# Finding the circumference of the loop in parametrics

1. Feb 10, 2010

### hangainlover

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. Feb 11, 2010

### 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)