# Math Help - Another Vector Problem

1. ## Another Vector Problem

The parametric form of the tangent line to the image of f(t)=2t^24/tt−5 at t=−1 is ?

I got the normal vector to the tangent, <4t,4/t^2, 1>.

Should I plug in the t=-1 to the original equation to find a point, then plug in t=-1, into the tangent line equation as well?

Thanks!

2. There is a mistake:

$f'(t)=(4t,-4/t^2,1)$ and so, $f'(-1)=(-4,-4,1),\;f(-1)=(2,-4,-6)$

The tangent line is:

$r \equiv\begin{Bmatrix}x=2-4t\\y=-4-4t\\z=-6+t\end{matrix}\quad (t\in{\mathbb{R}})$

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Fernando Revilla

3. Oh duh!

So I get,

<-4t,-4-4t,t-6>