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

Apr 26, 2016

Use algebra to get a length of $\frac{\pi}{2}$

#### Explanation:

$y = \sqrt{x - {x}^{2}}$ is equivalent to

${y}^{2} = x - {x}^{2}$ with the restriction $y \le 0$.

This is equivalent to ($y \ge 0$ on)

${x}^{2} - x + {y}^{2} = 0$.

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

${x}^{2} - x + \frac{1}{4} + {y}^{2} = \frac{1}{4}$, or, better yet

${\left(x - \frac{1}{2}\right)}^{2} + {y}^{2} = \frac{1}{4}$.

With $y \ge 0$, this is the upper semicircle centered at $\left(\frac{1}{2} , 0\right)$ with radius $r = \frac{1}{2}$

The length of the upper semicircle is half the circumference.

$\frac{1}{2} C = \frac{1}{2} \left(2 \pi r\right) = \pi \cdot \frac{1}{2} = \frac{\pi}{2}$

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.