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

1 Answer
Noah G
Feb 26, 2017

#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!

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