Question

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

Answer 1

`2` units.

Explanation:

The arc length of a continuous curve from `a` to `b` is given by `int_a^b sqrt(1+ (dy/dx)^2)`. Let's start by computing the derivative.

`y' = (1 - 2x)/(2sqrt(x - x^2)) + 1/(2sqrt(x - x^2)`

`y' = (1 - 2x + 1)/(2sqrt(x- x^2))`

`y' = (2 - 2x)/(2sqrt(x - x^2)`

`y' = (2(1 - x))/(2sqrt(x - x^2)`

`y' = (1 - x)/sqrt(x(1 - x))`

Now let's find the endpoints of the function `y`. The function `y = arcsinx` has domain `{x|-1 ≤ x ≤ 1, x in RR}`. However, since the value under the square root has to be positive, `y = arcsinsqrt(x)` has domain `{x| 0 ≤ x ≤ 1, x in RR}`.

The second part of the function, `y = sqrt(x - x^2)`, has the same domain as `y = arcsinsqrt(x)`. So, we can conclude our bounds of integration will be from `0` to `1`. Call the arc length `A`.

`A = int_0^1 sqrt(1 + ((1 - x)/sqrt(x(1 - x)))^2)dx`

`A = int_0^1 sqrt(1 + (1 - x)^2/(x(1 - x)))dx`

`A = int_0^1 sqrt(1 + (1 - x)/x) dx`

`A = int_0^1 sqrt(1 + 1/x - x/x)dx`

`A = int_0^1 sqrt(1 + 1/x - 1)dx`

`A = int_0^1 sqrt(x^-1)`

`A = int_0^1 (x^-1)^(1/2)`

`A = int_0^1 x^(-1/2)`

`A = [2x^(1/2)]_0^1`

`A = 2(1)^(1/2) - 2(0)^(1/2)`

`A = 2`

Hopefully this helps!