Parameterised Curve Proof Part 1

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Homework Help Overview

The discussion revolves around proving a relationship involving the arc length function \( s(t) \) for a parameterized curve \( \vec{r}(t) \). The original poster seeks guidance on how to demonstrate that the derivative of the arc length function \( s'(t) \) equals the norm of the derivative of the curve \( ||d\vec{r}(t)|| \).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants suggest looking up the Leibniz rule for differentiating integrals, with some questioning the application of this rule in the context of the problem. Others express confusion about integrating and differentiating implicit functions and the correct interpretation of limits in integrals.

Discussion Status

There are multiple lines of reasoning being explored, including the application of the Leibniz rule and the interpretation of the arc length function. Some participants have provided alternative perspectives on how to approach the differentiation of the integral, while others are still grappling with the foundational concepts involved.

Contextual Notes

Participants note that the original poster has also sought help on another forum without receiving replies, indicating a potential lack of resources or engagement on that platform. There are also discussions about the assumptions regarding the behavior of the derivative \( dr(t) \) and the implications of infinitesimals in the context of the problem.

  • #31
bugatti79 said:
ok,

||r'(t)||=v...?

Almost! :wink:

r'(t)=v
||r'(t)||=?
 
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  • #32
I like Serena said:
Almost! :wink:

r'(t)=v
||r'(t)||=?

The length of the vector is SRSS which is in this case v! No?
 
  • #33
bugatti79 said:
The length of the vector is SRSS which is in this case v! No?

Yes. You really need to start making distinctions between vectors, scalars, infinitesimal vectors, and infinitesimal scalars. :wink:

Anyway, your arc speed is ||r'(t)||=||v||.

For your reparameterisation you want it to be 1.

So perhaps s(t) is something like s(t)=c t for some constant c.
What will the arc speed become with this parameterisation?
 
  • #34
I like Serena said:
Yes. You really need to start making distinctions between vectors, scalars, infinitesimal vectors, and infinitesimal scalars. :wink:

Anyway, your arc speed is ||r'(t)||=||v||.

For your reparameterisation you want it to be 1.

So perhaps s(t) is something like s(t)=c t for some constant c.
What will the arc speed become with this parameterisation?

We are looking for r_1(s) = r(t(s)) such that ||r_1(s)||=1

IF s(t)=ct then s'(t)=c. Dont we need the reverse...t(s) etc to carry out the chain rule?
 
  • #35
bugatti79 said:
We are looking for r_1(s) = r(t(s)) such that ||r_1(s)||=1

Careful with the derivatives.
You want ||r_1'(s)||=1!


bugatti79 said:
IF s(t)=ct then s'(t)=c. Dont we need the reverse...t(s) etc to carry out the chain rule?

Yes...
 
  • #36
bugatti79 said:

Homework Statement



Suppose that \vec r(t) is a parameterised curve defined for a \le t\le b and

\displaystyle s(t)=\int_{a}^{t}\left \| d \vec r (t) \right \|dt is the arc length function measured from r(a)

a) Prove that s'(t) = || dr(t)||

How do I start this? It is easy to see that differentiating both sides will yield the proof but I don't know how to go about t. Any clues?

Note I have this also posted at MHF with no replies
http://www.mathhelpforum.com/math-help/f57/parameterised-curve-proof-part-1-a-191196.html"

We are integrating some vector valued function f(x), say.

So the integral of f(x) = F(x), which is evaluated from a to t, which is:
F(t) - F(a).
We different this then with respect to t:
dF(t)/dt - dF(a)/dt = f(t)
 
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  • #37
I like Serena said:
Careful with the derivatives.
You want ||r_1'(s)||=1!




Yes...

r'(t)=v and t'(s)=1/c implies r_1'(s)=r'(t)t'(s)=v/c but this is not a function of s anymore...?
 
  • #38
Norfonz said:
We are integrating some vector valued function f(x), say.

So the integral of f(x) = F(x), which is evaluated from a to t, which is:
F(t) - F(a).
We different this then with respect to t:
dF(t)/dt - dF(a)/dt = f(t)

Yes, we got this thanks. :-)
 
  • #39
bugatti79 said:
r'(t)=v and t'(s)=1/c implies r_1'(s)=r'(t)t'(s)=v/c but this is not a function of s anymore...?

Yes it is.
It is still a function of s.
It's just that there is no s in it any more.

How would you need to choose c to make sure the arc speed of r_1 is 1?
 
  • #40
I like Serena said:
Yes it is.
It is still a function of s.
It's just that there is no s in it any more.

How would you need to choose c to make sure the arc speed of r_1 is 1?

We need ||r_1'(s)||/||v|| =1 ie c=1...?
 
  • #41
bugatti79 said:
We need ||r_1'(s)||/||v|| =1 ie c=1...?

Nooo... you need ||r_1'(s)|| =1.

Consider the parameter s to be the distance along the curve - exactly.
 
  • #42
I like Serena said:
Nooo... you need ||r_1'(s)|| =1.

Consider the parameter s to be the distance along the curve - exactly.

Well if s=ct and s is the distance and t the time then c must be the velocity ...

but we already have a velocity r'(t)=v...
 
  • #43
Yes, what you say is true. So?

Perhaps I should remark that r'(t) ≠ r_1'(s).
 
  • #44
I like Serena said:
How would you need to choose c to make sure the arc speed of r_1 is 1?

ok, but I don't understand what you mean by this?
 
  • #45
Well, the arc speed of r_1 is ||r_1'(s)||.
This is the distance traveled along the curve r_1 if s increases by 1.

What is ||r_1'(s)||?
 
  • #46
I like Serena said:
Well, the arc speed of r_1 is ||r_1'(s)||.
This is the distance traveled along the curve r_1 if s increases by 1.

What is ||r_1'(s)||?

is = || r'(t) t'(s)||=||v/c||...where t(s)=s/c and r'(t)=v
 
  • #47
Good!

So what does c have to be to make it 1?
 
  • #48
I like Serena said:
Good!

So what does c have to be to make it 1?

The units would have to match therefore v must be equal c.
 
  • #49
Which units?
Anyway, the dimensions do not match, since v is a vector and c is a scalar.
So...?
 
  • #50
bugatti79 said:
Well if s=ct and s is the distance and t the time then c must be the velocity ...

but we already have a velocity r'(t)=v...

Thes units... c must be m/s and r'(t)=v which is velocity and henc also m/s...?
 
  • #51
Hmm, your problem statement said nothing about m/s.
So how did you assume it would be m/s?

Anyway, even assuming the speed is in m/s, what's stopping us from defining a new curve that has a unit-less speed of 1?
 
  • #52
I like Serena said:
Hmm, your problem statement said nothing about m/s.
So how did you assume it would be m/s?

Anyway, even assuming the speed is in m/s, what's stopping us from defining a new curve that has a unit-less speed of 1?

There was nothing said of m/s, I can't think of velocity being anything other than m/s!

I don't know how you would have a unit less speed. I give up :-)
 
  • #53
Well, let's make a side trip into physics.
Did you know that meters and seconds are actually the same unit with a conversion factor (speed of light in vacuum) between them?
In other words, in a sense m/s is also dimensionless!

This is however not really relevant to your problem.
But if you want to give up, that's up to you of course!
 
  • #54
I like Serena said:
Hmm, your problem statement said nothing about m/s.
So how did you assume it would be m/s?

Anyway, even assuming the speed is in m/s, what's stopping us from defining a new curve that has a unit-less speed of 1?

c would have to be 1 to make ||r_1(t)||=1 if v is dimensionless.
 
  • #55
You wrote before that ||r_1'(s)|| = ||v/c||
So if c = 1, then ||r_1'(s)|| = ||v||, which is presumably not 1.

Btw, you seem to have dropped a quote that should have been there to indicate the derivative.
And you also seem to have replaced the parameter s of r_1(s) by t.
Really!
 
  • #56
I like Serena said:
You wrote before that ||r_1'(s)|| = ||v/c||
So if c = 1, then ||r_1'(s)|| = ||v||, which is presumably not 1.

Btw, you seem to have dropped a quote that should have been there to indicate the derivative.
And you also seem to have replaced the parameter s of r_1(s) by t.
Really!

Well, I think I have exhausted all my possible answers based on my understanding. I don't know h0w ||r_1'(s)||=1 if c is not 1.
 
  • #57
Well, if v is a vector and c is a scalar, what if c=||v||?

And if you want, you can let v keep a unit of [m/s] and have c be dimensionless.
That should satisfy your sense of what speed should be.
 
  • #58
I like Serena said:
Well, if v is a vector and c is a scalar, what if c=||v||?

And if you want, you can let v keep a unit of [m/s] and have c be dimensionless.
That should satisfy your sense of what speed should be.

hmm... so if s(t) is the distance along the curve r_1(s) we would have s(t)=||v||t?

So in a nutshell we were finding a suitable expression for the distance s(t) such that we can maintain the requirement that ||r_1'(s)||=1 for all s?


Is it a similar approach for this last exercise?

reparameterise the curve r(t)=2cos(t)i+2cos(t)j+tk by it arc length. start at t=0
 
  • #59
bugatti79 said:
hmm... so if s(t) is the distance along the curve r_1(s) we would have s(t)=||v||t?

Uummm... s(t) depends on a parameter t that is not in r_1(s)...
But I think what you mean is probably right.

I'd say it something like this:

s is the length of the curve r_1 from 0 to s.

With s(t) = ||v||t, the curve r(t(s)) has this property.


bugatti79 said:
So in a nutshell we were finding a suitable expression for the distance s(t) such that we can maintain the requirement that ||r_1'(s)||=1 for all s?

Yes.

bugatti79 said:
Is it a similar approach for this last exercise?

reparameterise the curve r(t)=2cos(t)i+2cos(t)j+tk by it arc length. start at t=0

Yes.
 
  • #60
bugatti79 said:
Is it a similar approach for this last exercise?

reparameterise the curve r(t)=2cos(t)i+2cos(t)j+tk by it arc length. start at t=0

||r'(t)||=||-2sin(t)-2sin(t)+1||=(-4sin(t)+1)^(1/2).

If we let s(t)=ct then using the same idea we must have c=(-4sin(t)+1)^0.5 in order for ||r_1'(s)||=1...?
 

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