Question

Find the length of the curve defined by `y=18(4x^2−2ln(x)), x in[4,6]`?

Answer 1

`L~~1440-36ln(3/2)+1/576ln(143/63)` units is a first-order solution accurate to the 8th decimal place.

Explanation:

`y=18(4x^2−2lnx)=36(2x^2-lnx)`

`y'=36(4x-1/x)`

Arc length is given by:

`L=int_4^6sqrt(1+(y')^2)dx`

Rearrange:

`L=int_4^6y'sqrt(1+(y')^-2)dx`

Take the series expansion:

`L=int_4^6y'{sum_(n=0)^oo((1/2),(n))(y')^(-2n)}dx`

Isolate the `n=0` term and simplify:

`L=int_4^6y'dx+sum_(n=1)^oo((1/2),(n))int_4^6(36(4x-1/x))^(1-2n)dx`

Solve for the zeroth-order solution and rearrange:

`L=[y]_ 4^6+sum_(n=1)^oo((1/2),(n))1/6^(4n-2)int_4^6(x/(4x^2-1))^(2n-1)dx`

Isolate the `n=1` term and simplify:

`L=1440-36ln(3/2)+1/72int_4^6(x/(4x^2-1))dx+sum_(n=2)^oo((1/2),(n))1/12^(4n-2)int_4^6(1/(2x-1)+1/(2x+1))^(2n-1)dx`

Solve for the first-order correction:

`L=1440-36ln(3/2)+1/576[ln|4x^2-1|]_ 4^6+sum_(n=2)^oo((1/2),(n))1/12^(4n-2)int_4^6(1/(2x-1)+1/(2x+1))^(2n-1)dx`

Hence:

`L=1440-36ln(3/2)+1/576ln(143/63)+sum_(n=2)^oo((1/2),(n))1/12^(4n-2)int_4^6(1/(2x-1)+1/(2x+1))^(2n-1)dx`

Further work can be done to the desired level of accuracy.