Differentiation of Unit Tangent

[MichaelT]

So we are given T(t) = c'(t)/||c'(t)|| as well as ||T|| = 1

We also know T(t)dotT(t) = 1 and T'(t)dotT(t) = 0

The problem asks us to find T'(t)


I tried differentiating c'(t)/||c'(t)|| treating ||c'(t)|| as the square root of the dot product of c'(t) with itself. I used the product rule, chain rule, quotient rule, and ended up with some nasty terms, namely c'(t) dot c"(t).

I am pretty sure the answer we are looking for is T'(t) = c"(t). Therefore, if we can prove that T(t) = c'(t), then the answer T'(t) = c"(t) follows.

Please help! LOL Not being able to solve this has been bothering me big time!

 

[MichaelT]

Oh wait, I was wrong about something. T(t) will only equal c'(t) if it is parametrized by the arc length.

This is what I got, if anyone cares to check (please do!)

T'(t) = [||c'(t)||(c"(t)) - c'(t)(c'(t) dot c"(t))]/||c'(t)||³

 

[Dick]

I got c''(t)/|c'(t)|-c'(t)(c'(t).c''(t))/|c'(t)|^3. That's like yours but with some factors and parentheses moved around. You can do a quick check by testing whether T(t).T'(t)=0. Is it?

 

[MichaelT]

That is most definitely what I got when I just re-did the problem!

Yay! Thank you very much, I will go and check it now