Question

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

Answer 1

`"Arclength " approx 10.601267`

Explanation:

`f(t)=(t^2sqrt(t-1),t^2+t-1)`

Formula for parametric arclength: `color(blue)(L = int_a^b sqrt((dx/dt)^2+(dy/dt)^2)" "dt)`

First, find `f'(t)` by differentiating `f(t)`

`dx/dt=(t^2)(1/2(t-1)^(-1/2))+(2t)(t-1)^(1/2)`

`dx/dt=frac{t^2}{2sqrt(t-1)}+2tsqrt(t-1)=frac{t^2+4t(t-1)}{2sqrt(t-1)}`

`dy/dt=2t+1`

Plug in expressions for `dx/dt` and `dy/dt` into the arclength formula:
`L=int_2^3sqrt((frac{t^2+4t(t-1)}{2sqrt(t-1)})^2+(2t+1)^2)" "dt`

Use a calculator to evaluate:
`"Arclength " approx 10.601267`