What is the slope of f(t) = (t+1,t) at t =0?

Dec 10, 2017

The slope of the gradient of $f \left(t\right)$ is $1$ when $t = 0$

Explanation:

We have:

$f \left(t\right) = \left\{\begin{matrix}x \left(t\right) = t + 1 \\ y \left(t\right) = t\end{matrix}\right.$

Differentiating wrt $t$ we have:

$\frac{\mathrm{dx}}{\mathrm{dt}} = 1$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$

Then by the chain rule we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dy}}{\mathrm{dt}}\right) / \left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)$
$\setminus \setminus \setminus \setminus \setminus = 1$

So the slope of the gradient of $f \left(t\right)$ is $1$ when $t = 0$

We can also see this is the case if we consider isolate the parameter $t$ giving

$x = y + 1 \implies y = x - 1$

Which represents a straight line with gradient $1$.