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

Nov 10, 2016

The slope of $f \left(t\right)$ is infinite when $t = - 1$ (ie the tangent is vertical).

#### Explanation:

We have $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ Where $x \left(t\right) = {t}^{2} + 2 t$, $y \left(t\right) = 2 t - 3$

The by the chain rule, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$x \left(t\right) = {t}^{2} + 2 t \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 2 t + 2$
$y \left(t\right) = 2 t - 3 \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 2$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{2 t + 2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{t + 1}$

When $t = - 1 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{0} = \infty$, so the slope of $f \left(t\right)$ is infinite when $t = - 1$ (ie the tangent is vertical).

This can be visualised by looking at the graph of $f \left(t\right)$