What is the arclength of $f(t) = (e^(t)-e^(t^2)/t,t^2-1)$ on $t in [1,3]$ ?

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Answer 1 · 2018-06-23T12:32:57

$approx 2681.58$

With
$x(t)=e^t-e^(t^2)/t$ we get

$x'(t)=e^t-2e^(t^2)+e^(t^2)/t^2$

$y(t)=t^2-1$
so

$y'(t)=2t$
and we have to solve

$int_1^3sqrt((e^t-2e^(t^2)+e^(t^2)/t)^2+(2t)^2)dt approx 2681.58$

What is the arclength of f(t) = (e^(t)-e^(t^2)/t,t^2-1)f(t) = (e^(t)-e^(t^2)/t,t^2-1) on t in [1,3]t in [1,3]?