# What is the slope of f(t) = ((t-2)^3,2t) at t =-1?

Mar 28, 2016

Slope at $t = - 1$ is $- \frac{1}{3}$

#### Explanation:

At $t = - 1$, $f \left(t\right) = \left({\left(- 1 - 2\right)}^{2} , 2 \left(- 1\right)\right)$ or $f \left(t\right) = \left(9 , - 2\right)$

As $x = {\left(t - 2\right)}^{2}$, $\frac{\mathrm{dx}}{\mathrm{dt}} = 2 \left(t - 2\right) = 2 t - 4$

and $y = 2 t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 2$

Hence slope is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{2}{2 t - 4}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{2 \left(- 1\right) - 4} = \frac{2}{- 6} = - \frac{1}{3}$