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Arclength of exp from 0 to 1

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the arc length of y=e^x, from [0,1].


    2. Relevant equations



    3. The attempt at a solution
    [tex]s = \int_0^1 (1 + e^2^x)^(^1^/^2^)dx[/tex]

    I let t = e^x, dt=e^xdx; therefore dt/t=dx

    [tex] s = \int_1^e \frac{(1+t^2)^(^1^/^2^)}{t}\right) dx[/tex]

    Let t = tanT, dt = sec^2(T)dT (T for theta)

    [tex] s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{sec^3T}{tanT}\right)dT[/tex]

    [tex] s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{1/cos^3T}{sinT/cosT} \right) dT[/tex]

    [tex] s = \int_{\frac{Pi}{4}\right)}^{arctan(e)} \frac{1}{sinT}\right) * \frac{1}{cos^2T} \right)dT[/tex]

    Where do I go from here?

    Any help would be greatly appreciated. :)
     
    Last edited by a moderator: Dec 1, 2009
  2. jcsd
  3. Dec 1, 2009 #2
    A substitution t = sinh y might be a bit easier. It takes some work too though.
     
  4. Dec 1, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since that is an odd power of sin(T), a standard technique is to multiply both numerator and denominator by sin(T) to get
    [tex]s= \int_{\frac{pi}{4}}^{arctan(e)}\frac{cos^2(T)}{sin^2(T)} sin(T)dT[/tex]
    [tex]= \int_{\frac{pi}{4}}^{arctan(e)}\frac{cos^2(T)}{1- cos^2(T)} sin(T)dT[/tex]

    Now let u= cos(T) so du= -sin(T)dt and the integral becomes
    [tex]\int_{\frac{\sqrt{2}}{2}}^{\frac{1}{\sqrt{1+e^2}}}\frac{u^2}{1- u^2} du[/tex]
    Which you can do by "partial fractions".
     
  5. Dec 1, 2009 #4
    [tex]u=\sqrt{1+e^{2x}}[/tex]

    [tex]{\rm d}u=\frac{e^{2x}}{\sqrt{1+e^{2x}}}=\frac{u^2-1}{u} {\rm{d}x}[/tex]

    and

    [tex]s={\int\limits_{\sqrt{2}}^{\sqrt{1+e^2^}}} {\frac{u^2}{u^2-1^}} \quad {\rm d}u[/tex]

    ?
     
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