To find d²y/dx², we need to differentiate the equation for y with respect to x twice.
Given that x = a (cos t + log tan t/2) and y = a sin t, we can express y in terms of x and t by substituting the value of x into the equation for y:
y = a sin t
= a sin (arccos (x/a - log tan t/2))
To differentiate y with respect to x, we need to use the chain rule.
Let's start by finding dy/dt:
dy/dt = a cos (arccos (x/a - log tan t/2)) * (d/dt (arccos (x/a - log tan t/2)))
= a cos (arccos (x/a - log tan t/2)) * (d/dt (x/a - log tan t/2))
= a cos (arccos (x/a - log tan t/2)) * (1/a - (1/2) * sec²(t/2))
Now, let's differentiate dy/dt with respect to x using the chain rule:
(d²y/dx²) = (dy/dt) / (dx/dt)
= (dy/dt) / (d/dt (a (cos t + log tan t/2)))
To find dx/dt, let's differentiate x with respect to t:
dx/dt = d/dt (a (cos t + log tan t/2))
= -a sin t + (1/tan t/2) * (1/(cos² t/2)) * (1/2)
= -a sin t + (1/tan t/2) * (2/(cos t + 1)) * (1/2)
= -a sin t + (1/tan t/2) * (1/(cos t + 1))
= -a sin t + (1/tan t/2) * (1/(cos t + 1))
Now, let's substitute dy/dt and dx/dt into the expression for d²y/dx²:
(d²y/dx²) = (dy/dt) / (dx/dt)
= (a cos (arccos (x/a - log tan t/2)) * (1/a - (1/2) * sec²(t/2))) / (-a sin t + (1/tan t/2) * (1/(cos t + 1)))
To find d²y/dx² at t = π/3, substitute t = π/3 into the expression:
(d²y/dx²) = (a cos (arccos (x/a - log tan (π/3)/2)) * (1/a - (1/2) * sec²(π/6/2))) / (-a sin (π/3) + (1/tan (π/3)/2) * (1/(cos (π/3) + 1)))
Simplifying this expression will give you the value of d²y/dx² at t = π/3.