Show that the perimeter of the curve with equation x^2/3 + y^2/3 = 1 is 6.
How can I find this?
I forget the same of this particular shape, Maybe an astaroid or hypocycloid or something. but you can write it as
$\displaystyle y=(1-x^{\frac{2}{3}})^{\frac{3}{2}}$
Find the derivative and square it and you get:
$\displaystyle \frac{1-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}$
Add 1 and it becomes $\displaystyle \frac{1}{x^{\frac{2}{3}}}$
Now, use the arc length formula:
$\displaystyle \int\sqrt{\frac{1}{x^{\frac{2}{3}}}}dx$
Now, you simply have to integrate $\displaystyle 4\int_{0}^{1}x^{\frac{-1}{3}}dx$
Thanks, MrF.
Another way to do this is with parametrics.
The parametric equations for an astroid are
$\displaystyle x=cos^{3}(t), \;\ y=sin^{3}(t)$
Now, we can use the parametric arc length formula.
$\displaystyle \sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}=3si n(t)cos(t)$
$\displaystyle 6\int_{0}^{\frac{\pi}{2}}sin(2t)dt$
Gives same result as with the rectangular.