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

I used Wolframalpha to evaluate the integral.

Here is a reference for Arc length of parametric equations

From the reference, the equation for arclength is:

I used Wolframalpha to evaluate the integral:

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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]).

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