Answer:
You made it hard to answer so heres my answer :
A:
This equation holds when t = pi/4 or t = 5pi/4. These are the values of t where the speed is at a maximum.
B:
Since the cycloid has two arches, the total length of the cycloid is 2*pi = 8. This shows that one arch of the cycloid has length 8.
Explanation:
A:
a. To find the value of t where the speed is at a maximum, we can take the derivative of the parametrization r(t) with respect to t and set it equal to 0:
r'(t) = <1 - cos t, sin t>
The speed is given by the magnitude of the velocity vector, which is the square root of the sum of the squares of the components:
|r'(t)| = sqrt((1 - cos t)^2 + sin^2 t)
To find the value of t where the speed is at a maximum, we can take the derivative of this expression with respect to t and set it equal to 0:
(|r'(t)|^2)' = 0
((1 - cos t)^2 + sin^2 t)' = 0
2(1 - cos t)(-sin t) + 2sin t(cos t) = 0
2sin t - 2cos t = 0
sin t = cos t
B:
b. To show that one arch of the cycloid has length 8, we can use the parametrization to find the length of one arch by integrating the length of the velocity vector over the interval [0, 2*pi]:
L = integral from t=0 to t=2pi of |r'(t)| dt
= integral from t=0 to t=2pi of sqrt((1 - cos t)^2 + sin^2 t) dt
We can evaluate this integral using the substitution u = 1 - cos t, which gives us du = sin t dt. This gives us:
L = integral from u=1 to u=0 of sqrt(u^2 + (1 - u)^2) du
= integral from u=1 to u=0 of sqrt(2u^2 - 2u + 1) du
= integral from u=1 to u=0 of sqrt(u^2 - u + 1/2) du
We can now complete the square to rewrite the expression inside the square root in vertex form:
L = integral from u=1 to u=0 of sqrt((u - 1/4)^2 + 1/4) du
This integral is an arc length integral, and the expression inside the square root is the equation of a circle with radius 1/2 and center (1/4, 0). Thus, the integral is simply the circumference of this circle, which is 2pi(1/2) = pi.