How to find a particle's position after it off on the tangent?

Question

Problem:

A particle follows a trajectory given by $\sigma(t)=(t^{2},t^{3}-4t,0)$ and it off at a tangent at $t=2$. Find the particle's position at $t=3$. Make a draw of what is happening.

Solution:

And this is what I did to find the position at $t=2$:

First, to find the particle new position it's required to find a point $P$ and direction vector $\vec{v}$:

$P$: $\sigma(2)=(4,0,0)$

$\vec{v}$: $\sigma'(2)=(4,8,0)$

Hence, the equation to find the new particle position is:

$l(t)=P+(t-2)\vec{v}$

Is that correct? Does the equation model the particle's position at any given $t$ time?

Answer 1

Yes this is correct, expect for "Does the equation model the particle's position at any given $t$?" It is only valid for $t\geq 2$.

Answer 2

What you have obtained is indeed the correct form of the tangent line. Another way to obtain this is to expand component to first order in $t-2$:

\begin{align} \sigma(t) &=(t^2,t^3-4t,0)\\ &=([(t-2)+2]^{2},t(t-2)(t+2),0)\\ &\approx (4+4(t-2),8(t-2),0)\\ &=(4,0,0)+(4,8,0)(t-2). \end{align} This is exactly the form you found.