# Arc Length Formula Problem

1. Oct 28, 2007

### Lanza52

[SOLVED] Arc Length Problem

$$y=\sqrt{x^{3}}$$

So you plug it into the formula for arc length. (integral of the sqrt of 1+y'^2)

And it yields $$\int \sqrt{1+(\frac{3x^{2}}{2\sqrt{x^{3}}})^{2}dx$$

From there you would use trig substitution, 1+tan^2theta = sec^2theta. But converting the dx to dtheta is a complete pain. And from what I can tell, it looks like it gets ugly.

So the ugliness makes me think I am wrong. Can anybody check this up to this point?

Thanks.

2. Oct 28, 2007

### rock.freak667

well

$$\int \sqrt{1+(\frac{3x^{2}}{2\sqrt{x^{3}}})^{2}dx$$

can be simplified even more to give

$$\frac{1}{2}\int \sqrt{4+9x} dx$$

and from there it should become much easier

3. Oct 28, 2007

### Lanza52

Solved. Thanks =P