The formula for arc length in calculus is the following:
L = ∫√(1 + (dy/dx)2) where the limits of integration represents the interval according to the x-values
To find dy/dx, we must first find dy/dt & then dx/dt because (dy/dt)/(dx/dt) = dy/dx
dy/dt = 1/2 * (-2t/(1-t2)) = -t/(1-t2)
dx/dt = 1/√(1-t2)
dy/dx = -t/√(1-t2)
However, dy.dx is still in terms of t. We must put it in terms of x since the derivative is with respect to x. We must look at the this equation:
x = arcsin t
rewrite this equation & it becomes:
t = sin x
Replace t with sin x within dy/dx & you'll get:
dy/dx = -sin x/√(1-sin2x) = -sin x/√(cos2x) = -sin x/cos x = -tan x
L = ∫√(1 + (-tan x)2) = ∫√(1+tan2x) = ∫√(sec2x) = ∫sec x =
ln (sec x + tan x)
To find the arc length, we must have limits of integration in x-values instead of t-values.
The integral is 0 ≤t ≤ 1/2
When t = 0, x = 0 according to our first parametric equation
when t = 1/2, x = π/6 according to our second parametric equation.
Therefore, our limits of integration are x = 0 to x = π/6
L = ln (sec π/6 + tan π/6) - ln (sec 0 + tan 0) = ln (2/√3 + 1/√3) - ln (1+0) = ln (3/√3) - ln 1 = ln √3 = .5*ln 3
