Find parametric equations for the tangent line to the curve with the parametric equations $x=t, y=t^2, z=t^3$ at the point $(1, 1, 1)$.

For the Solution I know the method of solving it. But have a small problem in the procedure:

The parameter corresponding to the point $(1, 1, 1)$ is $t = 1$, so

And we have that $r(t)=(t,t^2,t^3)$
Thus $r'(1)=(1,2,3)$

And my doubt is after this step why do we write the parametric equation as

$x=1+t, y=1+2t, z=1+3t$?

  • $\begingroup$ Do you know about the Euler method? $\endgroup$
    – Toby Mak
    Feb 19, 2019 at 0:42

1 Answer 1


The (direction) vector $\mathbf{v}=(1,2,3)$ is the direction of your curve (and therefore the tangent line) at this (position) vector $\mathbf{p}=(1,1,1)$.

Then you can use that the (vector-valued) equation of the line through $p$ with direction $\mathbf{v}$ is given by $$(x(t),y(t),z(t))=L(t)=\mathbf{p}+\mathbf{v}t=(1,1,1)+(1,2,3)t=(1+t,1+2t,1+3t)$$ from which you can read off the conclusion you desire.

  • $\begingroup$ Ah that makes sense. Thank you! $\endgroup$
    – Charith
    Feb 19, 2019 at 4:06

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