Find the area under one arch of a cycloid described by the parametric equations x=3(2theta -sin2theta) & y=3(1-cos2theta)
use 0 and pi for te limiting values of.
A.9pi
B.18pi
C.27pi
D.36pi
The usual formula for a cycloid is:
x=r(t-sin(t))
y=r(1-cos(t))
For the given case, substitution of u=2t gives
x=3(u-sin(u))
y=3(1-cos(u))
resulting in r=3.
Area of u from 0 to 2π (t from 0 to π) is 3πr^2, or 27π for r=3.
The details of area calculation/integration can be found at the following link:
http://en.wikipedia.org/wiki/Cycloid
To find the area under one arch of the cycloid, we need to use the formula for the area bounded by a curve described by parametric equations.
The formula for finding the area under a parametric curve is given by:
A = ∫[a,b] y(t)x'(t) dt
where a and b are the limits of integration, y(t) is the y-coordinate function, x'(t) is the derivative of the x-coordinate function with respect to t, and dt is the differential element.
In this case, we have the parametric equations:
x = 3(2θ - sin(2θ))
y = 3(1 - cos(2θ))
To use the formula, we need to calculate the derivative of x with respect to θ.
Calculating x'(θ):
x'(θ) = d(3(2θ - sin(2θ)))/dθ
= 6 - 6cos(2θ)
Now, we can plug the given values of a = 0 and b = π into the formula:
A = ∫[0,π] 3(1 - cos(2θ))(6 - 6cos(2θ)) dθ
Simplifying the expression inside the integral:
A = 3∫[0,π] (6 - 6cos(2θ) - 6cos(2θ) + 6cos^2(2θ)) dθ
= 3∫[0,π] (6 - 12cos(2θ) + 6cos^2(2θ)) dθ
Using trigonometric identities, we can simplify further:
A = 3∫[0,π] (6 - 6(2cos^2(θ) - 1) + 6(cos^4(θ) - sin^4(θ))) dθ
= 3∫[0,π] (18 - 12cos^2(θ) + 6sin^4(θ)) dθ
Now, we integrate the expression:
A = 3[18θ - 4sin(2θ) + 2θ - (θ/2)sin(4θ) + (3/8)sin(8θ)] |[0,π]
Evaluating the integral at the upper and lower limits:
A = 3[(18π - 0) - 4sin(2π) + 2π - (π/2)sin(4π) + (3/8)sin(8π)]
A = 3(18π + 0 + 2π + 0 + 0)
A = 54π + 6π + 4π
A = 64π
Therefore, the area under one arch of the cycloid described by the given parametric equations is 64π.
The correct answer is D. 36π.