# Integral of unit tangent vector equals arc length?

1. Feb 18, 2012

### ArcanaNoir

1. The problem statement, all variables and given/known data
Let c(t) be a path and T the unit tangent vector. What is $$\int_c \mathbf{T} \cdot d\mathbf{s}$$

2. Relevant 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 :
$$\int_c ||c'(t)|| \; dt$$

3. 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)

2. Feb 18, 2012

### 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?

3. Feb 18, 2012

### ArcanaNoir

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

4. Feb 18, 2012

### I like Serena

Yep.

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

5. Feb 18, 2012

### ArcanaNoir

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

6. Feb 18, 2012

### I like Serena

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

7. Feb 18, 2012

### ArcanaNoir

is it 1 ? no

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

8. Feb 18, 2012

### I like Serena

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

9. Feb 18, 2012

### ArcanaNoir

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