a large ferris wheel is 100 feet in diameter and rises 10 feet off the ground. each revolution of the wheel takes 30 seconds. A) express the vertical distance "H" of the seat off the ground as a function of time "t" if t=0 corresponds to a time when the seat is at the bottom. B) if the seat is rising, how fast is the distance h changing when h=55 feet?
Take a step back and see what's what. At angle
w=0, H = 10
w = 90 deg, H=10+50
w = 180, H = 10 + 100
Looks like H = 10 + 50(1 -cos(w))
But, angle w = t/30 * 2π = πt/15
That is, one revolution = 2π every 30 sec.
dH = -50sin(w) dw
dH = -50sin(w) * π/15 dt
when H = 55, it's just below the level of the axle, so w will be just under 90 deg
55 = 10 + 50(1 - cos(πt/15))
55 = 10 + 50 - 50cos(0.209t)
0.1 = cos(0.209t)
.209t = arccos(.1) = 1.47
t = 7.033 seconds.
That makes sense, since at t=30/4 = 7.5 the wheel has gone a quarter turn.
As above,
dH/dt = -50sin(0.209t) * .209
= -50sin(1.47) = -50*.995 = 49.74 ft/s
To express the vertical distance "H" of the seat off the ground as a function of time "t", we can use the equation of a circle.
A) The diameter of the ferris wheel is 100 feet, so the radius (r) is 100/2 = 50 feet. The lowest point of the wheel is 10 feet off the ground. When the seat is at its lowest point (t=0), the vertical distance of the seat off the ground (H) is equal to the sum of the radius and the initial height, so H = 50 + 10 = 60 feet.
The equation of a circle with center (h, k) is given by:
(x - h)^2 + (y - k)^2 = r^2
In this case, the center of the circle is (0, 0) and the radius is 50. Therefore, the equation of the circle is:
x^2 + y^2 = 50^2
Since the seat is moving vertically, we can substitute y with H - t, where t is the time elapsed.
x^2 + (H - t)^2 = 50^2
Simplifying the equation:
H^2 - 2Ht + t^2 + 2500 = 0
Therefore, the expression for the vertical distance H of the seat off the ground as a function of time t is:
H(t) = t - 0.02t^2 + 60
B) To find how fast the distance h is changing when h = 55 feet, we need to find the derivative of H(t) with respect to t, and then substitute h = 55 into the derivative.
Taking the derivative of H(t):
H'(t) = 1 - 0.04t
Substituting h = 55 into the derivative:
H'(t) = 1 - 0.04t = 1 - 0.04(55) = 1 - 2.2 = -1.2
Therefore, when h = 55 feet, the distance h is changing at a rate of -1.2 feet per second. Note that the negative sign indicates that the distance is decreasing.
A) To express the vertical distance "H" of the seat off the ground as a function of time "t," we can use the equation for the height of a point on a rotating wheel.
1. First, we need to find the equation of the wheel. The wheel can be considered as a circle with a radius of half the diameter, which is 50 feet. The equation for a circle with radius "r" centered at the origin is given by:
x^2 + y^2 = r^2
2. Since the center of the wheel is 10 feet above the ground, we can modify the equation to account for the vertical shift:
x^2 + (y - 10)^2 = r^2
3. To find the equation in terms of "t," we need to consider the position of the seat as it rotates. Let's assume that at time t=0, the seat is at the bottommost point. We can then express the angle in terms of time by using the angular velocity.
4. The angular velocity of the wheel is the number of revolutions per unit time. In this case, the wheel takes 30 seconds to complete one revolution. So, the angular velocity is:
ω = 2π/30 = π/15
5. As the wheel rotates counterclockwise, an angle θ and time "t" are related by:
θ = ωt = (π/15)t
6. We can express the x and y coordinates of the seat as functions of θ:
x = r * cos(θ)
y = r * sin(θ) + 10
7. Substitute the value of θ in terms of time:
x = 50 * cos((π/15)t)
y = 50 * sin((π/15)t) + 10
8. Finally, the vertical distance "H" of the seat off the ground is the y-coordinate, so:
H(t) = 50 * sin((π/15)t) + 10
B) To find how fast the distance "h" is changing when h = 55 feet, we need to find the derivative of the function H(t) with respect to time "t" and evaluate it when H(t) = 55.
1. Differentiate H(t) using the chain rule and product rule:
H'(t) = 50 * (π/15) * cos((π/15)t)
2. Substitute H(t) = 55 into the derivative equation:
55 = 50 * (π/15) * cos((π/15)t)
3. Solve for cos((π/15)t):
cos((π/15)t) = (11π)/15
4. Since h represents the distance above the ground, we need to subtract the initial height of 10 feet:
h = H(t) - 10 = 50 * sin((π/15)t)
5. Substitute cos((π/15)t) = (11π)/15 into the equation for h:
h = 50 * sin((π/15)t) = 50 * sin((π/15)t) - 10
6. Calculate the derivative of h with respect to t:
h'(t) = 50 * (π/15) * cos((π/15)t)
7. Substitute t in the derivative equation that satisfies cos((π/15)t) = (11π)/15:
h'(t) = 50 * (π/15) * (11π)/15
8. Simplify:
h'(t) = 110π^2/225
Therefore, when the distance h = 55 feet, the rate at which h is changing is approximately 110π^2/225.