# Arc Length Problem

1. Feb 22, 2010

### 3.141592654

1. The problem statement, all variables and given/known data

Find the arc length of the curve:

$$y=\frac{x^5}{6}+\frac{1}{10x^3}$$

1$$\leq$$x$$\leq$$2

2. Relevant equations

$$ds=\sqrt{dx^2+dy^2}$$

$$ds=\sqrt{1+\frac{dy}{dx}^2}dx$$

3. The attempt at a solution

$$\frac{dy}{dx}=\frac{5}{6}x^4-\frac{3}{10x^4}$$

$$ds=\sqrt{1+(\frac{5}{6}x^4-\frac{3}{10x^4})^2}dx$$

If I use Trig. Sub, I have $$tan \theta=(\frac{5}{6}x^4-\frac{3}{10x^4})$$.

However, I don't know how to solve this for X. If this is the right path, can anyone help with the next steps, and is there a better way to do this problem?

Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 22, 2010

### Staff: Mentor

OK, hold it right there. Just carry out the multiplication, and add the 1. You'll see that you still have a perfect square so you can take its square root.

This is a perfect example of a problem that has been laboriously cooked-up so that it's doable. Most of the arc length problems come out with difficult integrands, so some effort has to go into one to make it not too difficult to integrate.