Proof of Arc Length Parameter

[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. Homework 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!

 

[micromass]

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

 

[yazz912]

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

 

[micromass]

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

Can you express that mathematically?

 

[yazz912]

Unfortunately no:/

 

[micromass]

How do you calculate the arc length of a curve?

 

[yazz912]

How do you calculate the arc length of a curve?

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

 

[micromass]

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

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

[tex]\int_0^s \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 + (z^\prime(t))^2}dt[/tex]

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

[tex]s = \int_0^s \sqrt{(x^\prime(t))^2 + (y^\prime(t))^2 + (z^\prime(t))^2}dt[/tex]

Now differentiate both sides.

 

[yazz912]

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

 

[HallsofIvy]

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

 

[yazz912]

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

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?

 

[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.

 

[pasmith]

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 ]

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

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

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