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Math Help - ARC Length Help!!

  1. #1
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    ARC Length Help!!

    hey, guys and gals,
    just wanted to see if any of you can direct me towards the correct way of
    approaching this problem.

    the problem asks to find the arc length of y=e^2x
    now ive obtained the derivative dy/dx to be
    2e^2x and when I use the formula for finding the
    arclength, I then have the integral:
    √(1+(2e^2x )) dx
    next I use
    u=√(1+(2e^2x ))
    so that if I square both sides I get:
    u=1+4e^4x then, to get du:
    2udu=16e^4x dx

    now i noticed that above in "u=1+4e^4x " i could solve for
    e^4x to =(u-1)/4 so that in the equation,
    2udu=16e^4x dx, i could also solve for e^4x dx=1/8udu, then substituting
    the e^4x=(u-1)/4 into e^4x dx=1/8udu to become
    (u-1)/4 dx=1/8udu then solving for dx, I found
    dx=u/2(u-1) du

    so my question is what do you do next? I substituted back u and du
    into the integral and get u/2(u-1) du but I dont know where to go from here. Can somebody please help?

    thanks in advance for your efforts,

    07 Yamaha R6
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  2. #2
    Math Engineering Student
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    Where're the bounds?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by 07 Yamaha R6 View Post
    hey, guys and gals,
    just wanted to see if any of you can direct me towards the correct way of
    approaching this problem.

    the problem asks to find the arc length of y=e^2x
    now ive obtained the derivative dy/dx to be
    2e^2x and when I use the formula for finding the
    arclength, I then have the integral:
    √(1+(2e^2x )) dx
    next I use
    u=√(1+(2e^2x ))
    so that if I square both sides I get:
    u=1+4e^4x then, to get du:
    2udu=16e^4x dx

    now i noticed that above in "u=1+4e^4x " i could solve for
    e^4x to =(u-1)/4 so that in the equation,
    2udu=16e^4x dx, i could also solve for e^4x dx=1/8udu, then substituting
    the e^4x=(u-1)/4 into e^4x dx=1/8udu to become
    (u-1)/4 dx=1/8udu then solving for dx, I found
    dx=u/2(u-1) du

    so my question is what do you do next? I substituted back u and du
    into the integral and get u/2(u-1) du but I dont know where to go from here. Can somebody please help?

    thanks in advance for your efforts,

    07 Yamaha R6
    \int\sqrt{1+(2e^{2x})^2}dx

    Let u=\sqrt{1+(2e^{2x})^2}\Rightarrow{\frac{1}{4}\ln(u  ^2-1)-\frac{\ln(2)}{2}=x}

    so dx=\frac{1}{2}\frac{u}{u^2-1}

    So we have

    \frac{1}{2}\int\frac{u^2}{u^2-1}du=\frac{1}{2}\bigg[\int\frac{u^2-1}{u^2-1}du+\int\frac{1}{u^2-1}du\bigg]

    integrating we get

    \frac{1}{2}\bigg[u-arctanh(u)\bigg]+C

    subbing back in we get

    \int\sqrt{1+(2e^{2x})^2}dx=\frac{1}{2}\bigg[\sqrt{1+(2e^{2x})^2}-arctahn\bigg(\sqrt{1+(2e^{2x})^2}\bigg)\bigg]+C
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  4. #4
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    right, the bounds are from 0 to 1
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int\sqrt{1+(2e^{2x})^2}dx

    Let u=\sqrt{1+(2e^{2x})^2}\Rightarrow{\frac{1}{4}\ln(u  ^2-1)-\frac{\ln(2)}{2}=x}

    so dx=\frac{1}{2}\frac{u}{u^2-1}

    So we have

    \frac{1}{2}\int\frac{u^2}{u^2-1}du=\frac{1}{2}\bigg[\int\frac{u^2-1}{u^2-1}du+\int\frac{1}{u^2-1}du\bigg]

    integrating we get

    \frac{1}{2}\bigg[u-arctanh(u)\bigg]+C

    subbing back in we get

    \int\sqrt{1+(2e^{2x})^2}dx=\frac{1}{2}\bigg[\sqrt{1+(2e^{2x})^2}-arctahn\bigg(\sqrt{1+(2e^{2x})^2}\bigg)\bigg]+C
    Quote Originally Posted by 07 Yamaha R6 View Post
    right, the bounds are from 0 to 1
    \int_0^{1}\sqrt{1+(2e^{2x})^2}dx=\frac{1}{2}\bigg[\sqrt{1+(2e^{2x})^2}-arctanh\bigg(\sqrt{1+(2e^{2x})^2}\bigg)\bigg]\bigg|_0^{1}\approx{6.5}
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  6. #6
    Eater of Worlds
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    Yes, where are the bounds?. But we can proceed and then you can use them.

    One possible sub is to let u=4e^{4x}

    Then, x=\frac{ln(u/4)}{4}, \;\ dx=\frac{1}{4u}du

    \frac{1}{4}\int\frac{\sqrt{1+u}}{u}du

    Now, a sub could be w=\sqrt{1+u}, \;\ w^{2}=1+u, \;\ 2wdw=du

    Making the subs and expanding gives:

    \frac{1}{2}+\frac{1}{4(w-1)}-\frac{1}{4(w+1)}

    Now, it is just a matter of ln.

    This is one possible approach.

    Resubbing we get \frac{ln(\sqrt{1+4e^{4x}}-1)}{2}+\frac{\sqrt{1+4e^{4x}}}{2}-x
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Yes, where are the bounds?. But we can proceed and then you can use them.

    One possible sub is to let u=4e^{4x}

    Then, x=\frac{ln(u/4)}{4}, \;\ dx=\frac{1}{4u}du

    \frac{1}{4}\int\frac{\sqrt{1+u}}{u}du

    Now, a sub could be w=\sqrt{1+u}, \;\ w^{2}=1+u, \;\ 2wdw=du

    Making the subs and expanding gives:

    \frac{1}{2}+\frac{1}{4(w-1)}-\frac{1}{4(w+1)}

    Now, it is just a matter of ln.

    This is one possible approach.

    Resubbing we get \frac{ln(\sqrt{1+4e^{4x}}-1)}{2}+\frac{\sqrt{1+4e^{4x}}}{2}-x
    Just to satisfy my curiousity, I have always when seeing a nested exponential defined u as the whole quantity due to the simplification qualities used when solving for x (because of ln(u(x))), but no matter where I look no one seems to do it that way,

    Is my way either invalid, overcomplicated, or just uncommon?
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  8. #8
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    mathstud,
    hey thanks for the help bra! thats right, my problem was
    not taking the 1/2 out of the integral, thats why I couldnt see
    the "solution" to the 1/u-1. The only question was that
    isnt 1/u-1=1/(2a)*ln|(u-a)/(u+a)| + C ?
    instead of the arctan?

    please advice,

    07 Yamaha R6
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by 07 Yamaha R6 View Post
    mathstud,
    hey thanks for the help bra! thats right, my problem was
    not taking the 1/2 out of the integral, thats why I couldnt see
    the "solution" to the 1/u-1. The only question was that
    isnt 1/u-1=1/(2a)*ln|(u-a)/(u+a)| + C ?
    instead of the arctan?

    please advice,

    07 Yamaha R6
    I did not put arctan(x), I put arctanH(x), which is hyperbolic arctangent, it is actually equal to what you said,

    arctanh(x)=\frac{1}{2}\ln\bigg(\frac{1+x}{1-x}\bigg)
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  10. #10
    Eater of Worlds
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    Whatever works.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by galactus View Post
    Whatever works.
    I found that post funny for some reason
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