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Math Help - Length of a curve

  1. #1
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    Length of a curve

    I have a task to calculate the length of the curve. Does anyone know outcome?

    I big idiot((

    http://www.sdilej.eu/pics/ee9c6c6833...d2329b4eea.jpg

    Could anyone help me please?
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  2. #2
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    It looks as though you are using the correct method, and you almost have the correct answer (except that you have crossed out the calculation at a crucial stage).

    The function is y = \frac13(x^{3/2} - 3x^{1/2}). Its derivative is y' = \frac12(x^{1/2}-x^{-1/2}). Then y'^2 = \frac14(x+x^{-1}-2), and 1+y'^{\,2} = \frac14(x+x^{-1}+2) = \bigl(\frac12(x^{1/2}+x^{-1/2})\bigr)^2. So the integral for the length is \displaystyle\int_0^3\tfrac12(x^{1/2}+x^{-1/2})dx = \Bigl[\tfrac13(x^{3/2} + 3x^{1/2})\Bigr]_0^3, from which you should get the answer 2\sqrt3.
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  3. #3
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    Hello, snopy!

    \text{Find the length of the curve: }\:y \:=\:\frac{1}{3}(x-3)\sqrt{x}\:\text{ from }x = 0\text{ to }x = 3

    The function is: . y \:=\:\frac{1}{3}(x-3)x^{\frac{1}{2}} \;=\;\frac{1}{3}\left(x^{\frac{3}{2}} - 3x^{\frac{1}{2}}\right)


    Differentiate: . y' \;=\;\frac{1}{3}\left(\frac{3}{2}x^{\frac{1}{2}} - \frac{3}{2}x^{-\frac{1}{2}}\right) \;=\;\frac{1}{2}\left(x^{\frac{1}{2}} - x^{-\frac{1}{2}}\right)


    Square: . (y')^2 \;=\;\frac{1}{4}(x - 2 + x^{-1}) \;=\;\frac{1}{4}x - \frac{1}{2} + \frac{1}{4}x^{-1}


    Add one: . 1 + (y')^2 \;=\;1 + \frac{1}{4}x - \frac{1}{2} + \frac{1}{4}x^{-1} \;=\;\frac{1}{4}x + \frac{1}{2} + \frac{1}{4}x^{-1}

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


    Square root: . \sqrt{1 + (y')^2} \;=\;\frac{1}{2}\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)


    Therefore: . \displaystyle L \;=\;\tfrac{1}{2}\int^3_0\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)dx

    Got it?

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  4. #4
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    thanks )))
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  5. #5
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    One thing that is generally true of "length" integrals is that they tend to be very hard if not impossible! That idea of choosing a function so that 1+ f'^2 turns out to be a perfect square is a favorite of Calculus teachers.
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