# Proof: Arc Length parameter

1. Mar 14, 2014

### yazz912

1. The problem statement, all variables and given/known

If C is a smooth curve given by
r(s)= x(s)i + y(s)j + z(s)k

Where s is the arc length parameter. Then

||r'(s)|| = 1.

My professor has stated that this is true for all cases the magnitude of r'(s) will always equal 1. But he wants me to PROVE it. ( of course not with an example)

2. Relevant equations

r(s)= x(s)i + y(s)j + z(s)k

r'(s)= x'(s)i + y'(s)j + z'(s)k

3. Attempt at the solution

To be quite honest, usually with math problems I will have some sort of attempt to try and solve it. But when it comes to proofs... I seem to get stuck.

Well I know I'm trying to prove that. ||r'(s)|| = 1

So the magnitude of r'(s)
Will be given by
SQRT[ x'(s)^2 + y'(s)^2 + z'(s)^2 ]

After this I don't know what I can do to make it equal 1. Any help will be greatly appreciated!

2. Mar 14, 2014

### micromass

You know $s$ is the arc-length parameter, so what does that mean?

3. Mar 14, 2014

### yazz912

Um, That it is measuring the length along the Curve ..

4. Mar 14, 2014

### micromass

Can you express that mathematically?

5. Mar 14, 2014

### yazz912

Unfortunately no:/

6. Mar 14, 2014

### micromass

How do you calculate the arc length of a curve?

7. Mar 17, 2014

### yazz912

by integrating from a to b
On the following
sqrt[ x'(t)^2 + y'(t)^2 + z'(t)^2 ] dt

8. Mar 17, 2014

### micromass

Right, so the length from $0$ to $s$ is defined as

$$\int_0^s \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 + (z^\prime(t))^2}dt$$

by definition of arc-lenght parameter, this is equal to $s$. So

$$s = \int_0^s \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 + (z^\prime(t))^2}dt$$

Now differentiate both sides.

9. Mar 17, 2014

### yazz912

Will differentiating on the side with the integral wipe out the integral completely? Or would I have to integrate and then differentiate?

10. Mar 17, 2014

### HallsofIvy

Differentiating on the side with the integral, use the fundamental theorem of Calculus:
$$\frac{d}{ds}\left(\int_a^s f(t)dt\right)= f(s)$$

11. Mar 17, 2014

### yazz912

Ok so by using that theorem, to find the derivative of an integral. I have to plug my upper limit back into the function.

Which would be
SQRT[ x'(s)^2 + y'(s)^2 + z'(s)^2 ]

How does this prove that ||r'(t)|| always =1 where t is any parameter?

12. Mar 17, 2014

### LCKurtz

Here's a hint for an alternate (easier) way. Use the chain rule for$$\frac d {ds} \vec r(t) = r'(t)\cdot (?)$$by the chain rule. Work that out and calculate its length. You don't have to consider components.

13. Mar 18, 2014

### pasmith

So far you have
$$s = \int_0^s \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\,dt$$
and have differentiated with respect to $s$ to find
$$1 = \sqrt{x'(s)^2 + y'(s)^2 + z'(s)^2}.$$
But the right hand side is exactly $\|r'(s)\|$!

It doesn't, becuase it's not true that $\|r'(t)\| = 1$ for any parameter: consider the line from $(0,0,0)$ to $(1,1,1)$ with the parametrization
$$(x,y,z) = (t,t,t),\qquad 0 \leq t \leq 1$$
and compare it with the parametrization
$$(x,y,z) = (2u,2u,2u),\qquad 0 \leq u \leq \tfrac 12.$$
The result $\|r'(s)\| = 1$ holds only when $s$ is arclength.