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

Question

Charles Anctil · Accepted Answer

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

#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.

Christopher Agresta · Answer

undefined To find the length of the curve $ y = \sqrt{x - x^2} $, you can use the arc length formula. The arc length formula for a curve given by $ y = f(x) $ from $ x = a $ to $ x = b $ is:

$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

For the given curve $ y = \sqrt{x - x^2} $, differentiate it to find $ \frac{dy}{dx} $:
$$ \frac{dy}{dx} = \frac{1 - 2x}{2\sqrt{x - x^2}} $$

Now, square this derivative and add 1 to it:
$$ \left(\frac{dy}{dx}\right)^2 + 1 = \frac{(1 - 2x)^2}{4(x - x^2)} + 1 $$

Integrate this expression from the lower limit of $ x $ to the upper limit to find the arc length. Since the function is symmetric about the midpoint of its domain, we can integrate from $ x = 0 $ to $ x = \frac{1}{2} $, and then double the result to find the total length.

$$ L = 2\int_{0}^{\frac{1}{2}} \sqrt{\frac{(1 - 2x)^2}{4(x - x^2)} + 1} \, dx $$

Evaluate this integral to find the length of the curve.