What is the arc length of f(x) = (x^2-1)^(3/2) f(x)=(x21)32 on x in [1,3] x[1,3]?

1 Answer
May 31, 2018

A first-order approximation gives L=23-5/(24sqrt2)ln((5+2sqrt2)/(5-2sqrt2))L=235242ln(5+22522) units.

Explanation:

f(x)=(x^2-1)^(3/2)f(x)=(x21)32

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