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

Apr 21, 2018

First, differentiate each individual function as you have been all year.

using the chain rule,

$x \left(t\right) = {\left(t - 2\right)}^{\frac{1}{2}}$

$\text{ "" } \implies \frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1}{2} {\left(t - 2\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dt}} \left(t - 2\right)$

$\text{ "" } \textcolor{w h i t e}{\implies \frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{1}{2} {\left(t - 2\right)}^{- \frac{1}{2}} \cdot 1$

" "" "color(white)(=>dx/dt)=color(red)(1/(2sqrt(t-2))

and, by the product rule,

$y \left(t\right) = {t}^{2} - 2 {t}^{3} {e}^{t}$

$\text{ "" } \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - \left[\left(\frac{d}{\mathrm{dt}} 2 {t}^{3}\right) {e}^{t} - 2 {t}^{3} \left(\frac{d}{\mathrm{dt}} {e}^{t}\right)\right]$

$\text{ "" } \textcolor{w h i t e}{\implies \frac{\mathrm{dy}}{\mathrm{dt}}} = 2 t - \left[6 {t}^{2} {e}^{t} - 2 {t}^{3} {e}^{t}\right]$

" "" "color(white)(=>dy/dt)=color(red)(2t(1-3te^t+t^2e^t)

if we want to know $\frac{\mathrm{dy}}{\mathrm{dx}}$, we see that:

dy/dx=(dy/dt)/(dx/dt)=(2t(1-3te^t+t^2e^t))/(1/(2sqrt(t-2)))=color(blue)(4tsqrt(t-2)(1-3te^t+t^2e^t)