What is the arclength of #f(t) = (tsqrt(lnt),t^3/(4-t))# on #t in [1,e]#?

Question

Charles Anctil · Accepted Answer

#\approx 16.0871# We get #x'(t)=1/(2sqrt(log(t)))+sqrt(log(t))# #y'(t)=3t^2/(4-t)+t^3/(4-t)^2# so we have the integral #int_{0}^{e}sqrt((1/(2*sqrt(log(t)))+sqrt(log(t)))^2+(3*t^2/(4-t)+t^3/(4-t)^2)^2)dt# Good luck!

Aurora Alvarado · Answer

undefined To find the arc length of the curve defined by $ f(t) = (t \sqrt{\ln t}, \frac{t^3}{4-t}) $ on the interval $ t \in [1,e] $, we use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dt}\right)^2} \, dt $$

We first find $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $:

$$ \frac{dx}{dt} = \frac{1}{2} \sqrt{\frac{\ln t}{t}} + \frac{t \cdot \frac{1}{t}}{2 \sqrt{\ln t}} = \frac{1}{2} \left(\frac{1}{\sqrt{t}} + \frac{1}{\sqrt{t \ln t}}\right) $$

$$ \frac{dy}{dt} = \frac{3t^2(4-t) - t^3(-1)}{(4-t)^2} = \frac{3t^2(4-t) + t^3}{(4-t)^2} $$

Now, we compute $ \sqrt{1 + \left(\frac{dy}{dt}\right)^2} $:

$$ \sqrt{1 + \left(\frac{dy}{dt}\right)^2} = \sqrt{1 + \frac{(3t^2(4-t) + t^3)^2}{(4-t)^4}} $$

Then, we integrate $ \sqrt{1 + \left(\frac{dy}{dt}\right)^2} $ over the interval $ t \in [1,e] $:

$$ L = \int_{1}^{e} \sqrt{1 + \left(\frac{dy}{dt}\right)^2} \, dt $$

After integrating, you will get the arc length of the curve.