Integral of unit tangent vector equals arc length?

[ArcanaNoir]

Homework Statement

Let c(t) be a path and T the unit tangent vector. What is \int_c \mathbf{T} \cdot d\mathbf{s}

Homework Equations

The unit tangent vector of c(t) is c'(t) over the magnitude of c'(t) :
\mathbf{T} = \frac{c'(t)}{||c'(t)||}

The length of c(t) can be represented by :
\int_c ||c'(t)|| \; dt

The Attempt at a Solution

\int_c \mathbf{T} \cdot d\mathbf{s} = \int_c \frac{c'(t)}{||c'(t)||} ... d-something. dt I suppose.

But this is clearly not quite the arc length integral. So what am I missing? (the book says the answer is the length of c)

 

[I like Serena]

Hi Arcana!

I guess you need that $\mathbf{s} = \mathbf{c}(t)$.
That is, $\mathbf{s}$ is a point on the curve that is parametrized by $\mathbf{c}(t)$.
What do you think $\textrm{d}\mathbf{s}$ is?

 

[ArcanaNoir]

ds=c'(t)dt ?
But then I still have 1/magnitude, so still not quite there.

 

[I like Serena]

ds=c'(t)dt ?

Yep.

But then I still have 1/magnitude, so still not quite there.

How so?
What do you get if you substitute ds=c'(t)dt?

 

[ArcanaNoir]

(c'(t))^2 / llc'(t)ll dt

 

[I like Serena]

Yes.
And what is $\mathbf{c}'(t) \cdot \mathbf{c}'(t)$?

 

[ArcanaNoir]

is it 1 ? no

it's the same thing as we have under the square root (calculating...)
okay I am stuck

 

[I like Serena]

is it 1 ?

No.
Suppose $\mathbf{v}$ is a vector.
Did you know that $\mathbf{v} \cdot \mathbf{v} = ||\mathbf{v}||^2$?

 

[ArcanaNoir]

oh. excellent observation. I see it now. thanks :)