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

Nov 12, 2016

The slope of $f \left(t\right)$ is $1$ when $t = 1$

Explanation:

Let $f \left(t\right) = \left\{\begin{matrix}x \left(t\right) = {t}^{2} - t & \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 2 t - 1 \\ y \left(t\right) = t + 3 & \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 1\end{matrix}\right.$

Then By the chain rule; $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} . \frac{\mathrm{dt}}{\mathrm{dx}}$

Alternatively, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 t - 1}$

When $t = 1 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 - 1} = 1$

Hence, the slope of $f \left(t\right)$ is $1$ when $t = 1$