A Ferris wheel with a radius of 5 m is rotating at a rate of one revolution every two minutes. Exactly how fast (in m/min) is a rider rising when his seat is 8 m above ground level?


To solve this, we need to calculate the rate at which the rider is rising when their seat is 8 meters above ground level.

### Given:
- Radius of the Ferris wheel \( r = 5 \) meters.
- Rate of rotation: 1 revolution every 2 minutes, so angular velocity \( \omega = \frac{2\pi}{2} = \pi \) radians per minute.
- The center of the Ferris wheel is 5 meters above the ground because the radius is 5 meters.
- The rider is 8 meters above the ground.

### Step 1: Equation of height as a function of time

The vertical position \( y(t) \) of the rider can be described by the equation:
\[
y(t) = 5 + 5\sin(\theta(t))
\]
where \( \theta(t) \) is the angle in radians and \( \theta(t) = \pi t \) (since \( \omega = \pi \) rad/min).

Thus, the height as a function of time is:
\[
y(t) = 5 + 5\sin(\pi t)
\]

### Step 2: Find the derivative of the height

To find the rate of change of height, we differentiate \( y(t) \) with respect to time:
\[
\frac{dy}{dt} = 5\cos(\pi t) \cdot \frac{d}{dt}(\pi t) = 5\pi \cos(\pi t)
\]
So the rate of change of height is:
\[
\frac{dy}{dt} = 5\pi \cos(\pi t)
\]

### Step 3: Find when the rider's height is 8 meters

We know that the rider's height is given by \( y(t) = 5 + 5\sin(\pi t) \), and we want to find when the rider is 8 meters above the ground:
\[
8 = 5 + 5\sin(\pi t)
\]
\[
\sin(\pi t) = \frac{3}{5}
\]
\[
\pi t = \arcsin\left(\frac{3}{5}\right)
\]
Using a calculator:
\[
\pi t \approx 0.6435 \text{ radians}
\]
\[
t \approx 0.2047 \text{ minutes}
\]

### Step 4: Calculate the rate of rising at this time

Substitute \( \pi t = 0.6435 \) into \( \frac{dy}{dt} = 5\pi \cos(\pi t) \):
\[
\frac{dy}{dt} = 5\pi \cos(0.6435)
\]
Using a calculator to find \( \cos(0.6435) \approx 0.8 \):
\[
\frac{dy}{dt} = 5\pi \times 0.8 \approx 12.566 \text{ m/min}
\]

Thus, the rider is rising at approximately **12.57 m/min** when they are 8 meters above the ground.

Would you like more details or have any questions?

### Related Questions:
1. How does angular velocity affect the speed of a rider at different heights?
2. What would be the rate of rising if the Ferris wheel's speed were doubled?
3. How does the Ferris wheel’s radius influence the maximum vertical speed of a rider?
4. What is the maximum possible rate at which the rider can rise on this Ferris wheel?
5. How would the calculation change if the ground level was defined differently?

**Tip:** When dealing with circular motion, always think about angular velocity and how it relates to linear velocity at different points in the cycle.

A Ferris wheel with a radius of 5 m is rotating at a rate of one revolution every two minutes. Exactly how fast (in m/min) is a rider rising when his seat is 8 m above ground level?

Calculate how fast a rider is rising on a Ferris wheel with a 5 m radius when the seat is 8 m above ground level. This problem involves trigonometry, circular motion, and related rates, suitable for students in Grades 10-12.

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