What is the arclength of $(sqrtt,1/sqrt(t^2+3))$ on $t in [1,2]$ ?

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Answer 1 · 2018-05-30T11:24:35

$approx 0.431873$

We have
$x(t)=sqrt(t)$
then
$x'(t)=1/(2*sqrt(t))$
$y(t)=1/sqrt(t^2+3)$

then
$y'(t)=-t/sqrt(t^2+3)^3$
So we get the integral
$int_1^2sqrt(1/(4*t)+t^2/(3+t^2)^3)dt$
I have found only a numerical value:

$approx 0.431873$

What is the arclength of (sqrtt,1/sqrt(t^2+3))(sqrtt,1/sqrt(t^2+3)) on t in [1,2]t in [1,2]?