# Arc Length Question

1. Nov 24, 2008

### kevtimc

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

y = (2/3) * (x^2 - 1) ^ (3/2) 1 <= x <= 3
Length = ?

2. Relevant equations

L = $$\int\sqrt{1 + (dy/dx)^2} dx$$

3. The attempt at a solution

dy/dx y = (2/3) * (x^2 - 1) ^ (3/2) =
2x * sqrt(x - 1)

Any ideas for a proper substitution? The answer on wolfram seems ridiculous.

Last edited: Nov 24, 2008
2. Nov 24, 2008

### Citan Uzuki

Hmm... it seems wolfram is being ridiculous on this problem. But you don't need a substitution here. Observe that the derivative is 2x√(x² - 1), not 2x√(x - 1). So once you go back and do the arithmetic correctly, you will find that 1 + (dy/dx)² is the square of a polynomial (specifically, it is (2x² - 1)²). After that, this problem should be a piece of cake.

3. Nov 24, 2008

### kevtimc

I actually knew it was x2, just a typing error. I actually got (2x2 - 1)2
the first time I calculated it, but I didn't get the right answer for the definite integral at 1 through 3 (15.333).

My solution:

(2x)^2 * (x^2 - 1) + 1 = 4x^4 - 4x^2 + 1 = (2x^2 - 1)^2

= (2/3) * x3 -x + C ] (1-3) = 15.333 Nevermind . . .

Last edited: Nov 24, 2008
4. Nov 24, 2008

### Citan Uzuki

Glad you figured it out. It looks like you have the theory right, and were just getting caught up with typos.

5. Nov 25, 2008

### kevtimc

Story of my mathematical life.