldots \ldots \ldots \left(2\right) .$

Now, by the Rule for Parametric Diffn., we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y ' \left(t\right)}{x ' \left(t\right)} , x ' \left(t\right) \ne 0. \ldots \ldots \ldots . \left(\star\right) .$

Now, $x ' \left(t\right) = 0 \Rightarrow \left(1 + 7 t + 3 {t}^{2}\right) {e}^{t} = 0$

$\Rightarrow 1 + 7 t + 3 {t}^{2} = 0 , \because , {e}^{t} \ne 0 , \forall t \in \mathbb{R} .$

$\Rightarrow t = \frac{- 7 \pm \sqrt{49 - 4 \left(1\right) \left(3\right)}}{2} = \frac{- 7 \pm \sqrt{37}}{2.}$

$\Rightarrow t \ne \frac{- 7 \pm \sqrt{37}}{2} , x ' \left(t\right) \ne 0.$

$\therefore , b y \left(\star\right) , \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - 3 t\right) {e}^{- 3 t}}{\left(1 + 7 t + 3 {t}^{2}\right) {e}^{t}} , \mathmr{and} ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 3 t}{\left(1 + 7 t + 3 {t}^{2}\right) {e}^{4 t}} , t \ne \frac{- 7 \pm \sqrt{37}}{2.}$

Enjoy Maths.!