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

Aug 21, 2017

$f \left(t\right) = \left({t}^{2} \ln t , t - \ln t\right)$

$f ' \left(t\right) = \left(2 t \ln t + t , t - \frac{1}{t}\right)$

For a parametric function $f \left(t\right) = \left(x , y\right)$, the arc length for $t \in \left[a , b\right]$ is given by:

$s = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

Thus, we have an arc length of:

color(blue)(s=int_1^2sqrt((2tlnt+t)^2+(t-1/t)^2)dtapprox2.8914

Find the approximation using a calculator or a website like WolframAlpha.