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Math Help - lenght of a curve

  1. #1
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    lenght of a curve

    Find the length of the curve defined by
    y=3x3211
    from x=2 to x=10.

    ok i think that

    L= integral from 2 to 10 sqrt((dx/dt)^2+(dy/dt)^2)dt

    dx/dt would be 3x^3/2-11

    do i get dy/dt by just solving for x then plug them both in because it doesnt seem to be working for me
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  2. #2
    Member eXist's Avatar
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    Try this:

    L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2}
    Last edited by eXist; August 10th 2009 at 11:31 AM.
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  3. #3
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    I'm assuming you want the length of curve of y=3x^\frac {3}{2}-11 from 2 to 10?

    This is the formula to find the length of the curve.

    L=\int_a^b \sqrt {1+\bigg( \frac{dy}{dx}\bigg )^2}dx
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  4. #4
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    Quote Originally Posted by chengbin View Post
    I'm assuming you want the length of curve of y=3x^\frac {3}{2}-11 from 2 to 10?

    This is the formula to find the length of the curve.

    L=\int_a^b \sqrt {1+\bigg( \frac{dy}{dx}\bigg )^2}dx

    so its intergral from 2 to 10 sqrt(1+(3x^3/2-11)^2)
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  5. #5
    Senior Member DeMath's Avatar
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    L = \int\limits_2^{10} {\sqrt {1 + {{\left( {\frac{d}{{dx}}\left( {3{x^{3/2}} - 11} \right)} \right)}^2}}\, dx}  = \int\limits_2^{10} {\sqrt {1 + {{\left( {\frac{9}{2}{x^{1/2}}} \right)}^2}}\, dx}  = \int\limits_2^{10} {\sqrt {1 + \frac{{81}}{4}x}\, dx}  =

    = \frac{1}{2}\int\limits_2^{10} {\sqrt {4 + 81x}\, dx}  = \frac{1}{{162}}\int\limits_2^{10} {\sqrt {4 + 81x}\, d\left( {4 + 81x} \right)}  = \left. {\frac{1}{{162}} \cdot \frac{2}{3}\sqrt {{{\left( {4 + 81x} \right)}^3}} } \right|_2^{10} =

    = \frac{1}{{243}}\left( {\sqrt {{{814}^3}}  - \sqrt {{{166}^3}} } \right) = \frac{2}{{243}}\left( {407\sqrt {814}  - 83\sqrt {166} } \right) \approx 86.77044{\text{ }}\left( {{\text{linear units}}} \right).
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