- #1
Lanza52
- 63
- 0
[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.
[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.