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=cos³ t, y=sin³ t.

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

So, the perimeter is

4ò₀^p/2((dx/dt)²+(dy/dt)²)^1/2 dt=12ò₀^p/2 (cos² tsin² t(cos² t+sin² t))^1/2t dt=12ò₀^p/2costsint dt = 6

Kerwin