Finding the circumference of the loop in parametrics

[hangainlover]

Homework Statement

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

Homework Equations

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

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...

 

[Mark44]

Homework Statement

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

It's a curve, but not a loop. A loop closes back on itself.

Homework Equations

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

The Attempt at a Solution

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

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)y^3/2. Since you want the arclength between (0,0) and (2/3, 1), x will be >= 0, so the equation simplifies to x = +(2/3)y^3/2. The resulting integral can be done with an ordinary substitution.

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...

Are you sure that's the right answer? I get (2/3)(2sqrt(2) - 1)