1. ## Unit vectors question

The following problem was confusing me so any help would be greatly appreciated!

Find the unit vectors T and N for the given curve at the indicated point.

r(t) = <t, t^2, t^3> at (1, 1, 1)

2. Originally Posted by meebo0129
The following problem was confusing me so any help would be greatly appreciated!

Find the unit vectors T and N for the given curve at the indicated point.

r(t) = <t, t^2, t^3> at (1, 1, 1)
T and N are tangent and normal vectors, yes?

I can get you T, but I don't have a specific definition for N: the best I can do is tell you how to find one of an infinite number of normal vectors.

Anyway, for T:
$\displaystyle \nabla r(t) = <1, 2t, 3t^2>$

At the point (1, 1, 1) we have for an unnormalized T = <1, 2, 3>. This vector has a length of $\displaystyle \sqrt{1^2 + 2^2 + 3^3} = \sqrt{14}$, so your answer for the unit vector in the direction of T is:

$\displaystyle \frac{<1, 2, 3>}{\sqrt{14}}$

Someone else is going to have to show you how to get N. (I can find a vector perpendicular to T easily enough using $\displaystyle T \cdot N = 0$, but there will be an infinite number of possible N that can do this. You obviously have a specific property defined that picks out a specific N.)

-Dan

3. $\displaystyle r(t)=ti+t^{2}j+t^{3}k$

$\displaystyle f'(t)=i+2tj+3t^{2}k$

$\displaystyle T=\frac{i+2tj+3t^{2}k}{\sqrt{14}}=\frac{1}{\sqrt{1 4}}i+\frac{2}{\sqrt{14}}j+\frac{3}{\sqrt{14}}k$

Now, $\displaystyle N(t)=\frac{T'(t)}{||T||}$

If I remember correctly, this is how T and N are found. It's been a while.