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

Jul 28, 2017

Slope is $3$

#### Explanation:

The slope of $f \left(t\right) - \left(x \left(t\right) , y \left(t\right)\right)$ is given by $\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here $y = {t}^{3}$ hence $\frac{\mathrm{dy}}{\mathrm{dt}} = 3 {t}^{2}$

and $x = t - 1$ and hence $\frac{\mathrm{dx}}{\mathrm{dt}} = 1$

Hence slope is $\frac{3 {t}^{2}}{1} = 3 {t}^{2}$

and at $t = 1$, it is $3 \times {1}^{2} = 3$

one can see it in the following graph as $t = 1$ means slope at $\left(0 , 1\right)$

graph{y=(x+1)^3 [-5, 5, -2.5, 2.5]}