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Finding the circumference of the loop in parametrics

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data

    find the cumference of the loop 9x^2=4y^3 from (0,0) to (2/3, 1)


    2. Relevant equations

    length of curve square root (1+(dx/dy)^2)dy or square root ((dx/dt)^2 + (dy/dt)^2) dt

    3. The attempt at a solution

    i said x=9t/4 and y= (9t^(2/3))/4

    then i found that 0≤t≤8/27

    so after finding dy/dt and dx/dt i pluged that into that lengh of curve formula
    the answer should be (4(square root 3))/3

    But i m not getting the answer....
     
  2. jcsd
  3. Feb 11, 2010 #2

    Mark44

    Staff: Mentor

    It's a curve, but not a loop. A loop closes back on itself.
    I found it easier to solve for x as a function of y, and then use the first formula you have above. From your given equation, x = +/-(2/3)y3/2. Since you want the arclength between (0,0) and (2/3, 1), x will be >= 0, so the equation simplifies to x = +(2/3)y3/2. The resulting integral can be done with an ordinary substitution.
    Are you sure that's the right answer? I get (2/3)(2sqrt(2) - 1)
     
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