Find the exact length of the curve. y= (x-x^2)^1/2+ sin^-1(x)^1/2   I get confused after I try to square y'. I would appreciate if all steps are clearly shown.

Expert Answers
sciencesolve eNotes educator| Certified Educator

You should evaluate the length of the curve using the following formula such that:

`int_a^b sqrt(1 + ((dy)/(dx))^2)dx `

You need to differentiate the given function with respect to x, using the chain rule such that:

`(dy)/(dx) = (1-2x)/(2sqrt(x - x^2)) + 1/(sqrt(1 - (sqrt x)^2))*(1/(2sqrt x))`

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

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

`(dy)/(dx) = (2-2x)/(2sqrt(x - x^2))`

Factoring out 2 yields:

`(dy)/(dx) = (2(1-x))/(2sqrtx*sqrt(1 - x))`

`(dy)/(dx) = (sqrt(1-x))/(sqrt x) => (dy)/(dx) = sqrt((1-x)/x)`

Raising to square yields:

`((dy)/(dx))^2 = ((1-x)/x)`

`int_a^b sqrt(1 + (1-x)/x)dx => int_a^b sqrt((x+1-x)/x)`

`int_a^b sqrt(1/x) dx = int_a^b (1/x)^(1/2)dx`

`int_a^b (x^(-1))^(1/2)dx = x^(-1/2+1)/(-1/2+1)|_a^b`

`int_a^b (x^(-1))^(1/2)dx = x^(1/2)/(1/2)|_a^b`

`int_a^b (x^(-1))^(1/2)dx = 2sqrt x|_a^b`

`int_a^b (x^(-1))^(1/2)dx = 2(sqrt b - sqrt a)`

Hence, evaluating the length of the given curve over the interval [a,b] yields `int_a^b sqrt(1 + (1-x)/x)dx = 2(sqrt b - sqrt a).`