How do you find the length of the curve #y=sqrt(x-x^2)#?

1 Answer
Apr 26, 2016

Use algebra to get a length of #pi/2#

Explanation:

#y=sqrt(x-x^2)# is equivalent to

#y^2 = x-x^2# with the restriction #y <= 0#.

This is equivalent to (#y >= 0# on)

#x^2-x+y^2 = 0#.

This is the equation of a circle. Complete the square to get

#x^2-x+1/4+y^2 = 1/4#, or, better yet

#(x-1/2)^2 + y^2 = 1/4#.

With #y >= 0#, this is the upper semicircle centered at #(1/2,0)# with radius #r = 1/2#

The length of the upper semicircle is half the circumference.

#1/2C = 1/2(2pir) = pi*1/2 = pi/2#

Note
I also tried to do this using the integral, but it became too complicated for me to continue when there was a much cleaner solution available.