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