Question

What is the arclength of f(t)=(3t^2-1, 3t^3-t) on t∈[-1/sqrt(3),1/sqrt(3)]?

Answer 1

`{4sqrt3}/3`

Explanation:

For the given curve, we have

`x(t) = 3t^2-1,qquad y(t) = 3t^3-t`

Hence

`dx/dt = 6t,qquad dy/dt = 9t^2-1`

Thus, the infinitesimal arc length is

`ds = sqrt{(dx/dt)^2+(dy/dt)^2}dt`
`qquad = sqrt{36t^2+(9t^2-1)^2}dt`
`qquad = sqrt{(9t^2+1)^2}dt`
`qquad = (9t^2+1)dt`

Thus, the arc length required is

`int_{-1/sqrt3}^{+1/sqrt3}(9t^2+1)dt = (3t^3+t)_{-1/sqrt3}^{+1/sqrt3}`

`qquad qquad = {t(3t^2+1)}_{-1/sqrt3}^{+1/sqrt3}=4/sqrt3`