Length of a curve - A calculus approach

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Now i'm a 16-year old high school student, and as some of you might know, i like math :) I have been studying the integration by riemann sums lately, and i truly love the logical concept. So I decided to create my own formula for calculation of graph curve length (without looking at the present solutions - this i consider cheating, and i only do this if i find the task impossible .. :) and after 30 min i ended up with a formula, that works (i think). I can use it to calculate curve lengtt manually by doing it in hand, and choose a certain number of rectangles n=some number. But after i confirmed that it works, i wanted to evaluate the limit for n-->infinity, so the true length shone through. This was my original thought with the formula .. I ended up hitting a dead end though. I am simply not able to evaluate the limit....

I don't know if it's because my skills are good enough, or if it's because it's impossible to do so :) So i need a bit of help. This is how the formula looks like;
(it's enclosed in the word document) - i would appreciate if anyone would write in a post, or tell me how to write in the post .. :) It's annoying for you to download the formula

I have tried evaluating for a both a linear function and a non-linear .. I can't seem to find the limit. Maybe it doesn't exist? Maybe my formula is completely wrong? I don't know :P

thx
 

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  • #2
jambaugh
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Your formula looks good. (Nice work on your own).

As far as finding the limit, its going to be rather messy. Remember in the Riemann integrals the differential [dx] term corresponds to the interval lengths [Delta x] corresponding to your (x2-x1)/n factors.

You can factor their squares out of the sum inside the radical and then taking the square root you'd get something like:

[tex] \sum_i \sqrt{1+ [f'(x_i)]^2} \Delta x[/tex]
which becomes in the limit:
[tex] \int_{x_1}^{x_2} \sqrt{1+[f'(x)]^2} dx[/tex]
That's the textbook formula you find for arc length of a function's curve.

You may find this makes more sense, and generalizes nicely to parametric curves if you work in terms of differentials.
 
  • #3
mathman
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The expression you have when taken to the limit can be expressed as:
∫(dx2 + dy2)1/2 = ∫(1 + (dy/dx)2)1/2dx
where y=f(x). This is the usual expression for curve length.
 
  • #4
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Your formula looks good. (Nice work on your own).

As far as finding the limit, its going to be rather messy. Remember in the Riemann integrals the differential [dx] term corresponds to the interval lengths [Delta x] corresponding to your (x2-x1)/n factors.

You can factor their squares out of the sum inside the radical and then taking the square root you'd get something like:

[tex] \sum_i \sqrt{1+ [f'(x_i)]^2} \Delta x[/tex]
which becomes in the limit:
[tex] \int_{x_1}^{x_2} \sqrt{1+[f'(x)]^2} dx[/tex]
That's the textbook formula you find for arc length of a function's curve.

You may find this makes more sense, and generalizes nicely to parametric curves if you work in terms of differentials.
well, i feel very, very stupid right now - how am i suppose to factor out the the dx term ? :P I have only received 3 months worth of lessons in very basic algebra, since im fresh out of 9nth grade :) I haven't even had a single lesson about functions lol. I don't see how i can factor out the dx term?

And yea i know my ((x2-x1)/n) term is equivalent to dx in the summs, but i prefer to write it the other way to give it a personal touch i guess :)

I apologise for any possible errors in notation, but since all of this is self-study, it takes a while to get familiar with the correct notations :)

But if that's the original formular for arch length, then i have actually re-invented the formula ?:P
 
  • #5
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hey if your looking to get ahead of your class a bit a great site for you is www.khanacademy.org ... goes from secondary school maths right up to 1 year of college maths.
 
  • #6
192
3
I teach Calculus
Yes, Khan Academy is a great site for learning online by video.
Also Google for "calculus university houston" and find 53 excellent Calc videos

But be careful. You need to be able to learn by reading from a text too.
Videos do not exist [yet] for all topics or all subjects.

I wish my 9th graders had half the initiative and potential you clearly have.

Good luck
 
  • #7
724
0
I teach Calculus
Yes, Khan Academy is a great site for learning online by video.
Also Google for "calculus university houston" and find 53 excellent Calc videos

But be careful. You need to be able to learn by reading from a text too.
Videos do not exist [yet] for all topics or all subjects.

I wish my 9th graders had half the initiative and potential you clearly have.

Good luck
You teach Calc to 9th graders?
 
  • #8
jambaugh
Science Advisor
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well, i feel very, very stupid right now - how am i suppose to factor out the the dx term ? :P I have only received 3 months worth of lessons in very basic algebra, since im fresh out of 9nth grade :) I haven't even had a single lesson about functions lol. I don't see how i can factor out the dx term?

And yea i know my ((x2-x1)/n) term is equivalent to dx in the summs, but i prefer to write it the other way to give it a personal touch i guess :)

I apologise for any possible errors in notation, but since all of this is self-study, it takes a while to get familiar with the correct notations :)

But if that's the original formular for arch length, then i have actually re-invented the formula ?:P
There are a few details I'd like to mention. The algebra of it is:
[tex] \sqrt{X^2A + X^2B} = \sqrt{X^2(A+B)} = \sqrt{X^2} \cdot \sqrt{A+B}[/tex]
one uses the algebra fact the square root of a product is the product of the square roots, (and of course the distributive law to factor the sum).

Here's the tiny detail with which one should be careful:
[tex]\sqrt{X^2} = |X| [/tex]
not simply X.

What this means is you should get the right arc-length as long as you integrate in the positive direction i.e. x2 > x1. But if you reverse the integration (which changes the sign) the dx (incorrectly) becomes negative giving you a negative arclength. We should be more careful to put in the absolute value |dx| in the integral. But |dx| = dx if dx>0 and |dx| = -dx if dx<0. So we break it up into cases. If we integrate from larger x value to smaller (which gives us a negative value) we change the sign (to get back the positive arclength).

Or we can simply write:
[tex]S =\left| \int_{x_1}^{x_2}\sqrt{1+f'^2(x)}dx \right| [/tex]

Most texts don't bother but there's the chance a student will occasionally integrate in a negative direction (say by using a substitution) and wonder why they keep getting a sign error. It's really an error in the formula in neglecting this |dx| business.

Again, good work on that derivation. Keep it up and you'll go far. Oh, and there were no errors in notation, only that you choice not to use it.
 
  • #9
192
3
You teach Calc to 9th graders?
No but I can see how you concluded that.
I teach 12th graders Calculus.
My 9th graders are about the same age as the OP author.
I wish more of them were as curious as this fellow is.
Unfortunately it is an expensive private school and many
come from wealthy homes and are not motivated much less curious.
Sigh .....
 
  • #10
43
0
There are a few details I'd like to mention. The algebra of it is:
[tex] \sqrt{X^2A + X^2B} = \sqrt{X^2(A+B)} = \sqrt{X^2} \cdot \sqrt{A+B}[/tex]
one uses the algebra fact the square root of a product is the product of the square roots, (and of course the distributive law to factor the sum).

Here's the tiny detail with which one should be careful:
[tex]\sqrt{X^2} = |X| [/tex]
not simply X.

What this means is you should get the right arc-length as long as you integrate in the positive direction i.e. x2 > x1. But if you reverse the integration (which changes the sign) the dx (incorrectly) becomes negative giving you a negative arclength. We should be more careful to put in the absolute value |dx| in the integral. But |dx| = dx if dx>0 and |dx| = -dx if dx<0. So we break it up into cases. If we integrate from larger x value to smaller (which gives us a negative value) we change the sign (to get back the positive arclength).

Or we can simply write:
[tex]S =\left| \int_{x_1}^{x_2}\sqrt{1+f'^2(x)}dx \right| [/tex]

Most texts don't bother but there's the chance a student will occasionally integrate in a negative direction (say by using a substitution) and wonder why they keep getting a sign error. It's really an error in the formula in neglecting this |dx| business.

Again, good work on that derivation. Keep it up and you'll go far. Oh, and there were no errors in notation, only that you choice not to use it.
Haha thats rather obvius .. weird i didn't see the fact that i could just factor out dx :)

Now i tried to evaluate the limit of f(x)=x^2, from 0-5. I ended up this far;

(in the document .. sorry :)

And when i solved it on maple, it seemed correct. Hmm maybe i'm missing something elementary ? :) Its hard to study up on everything when one has to do it on their own.

Any suggestions ?
 

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  • #11
43
0
I teach Calculus
Yes, Khan Academy is a great site for learning online by video.
Also Google for "calculus university houston" and find 53 excellent Calc videos

But be careful. You need to be able to learn by reading from a text too.
Videos do not exist [yet] for all topics or all subjects.

I wish my 9th graders had half the initiative and potential you clearly have.

Good luck
yeah i do prefer books aswell - i think watching a lesson on video is very boring :) I would either be there in person or just read a book. It's a good skill to master too, since the majority of all the wisdom is this world is written in books hehe.

and thx for the comment - although my intelligence level is rather average actually :)
 

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