Calculus 2: Finding Arc Length | Florida A&M Univ.

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    Arc Arc length Length
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SUMMARY

The discussion focuses on the complexities of finding arc length in Calculus 2, specifically using the integral formula ∫[a to b] √[1 + (f'(x))²] dx. Participants highlight that while the algebra can be tedious, many functions are designed to simplify the process, often referred to as "rigged" functions. The conversation also touches on the transition from Cartesian to polar coordinates for easier integration. Key insights include the importance of understanding basic geometric shapes, such as ellipses, to facilitate learning in calculus.

PREREQUISITES
  • Understanding of integral calculus, specifically arc length calculations
  • Familiarity with derivatives and their notation
  • Basic knowledge of geometric shapes, particularly ellipses
  • Concept of "rigged" functions in mathematical contexts
NEXT STEPS
  • Study the integral formula for arc length in greater detail
  • Explore examples of "rigged" functions and their properties
  • Learn about polar coordinates and their application in integration
  • Review geometric concepts related to ellipses and their equations
USEFUL FOR

Students in Calculus 2, particularly those struggling with arc length problems, as well as educators seeking to clarify the concept of "rigged" functions and their applications in calculus.

Techman07
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The arc length...

Hello all, this is my first post. I am a Computer Engineering Student at Florida A&M University taking Calculus 2 over the summer semester.

Should finding the arc length be so extensive?? Are their shortcuts that I am missing?

If you don't know the formula, the arc length is the integral from a to b of


[1 + (f'(x))^2]^1/2

If finding the arc length IS this difficult, can someone direct me towards some insight. I get caught up in the VERY tedious algebra involved.

thank you.
 
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Well,the algebra should be a snack.I mean,it should be the least worrying.

The integral \int_{a}^{b} \sqrt{1+\left[f'(x)\right]^{2}} \ dx is on rare occasions expressible in terms of elementary functions.

Think the ellipse.Try to compute the length of the ellipse...:wink: You'll see that algebra won't be a problem at all.

Daniel.

P.S.Surely,that formula is valid in cartesian coordinates.You can Switch to polar,if u find the integration to be easier.
 
Last edited:
You have the wrong sign on your f(x) derivative dex.
OP: You'll mostly find the functions rigged so that the algebra is really simple. I remember mostly arclength of sin functions in my class.
 
Thanks for pointing it out.What does this " You'll mostly find the functions rigged "mean...?(Sorry,it's a bit offtopic,i guess,but i have no idea what that means).

Daniel.
 
Rigging something means setting it up so that something specific happens. For example, for the arclength of the cos function you get \sqrt{1-sin^2x} which is just cos(x).

Are you belgian native?
 
Nope,i'm Romanian and for about a week I'm wondering how should i translate "rigged Hilbert space" into Romanian.This word "rigged" has been obssessing me :-p

Daniel.
 
Examples...

I guess that i should give an example, but what is an ellipse?
 
You have to be kidding,right...?Tell me you are.


Daniel.
 
Nope, not kidding, never used the term in calculus I, haven't seen it yet in cal II
 
  • #10
You should know what an ellipse is from geometry.

Dex, I'd help you out, but first I'd have to know what a Hilbert space is, wouldn't I? :P
 
  • #11
Nope, not kidding, never used the term in calculus I, haven't seen it yet in cal II

lol, that's because you first encountered the term in pre-school or 1st grade (whenever you begin learning about basic shapes.)
 
  • #12
well, thanks for input.
 
  • #13
whozum said:
You should know what an ellipse is from geometry.

Dex, I'd help you out, but first I'd have to know what a Hilbert space is, wouldn't I? :P

It's a complete preHilbert space.What does "rigged" mean...?

Daniel.
 
  • #14
Rigged

Rigging something means setting it up so that something specific happens. For example, the rigged equation above is set up so that once you solve the correct problem, the result is really simple to evaluate. It simplifies easily due to the properties of the function. If it wasnt rigged, fro example if f(x) = 3x^3, then the integrand is considerably more difficult than if it was just f(x) = cos(x).


Rigged hilbert Space

I don't know anything about hilbert spaces, I haven't began quantum mechanics yet, but I'm sure this will help you out:

http://en.wikipedia.org/wiki/Rigged_Hilbert_space
 

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