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Math Help - Arc Length Problem - 2

  1. #1
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    Arc Length Problem - 2

    x = \sqrt{y} - 8y evaluated from upper bound 4 to lower bound 1

    x'(\sqrt{y} - 8y) = \dfrac{1}{2}y^{-1/2} - 8

     L = \int_{1}^{4} \sqrt{1 + (\dfrac{1}{2}y^{-1/2} - 8)^{2}}

    (\dfrac{1}{2}y^{-1/2} - 8)^{2} =

    (\dfrac{1}{2}y^{-1/2} - 8)(\dfrac{1}{2}y^{-1/2} - 8)

     = \dfrac{1}{4}y^{-1} - 4y^{-1/2 }- 4y^{-1/2} + 64

    \dfrac{1}{4}y^{-1} - 8y^{-1/2 }+ 64

     L = \int_{1}^{4} \sqrt{1 + \dfrac{1}{4}y^{-1} - 8y^{-1/2 }+ 64}

     L = \int_{1}^{4} (1 + \dfrac{1}{4}y^{-1} - 8y^{-1/2 }+ 64)^{1/2}

     L = \int_{1}^{4} (u)^{1/2}

    u = 1 + \dfrac{1}{4}y^{-1} - 8y^{-1/2 }+ 64

    du = -\dfrac{1}{4} - 4y^{-3/2}

    Next step here?
    Last edited by Jason76; July 15th 2014 at 08:52 PM.
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  2. #2
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    Re: Arc Length Problem - 2

    That is correct so far (don't forget your dy on the integral)...

    $\displaystyle \begin{align*} L &= \int_1^4{\sqrt{ 1 + \frac{1}{4}y^{-1} - 8y^{-\frac{1}{2}} + 64}\,\mathrm{d}y } \\ &= \int_1^4{ \sqrt{ 65 - \frac{8}{\sqrt{y}} + \frac{1}{4y} } \, \mathrm{d}y } \\ &= \int_1^4{\sqrt{ \frac{260y - 32\sqrt{y} + 1}{4y}}\,\mathrm{d}y } \\ &= \int_1^4{ \frac{\sqrt{260y - 32\sqrt{y} + 1}}{\sqrt{4y}}\,\mathrm{d}y } \\ &= \int_1^4{ \sqrt{260 \left( \sqrt{y} \right) ^2 - 32\sqrt{y} + 1 } \, \left( \frac{1}{2\sqrt{y}} \right) \, \mathrm{d}y } \end{align*}$

    So now substitute $\displaystyle \begin{align*} u = \sqrt{y} \implies \mathrm{d}u = \frac{1}{2\sqrt{y}}\,\mathrm{d}y \end{align*}$ and noting that $\displaystyle \begin{align*} u(1) = 1 \end{align*}$ and $\displaystyle \begin{align*} u(4) = 2 \end{align*}$, the integral becomes

    $\displaystyle \begin{align*} \int_1^4{ \sqrt{ 260 \left( \sqrt{y} \right) ^2 - 32\sqrt{y} + 1 } \, \left( \frac{1}{2\sqrt{y}} \right) \, \mathrm{d}y } &= \int_1^2{ \sqrt{ 260u^2 - 32u + 1} \, \mathrm{d}u } \\ &= \int_1^2{ \sqrt{ 260 \left( u^2 - \frac{8}{65} u + \frac{1}{260} \right) } \, \mathrm{d}u } \\ &= \sqrt{260} \int_1^2{ \sqrt{ u^2 - \frac{8}{65}u + \frac{1}{260} } \, \mathrm{d}u } \\ &= 2\sqrt{65} \int_1^2{ \sqrt{ u^2 - \frac{8}{65}u + \left( -\frac{4}{65} \right) ^2 - \left( - \frac{4}{65} \right) ^2 + \frac{1}{260 } }\,\mathrm{d}u } \\ &= 2\sqrt{65} \int_1^2{\sqrt{ \left( u - \frac{4}{65} \right) ^2 + \frac{1}{16900} } \, \mathrm{d}u} \end{align*}$

    So now you can proceed by substituting $\displaystyle \begin{align*} u - \frac{4}{65} = \frac{1}{130}\tan{ \left( \theta \right) } \implies \mathrm{d}u = \frac{1}{130}\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$.
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    Re: Arc Length Problem - 2

    Prove It, I don't understand the third line from the top on your work.
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    Re: Arc Length Problem - 2

    Surely you know that to add fractions you need a common denominator...
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    Re: Arc Length Problem - 2

    Ok, makes sense.

    Quote Originally Posted by Prove It View Post
    Surely you know that to add fractions you need a common denominator...
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    Re: Arc Length Problem - 2

    x = \sqrt{y} - 8y

    1 \leq y \leq 4

    x = y^{1/2} - 8y

    x' = \dfrac{1}{2}y^{-/1/2} - 8

    L = \int_{1}^{4} \sqrt{1 + (\dfrac{1}{2}y^{-/1/2} - 8   )^{2}}


    Foiling out the derivative you get

    \dfrac{1}{4}y^{-1} - 8y^{-1/2} + 64

    L = \int_{1}^{4} \sqrt{\dfrac{1}{4}y^{-1} - 8y^{-1/2} + 65}

    L = \int_{1}^{4} \sqrt{\dfrac{1}{4}y^{-1} -\dfrac{32y^{-1/2}}{4} + \dfrac{260}{4}} What now?
    Last edited by Jason76; July 18th 2014 at 06:51 PM.
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    Re: Arc Length Problem - 2

    What should we go for u substitution (doesn't look like it), trig sub, partial fractions?
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    Re: Arc Length Problem - 2

    I already laid out the problem step by step. Re-read my previous post.
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    Re: Arc Length Problem - 2

    My professor said this problem could only be solved by approximation methods. However, don't try to solve this, cause I already got the answer.
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