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Definition of Arc Length Function

  1. Mar 30, 2004 #1

    I am having trouble remembering some of the material required for my current calculus course so I am reviewing some of the previous material that I have forgotten.

    I am having trouble following the definition of The Arc Length Function as presented in James Stewart's "Calculus: Fourth Ed." page 579.

    I already follow how to derive the formula for arc length. But I am having problems with the concept of the arc length function. I am given the following:

    Now I am not following the purpose of this section on arc length function. I can already figure out (5) with ds being equal to (3). So what is the purpose of this section? And I am also not following what the text is doing at the beginning by replacing the variable of integration with t and then differentiating and re-replacing t with x again. What the hey? I would really like to know what they are doing in this step.

    Any help is appreciated. Thankyou.
  2. jcsd
  3. Mar 30, 2004 #2


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    Well the purpose of the function is simply to work out what the length of a curve is between two points. Imagine the curve [itex]y = x^2[/itex] from the point 0 to 1. There is no mathematical way to accurately work out the length of the curve that you have probably come across, you could draw it very accurately and use a string and a ruler but that is hardly mathematically sound.

    Imagine a very small section of the curve, the horizontal distance being a very small section of x, [itex]\delta x[/itex] the vertical distance being a very small section of y, [itex]\delta y[/itex] and the length of this very small section of curve being [itex]\delta s[/itex]/

    Now because you have such a small section of the curve it is almost a right angle triangle. So we can say:

    [tex](\delta s)^2 \approx (\delta x)^2 + (\delta y)^2[/tex]

    Dividing both sides by [itex](\delta x)^2[/itex] we get:

    [tex]{\left( \frac{\delta s}{\delta x} \right)}^2 \approx 1 + {\left( \frac{\delta y}{\delta x} \right)}^2[/tex]

    If we let [itex]\delta x[/itex] approach 0 we can say that:

    [tex]{\left( \frac{ds}{dx} \right)}^2 = 1 + {\left( \frac{dy}{dx} \right)}^2[/tex]

    And the rest is manipulation, hope that helps.
  4. Mar 30, 2004 #3

    Perhaps my question is not a good question - too trivial?

    See, I completely understand what you are saying. I have no problem with that. In fact I have no problem deriving L. What I don't follow is the way that they are presenting the first half of the above passage.

    What is going on with the replacement of x with t and then back to x again?

    Also, am I to see that

    L = s(x) = integral ds = integral (1+ (dy/dx)^2)^1/2 dx?

    If so, I can see it. But like I said again I am not following the replacement of integrating variable x with t then back again. What are they doing here in the above passage?

    I know this is extremely trivial but it is confusing me. Perhaps it is inconsequential since I can follow the derviation of arc length. But I would like to know how they go from (1) to (2).
  5. Mar 30, 2004 #4


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    You might be having trouble because (1) is wrong! The formula for s is:

    s(x) := \int_a^x \sqrt{1 + f'(t)^2} \, dt
  6. Mar 30, 2004 #5
    Nope. That is not my trouble. Simply a typo.


    What are they doing with the exchange of variables back and forth?
  7. Mar 30, 2004 #6


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    Do you see the trouble inherent in this expression:

    g(x) = \int_a^x f(x) \, dx

  8. Mar 30, 2004 #7
    No I don't. I don't see the problem. What am I missing here? It's been a while since my last calculus course. And a while before that one as well, so I have gaps in my foundation. And it is causing me trouble.

    g(x) is dependent on the limits [a,x].

    But it is not the same as f(x). Let y = f(x). Then y changes as x changes. x in this case will change over the interval [a,x]. And there lies the problem. x is changing while x in the interval does not.


    I am sort of confused.

    I would have been more comfortable to give the limit a different variable say t then the other way around. That is

    [tex]g(t) = \int_a^t f(x) \, dx[/tex]

    What the hey. I am still confused.
  9. Mar 30, 2004 #8
    I think you caught what was wrong.

    Hurkyl's expression for g(x) has x in both the boundaries of integration and the integrand, which is absurd. The expression you made for g(t) is better.

  10. Mar 30, 2004 #9
    Yes. This is a problem for me. I can see it. But I can't explain it properly. It's causing me problems in my current course. There are small parts of concepts that I am not entirely clear on but I still seem to get the whole of concept.
  11. Mar 30, 2004 #10
    You probably can't explain the meaning of a definite integral with the same variable in both the boundaries and the integrand because it doesn't make any sense anyway. =]

    The variable in the integrand for indefinite integrals is just a dummy variable. It doesn't matter what it is. You can make it whatever you want, like calling it "cookiemonster" for all anybody cares. The true variable is the upper boundary of the integral.

  12. Apr 7, 2010 #11
    could some genius post a worked example of arc length calculation please.What is arc length of curve y=x^2 from x=0 to x=3? ?
    Last edited: Apr 7, 2010
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