Answer: (a) Let the circle roll along the x-axis with the center at the origin. Then, the parametric equations of the cycloid are given by:
x = t - sin(t)
y = 1 - cos(t)
where t is the angle in radians that the circle has rotated from its initial position.
(b) The arclength of one period of the cycloid is given by the integral:
L = ∫(0,2π) √[dx/dt]^2 + [dy/dt]^2 dt
Substituting the parametric equations from part (a), we get:
dx/dt = 1 - cos(t)
dy/dt = sin(t)
Therefore,
[dx/dt]^2 + [dy/dt]^2 = 2 - 2cos(t)
Substituting back into the arclength integral, we get:
L = ∫(0,2π) √(2 - 2cos(t)) dt
This integral can be evaluated using the half-angle formula:
cos(t) = 1 - 2sin^2(t/2)
Substituting this into the integral and simplifying, we get:
L = 8∫(0,π/2) √(sin^3(t/2)) dt
This integral can be evaluated using a substitution u = cos(t/2), du = -sin(t/2)dt:
L = 16∫(0,1) √[(1-u^2)^3] du
This integral can be further simplified by making the substitution u = sin(θ):
L = 16∫(0,π/2) cos^4(θ) dθ
This integral can be evaluated using integration by parts and trigonometric identities. After some algebraic manipulations, we get:
L = 8π
Explanation: