
Another Vector Problem
The parametric form of the tangent line to the image of f(t)=http://omega.math.union.edu/webwork2...144/char0A.png2t^2http://omega.math.union.edu/webwork2...144/char3B.png4/thttp://omega.math.union.edu/webwork2...144/char3B.pngt−5http://omega.math.union.edu/webwork2...144/char0B.png 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!

There is a mistake:
$\displaystyle f'(t)=(4t,4/t^2,1)$ and so, $\displaystyle f'(1)=(4,4,1),\;f(1)=(2,4,6)$
The tangent line is:
$\displaystyle r \equiv\begin{Bmatrix}x=24t\\y=44t\\z=6+t\end{matrix}\quad (t\in{\mathbb{R}})$

Fernando Revilla

Oh duh!
So I get,
<4t,44t,t6>