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Math Help - Arclength

  1. #1
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    Arclength

    I am getting nowhere with this problem. I would appreciate it if one could show the work, including the answer. thank you.

    y = 1/3x^(3/2) - x^(1/2) , [2,8]

    Thanks again!
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  2. #2
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    Quote Originally Posted by ch2kb0x View Post
    I am getting nowhere with this problem. I would appreciate it if one could show the work, including the answer. thank you.

    y = 1/3x^(3/2) - x^(1/2) , [2,8]

    Thanks again!
    According to your headline I assume that you want to calculate the length of the curve of the function from 2 to 8 (?).

    You know that the length l of the arc can be calculated by:

    l = \int \sqrt{1+(y')^2}\ dx

    1. Determine the first derivative. I've got: y'=\frac12 \cdot x^{\frac12}-\frac12 \cdot x^{-\frac12}

    2. Calculate (y')

    3. Calculate 1 + (y')^2= \frac14 x + \frac12 + \frac14 x^{-1} = \left( \frac12x^{\frac12} + \frac12 x^{-\frac12} \right)^2

    4. Now calculate the arc length:

    l = \int_2^8 \sqrt{\left( \frac12x^{\frac12} + \frac12 x^{-\frac12} \right)^2}\ dx

    5. I've got: l = \frac{17}3 \cdot \sqrt{2} \approx 8.01...
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