# Parameterised Curve Proof Part 1

## 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 dont 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"

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LCKurtz
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Gold Member
Look up Leibniz rule in your calculus book. It will tell you how do differentiate an integral as a function of the upper limit.

I like Serena
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Hi bugatti79!

Funny that you'd get no reply at MHF.
I'm not familiar with MHF, but can you compare PF to MHF?
And what is that avatar?
http://www.mathhelpforum.com/math-help/avatars/bugatti79.gif [Broken]

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Hi bugatti79!

Funny that you'd get no reply at MHF.
I'm not familiar with MHF, but can you compare PF to MHF?
And what is that avatar?
http://www.mathhelpforum.com/math-help/avatars/bugatti79.gif [Broken]

The MHF website was down for a few days so there are not many people checking the forum as the minute. I think its a very good website, very responsive.
Its just some fractals jpeg I got off the web. :-)

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Look up Leibniz rule in your calculus book. It will tell you how do differentiate an integral as a function of the upper limit.

My interpretation of Leibniz rule for this problem is not correct which is

$\displaystyle \frac{d}{dt} \int_{u(x)=a}^{v(x)=t} || dr(t)||dt=f(v(x))\frac{dv}{dx}-f(u(x))\frac{du}{dx}$............?

I like Serena
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My interpretation of Leibniz rule for this problem is not correct which is

$\displaystyle \frac{d}{dt} \int_{u(x)=a}^{v(x)=t} || dr(t)||dt=f(v(x))\frac{dv}{dx}-f(u(x))\frac{du}{dx}$............?

Hmm, you did not apply the rule correctly...

But I'll give you another way to look at it.

Suppose D(t) is the anti-derivative of $||d\vec r(t)||$.
Then D'(t)=$||d\vec r(t)||$.

Furthermore s(t)=D(t)-D(a).
Can you take the derivative on both sides with respect to t?

LCKurtz
Homework Helper
Gold Member
$$\frac d {dt}\int_a^t f(u)\,du = f(t)$$

Hmm, you did not apply the rule correctly...

But I'll give you another way to look at it.

Suppose D(t) is the anti-derivative of $||d\vec r(t)||$.
Then D'(t)=$||d\vec r(t)||$.

Furthermore s(t)=D(t)-D(a).
Can you take the derivative on both sides with respect to t?

No, I dont know how to differentiate this $\displaystyle s(t)=\int_{a}^{t}\left \| d \vec r (t) \right \|dt$ on the RHS...? THe LHS is just s'(t)

LCKurtz
Homework Helper
Gold Member
No, I dont know how to differentiate this $\displaystyle s(t)=\int_{a}^{t}\left \| d \vec r (t) \right \|dt$ on the RHS...? THe LHS is just s'(t)

Write it like this:
$$s(t)=\int_{a}^{t}\left \| \vec r' (u) \right \|du$$
and look at post #7.

$$\frac d {dt}\int_a^t f(u)\,du = f(t)$$

What is throwing me off is how to integrate an implicit function and put in the the limits and then differentiate it.

Ie, one cant write f(u)^2/2|^t_a etc....I dont believe this is correct.

I can see how integrating the function yields 2 terms..one a function of t and the other a function of a. Then differentiating wrt t takes the term a function of a go to 0 etc.

LCKurtz
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Gold Member
$$\frac d {dt}\int_a^t f(u)\,du = f(t)$$

What is throwing me off is how to integrate an implicit function and put in the the limits and then differentiate it.

Ie, one cant write f(u)^2/2|^t_a etc....I dont believe this is correct.
Of course that isn't correct.
I can see how integrating the function yields 2 terms..one a function of t and the other a function of a. Then differentiating wrt t takes the term a function of a go to 0 etc.

Suppose the antiderivative of f is F, whatever formula that may be. Then the integral gives you F(t) - F(a). What happens if you differentiate that?

I like Serena
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Hmm, you did not apply the rule correctly...

But I'll give you another way to look at it.

Suppose D(t) is the anti-derivative of $||d\vec r(t)||$.
Then D'(t)=$||d\vec r(t)||$.

Furthermore s(t)=D(t)-D(a).
Can you take the derivative on both sides with respect to t?

No, I dont know how to differentiate this $\displaystyle s(t)=\int_{a}^{t}\left \| d \vec r (t) \right \|dt$ on the RHS...? THe LHS is just s'(t)

If you take the derivative on both sides you get:
s'(t)=D'(t)-0
And because of our definition of D we have: D'(t)=$||d\vec r(t)||$.
So:
$s'(t)=||d\vec r(t)||$.

Of course that isn't correct.

Suppose the antiderivative of f is F, whatever formula that may be. Then the integral gives you F(t) - F(a). What happens if you differentiate that?

You are just left with s'(t)=F'(t) where the RHS is ||r'(t)|| etc. But how do you prove it? THat is the original question.

I like Serena
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You are just left with s'(t)=F'(t) where the RHS is ||r'(t)|| etc. But how do you prove it? THat is the original question.

This is the proof.
Note that you only used the definitions of the anti-derivative and the integral.

LCKurtz
Homework Helper
Gold Member
You are just left with s'(t)=F'(t) where the RHS is ||r'(t)|| etc. But how do you prove it? THat is the original question.

If what you are asking is how to prove Leibnitz Rule then, like I said in my very first post, look in your calculus book for a proof. Although ILS and I have both outlined a version of a proof in this thread.

OK, the next bit.

Assume dr (t) is never 0. Then ds(t) >0 for all t so we can in principle express t in terms of s and then obtain the new parameterisation r_1 (s)= r(t(s)) where 0 <= s<=L(=\int_{a}^{b} ||dr(t)||dt).

a) Prove ||dr(s)||=1.

My attempt.

r_1(s)=dr(t).dt(s). We know s'(t)= ||dr(t)|| therefore t'(s)=1/||dr(t)||

then r_1(s)=dr(t)/||dr(t)||=1

Correct?

I like Serena
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OK, the next bit.

Assume dr (t) is never 0. Then ds(t) >0 for all t so we can in principle express t in terms of s and then obtain the new parameterisation r_1 (s)= r(t(s)) where 0 <= s<=L(=\int_{a}^{b} ||dr(t)||dt).

It should be \int_{a}^{b} ||dr(t)||.
That is, without the dt.
There can (effectively) be only 1 infinitesimal in an integral.

a) Prove ||dr(s)||=1.

This cannot be true, since the left hand side (LHS) is an infinitesimal, while the right hand side (RHS) is not.

Perhaps you meant: ||dr(s)||=ds?

My attempt.

r_1(s)=dr(t).dt(s).

I'm afraid you have another infinitesimal problem here.
No infinitesimals on the LHS, and a product of 2 at the RHS is simply not possible.

You should define r_1 in terms of r though.
This would be:
r_1(s) = r(t(s))

We know s'(t)= ||dr(t)|| therefore t'(s)=1/||dr(t)||

Good!
You will want to use this.

then r_1(s)=dr(t)/||dr(t)||=1

Correct?

I'm afraid you have a vector at the LHS, and in the middle, but a scalar at the RHS.
Not possible.

r_1(s) = r(t(s))

Convert first both sides to vector-infinitesimals.
And then take the norm of the vector-infinitesimals.

It should be \int_{a}^{b} ||dr(t)||.
That is, without the dt.
There can (effectively) be only 1 infinitesimal in an integral.

This cannot be true, since the left hand side (LHS) is an infinitesimal, while the right hand side (RHS) is not.

Perhaps you meant: ||dr(s)||=ds?

it says in the handout ||dr_1(s)||=1 for all s....

I'm afraid you have another infinitesimal problem here.

I think I should have written it

[ITEX]r_1(s)=\frac{dr(t)}{dt}\frac{dt}{ds}....[/ITEX] ie chain rule....?

No infinitesimals on the LHS, and a product of 2 at the RHS is simply not possible.

You should define r_1 in terms of r though.
This would be:
r_1(s) = r(t(s))

Good!
You will want to use this.

I'm afraid you have a vector at the LHS, and in the middle, but a scalar at the RHS.
Not possible.

r_1(s) = r(t(s))

Convert first both sides to vector-infinitesimals.
And then take the norm of the vector-infinitesimals.

HallsofIvy
Homework Helper
Since, for this problem, the lower limit is a constant and the upper limit is t, you don't really need the full Leibniz rule. Just the "Fundamental Theorem of Calculus:
The derivative of
$$\frac{d}{dx}\int_a^x f(t) dt= f(x)$$

It should be \int_{a}^{b} ||dr(t)||.
That is, without the dt.
There can (effectively) be only 1 infinitesimal in an integral.

This cannot be true, since the left hand side (LHS) is an infinitesimal, while the right hand side (RHS) is not.

Perhaps you meant: ||dr(s)||=ds?

I'm afraid you have another infinitesimal problem here.
No infinitesimals on the LHS, and a product of 2 at the RHS is simply not possible.

You should define r_1 in terms of r though.
This would be:
r_1(s) = r(t(s))

Good!
You will want to use this.

I'm afraid you have a vector at the LHS, and in the middle, but a scalar at the RHS.
Not possible.

r_1(s) = r(t(s))

Convert first both sides to vector-infinitesimals.
And then take the norm of the vector-infinitesimals.

Since, for this problem, the lower limit is a constant and the upper limit is t, you don't really need the full Leibniz rule. Just the "Fundamental Theorem of Calculus:
The derivative of
$$\frac{d}{dx}\int_a^x f(t) dt= f(x)$$

Yes, I have that thanks. Its the next bit I wanna work on

Assume r' (t) is never 0. Then s'(t) >0 for all t so we can in principle express t in terms of s and then obtain the new parameterisation r_1 (s)= r(t(s)) where 0 <= s<=L(=\int_{a}^{b} ||r'(t)||).

a) Prove ||r_1(s)||=1 for all s

My attempt.

$r_1(s)=\frac{dr}{dt}*\frac{dt}{ds}$.

We know s'(t)= ||r'(t)|| therefore t'(s)=1/||r'(t)||

then r_1(s)=r'(t)/||r'(t)||=1....?

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I like Serena
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Yes, I have that thanks. Its the next bit I wanna work on

Assume r' (t) is never 0. Then s'(t) >0 for all t so we can in principle express t in terms of s and then obtain the new parameterisation r_1 (s)= r(t(s)) where 0 <= s<=L(=\int_{a}^{b} ||r'(t)||).

Hmm, are you aware that $r'(t)={dr \over dt}$?

$$L=\int_{a}^{b} ||r'(t)|| dt$$
Or:
$$L=\int_{a}^{b} ||dr(t)||$$

a) Prove ||r_1(s)||=1 for all s

No, this still won't work.

Do you perhaps mean: ||r_1'(s)||=1 for all s?

My attempt.

$r_1(s)=\frac{dr}{dt}*\frac{dt}{ds}$.

We know s'(t)= ||r'(t)|| therefore t'(s)=1/||r'(t)||

then r_1(s)=r'(t)/||r'(t)||=1....?

Slight modifications:

$r_1'(s)=\frac{dr}{dt}*\frac{dt}{ds}$

We know s'(t)= ||r'(t)|| therefore t'(s)=1/||r'(t)||

then r_1'(s)=r'(t)/||r'(t)||.

And it follows that ||r_1'(s)||=1.

QED.

Note the additions of a couple of derivative-quotes.
And your last statement is about vectors, and a vector cannot be 1.
But the norm of a vector can be 1.

Ok, thanks! Thats makes sense now. The next bit
Reparameterise the curve r(t)=u+tv by its arc length. Start at t=0. My notes gave a definition of reparameterisation but not an example. How do I approach this?

A function h:[a_1,b_1] to [a,b] is called a reparameterisation if:

h is bijective, ii h'(t) exist for all t and its derivatives, iii h'(t) is never 0....

I like Serena
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Well, what is the current "arc-speed"?
That is, what is ||r'(t)||?
Can you think up an h(t), such that r_1(h) = r(t(h)) has ||r_1'(h)|| = 1?

Well, what is the current "arc-speed"?
That is, what is ||r'(t)||?
Can you think up an h(t), such that r_1(h) = r(t(h)) has ||r_1'(h)|| = 1?

The arc speed or ||r'(t)|| is ||dr(t)/dt|| which we know is s'(t). I dont intuitively understand what we are doing?

You have just replaced s with h in your last line....?

I like Serena
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The arc speed or ||r'(t)|| is ||dr(t)/dt|| which we know is s'(t). I dont intuitively understand what we are doing?

The arc speed is the distance r(t) covers if t increases by 1.
Since this is a new problem, s(t) is as yet undefined.
r'(t) is the derivative of r(t) which you defined in your new problem.

What your sequence of problems is trying to do, is explain how you can reparametrize a curve, such that its arc speed is 1.
That is, when you increase s by 1, the distance covered is also 1.

This in turn is an intro in what is to come for the definition of curvature.
(Actually I'm surprised that you are learning this stuff.)

You have just replaced s with h in your last line....?

I didn't, you did!
You started a new problem, effectively discarding the definition of s(t).
And then you introduced h(t) as a parameterisation, which is exactly the same as what s(t) represented in the previous problem.