Question

What is the arc length of the curve given by `x =sqrt( t^2+t+1)` and `y= 1/(t^2+1) -t`, for `0<t<1`?

Answer 1

`approx 1.66982`

Explanation:

With
`x(t)=sqrt(t^2+t+1)`
we get
`x'(t)=(1+2t)/sqrt(t^2+t+1)`
`y(t)=1&(t^2+1)-t`
so
`y'(t)=-1-2t/(1+t^2)^2`
So we have to solve
`int_0^1sqrt((1+2t)^2/(4(1+t+t^2)^2)+(1+2t/(1+t^2)^2))dt`
I only found an approximate value.