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

Question

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

Scarlett Allison · Accepted Answer

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

#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#

Adam Adkins · Answer

undefined To find the arc length of $ f(x) = (x^2-1)^{3/2} $ on $ x $ in the interval $[1,3]$:

1. Calculate the first derivative of $ f(x) $.
2. Compute the square of the derivative and add 1.
3. Integrate the square root of the expression obtained in step 2 over the interval $[1,3]$.
4. The result of the integration is the arc length of the function on the specified interval.