Question

What is the arc length of the curve given by `r(t)= (ln(1/t),t^2,t)` on `t in [1, 10]`?

Answer 1

The arc length is `99+ln10-3/4(ln10+sum_(m=1)^oo((-1),(m))(1/2)^m(1/(2m))(1-1/100^m))+sum_(n=2)^oosum_(m=0)^oo((1/2),(n))((1-2n),(m))(-3/4)^n(1/2)^m(1/(n+m))(1-1/100^(n+m))` units.

Explanation:

`r(t)=(ln(1/t),t^2,t)=(-lnt,t^2,t)`

`r'(t)=(-1/t,2t,1)`

Arc length is given by:

`L=int_1^10sqrt(1/t^2+4t^2+1)dt`

Complete the square in the square root:

`L=int_1^10sqrt((2t+1/t)^2-3)dt`

Factor out the larger piece:

`L=int_1^10(2t+1/t)sqrt(1-3/(2t+1/t)^2)dt`

For `t in [1,10]`, `3/(2t+1/t)^2<1`. Take the series expansion of the square root:

`L=int_1^10(2t+1/t){sum_(n=0)^oo((1/2),(n))(-3/(2t+1/t)^2)^n}dt`

Isolate the `n=0` term and simplify:

`L=int_1^10(2t+1/t)dt+sum_(n=1)^oo((1/2),(n))(-3)^nint_1^10(2t+1/t)^(1-2n)dt`

Rearrange:

`L=[t^2+lnt]_ 1^10+2sum_(n=1)^oo((1/2),(n))(-3/4)^nint_1^10t^(1-2n)(1+1/(2t^2))^(1-2n)dt`

Take a second series expansion:

`L=99+ln10+2sum_(n=1)^oo((1/2),(n))(-3/4)^nint_1^10t^(1-2n){sum_(m=0)^oo((1-2n),(m))(1/(2t^2))^m}dt`

Simplify:

`L=99+ln10+2sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((1-2n),(m))(-3/4)^n(1/2)^m int_1^10t^(1-2n-2m)dt`

Isolate the `n=1` case:

`L=99+ln10-3/4sum_(m=0)^oo((-1),(m))(1/2)^m int_1^10t^(-1-2m)dt+2sum_(n=2)^oosum_(m=0)^oo((1/2),(n))((1-2n),(m))(-3/4)^n(1/2)^m int_1^10t^(1-2n-2m)dt`

Isolate the `n=1,m=0` case:

`L=99+ln10-3/4(int_1^10 1/tdt+sum_(m=1)^oo((-1),(m))(1/2)^m int_1^10t^(-1-2m)dt)+2sum_(n=2)^oosum_(m=0)^oo((1/2),(n))((1-2n),(m))(-3/4)^n(1/2)^m int_1^10t^(1-2n-2m)dt`

The remaining integrals are trivial:

`L=99+ln10-3/4([lnt]_ 1^10+sum_(m=1)^oo((-1),(m))(1/2)^m [t^(-2m)]_ 1^10/(-2m))+2sum_(n=2)^oosum_(m=0)^oo((1/2),(n))((1-2n),(m))(-3/4)^n(1/2)^m[t^(-2n-2m)]_1^10/(-2n-2m)`

Insert the limits of integration and simplify:

`L=99+ln10-3/4(ln10+sum_(m=1)^oo((-1),(m))(1/2)^m(1/(2m))(1-1/100^m))+sum_(n=2)^oosum_(m=0)^oo((1/2),(n))((1-2n),(m))(-3/4)^n(1/2)^m(1/(n+m))(1-1/100^(n+m))`