# Motion defined by a parametric eqn

1. Oct 19, 2011

### Locoism

1. The problem statement, all variables and given/known data
For the curve:
r(t) = ⟨$\frac{1}{2}$t5, $\frac{1}{3}$t5, $\frac{1}{6}$t5⟩,

Find the arc length s(t)
Find the unit tangent T(t) and T(1)
Find the principle unit normal N(t) and N(1)
Find the binormal vector B(t) and B(1)

2. Relevant equations

T(t) = $\frac{r'(t)}{|r'(t)|}$

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

3. The attempt at a solution
I found s(t) as $\frac{1}{6}$t5$\sqrt{10}$

and r'(t) = <$\frac{5}{2}$t4, $\frac{5}{3}$t4, $\frac{5}{6}$t4>

But now if I calculate T(t) I get $\frac{1}{\sqrt{10}}$<3, 2, 1>
I'm sure this is wrong because first of all the question wouldn't ask for T(1), and secondly because now T'(t) is a zero vector, which makes N(t) a zero vector, and B(t) likewise.
Have I made a mistake or is the question just asking for some really trivial stuff?

2. Oct 19, 2011

### Dick

Well, the curve r(t) is a straight line, right? So it would make sense T(t) is a constant. And N(t) is going to be undefined, not zero. So that does make it a pretty strange question. Maybe there's a typo in r(t)??

3. Oct 19, 2011

### Locoism

Hm ok then, maybe the question has a mistake... But why would N be undefined and not zero?
Also the equation of the osculating plane at t=1 would also be undefined?
Thank you

4. Oct 19, 2011

### Dick

If T'(t)=0 then N(t)=T'(t)/|T'(t)| is 0/0. That's undefined. There's no unique normal. ANY vector perpendicular to the line is a normal. Nope, no osculating plane either.

5. Oct 19, 2011

### Locoism

Ah ok thanks a lot!