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Arc Length: Definite and Indefinite Integration

  1. Dec 25, 2014 #1
    Several authors state the formula for finding the arc length of a curve defined by ##y = f(x)## from ##x=a## to ##x=b## as:

    $$\int ds = \int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx$$

    Isn't this notation technically wrong, since the RHS is a definite integral, and the LHS is an indefinite integral (family of antiderivatives)?

    I understand that no limits are placed on the integral of ##ds## since ##ds## can be defined in several equivalent ways: y as the independent variable, x as the independent variable, parametrically, or in terms of polar coordinates.

    But why can't we write the integral as:

    $$\int_{x=a}^{x=b} ds$$

    Writing it this way makes it explicit that the variable ##x## is changing from ##a## to ##b##.

    Alternatively, can't we also write it as:

    $$\int_P ds$$

    Where P is the path I have defined above.
     
    Last edited: Dec 25, 2014
  2. jcsd
  3. Dec 25, 2014 #2

    lavinia

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    You can not put in integration bounds for ds before you know the integral But you could put the subscript P if you like but that is clear anyway.

    The integration bounds x = a to x = b are incorrect because you are integrating with respect to s not with respect to x.
     
  4. Dec 26, 2014 #3
    What does carrying out indefinite integration with respect to s mean?

    $$\int ds = s + C$$

    For ##C ∈ ℝ##

    What does the right hand side represent?
     
  5. Dec 26, 2014 #4

    pasmith

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    Yes. The left hand side should be [tex]
    L = \int_0^L\,ds = \int_a^b \frac{ds}{dx}\,dx[/tex] where [itex]L[/itex] is the length of the curve in question.
     
  6. Dec 26, 2014 #5

    lavinia

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    All I am saying is that the left hand side does not mean an indefinite integral. Perhaps it is an abuse of notation but the integration bounds are implicit. You do not know the integral of ds until you do the integration. You could put it in as an unknown as pasmith did but in practice you can not do this integral with respect to ds. You need a change of parameter where you do know the bounds. But the thing to keep in mind is that an indefinite integral is not meant here.
     
  7. Dec 26, 2014 #6
    This makes sense. But if L is the arc length of the curve, what exactly does the variable s represent?
     
  8. Dec 26, 2014 #7
    Exactly. Abuse of notation.
    This is done over and over again in several texts.
     
  9. Dec 26, 2014 #8

    pasmith

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    If [itex]a \leq x \leq b[/itex] then [itex]s(x)[/itex] is the length of the curve from [itex](a,f(a))[/itex] to [itex](x, f(x))[/itex]. We have by definition [itex]s(a) = 0[/itex] and [itex]s(b) = L[/itex].
     
  10. Dec 26, 2014 #9
    I get it now. It's the same logic used when deriving the area of a circle right?

    $$dA = 2πr dr$$
    $$\int_0^A dA' = \int_0^r 2πr' dr'$$
    $$A = πr^2$$
     
  11. Dec 26, 2014 #10

    Stephen Tashi

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    The variable [itex] s [/itex] must represent distance along the path of the curve, measured from some arbitrary starting pont. For example, there are occasions when you see a curve "parameterized by arc length" in the form (x(s),y(s)).
     
  12. Dec 27, 2014 #11
    Remember that the integral sign is a glorified plus sign. ds is an infinitesimal arc length. So when you take the integral of ds, you are adding up all the tiny infinitesimal lengths. the integral of ds without limits is pointless, because if you wanted to restate that in words, it would go like this: add up all those infinitesimal arc lengths. When do I stop? Do I keep going? Limits are there do tell you where to start adding and when to stop adding.

    To answer your question, the integral of ds is pointless, or rather a symbolic notation. That is why there are no limits. You have to define what ds is first, and then introduce the limits
     
  13. Dec 31, 2014 #12
    It is implied that the LHS has lower and upper limit as 0 and l respectively.
     
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