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Homework Help: Integral problem.(arc length)

  1. May 25, 2005 #1
    my problem is to find the arc length of [tex]y=e^x[/tex] between 0 and 1

    what i've got is [tex]\int_{0}^{1} \sqrt{1+(e^x)^2}dx[/tex] which i then substitute [tex]u=e^x[/tex] giving [tex]\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du[/tex] which i then sub in v=[tex]\sqrt{1+u^2}[/tex] giving [tex]\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v-1)(v+1)} dv[/tex] which then needs to be integrated using partial fractions, and this is where i run into problems

    i get [tex]\frac{1}{2} \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{1}{v-1}+\frac{1}{v+1}dv[/tex] which results in my getting an answer of 1.

    the book on the other hand gets [tex]\int_{\sqrt{2}}^{\sqrt{1+e^2}}1+ \frac{\frac{1}{2}}{v-1}-\frac{\frac{1}{2}}{v+1}dv[/tex] but i cant see how they got the 1, or the negative sign on the second fraction
     
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  3. May 25, 2005 #2

    dextercioby

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    I think you have a problem there.EDIT;It doesn't work.Well,when u got rid of the square in the variable "v".

    Daniel.
     
  4. May 25, 2005 #3

    arildno

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    This equality is wrong:
    [tex]\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv[/tex]
     
  5. May 25, 2005 #4
    yeah, i am seeing that...the book jumps from [tex]\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du[/tex] to [tex]\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv[/tex] and (without checking) i figured [tex]\frac{x^2}{y^2}=\frac{x}{y}[/tex] my mistake. now i'm really confused, because i see how they get [tex]\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du[/tex] but dont know where to go from there now.
     
  6. May 25, 2005 #5

    dextercioby

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    [tex] \int_{1}^{e} \sqrt{\frac{1}{u^{2}}+1} \ du [/tex]

    Make the sub.

    [tex]\frac{1}{u} =\sinh t [/tex]

    Daniel.
     
  7. May 25, 2005 #6

    GCT

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    From here, [tex]\int_{1}^{e} \frac{\sqrt{1+u^2}}{u} du[/tex][/quote]

    try the substitution [tex]u=1/t[/tex]
     
  8. May 25, 2005 #7
    can anyone see what the book did(i would like to get it solved the same way they did it.) up to the point where it subs in v i can see what they did, but when i sub in v i get [tex]\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv [/tex] and dont know how they got the roots of it.
    and even then i dont see how they got the 1 and the negative in the partial fraction.
     
  9. May 25, 2005 #8

    dextercioby

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    GCT,you're not being original...:wink:

    Daniel.
     
  10. May 26, 2005 #9

    GCT

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  11. May 27, 2005 #10

    OlderDan

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    If the book makes that jump, it is a typographical error. It should be

    [tex]\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)} dv[/tex]

    And their result follows. The u in this equation is just the original y, and v is as you defined it. Taking it from the top

    [tex]y = e^x [/tex]

    [tex] dy = e^x dx = ydx \Rightarrow dx = \frac{{dy}}{y}[/tex]

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

    [tex] v = \sqrt {1 + y^2 } \Rightarrow dv = \frac{1}{{2v}}2ydy = \frac{{ydy}}{v} \Rightarrow dy = \frac{v}{y}dv [/tex]

    [tex] v^2 = 1 + y^2 \Rightarrow y^2 = v^2 - 1 [/tex]

    [tex] ds = \frac{{\sqrt {1 + y^2 } }}{y}dy = \frac{v}{y} \cdot \frac{v}{y}dv = \frac{{v^2 }}{{y^2 }}dv = \frac{{v^2 }}{{v^2 - 1}}dv [/tex]

    [tex] ds = \left( {1 + \frac{1}{{v^2 - 1}}} \right)dv = \left( {1 + \frac{1}{{v - 1}} \cdot \frac{1}{{v + 1}}} \right)dv = \left( {1 + \frac{{\frac{1}{2}}}{{v - 1}} - \frac{{\frac{1}{2}}}{{v + 1}}} \right)dv [/tex]
     
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