# 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.

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## Answers and Replies

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 . . .

Thanks for your help :]

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

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