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Math Help - Arc length

  1. #1
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    Arc length

    Hey guys and gals,
    I just wanted to see if any one can explain to me how to derive the arc length of y=e^x; 0<x<1.

    I know the formula to finding the arc length is the square root of 1 + (dy/dx)^2 in the integrand. my question is how do you go from √(1 + u)/u du to √(1 + u)/u. Can some one help out?

    thanks in advance,

    Jose
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  2. #2
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    Quote Originally Posted by 07 Yamaha R6 View Post
    Hey guys and gals,
    I just wanted to see if any one can explain to me how to derive the arc length of y=e^x; 0<x<1.

    I know the formula to finding the arc length is the square root of 1 + (dy/dx)^2 in the integrand. my question is how do you go from √(1 + u)/u du to √(1 + u)/u. Can some one help out?

    thanks in advance,

    Jose
    Are you asking how \frac{\sqrt{1 + u^2}}{u} becomes \sqrt{\frac{1 + u^2}{u^2}} ?
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    arc length of y=e^x; 0<x<1.

    Yes sir!
    I understand we have to use the technique of using the ln to get rid of e^x so that we can find x, but im having trouble understanding that.
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  4. #4
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    If

    your asking how \frac{\sqrt{u^2+1}}{u} becomes \sqrt{\frac{u^2+1}{u^2}} all you do is since powers in this case \frac{1}{2} distribute if you distributed the squareroot of the second one you would get |u| but in this case disregard the ||
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Are you asking how \frac{\sqrt{1 + u^2}}{u} becomes \sqrt{\frac{1 + u^2}{u^2}} ?
    Quote Originally Posted by 07 Yamaha R6 View Post
    Yes sir!
    I understand we have to use the technique of using the ln to get rid of e^x so that we can find x, but im having trouble understanding that.
    \frac{\sqrt{1 + u^2}}{u} = \frac{\sqrt{1 + u^2}}{\sqrt{u^2}} = \sqrt{\frac{1 + u^2}{u^2}}.
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    sorry mr fantastic,

    first of all let me thank you for taking the time in helping me out, I appreciate a lot.

    second,
    I am trying to derive the arc length of y = e^x from 0 to 1

    first step is to get dy/dx

    that is simply e^x dx

    next I have ∫√(1+(e^x)) dx

    then I use u substitution, where u=e^x and du=e^x dx

    next I plug the values in and I get ∫√(1+(u)) du/u or ∫(√(1+(u)))/u du

    now, I have been helped at school and was told that the next step to this would be to go from ∫(√(1+(u)))/u du to ∫(√(1+(u)))/u du,

    my question is what did they do to get there and what are the next steps?
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  7. #7
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    Quote Originally Posted by 07 Yamaha R6 View Post
    sorry mr fantastic,

    first of all let me thank you for taking the time in helping me out, I appreciate a lot.

    second,
    I am trying to derive the arc length of y = e^x from 0 to 1

    first step is to get dy/dx

    that is simply e^x dx

    next I have ∫√(1+(e^x)) dx

    then I use u substitution, where u=e^x and du=e^x dx

    next I plug the values in and I get ∫√(1+(u)) du/u or ∫(√(1+(u)))/u du

    now, I have been helped at school and was told that the next step to this would be to go from ∫(√(1+(u)))/u du to ∫(√(1+(u)))/u du,

    my question is what did they do to get there and what are the next steps?
    I have posted the answer to what they did to get there ......

    The substitution u = e^x is not a good one to use. A much better substitution is u^2 = 1 + e^{2x}.

    Then 2u \frac{du}{dx} = 2 e^{2x} = 2 (u^2 - 1) \Rightarrow \frac{du}{dx} = \frac{u^2 - 1}{u} \Rightarrow dx = \frac{u}{u^2 - 1}.

    Then:

     \int_{0}^{1} \sqrt{1 + e^{2x}} \, dx becomes \int_{\sqrt{2}}^{\sqrt{1 + e^2}} \frac{u^2}{u^2 - 1} \, du = \int_{\sqrt{2}}^{\sqrt{1 + e^2}} \frac{(u^2 - 1) + 1}{u^2 - 1} \, du = \int_{\sqrt{2}}^{\sqrt{1 + e^2}} 1 + \frac{1}{u^2 - 1} \, du .
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    hey mr fantastic,

    thanks again,

    that substitution really makes it easier. Just to clarify though, when you substitute u=√(1+e^2x), I see how you have to square u to get u^2=1+e^2x what I dont understand is how you isolated du to be equal to u/u^2-1

    Can you clarify?
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  9. #9
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    Quote Originally Posted by 07 Yamaha R6 View Post
    hey mr fantastic,

    thanks again,

    that substitution really makes it easier. Just to clarify though, when you substitute u=√(1+e^2x), I see how you have to square u to get u^2=1+e^2x what I dont understand is how you isolated du to be equal to u/u^2-1

    Can you clarify?
    Do you understand where 2u \frac{du}{dx} = 2 e^{2x} = 2 (u^2 - 1) has come from?

    Then divide both sides by 2u:

    \frac{du}{dx} = \frac{u^2 - 1}{u} \Rightarrow \frac{dx}{du} = \frac{u}{u^2 - 1} \Rightarrow dx = .....
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    Got It!!

    Ok, Fantastic,

    I got a break through,
    feels good too!! anyways, so I see how you got u^2-1. that is, you set e^2x=u^2-1 right?

    got it!!!

    So, let me ask you this, this would be the same for y=e^3x or y=e^9x? I mean the technique and substitution we used for y=e^x?

    Please tell me it is so!

    Thanks in advance for your efforts,

    Jose
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  11. #11
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    Quote Originally Posted by 07 Yamaha R6 View Post
    Ok, Fantastic,

    I got a break through,
    feels good too!! anyways, so I see how you got u^2-1. that is, you set e^2x=u^2-1 right?

    got it!!!

    So, let me ask you this, this would be the same for y=e^3x or y=e^9x? I mean the technique and substitution we used for y=e^x?

    Please tell me it is so!

    Thanks in advance for your efforts,

    Jose
    Yes, it is so.
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