I just read that given $\alpha : I \rightarrow R^3$ a parametrised differentiable curve, for each $t \in I$ where $\alpha '(t) \neq 0$, there is a well-defined straight line that contains the point $\alpha (t)$ and the vector $\alpha ' (t)$. I understand this line is called the tangent line to $\alpha $ at $t$. Could somebody give me an example of how the equation of this tangent line is derived? For example, how could I find the tangent line of the tractrix at $t=t_0$?
Thanks.
On
Given a curve $\alpha : I \to \mathbb{R}^3$, its velocity vector at a point $t_0$ is $\alpha'(t_0)$. The line tangent to the curve that this point is then formed as follows: you start at the point $\alpha(t_0)$, and then you travel parallel to the velocity vector. So it would be the line parametrized in $s$ by $$ \alpha(t_0) + s\alpha'(t_0), \ \ \ \ \ \ \ \ \ s \in \mathbb{R}. $$
Example
$$x=t, y= t^2+1, z= 2-3t^3 $$
Find the tangent line at $t=1$
First you notice that at $t=1$, we have $$ (x,y,z) = (1,2,-1)$$
On the other hand $$ (x',y',z') = (1,2t,-9t^2)=(1,2,-9)$$
Thus the equation of tangent line is
$$ x=1+t\\ y=2+2t\\z=-1-9t $$