Perimeter of curve x^2/3 + y^2/3 = 1

By Anonymous on Friday, March 9, 2001 - 02:28 pm :

If x^2/3 + y^2/3 = 1, how do you show that the perimeter is 6?

By Kerwin Hui (Kwkh2) on Friday, March 9, 2001 - 03:27 pm :

Parametrise the curve by

x = \cos^{3} t

y = \sin^{3} t

Now, we know that the curve is symmetric about the $x$ and $y$ axes.

So, the perimeter is

$4 \int_{0}^{π / 2} ((dx / dt)^{2} + (dy / dt)^{2})^{1 / 2} dt = 12 \int_{0}^{π / 2} (\cos^{2} t \sin^{2} t (\cos^{2} t + \sin^{2} t {))}^{1 / 2} t dt = 12 \int_{0}^{π / 2} \cos t \sin t dt = 6$

Kerwin