# How do you differentiate the following parametric equation:  x(t)=3(t+1)^2+2e^t, y(t)= (t+2)^2+t^2?

Dec 3, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \left(t + 1\right)}{2 t + 1}$

#### Explanation:

Step 1: Find $\frac{\mathrm{dx}}{\mathrm{dt}}$
Step 2: Find $\frac{\mathrm{dy}}{\mathrm{dt}}$
3) Step 3: $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\mathrm{dx}}{\mathrm{dt}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \Leftrightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Step 1: Given $x \left(t\right) = 3 {\left(t + 1\right)}^{2}$
$x ' \left(t\right) = \frac{\mathrm{dx}}{\mathrm{dt}} = 3 \left(2\right) \left(t + 1\right) \left(1\right)$

$\textcolor{red}{\frac{\mathrm{dx}}{\mathrm{dt}} = 6 \left(t + 1\right)}$

Step 2: Given $y \left(t\right) = {\left(t + 2\right)}^{2} + {t}^{2}$
$y ' \left(t\right) = \frac{\mathrm{dy}}{\mathrm{dt}} = 2 \left(t + 2\right) \left(1\right) + 2 t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t + 4 + 2 t = 4 t + 2$
$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dt}} = 4 t + 2 = 2 \left(2 t + 1\right)}$

Step 3: $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\mathrm{dx}}{\mathrm{dt}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \Leftrightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\textcolor{red}{6 \left(t + 1\right)}}{\textcolor{b l u e}{2 \left(2 t + 1\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \left(t + 1\right)}{2 t + 1}$