A curve is given by $x = t^{2} + 1$, $y = t^{3}$. What is its slope $\frac{dy}{dx}$ at $t = 2$?

Explanation

1. 01

   $\frac{dx}{dt} = 2 t$ and $\frac{dy}{dt} = 3 t^{2}$.

2. 02

   $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 t}{2}$.

3. 03

   At $t = 2$ the slope is $3$.