Question

What is the arc length of `f(x) = (x^2-1)^(3/2)` on `x in [1,3]`?

Answer 1

A first-order approximation gives `L=23-5/(24sqrt2)ln((5+2sqrt2)/(5-2sqrt2))` units.

Explanation:

`f(x)=(x^2-1)^(3/2)`

`f'(x)=3xsqrt(x^2-1)`

Arc length is given by:

`L=int_1^3sqrt(1+9x^2(x^2-1))dx`

Expand:

`L=int_1^3sqrt(9x^4-9x^2+1)dx`

Complete the square:

`L=1/2int_1^3sqrt(9(2x^2-1)^2-5)dx`

Factorize:

`L=3/2int_1^3(2x^2-1)sqrt(1-5/(9(2x^2-1)^2))dx`

For `x in [1,3]`, `5/(9(2x^2-1)^2<1`. Take the series expansion of the square root:

`L=3/2int_1^3(2x^2-1){sum_(n=0)^oo((1/2),(n))(-5/(9(2x^2-1)^2))^n}dx`

Isolate the `n=0` term and simplify:

`L=3/2int_1^3(2x^2-1)dx+3/2sum_(n=1)^oo((1/2),(n))(-5/9)^nint_1^3(1/(2x^2-1))^(2n-1)dx`

Apply the difference of squares:

`L=3/2[2/3x^3-x]_ 1^3+3/2sum_(n=1)^oo((1/2),(n))(-5/9)^nint_1^3 (1/((sqrt2x-1)(sqrt2x+1)))^(2n-1)dx`

Apply partial fraction decomposition:

`L=23+3sum_(n=1)^oo((1/2),(n))(-5/36)^nint_1^3 (1/(sqrt2x-1)-1/(sqrt2x+1))^(2n-1)dx`

Isolate the `n=1` term and simplify:

`L=23-5/24int_1^3 (1/(sqrt2x-1)-1/(sqrt2x+1))dx+3sum_(n=2)^oo((1/2),(n))(-5/36)^nint_1^3 (1/(sqrt2x-1)-1/(sqrt2x+1))^(2n-1)dx`

Hence

`L=23-5/(24sqrt2)[ln|sqrt2x-1|-ln|sqrt2x+1|]_ 1^3+3sum_(n=2)^oo((1/2),(n))(-5/36)^nint_1^3 (1/(sqrt2x-1)-1/(sqrt2x+1))^(2n-1)dx`

Giving:

`L=23-5/(24sqrt2)ln((5+2sqrt2)/(5-2sqrt2))+3sum_(n=2)^oo((1/2),(n))(-5/36)^nint_1^3 (1/(sqrt2x-1)-1/(sqrt2x+1))^(2n-1)dx`