Deriving T'(t) for T(t) = c'(t)/||c'(t)|| with Norm and Dot Product

[MichaelT]

I am trying to derivate the equation T(t) = c'(t)/||c'(t)||

||c'(t)|| being the norm of c'(t). ||T(t)||= 1 and I need to solve for T'(t)

I keep getting stuck with some nasty terms. I assume that I need to deal with the norm as the square root of the dot product of c'(t) with itself.

Any help would be much appreciated. I have a feeling there is a simpler way to solve this than the way I have been trying.

 

[tiny-tim]

I am trying to derivate the equation T(t) = c'(t)/||c'(t)||

||c'(t)|| being the norm of c'(t). ||T(t)||= 1 and I need to solve for T'(t)

I keep getting stuck with some nasty terms. I assume that I need to deal with the norm as the square root of the dot product of c'(t) with itself.

Any help would be much appreciated. I have a feeling there is a simpler way to solve this than the way I have been trying.

Hi MichaelT!

I don't understand what you're trying to do.

If c(t) is a curve, then c'(t)/||c'(t)|| is the unit vector tangent to the curve … what is the problem?

 

[MichaelT]

So T(t) = c'(t)/||c'(t)||. We are asked to find a formula for T'(t) in terms of c.

Sorry, looking back at my original post I realize I was not so clear.

Thank you!

 

[tiny-tim]

So T(t) = c'(t)/||c'(t)||. We are asked to find a formula for T'(t) in terms of c.

Sorry, looking back at my original post I realize I was not so clear.

Thank you!

Hi MichaelT! Thanks for your PM.

As you said:

I assume that I need to deal with the norm as the square root of the dot product of c'(t) with itself.

So let's write it out in full …

||c'(t)|| = √(c'(t).c'(t)) …

use the product rule (it works for vectors just as it does for scalars) to differentiate the inside of the √, and then use the chain rule to differentiate the whole thing.

(btw … no such word as "derivate" … and "derviation" would be completely different from "differentiation" )

 

[MichaelT]

Ok, I think I have another way to approach the problem. From subsequent problems I have learned that T'(t) = c"(t)

Therefore, if I can prove that T(t) = c'(t), then I am set. Can you give me any direction on doing that? I've tried with no luck, though I do know it is true!

 

[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

 

[tiny-tim]

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

Hi MichaelT!

Aren't you missing a ||c'(t)|| at the beginning?

 

[MichaelT]

yeah it looks like I did miss something! Does it look like I am approaching this the right way?

Just re-did the problem, and this is what I got

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

 

[tiny-tim]

yeah it looks like I did miss something! Does it look like I am approaching this the right way?

Just re-did the problem, and this is what I got

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

Yes, that is the right way, and the right result this time!

btw, in your previous single-fraction answer, which should have been

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

the numerator is in the form A(B.C) + B(A.C), so you could (probably not very usefully ) write it as … ? ​