# How do you differentiate the following parametric equation:  x(t)=(t-1)^2-e^t, y(t)= (t+2)^2+t^2?

Sep 10, 2017

To find derivatives with respect to t, differentiate normally. To find $\frac{\mathrm{dy}}{\mathrm{dx}}$, use $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$. Solutions below.

#### Explanation:

Provided you mean finding $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$ , we do it the same way that we would if t were x and x (t) and y (t) were both functions of x. The usual rules apply, including the chain rule ($f \left(x\right) = g \left(h \left(x\right)\right) \to f ' \left(x\right) = h ' \left(x\right) g ' \left(h\right)$. Thus...

$x \left(t\right) = {\left(t - 1\right)}^{2} - {e}^{t} \to \frac{\mathrm{dx}}{\mathrm{dt}} = 2 \left(t - 1\right) - {e}^{t}$

And

$y \left(t\right) = {\left(t + 2\right)}^{2} + {t}^{2} \to \frac{\mathrm{dy}}{\mathrm{dt}} = 2 \left(t + 2\right) + 2 t = 4 t + 2$

If instead you want $\frac{\mathrm{dy}}{\mathrm{dx}}$, we can solve that as follows:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\mathrm{dy}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}}$ (chain rule).

Rearranging, we can get...

(dy)/dx = ((dy)/dt)/((dx)/dt) -> (dy)/dx =(4t+2)/(2(t-1)-e^t