Homework Help: Parameterised Curve Proof Part 1

1. Nov 5, 2011

bugatti79

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

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"

Last edited by a moderator: Apr 26, 2017
2. Nov 5, 2011

LCKurtz

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

3. Nov 5, 2011

I like Serena

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]

Last edited by a moderator: May 5, 2017
4. Nov 5, 2011

bugatti79

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. :-)

Last edited by a moderator: May 5, 2017
5. Nov 5, 2011

bugatti79

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

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

6. Nov 5, 2011

I like Serena

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?

7. Nov 5, 2011

LCKurtz

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

8. Nov 11, 2011

bugatti79

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)

9. Nov 11, 2011

LCKurtz

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

10. Nov 11, 2011

bugatti79

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.

11. Nov 11, 2011

LCKurtz

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?

12. Nov 11, 2011

I like Serena

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:
$s'(t)=||d\vec r(t)||$.

13. Nov 11, 2011

bugatti79

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.

14. Nov 11, 2011

I like Serena

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

15. Nov 11, 2011

LCKurtz

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.

16. Nov 18, 2011

bugatti79

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?

17. Nov 18, 2011

I like Serena

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.

18. Nov 19, 2011

bugatti79

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

19. Nov 19, 2011

HallsofIvy

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
$$\frac{d}{dx}\int_a^x f(t) dt= f(x)$$

20. Nov 19, 2011

bugatti79

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....?

Last edited by a moderator: Nov 22, 2011