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Arc Length formula

  1. Mar 10, 2004 #1
    Find a function F(x) whose arc length L(x) from (1,1/2) to (x,F(x)), x>1 is (1/2)x^2 + (1/4)Ln(x).

    First some short hand notation.

    Int[f(x),dx], means the indefinite integral of the function f(x).

    Int[f(x),dx,a,b], means the definite integral where a is the lower bound and b is the upper bound.

    d/(dx)[f(x)], is the derivate of f(x).

    Sqrt[f(x)], is the square root of f(x).

    Basically I am trying to follow Mathematica conventions and adding a touch of Leibniz to it.

    Ok, now here is what I have done and where I am at thus far.

    Arc length formula is Int[Sqrt[1+(d/dt)[f(t)]^2],dt,a,x]

    So I have Int[Sqrt[1+((d/dt)[f(t)])^2],dt,1,x] = (1/2)x^2 + (1/4)Ln(x)

    So by the FTC I can say that

    d/(dx)[(1/2)x^2 + (1/4)Ln(x)] = Sqrt[1+(d/dt)[f(t)]^2]

    Which is

    x + 1/(4x) = Sqrt[1+(d/dt)[f(t)]^2]

    Squaring both sides I have

    (x + 1/(4x))^2 = abs[1+(d/dt)[f(t)]^2]

    I drop the absolute value here so I have.

    (x + 1/(4x))^2 = 1+(d/dt)[f(t)]^2

    x^2 + 1/2 + (1/(4x))^2 = 1+(d/dt)[f(t)]^2

    Subtracting 1 from both sides and factoring the RHS

    (x - 1/(4x))^2 = ((d/dt)[f(t)])^2

    Taking square root of both sides and discarding the absolute value I have.

    (d/dt)[f(t)] = x - 1/(4x)

    Integrating both sides I get.

    f(x) =(1/2)x^2 - (1/4)Ln(x) +c

    I use the point (1,(1/2)) that is given to find c. Thus c = 0.

    Seems ok, right?

    Well that means that if I put f(x) back into my arc length formula I should get back to the given function.

    Ok.... Did that and now I have an extra -(1/2) on my function? What am I missing here? One thing to notice is

    Int[Sqrt[1+((d/dt)[f(t)])^2],dt,1,x] = (1/2)x^2 + (1/4)Ln(x)

    When x=1 you have 0 = 1/2

    Of course the restriction was x>1

    Any help would be great, thanks.

  2. jcsd
  3. Mar 10, 2004 #2
    Interesting. I'm getting the same thing.

    You'll notice that the method fails right after applying the FTCs. You get an expression for Sqrt[1+f'[t]^2] that, if you integrate from 1 to x with respect to t, has the extra -1/2. Additionally, maybe I just can't use my calculator, but whenever I asked it to find the arclength numerically between two finite points for x^2/2 + 1/(4x), the answer did not agree with x + Log[x]/4 - 1/2.

    Sorry I'm not more help, but mathematical rigor is most definitely not my strong point.

  4. Mar 11, 2004 #3
    i'm tempted to think there is no solution because if x=1, the arc length formula would give 1/2. the restriction x>1 may be the resolution to that little dilemma. seems odd though that the arc length function L(x) is discontinuous at x=1, jumping from 0 to 1/2. L(1) should be 0.
  5. Mar 11, 2004 #4
    I don't see the problem, I don't get an extra 1/2 at the end. I get the same result as you do finding the function but I don't have any problem inputing back into the equation.

    [tex]f(x) = x^2 - \frac{ln|x|}{4}[/tex]

    [tex]\frac{df(x)}{dx} = x - \frac{1}{4x}[/tex]

    [tex]\frac{df(x)}{dx}^2 = x^2 - \frac{1}{2} + \frac{1}{16x^2}[/tex]

    [tex]\frac{df(x)}{dx}^2 + 1 = x^2 + \frac{1}{2} + \frac{1}{16x^2} = (x + \frac{1}{4x})^2[/tex]

    [tex]L = \int_{1}^{x} \sqrt(x + \frac{1}{4x})^2 dx = \int_{1}^{x} (x + \frac{1}{4x}) dx[/tex]

    [tex]L = [ \frac{x^2}{2} + \frac{ln|x|}{4} ]_{1}^{x}[/tex]

    Like I said, unless I missed something in your post, I don't see any problem.
  6. Mar 11, 2004 #5
    in [tex]L = [ \frac{x^2}{2} + \frac{ln|x|}{4} ]_{1}^{x}[/tex], when you subtract off the result of plugging in the 1, you get -1/2.
  7. Mar 11, 2004 #6
    Oh, I see. That's odd. Are you sure that in the text they didn't mean that the result from the arclength formula should be x^2/2 + lnx/4 before you put in the limits?
  8. Mar 11, 2004 #7
    or that the initial point might be (1,0)? if it were (1,0), then x^2/2+(logx)/4 would work.

    kinda neat, actually. if A(f)(x) gives the arc length of f as a function of x and the initial point was (1,0), then this particular function x^2/2+(logx)/4 is a fixed "point" of the operator A.
  9. Mar 12, 2004 #8


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    Dearly Missed

    Your arclength formula is wrong. Starting with the derivative of the arclength, you have:
    Integrating from 1 to x yields:
    When L(1) is non-zero, as in your case, it simply means that the graph (t,f(t)) (on 1<t<x) represents a segment of a curve starting at some point (x0,y0) such that the arclength from (x0,y0) to (1,1/2) is L(1)..
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