# Arc Length Question

## Homework Statement

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

## Homework Equations

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

## 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:

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.

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.

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 . . .