Find the length of the curve?

Question

x=2/3*t^3, y=3t^2 on the interval from 0 is less than or equal to t is less than or equal to 4. I am getting stuck trying to integrate the sqrt of 4t^4+324t^2...

Answer 1

Where did the 324 come from?

dx = 2t^2dt

dy = 6tdt

ds = ((4t^4 + 36t^2)^(1/2))dt

ds = 2t((t^2 + 9)^(1/2))dt

let u = t^2 + 9, du = 2tdt

ds = [u^(1/2)]du

s = (2/3)u^(3/2) = (2/3)(t^2 + 9)^(3/2) from t=0 to t=4

s = (2/3)[(16 + 9)^(3/2) - (0 + 9)^(3/2)]

s = (2/3)[25^(3/2) - 9^(3/2)]

s = (2/3)(125 - 27)

s = (2/3)(98)

s ≈ 65.333 or 65 1/3