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Arc Length (Astroid)

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    the graph of the equation x^(2/3) + y^(2/3) = 7^(2/3) is one of the family of curves called astroids.


    2. Relevant equations
    find the length of first-quadrant and multiply by 8.
    1. y=(7^(2/3) - x^(2/3))^(3/2)) ; 7sqrt(2)/4 <= x <= 7



    3. The attempt at a solution
    1. i found dy/dx, (dy/dx)^2, and now i'm at the length integral.
    2. L = integral sqrt(1+7^(2/3) * x^(2/3) - x^(4/3))

    i'm not sure how to evaluate this integral (#2) without using a calculator, but that doesn't give me the exact answer that i need.
     
  2. jcsd
  3. Feb 3, 2013 #2

    LCKurtz

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    I get a simpler expression for ##1 + y'^2##. Please show your work.
     
  4. Feb 3, 2013 #3
    dy/dx = -x^1/3 (7^2/3 - x^2/3)^(1/2)
    (dy/dx)^2 = x^(2/3) * (7^(2/3) - X^(2/3)V ----> 7^(2/3)* X^(2/3) - X^(4/3)

    L = ∫ (x=a to x=b) SQRT(1 + 7^(2/3) * x^(2/3) - x^(4/3))
     
    Last edited: Feb 3, 2013
  5. Feb 3, 2013 #4

    haruspex

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    Instead of converting the equation to the form y = f(x), just differentiate the equation as given. It's less messy, so less prone to error.
     
  6. Feb 4, 2013 #5
    it's already solved for y thats how the want you to solve it.
     
  7. Feb 4, 2013 #6

    LCKurtz

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    That is not the easy way to solve it and is probably why you are having trouble. Differentiate the equation ##x^{2/3} + y^{2/3} = 7^{2/3}## implicitly with respect to ##x## to get ##y'## and calculate ##1 + y'^2##. Show us what you get.
     
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