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

Question

Ezra Adams · Accepted Answer

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

Emily Ammerman · Answer

undefined To find the arc length of the curve $ f(t) = (t - \sqrt{t - 1}, \frac{t^2}{t^2 - 1}) $ on the interval $ t \in [2,3] $, we use the formula for arc length:

$$
L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt
$$

For $ f(t) $, we first find $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $:

$$
\frac{dx}{dt} = 1 - \frac{1}{2\sqrt{t - 1}}
$$
$$
\frac{dy}{dt} = \frac{2t}{(t^2 - 1)^2}
$$

Now, we substitute these derivatives into the arc length formula and integrate over the interval $ t \in [2,3] $:

$$
L = \int_{2}^{3} \sqrt{\left(1 - \frac{1}{2\sqrt{t - 1}}\right)^2 + \left(\frac{2t}{(t^2 - 1)^2}\right)^2} \, dt
$$

This integral may not have a simple closed-form solution and may need to be evaluated numerically.