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

1 Answer
May 31, 2018

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#