# Parametric equation for the tangent curve

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

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

• Do you know about the Euler method? Commented Feb 19, 2019 at 0:42

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.