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

Apr 8, 2018

The derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ of the parametric equation is $\frac{3 {t}^{2} - 2 t + 3}{2 t - 1}$.

Explanation:

To find $\frac{\mathrm{dy}}{\mathrm{dx}}$ of the parametric equation $\left({t}^{2} - t , {t}^{3} - {t}^{2} + 3 t\right)$, find $\mathrm{dx}$ from $x \left(t\right)$ and $\mathrm{dy}$ from $y \left(t\right)$ in terms of $\mathrm{dt}$:

$\textcolor{w h i t e}{\implies} x \left(t\right) = {t}^{2} - t$

$\implies \mathrm{dx} = \left(2 t - 1\right) \mathrm{dt}$

and

$\textcolor{w h i t e}{\implies} y \left(t\right) = {t}^{3} - {t}^{2} + 3 t$

$\implies \mathrm{dy} = \left(3 {t}^{2} - 2 t + 3\right) \mathrm{dt}$

Now, $\frac{\mathrm{dy}}{\mathrm{dx}}$ will be the quotient of these two:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(3 {t}^{2} - 2 t + 3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\mathrm{dt}}}}}{\left(2 t - 1\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\mathrm{dt}}}}} = \frac{3 {t}^{2} - 2 t + 3}{2 t - 1}$

That's the derivative. Hope this helped!