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Math Help - Find the length of a graph

  1. #1
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    Find the length of a graph

    Find the length of the graph of y= from(1, - ) to (4, )
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by evansf View Post
    Find the length of the graph of y= from(1, - ) to (4, )
    y = \frac{1}{3}x^{3/2} - x^{1/2}


    The arc length formula is
    L = \int_a^b \sqrt{1 + \left ( y^{\prime } \right ) ^2 }~dx

    So
    L = \int_1^4 \sqrt{1 + \left ( \frac{1}{2}x^{1/2} - \frac{1}{2}x^{-1/2} \right ) ^2 }~ dx

    L = \int_1^4 \sqrt{1 + \frac{1}{4} \cdot \left ( x - 2 + \frac{1}{x} \right ) }~ dx

    L = \int_1^4 \sqrt{\frac{1}{2} + \frac{1}{4} \cdot \left ( x + \frac{1}{x} \right ) }~ dx

    L = \int_1^4 \sqrt{\frac{1}{2} + \frac{1}{4} \cdot \left ( x + \frac{1}{x} \right ) }~ dx

    Unfortunately, this form shows that the integral is going to turn out to be an elliptic integral. It cannot be calculated by hand, only approximated.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    L = \int_1^4 \sqrt{\frac{1}{2} + \frac{1}{4} \cdot \left ( x + \frac{1}{x} \right ) }~ dx

    Unfortunately, this form shows that the integral is going to turn out to be an elliptic integral. It cannot be calculated by hand, only approximated.
    ?

    It can be computed
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  4. #4
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    Hello, evansf!

    Find the length of the graph of: .  y \:= \:\frac{1}{3}x^{\frac{3}{2}} - x^{\frac{1}{2}} . from \left(1,\,-\frac{2}{3}\right) to \left(4,\,\frac{2}{3}\right)
    Arc Length Formula: . \boxed{L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx}

    We have: . y \;=\;\frac{1}{3}x^{\frac{3}{2}} - x^{\frac{1}{2}}

    Then: . \frac{dy}{dx}\;=\;\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}<br />

    And: . 1 + \left(\frac{dy}{dx}\right)^2\;=\;1 + \left(\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}\right)^2 \;=\;1 + \frac{1}{4}x^{\frac{1}{4}} - \frac{1}{2} + \frac{1}{4}x^{-\frac{1}{4}}

    . . . . . . =\;\frac{1}{4}x^{\frac{1}{4}} + \frac{1}{2} + \frac{1}{4}x^{-\frac{1}{4}} \;=\;\left(\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}\right)^2

    Hence: . \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\; \sqrt{\left(\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}}\right)^2 } \;=\;\frac{1}{2}x^{\frac{1}{2}} + \frac{1}{2}x^{-\frac{1}{2}} \;=\;\frac{1}{2}\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)<br />


    Therefore, we have: . L \;=\;\frac{1}{2}\int^4_1\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)\,dx


    . . . Can you finish it?

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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post

    And: . 1 + \left(\frac{dy}{dx}\right)^2\;=\;1 + \left(\frac{1}{2}x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}}\right)^2 \;=\;1 + \frac{1}{4}x^{\frac{1}{4}} - \frac{1}{2} + \frac{1}{4}x^{-\frac{1}{4}}
    You might want to take a look at your FOILing here.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    ?

    It can be computed
    Are you are saying that I goofed and it is not an elliptic integral, or are you merely saying that it can be approximated (fairly) easily?

    -Dan
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  7. #7
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    Put it at the integrator

    --

    Ohhh, Soroban made a mistake there
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