Question

What is the arclength of `f(t) = (t^3-t+55,t^2-1)` on `t in [2,3]`?

Answer 1

`L=56/3+2/9sum_(n=1)^oo((1/2),(n))2^nint_6^9(1/(u-1)-1/(u+1))^(2n-1)du` units.

Explanation:

`f(t)=(t^3-t+55,t^2-1)`

`f'(t)=(3t^2-1,2t)`

Arclength is given by:

`L=int_2^3sqrt((3t^2-1)^2+(2t)^2)dt`

Expand the squares:

`L=int_2^3sqrt(9t^4-2t^2+1)dt`

Complete the square in the square root:

`L=1/3int_2^3sqrt((9t^2-1)^2+8)dt`

Rearrange:

`L=1/3int_2^3(9t^2-1)sqrt(1+8/(9t^2-1)^2)dt`

For `t in [2,3]`, `8/(9t^2-1)^2<1`. Take the series expansion of the square root:

`L=1/3int_2^3(9t^2-1){sum_(n=0)^oo((1/2),(n))(8/(9t^2-1)^2)^n}dt`

Isolate the `n=0` term and simplify:

`L=1/3int_2^3(9t^2-1)dt+1/3sum_(n=1)^oo((1/2),(n))8^nint_2^3 1/(9t^2-1)^(2n-1)dt`

Apply partial fraction decomposition:

`L=1/3[3t^3-t]_ 2^3+2/3sum_(n=1)^oo((1/2),(n))2^nint_2^3(1/(3t-1)-1/(3t+1))^(2n-1)dt`

For ease of reading, apply the substitution `3t=u`:

`L=56/3+2/9sum_(n=1)^oo((1/2),(n))2^nint_6^9(1/(u-1)-1/(u+1))^(2n-1)du`

The `n=1` case is trivial.