What is the arclength of #(t^2lnt,(lnt)^2)# on #t in [1,2]#?

Question

Cameron Alexander · Accepted Answer

I used Wolframalpha to evaluate the integral.

#L = 2.815# Here is a reference for Arc length of parametric equations From the reference, the equation for arclength is: #L = int_alpha^beta sqrt((dx/dt)^2 + (dy/dt)^2)dt# #dx/dt = (d[t^2ln(t)])/dt = 2tln(t) + t# #dy/dt = (d[(ln(t))^2])/dt = (2ln(t))/t# I used Wolframalpha to evaluate the integral: #L = int_1^2 sqrt((2tln(t) + t)^2 + ((2ln(t))/t)^2)dt = 2.815#

Charles Arocha · Answer

undefined To find the arc length of the curve $(t^2 \ln t, (\ln t)^2)$ on the interval $[1, 2]$, we can use the formula for arc length:

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

where $ a = 1 $, $ b = 2 $, $ x(t) = t^2 \ln t $, and $ y(t) = (\ln t)^2 $. Differentiate $ x(t) $ and $ y(t) $ with respect to $ t $ to find $ dx/dt $ and $ dy/dt $, then substitute them into the formula and integrate over the given interval.

$$ dx/dt = 2t\ln t + t $$
$$ dy/dt = \frac{2 \ln t}{t} $$

Substituting these into the formula, we get:

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

After calculating this integral, you'll get the arc length of the curve on the interval $[1, 2]$.