# Troublesome Arc Length Problem

1. May 4, 2014

### defecritus

1.) The problem is:

Find the arc length of f(x)= x^3/3-1/(4x) from x=1 to 2

2.) Relevant formulas:

ds = √(1+(dy/dx))

abs(L) = ∫ds

3.) My work so far:

f'(x)= x^2+1/(4x^2)

abs(L) = ∫(from 1 to 2) √(1+(x^2+1/(4x^2))^2 dx
= ∫(from 1 to 2) √(1+(x^4+1/2+1/(16x^4)) dx
= ∫(from 1 to 2) √(x^4+3/2+1/(16x^4)) dx
= ∫(from 1 to 2) √((16x^8+24x^4+1)/(16x^4) dx
= 1/4∫(from 1 to 2) √((16x^8+24x^4+1)/(x^4) dx
= 1/4∫(from 1 to 2) √(16x^8+24x^4+1)/(x^2) dx

Generally, I would suspect the numerator would be able to be factored and that would make this integral much easier to solve. I'm stumped after this part, because it seems that the numerator is irreducible.

I'm up for any help, I'm not asking for the answer. I really just need guidance from this point...that includes the possibility that the professor may have made a typo. I really wanted to consider that I may just not be aware of the next step before accusing the professor of fat fingering this. Thanks for the help.

2. May 4, 2014

### Staff: Mentor

Are you sure this is the right function? Things would work out if the above was a sum, not a difference.
Your formula is neither relevant nor correct. Fortunately you used the right formula in your work.

$ds = \sqrt{1 + (dy/dx)^2}dx$
I would contact the instructor to confirm that the function you wrote is what he intended. Because of the radical in the integral, arc length problems are generally to difficult to integrate using the techniques that are taught in beginning calculus courses. The functions have to be "cooked up" so that 1 + (f'(x))2 is a perfect square, so that its square root can be simplified.