We need to model your position on the Ferris wheel, which involves periodic motion. Here's the breakdown:

Key Information:

• The diameter of the Ferris wheel is 20 meters, so the radius is half of that:
  $r = 10$ meters.
• You are 15 meters high, meaning you are 15 meters above the ground.
• The Ferris wheel completes one full rotation every 3 minutes.
• You boarded at the 6:00 position, which is on the ground.

Step 1: Modeling the Motion

Since the Ferris wheel makes a full rotation every 3 minutes, we can model the height $h(t)$ of your position as a function of time using the sine function. The general form of the vertical position on a Ferris wheel is:

$h(t) = r \cdot \sin(\omega t + \phi) + C$

where:

• $r = 10$ meters (radius),
• $\omega = \frac{2\pi}{3}$ radians per minute (angular velocity, since it takes 3 minutes for one full rotation),
• $C = 10$ meters (the center of the wheel is 10 meters above the ground),
• $\phi$ is the phase shift, which we will determine,
• $h(t)$ is the height at time $t$.

Step 2: Find the Height Function

At $t = 0$ (the moment you board), your height is at ground level, so:

$h(0) = 10 \cdot \sin(0 + \phi) + 10 = 0$

Solving for $\phi$, we get:

$\sin(\phi) = -1 \quad \Rightarrow \phi = -\frac{\pi}{2}$

Thus, the height function becomes:

$h(t) = 10 \cdot \sin\left( \frac{2\pi}{3} t - \frac{\pi}{2} \right) + 10$

Step 3: Solve for Time

You are 15 meters high at some point on the Ferris wheel. Therefore, we set $h(t) = 15$:

$15 = 10 \cdot \sin\left( \frac{2\pi}{3} t - \frac{\pi}{2} \right) + 10$

Simplifying:

$5 = 10 \cdot \sin\left( \frac{2\pi}{3} t - \frac{\pi}{2} \right)$

$\frac{1}{2} = \sin\left( \frac{2\pi}{3} t - \frac{\pi}{2} \right)$

The sine of $\frac{1}{2}$ occurs at $\frac{\pi}{6}$ and $\frac{5\pi}{6}$. Thus, we have two possible solutions for $\frac{2\pi}{3} t - \frac{\pi}{2}$:

1. $\frac{2\pi}{3} t - \frac{\pi}{2} = \frac{\pi}{6}$
2. $\frac{2\pi}{3} t - \frac{\pi}{2} = \frac{5\pi}{6}$

Step 4: Solve for $t$

Case 1: $\frac{2\pi}{3} t - \frac{\pi}{2} = \frac{\pi}{6}$

$\frac{2\pi}{3} t = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}$

$t = \frac{2\pi}{3} \times \frac{3}{2\pi} = 1 \text{ minute}$

Case 2: $\frac{2\pi}{3} t - \frac{\pi}{2} = \frac{5\pi}{6}$

$\frac{2\pi}{3} t = \frac{5\pi}{6} + \frac{\pi}{2} = \frac{8\pi}{6} = \frac{4\pi}{3}$

$t = \frac{4\pi}{3} \times \frac{3}{2\pi} = 2 \text{ minutes}$

Step 5: Conclusion

You could have been on the Ferris wheel for either 1 minute or 2 minutes. Both are valid solutions for the given height of 15 meters.

Would you like more details or have any further questions about the solution?

Additional Questions:

1. How would the position of a person on the Ferris wheel change if the wheel rotated faster or slower?
2. What would the height function look like if you started at the highest point (12:00 position)?
3. How do the phase shifts in the equation relate to the starting position of the person on the Ferris wheel?
4. What other heights can you calculate for specific times during the rotation?
5. How could this problem change if the Ferris wheel were not perfectly circular?

Tip:

In problems involving periodic motion, using sine or cosine functions helps model the motion based on known characteristics like amplitude (radius) and frequency (how fast the rotation is).