What is the slope of f(t) = (t-2t+4,t^2) at t =0?

Nov 22, 2016

0

Explanation:

Define $x \left(t\right) , y \left(t\right)$ as follows

$\left\{\begin{matrix}x \left(t\right) = t - 2 t + 4 = 4 - t \\ y \left(t\right) = {t}^{2}\end{matrix}\right. \implies f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$

Differentiating $x \left(t\right)$ wrt $t$ we get:

$\dot{x} \left(t\right) = \frac{\mathrm{dx}}{\mathrm{dt}} = - 1$

Differentiating $y \left(t\right)$ wrt $t$ we get:

$\dot{y} \left(t\right) = \frac{\mathrm{dy}}{\mathrm{dt}} = 2 t$

Then the applying the chain rule;

$f ' \left(t\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
$\therefore f ' \left(t\right) = \frac{2 t}{- 1}$
$\therefore f ' \left(t\right) = - 2 t$

So when $t = 0$ we get:

$f ' \left(0\right) = 0$

Hence, The slope of $f \left(t\right)$ is $0$ at $t = 0$