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

Feb 19, 2016

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 $f \left(t\right) = \left[x \left(t\right) , y \left(t\right)\right]$, deﬁned on an interval a ≤ t ≤ b, with the properties that $x ' \left(t\right) , y ' \left(t\right)$ are continuous on the interval a ≤ t ≤ b and the path represented by $f \left(t\right)$ is traversed exactly once on this interval. Then the arc length of this path is given by the formula:
$L = {\int}_{a}^{b} | f ' \left(t\right) | \mathrm{dt}$ ====> (1)
with $r \left(t\right) = \left[x \left(t\right) , y \left(t\right)\right]$
$| r ' \left(t\right) | = \sqrt{{\left(x ' \left(t\right)\right)}^{2} + {\left(y ' \left(t\right)\right)}^{2}}$
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: $L = 0.6301966330154752$