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

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Answer 1 · 2017-08-21T20:47:47

$f(t)=(t^2lnt,t-lnt)$

$f'(t)=(2tlnt+t,t-1/t)$

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

$s=int_a^bsqrt((dx/dt)^2+(dy/dt)^2)dt$

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.

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