The answer turned out to be 2pi2π units.
The arc length is essentially the usage of the distance formula with a small, independent change in x and a small change in y, where y changes according to f(x)f(x). Thus:
D = sqrt((Deltax)^2 + (Deltay)^2)
= sqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2*(Deltax)^2)
= sqrt(1 + (Deltay)^2/(Deltax)^2)Deltax
Make it a sum, and you've got the arc length over an interval [a,b].
s = sum_a^b sqrt(1 + (Deltay)^2/(Deltax)^2)Deltax
Turn it into an integral expression to get:
s = int_a^b sqrt(1 + ((dy)/(dx))^2)dx
So, we could take the derivative first to get:
(dy)/(dx) = 1/(2sqrt(4-x^2))*-2x = (-x)/(sqrt(4-x^2))
Then, let's square it and plug it in.
((dy)/(dx))^2 = x^2/(4-x^2)
-> s = int_a^b sqrt(1 + x^2/(4-x^2))dx
Cross-multiply:
= int_a^b sqrt((4-x^2 + x^2)/(4-x^2))dx
= int_a^b sqrt((4)/(4-x^2))dx
= 2int_a^b 1/sqrt(4-x^2)dx
And we've got ourselves a u-substitution, if you know that d/(dx)[arcsin(x)] = 1/sqrt(1-x^2).
= 2int_a^b 1/sqrt4*1/sqrt(1-x^2/4)dx
= int_a^b 1/sqrt(1-(x/2)^2)dx
Let u = x/2 and you get:
dx = 2du
= 2int_a^b 1/sqrt(1-u^2)du
= [2 arcsin(x/2)]|_(-2)^(2)
= 2 arcsin((2)/2) - 2 arcsin((-2)/2)
= 2 arcsin(1) - 2 arcsin(-1)
Let:
y = arcsin(1) -> siny = 1 -> y = pi/2
y = arcsin(-1) -> siny = -1 -> y = (3pi)/2
-> 2(pi/2) - 2((3pi)/2) -> 2(pi/2) - 2((-pi)/2)
(so that our answer is positive)
= pi + pi = color(blue)(2pi " u")
