Question

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

Answer 1

Result obtained using numerical integration, can't find closed form antiderivative:
`L = 0.6302`

Explanation:

First some definition and principles that will need to solve this:
Consider a vector function `f(t) =[x(t),y(t)]`, defined on an interval `a ≤ t ≤ b`, with the properties that `x'(t), y'(t)` are continuous on the interval a ≤ t ≤ b and the path represented by `f(t)` is traversed exactly once on this interval. Then the arc length of this path is given by the formula:
`L =int_a^b |f'(t)|dt` ====> (1)
with `r(t) =[x(t),y(t)]`
`|r'(t)| = sqrt((x'(t))^2+(y'(t))^2)`
Knowing this we are asked to find the length of the parametric curve given by
`f(t) = [t-sqrt(t-1), t^2/(t^2-1)]; " for " t: t in 2 ≤ t ≤ 3`
let's take the derivative of f(t):`f'(t)=[1−1/(2sqrt(t−1)), -(2t)/(t^2-1)^2]`
Now square each term and take the square root and integrate:
`L =int_2^3sqrt((1-1/(2sqrt(t-1)))^2 + (4t^2)/(t^2-1)^4`
This integral you have to integrate numerically, it does not have a close form antiderivative. You can us any online or fancy calculator"
My approximation is: `L = 0.6301966330154752`