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Homework Help: Finding length along ln(x)

  1. May 26, 2010 #1
    I'm currently trying to find the length along function of ln(x) for the heck of it.

    I set up this integral for length

    L= int/ sqrt(1+(y')^2)

    so y'=1/x

    so the integral becomes

    int/ sqrt(1+(1/x^2)) = int/ sqrt(x^2+1)/x

    So I used trig substitution. I set
    tanA=x
    secA=sqrt(x^2+1)
    dx = sec^2(A) dA

    so the integral becomes

    int/ secA/tanA * sec^2A dA= int/ sec^2A cscA dA

    d/dA cscA = -cotAcscA rearrange this and get -dsecA = cscA

    so the integral becomes - int/ sec^2A dsecA

    set u=secA

    so the integral becomes - int/ u^2 du

    so -(u^3/3) evaluated from limit a to b

    let the original limits be x=1 and x=5
    therefore they become A=arctan1 and A=arctan5
    and u=sec(arctan(1)) and u=sec(arctan(5))
    where sec(arctan(1))=sqrt(2) and sec(arctan(5)=sqrt(26)

    However, if you plug these limits in you get a negative answer. Where did I go wrong?
     
  2. jcsd
  3. May 26, 2010 #2
    Sorry, but this is meaningless! You need to learn better about differential forms. In reality, we have

    [tex]-d\,\sec A=-\sec A\tan A \,dA.[/tex]
     
  4. May 26, 2010 #3
    d/dA cscA = -cotAcscA

    so dcscA = -cotAcscA dA

    d cscA/-cotA = cscA dA

    cscA/cotA = secA

    -dsecA = cscA dA

    Which step is wrong?
     
  5. May 26, 2010 #4
    This is step is wrong because [tex]-\cot A[/tex] is not a constant with respect to A. So you cannot divide by [tex]-\cot A[/tex] in the differential form.
     
  6. May 26, 2010 #5
    Oh ok, I see that now. Thanks. I'll try a different way of solving this problem.
     
  7. May 26, 2010 #6
    I think this integral cannot be evaluated at all except by numerical methods. Ask your teacher about it. I guess there's a theorem in Differential Algebra that can decide whether an integral may be evaluated in terms of elementary functions or not.
     
  8. May 26, 2010 #7
    I did manage to get this far

    int/ sec^2A cscA
    = int/ (tan^2A + 1) cscA
    = int/ tanAsecA + int/ cscA
    = secA + int/ cscA

    is this right?

    isn't the integral of cscA equal to ln| tan x/2 | ?
     
  9. May 26, 2010 #8
    I think you've got a point. And you can write

    [tex]\int\frac{\sqrt{1+x^2}}{x}dx=\sqrt{1+x^2}-\ln\left|\frac{\sqrt{1+x^2}+1}{x}\right|[/tex]

    And, if I haven't done any typo,

    [tex]\int_1^5\frac{\sqrt{1+x^2}}{x}dx=4.3674894278899,[/tex]

    which is the length you desired. Yeah, two heads do think better than one. Mission accomplished! Now, please check my results, noting also that

    [tex]\int\csc A\, dA=-\ln|\csc A+\cot A|=\ln|\tan A/2|.[/tex]
     
  10. May 26, 2010 #9
    Yeah, looks right to me. Thanks for the help.
     
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