# Thread: perimeter of the curve

1. ## perimeter of the curve

Show that the perimeter of the curve with equation x^2/3 + y^2/3 = 1 is 6.

How can I find this?

2. I forget the same of this particular shape, Maybe an astaroid or hypocycloid or something. but you can write it as

$y=(1-x^{\frac{2}{3}})^{\frac{3}{2}}$

Find the derivative and square it and you get:

$\frac{1-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}$

Add 1 and it becomes $\frac{1}{x^{\frac{2}{3}}}$

Now, use the arc length formula:

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

Now, you simply have to integrate $4\int_{0}^{1}x^{\frac{-1}{3}}dx$

3. Originally Posted by galactus
I forget the same of this particular shape, Maybe an astaroid or hypocycloid or something. [snip]
Astroid - Wikipedia, the free encyclopedia

4. Thanks, MrF.

Another way to do this is with parametrics.

The parametric equations for an astroid are

$x=cos^{3}(t), \;\ y=sin^{3}(t)$

Now, we can use the parametric arc length formula.

$\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}=3si n(t)cos(t)$

$6\int_{0}^{\frac{\pi}{2}}sin(2t)dt$

Gives same result as with the rectangular.

5. Thank you so much galactus