How Do You Calculate the Arc Length of y=sqrt(x^3)?

[Lanza52]

[SOLVED] Arc Length Problem

[tex]y=\sqrt{x^{3}}[/tex]

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

And it yields [tex]\int \sqrt{1+(\frac{3x^{2}}{2\sqrt{x^{3}}})^{2}dx[/tex]

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.

 

[rock.freak667]

well

[tex]\int \sqrt{1+(\frac{3x^{2}}{2\sqrt{x^{3}}})^{2}dx[/tex]

can be simplified even more to give

[tex]\frac{1}{2}\int \sqrt{4+9x} dx[/tex]

and from there it should become much easier

 

[Lanza52]

Solved. Thanks =P