A Ferris wheel with a radius of 13 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when the rider is 25 m above ground level?
Well, isn't that a phe-nominal question! To solve this, let's assume our rider is at the top of the Ferris wheel, at a height of 25 m. We know that the wheel has a radius of 13 m and completes one full revolution every 2 minutes. So, the time it takes to complete half a revolution (180 degrees) is 1 minute.
Now, imagine our rider is drawing an imaginary triangle with the center of the Ferris wheel, the rider's current position, and the rider's future position. The side opposite to the right angle of this triangle represents the height the rider is rising.
Since we have half a revolution in 1 minute, we can say that the rider weeeeee-reases in height by the wheel's diameter in that time. The diameter of the Ferris wheel is twice the radius, which is 26 m. Therefore, in 1 minute, our rider rises by 26 m, but we only need the rate at the particular moment when our rider is 25 m above the ground.
So, if every minute the rider rises by 26 m, we can set up a proportion:
26 m -------- 1 minute
25 m -------- x minutes
Solve this proportion and we'll know how fast our rider is rising at 25 m above the ground. I hope this Ferris wheel of knowledge gave you a spin of fun!
To solve this problem, we can use the concept of related rates. We need to find the rate at which the rider is rising.
Let's start by defining some variables:
- r: radius of the Ferris wheel (given) = 13 m
- h: height of the rider above the ground (given at a specific moment) = 25 m
- θ: angle formed between the horizontal line and the line connecting the rider with the center of the Ferris wheel (unknown)
- t: time (unknown)
To find the relationship between the height of the rider (h) and the angle (θ), we can use the formula for the circumference of a circle:
C = 2πr
At any given moment, the arc length covered by the rider (s) is equal to the angle (θ) measured in radians multiplied by the radius (r):
s = θr
Since there is a relationship between s and h, we can differentiate both sides of the equation with respect to time (t):
ds/dt = d(θr)/dt
The derivative of s with respect to t is the rate at which the arc length is changing (ds/dt), and d(θr)/dt is the rate at which the angle is changing (dθ/dt):
ds/dt = r(dθ/dt)
Now, let's find the value of ds/dt. We are given that the Ferris wheel completes one revolution every 2 minutes, which means that in 2 minutes, the angle θ will change by 2π radians:
dθ/dt = (2π radians) / (2 minutes) = π radians/minute
Substituting this value and the value of r into the equation, we get:
ds/dt = (13 m)(π radians/minute)
ds/dt = 13π m/minute
Therefore, the rate at which the rider is rising when the rider is 25 m above ground level is 13π m/minute.
To find how fast the rider is rising, we need to determine the derivative of the rider's height with respect to time, which will give us the velocity of the rider.
Let's denote the height of the rider above ground level as "h" and the time as "t".
Given:
- The radius of the Ferris wheel, r = 13 m
- The time for one revolution, T = 2 minutes
First, we need to determine the equation that relates the height of the rider to time.
The height of the rider, h, can be represented by a trigonometric function, such as sine or cosine. In the case of a Ferris wheel, it is often more convenient to use the cosine function since the height starts at its maximum when the angle starts at 0 degrees (or 0 radians).
So, we can express the height of the rider as:
h = r * cos(theta)
Next, we need to determine the angular velocity, which is the rate at which the angle changes with respect to time. Since the Ferris wheel completes one revolution every 2 minutes, the angular velocity, omega (ω), can be calculated as:
ω = 2π / T
Substituting the values, we have:
ω = 2π / 2 = π
Now, we can find the derivative of h with respect to time (dh/dt) to determine the velocity of the rider:
dh/dt = d(r*cos(theta))/dt
To differentiate cos(theta) with respect to time, we need to use the chain rule. The chain rule states that the derivative of a composition of functions is the derivative of the outer function multiplied by the derivative of the inner function.
Let's consider that theta is a function of time, i.e., theta = f(t).
Then, the derivative of cos(theta) with respect to time is:
d(cos(theta))/dt = d(cos(f(t)))/d(f(t)) * d(f(t))/dt
To find d(cos(f(t)))/d(f(t)), we differentiate cos(f(t)) with respect to f(t), treating f(t) as a constant:
d(cos(f(t)))/d(f(t)) = -sin(f(t))
To find d(f(t))/dt, we differentiate theta with respect to t:
d(f(t))/dt = d(theta)/dt
Now, let's substitute these values into the expression for dh/dt:
dh/dt = d(r*cos(theta))/dt = d(r*cos(f(t)))/dt = -r*sin(f(t)) * d(theta)/dt
Recall that ω = d(theta)/dt, so we can simplify the expression as:
dh/dt = -r*sin(theta) * ω
Finally, we can substitute the given values into the equation and solve for dh/dt when the rider is 25 m above the ground level.
Given:
- r = 13 m
- h = 25 m
- ω = π
dh/dt = -r*sin(theta) * ω = -13*sin(theta) * π
To find theta, we need to relate it to the height of the rider. Since the rider is 25 m above ground level, we can use the equation:
h = r * cos(theta)
Substituting the values of h and r, we have:
25 = 13 * cos(theta)
Now we can solve for theta:
cos(theta) = 25/13
theta ≈ 1.114 radians
Plugging in the value of theta into the expression for dh/dt:
dh/dt = -13 * sin(theta) * π
dh/dt = -13 * sin(1.114) * π
Calculating this expression will give you the velocity (rate of change of height) at that specific height.
Consider the axle to be at (0,13)
Then y=13+13sin(πt)
when y=25,
13sin(πt) = 12
sin πt = 12/13
Looks like the familiar 5-12-13 triangle, so cos(πt) = 5/13
dy/dt = 13πcos(πt) = 13π*5/13 = 5π m/min