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Math Help - arc length

  1. #1
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    arc length

    What is the arc length of y=2x^(3/2), from x=1 to x=3?
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  2. #2
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    Formula is L=Intagral from a to b (square root of 1+ derivative of f(x) squared dx.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by chevy900ss View Post
    What is the arc length of y=2x^(3/2), from x=1 to x=3?
    {L_y} = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx}  = \int\limits_1^3 {\sqrt {1 + {{\left( {\frac{d}<br />
{{dx}}2{x^{3/2}}} \right)}^2}} dx}  = \int\limits_1^3 {\sqrt {1 + {{\left( {3{x^{1/2}}} \right)}^2}} dx}  =

    = \int\limits_1^3 {\sqrt {1 + 9x} dx}  = \frac{1}{9}\int\limits_1^3 {{{\left( {1 + 9x} \right)}^{1/2}}d\left( {1 + 9x} \right)}  = \left. {\frac{1}<br />
{9}\left( {\frac{2}{3}{{\left( {1 + 9x} \right)}^{3/2}}} \right)} \right|_1^3 =

    = \left. {\frac{2}{{27}}{{\left( {1 + 9x} \right)}^{3/2}}} \right|_1^3 = \frac{2}{{27}}\left( {\sqrt {{{28}^3}}  - \sqrt {{{10}^3}} } \right) = \frac{2}{{27}}\left( {\sqrt {{{28}^2} \cdot 28}  - \sqrt {{{10}^3}} } \right) =

    = \frac{2}{{27}}\left( {\sqrt {{{28}^2} \cdot {2^2} \cdot 7}  - \sqrt {{{10}^2} \cdot 10} } \right) = \frac{2}{{27}}\left( {56\sqrt 7  - 10\sqrt {10} } \right) =

    = \frac{4}{{27}}\left( {28\sqrt 7  - 5\sqrt {10} } \right) \approx 8.632540504
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  4. #4
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    thats awesome, i was on the right track. I also have 3 more questions. Should i post them on this same forum or make a new post for them. Or is there an email that i can email them to someone and they can help me. They are basically having to do with the same thing.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by DeMath View Post
    {L_y} = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} = \int\limits_1^3 {\sqrt {1 + {{\left( {\frac{d}<br />
{{dx}}2{x^{3/2}}} \right)}^2}} dx} = \int\limits_1^3 {\sqrt {1 + {{\left( {3{x^{1/2}}} \right)}^2}} dx} =

    = \int\limits_1^3 {\sqrt {1 + 9x} dx} = \frac{1}{9}\int\limits_1^3 {{{\left( {1 + 9x} \right)}^{1/2}}d\left( {1 + 9x} \right)} = \left. {\frac{1}<br />
{9}\left( {\frac{2}{3}{{\left( {1 + 9x} \right)}^{3/2}}} \right)} \right|_1^3 =

    = \left. {\frac{2}{{27}}{{\left( {1 + 9x} \right)}^{3/2}}} \right|_1^3 = \frac{2}{{27}}\left( {\sqrt {{{28}^3}} - \sqrt {{{10}^3}} } \right) = \frac{2}{{27}}\left( {\sqrt {{{28}^2} \cdot 28} - \sqrt {{{10}^3}} } \right) =

    = \frac{2}{{27}}\left( {\sqrt {{{28}^2} \cdot {2^2} \cdot 7} - \sqrt {{{10}^2} \cdot 10} } \right) = \frac{2}{{27}}\left( {56\sqrt 7 - 10\sqrt {10} } \right) =

    = \frac{4}{{27}}\left( {28\sqrt 7 - 5\sqrt {10} } \right) \approx 8.632540504
    De Math, I think understand the priciples behind arc length.

    1. partition

    2. sum the distances between each f(x_i) and f(x_{i-1})

    3. take the limit

    4. Define the limit as an integral

    So, I have some questions:

    1. is what I stated above correct?

    2. If it is, could you provide a complete worked example?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chevy900ss View Post
    . I also have 3 more questions.
    Generally, we start a new thread if the questions doesn't relate the this one specifically.
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