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Math Help - Arc Length Problem

  1. #1
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    Arc Length Problem

     y = \sqrt{2 - x^{2}} evaluated at upper bound 1 and lower bound 0

     y'(\sqrt{2 - x^{2}}) = -2x

     L = \int_{0}^{1} \sqrt{1 + (-2x)^{2}}

    L = \int_{0}^{1} \sqrt{1 + 4x^{2}}

     L = \int_{0}^{1} (1 + 4x^{2})^{1/2}

     L = \int_{0}^{1} (u)^{1/2}

     u = 1 + 4x^{2}

    du = 8x

    \dfrac{1}{8}du = dx

     L = \int_{0}^{1} (u)^{1/2}

     L = \dfrac{1}{8} \int_{0}^{1} ((u)^{1/2}

    (\dfrac{2}{3})(\dfrac{1}{8})(u)^{3/2} evaluated at upper bound 1 and lower bound 0

    (\dfrac{1}{12})(u)^{3/2} evaluated at upper bound 1 and lower bound 0

    (\dfrac{1}{12})(1 + 4x^{2})^{3/2} evaluated at upper bound 1 and lower bound 0

    [(\dfrac{1}{12})(1 + 4(1)^{2})^{3/2}] - (\dfrac{1}{12})(1 + 4(0)^{2})^{3/2} - Is this right?
    Last edited by Jason76; July 15th 2014 at 08:23 PM.
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  2. #2
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    Re: Arc Length Problem

    Why have you written your derivative in such an odd format?

    $\displaystyle \begin{align*} y &= \sqrt{ 2 - x^2} \\ \frac{ \mathrm{d}y }{\mathrm{d}x} &= -\frac{2x}{2 \sqrt{2 - x^2} } \\ &= -\frac{x}{\sqrt{2 - x^2}} \end{align*}$

    so your arclength is

    $\displaystyle \begin{align*} L &= \int_a^b{ \sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 } \, \mathrm{d}x } \\ &= \int_0^1{ \sqrt{ 1 + \left( -\frac{x}{\sqrt{2 - x^2} } \right) ^2 } \, \mathrm{d}x } \end{align*}$

    Go from here...
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  3. #3
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    Re: Arc Length Problem

    Sorry for delay in responding.

    \sqrt{2 - x^{2}}

    0\leq x \leq 1

    y = (2 - x^{2})^{2}

    y' = -x(2 - x^{2})^{-1/2}

    \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}}

    \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x}}

    \int_{0}^{1} \dfrac{\sqrt{2}}{\sqrt{2 - x^{2}}}

    Using trig substitution you get to this point:

    a = 2

    x = 2\sin\theta

    dx = 3\cos\theta

    4 - (2\sin\theta)^{2} = 4 - 4 \sin^{2}\theta = 4(1 - \sin^{2}\theta) = 4 cos^{2}\theta

    \int_{0}^{1}\dfrac{\sqrt{2} (2\cos \theta)}{\sqrt {4 cos^{2}\theta}} - What next it seems like all the thetas cancel out, so how can we evaluate it?

    \int_{0}^{1}\sqrt{2}
    Last edited by Jason76; July 18th 2014 at 06:15 PM.
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    Re: Arc Length Problem

    Quote Originally Posted by Jason76 View Post
    Sorry for delay in responding.

    \sqrt{2 - x^{2}}

    0\leq x \leq 1

    y = (2 - x^{2})^{2}

    y' = -x(2 - x^{2})^{-1/2}

    \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}}

    \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x^2}}
    Fixed that step

    \int_{0}^{1} \dfrac{\sqrt{2}}{\sqrt{2 - x^{2}}}

    Using trig substitution you get to this point:

    a = 2
    NO! Here your $\displaystyle \begin{align*} a = \sqrt{2} \end{align*}$ so the substitution you must make is $\displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \implies \mathrm{d}x = \sqrt{2}\cos{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$

    x = 2\sin\theta

    dx = 3\cos\theta

    4 - (2\sin\theta)^{2} = 4 - 4 \sin^{2}\theta = 4(1 - \sin^{2}\theta) = 4 cos^{2}\theta

    \int_{0}^{1}\dfrac{\sqrt{2} (2\cos \theta)}{\sqrt {4 cos^{2}\theta}} - What next it seems like all the thetas cancel out, so how can we evaluate it?
    Even though you are out by a significant factor from the above mistake, you come to the right conclusion that the "functional" part cancels out. So how would you evaluate $\displaystyle \begin{align*} \int{ \mathrm{d}x } \end{align*}$? Hint: $\displaystyle \begin{align*} \int{ \mathrm{d}x} = \int{ 1 \, \mathrm{d}x } \end{align*}$...
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    Re: Arc Length Problem

    Would the answer be \sqrt{2} \theta evaluated at 1 upper bound and 0 lower bound?
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    Re: Arc Length Problem

    Well, if you mean evaluated where x is between 0 and 1, then yes.

    But you will be better off changes your bounds to $\displaystyle \begin{align*} \theta \end{align*}$ bounds, so that you don't have to convert back to a function of x.
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    Re: Arc Length Problem

    How could I convert back to a function of x without all the trig functions? If I had those, then all I would have to do is make a right triangle.
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    Re: Arc Length Problem

    You NEED the trig functions to convert back to a function of x.

    Convert your bounds, it's a much easier process.
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    Re: Arc Length Problem

    Quote Originally Posted by Prove It View Post
    You NEED the trig functions to convert back to a function of x.

    Convert your bounds, it's a much easier process.
    how to convert bounds? hint?
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    Re: Arc Length Problem

    You know that $\displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \end{align*}$. If x = 0 then what is $\displaystyle \begin{align*} \theta \end{align*}$? If x = 1 then what is $\displaystyle \begin{align*} \theta \end{align*}$?
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    Re: Arc Length Problem

    Quote Originally Posted by Prove It View Post
    You know that $\displaystyle \begin{align*} x = \sqrt{2}\sin{ \left( \theta \right) } \end{align*}$. If x = 0 then what is $\displaystyle \begin{align*} \theta \end{align*}$? If x = 1 then what is $\displaystyle \begin{align*} \theta \end{align*}$?
    right I see. The sin of 0 is 0 and the sin of 1 is ?? Is it \dfrac{\pi}{2}

    Answer would be \sqrt{2}\sin(1) - \sqrt{2}\sin(0) =  \sqrt{2}(\dfrac{\pi}{2}) - \sqrt{2}(0) = \sqrt{2}\dfrac{\pi}{2} ??
    Last edited by Jason76; July 19th 2014 at 10:22 AM.
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    Re: Arc Length Problem

    NO! You are trying to figure out which ANGLES give those sine values! What angle gives the sine value as 0? (You have correctly identified it as 0). What angle gives $\displaystyle \begin{align*} \sqrt{2}\sin{ \left( \theta \right) } = 1 \implies \sin{ \left( \theta \right) } = \frac{1}{\sqrt{2}} \end{align*}$. This is another one that you should have committed to memory.
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    Re: Arc Length Problem

    Wouldn't it be?

    \sqrt{2}\sin(1) - \sqrt{2}\sin(0) =  \sqrt{2}(\dfrac{\pi}{2}) - \sqrt{2}(0) = \sqrt{2}\dfrac{\pi}{2}
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    Re: Arc Length Problem

    Don't listen to me then...
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    Re: Arc Length Problem

    \sqrt{2}\dfrac{\pi}{2} is not the answer on homework.
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