Question

How do you find the arc length of the curve `y=x^3` over the interval [0,2]?

Answer 1

A first-order approximation to the arc length gives `8+tan^(-1)(2sqrt3)` units.

Explanation:

`y=x^3`

`y'=3x^2`

Arc length is given by:

`L=int_0^2sqrt(1+9x^4)dx`

Apply the substitution `sqrt3x=u`:

`L=1/sqrt3int_0^(2sqrt3)sqrt(1+u^4)du`

Complete the square in the square root:

`L=1/sqrt3int_0^(2sqrt3)sqrt((u^2+1)^2-2u^2)du`

Factor out the larger piece:

`L=1/sqrt3int_0^(2sqrt3)(u^2+1)sqrt(1-(2u^2)/(u^2+1)^2)du`

For `x in [0,2sqrt3]`, `(2u^2)/(u^2+1)^2<1`. Take the series expansion of the square root:

`L=1/sqrt3int_0^(2sqrt3)(u^2+1){sum_(n=0)^oo((1/2),(n))(-(2u^2)/(u^2+1)^2)^n}du`

Isolate the `n=0` and `n=1` cases:

`L=1/sqrt3int_0^(2sqrt3){(u^2+1)-u^2/(u^2+1)}du+1/sqrt3sum_(n=2)^oo((1/2),(n))(-2)^nint_0^(2sqrt3)u^(2n)/(u^2+1)^(2n-1)du`

Hence

`L=1/sqrt3int_0^(2sqrt3)(u^2+1/(u^2+1))du+1/sqrt3sum_(n=2)^oo((1/2),(n))(-2)^nint_0^(2sqrt3)u^(2n)/(u^2+1)^(2n-1)du`

Ignoring the higher-order terms, a first-order approximation gives:

`L~~1/sqrt3[1/3u^3+tan^(-1)u]_ 0^(2sqrt3)`

Insert the limits of integration:

`L~~8+tan^(-1)(2sqrt3)`