Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

A Ferris wheel of radius 20 feet is rotating counterclockwise with an angular velocity of 1 radian per second. One seat on the rim is at \((20,0)\) at time \(t=0\). (a) What are its coordinates at \(t=\pi / 6 ?\) (b) How fast is it rising (vertically) at \(t=\pi / 6 ?\) (c) How fast is it rising when it is rising at the fastest rate?

Step-by-Step Solution

Answer

(a) \((10\sqrt{3}, 10)\); (b) \(10\sqrt{3}\) ft/s; (c) 20 ft/s.

Step 1: Understanding the Problem

We need to calculate the position and vertical velocity of a point on the Ferris wheel after being at initial position \((20,0)\) and rotating counterclockwise. We can represent this point's coordinates using trigonometric functions involving the angle from the x-axis.

Step 2: Calculating Coordinates at \(t=\pi/6\)

The angle \(\theta\) at time \(t\) is given by \(\theta = \omega t\), where \(\omega = 1\) rad/s is the angular velocity. So, at \(t = \pi/6\), \(\theta = \pi/6\). The x-coordinate is given by \(x = r \cos\theta = 20 \cos(\pi/6)\). The y-coordinate by \(y = r \sin\theta = 20 \sin(\pi/6)\). Calculating these gives us the coordinates.

Step 3: Plug in Values to Sine and Cosine

\(\cos(\pi/6) = \sqrt{3}/2\) and \(\sin(\pi/6) = 1/2\). Thus, the coordinates \(x, y\) at \(t = \pi/6\) are \(x = 20 \cdot \sqrt{3}/2 = 10\sqrt{3}\) and \(y = 20 \cdot 1/2 = 10\). So, the coordinates at \(t=\pi/6\) are \((10\sqrt{3}, 10)\).

Step 4: Angular Velocity and Vertical Speed Formula

The vertical (upward) speed can be derived from the derivative of \(y = r \sin(\theta)\) with respect to time \(t\). We have \(\frac{dy}{dt} = r \cos(\theta) \cdot \omega\), which is the component of velocity in the vertical direction.

Step 5: Calculating Vertical Speed at \(t=\pi/6\)

At \(t=\pi/6\), we substitute \(\theta = \pi/6\) into the expression for the vertical speed: \(\frac{dy}{dt} = 20 \cos(\pi/6) \cdot 1 = 20 \cdot \sqrt{3}/2 = 10\sqrt{3}\). Therefore, the vertical speed at \(t=\pi/6\) is \(10\sqrt{3}\) feet per second.

Step 6: Determining Maximum Rise Speed

The vertical speed is maximized when \(\cos(\theta)\) is maximized, which occurs at \(\theta = 0\) (as the wheel begins). Thus, the maximum speed is when \(\cos(0) = 1\). Substituting into the formula, the maximum rise speed is \(r \cdot \omega = 20 \cdot 1 = 20\).

Step 7: Summarizing Results

Using our calculations, the coordinates at \(t=\pi/6\) are \((10\sqrt{3}, 10)\), the vertical speed at \(t=\pi/6\) is \(10\sqrt{3}\) feet per second, and the maximum vertical speed is 20 feet per second.

Key Concepts

Angular Velocity

Angular velocity is a fundamental concept in rotational motion, representing how fast an object rotates around a central point. It is often denoted by the Greek letter \( \omega \) (omega) and is measured in radians per second. Consider a Ferris wheel turning at a steady pace, like in our exercise problem. Here, the angular velocity is consistently 1 radian per second.

This means that every second, the Ferris wheel rotates by one radian. Since there are \( 2\pi \) radians in a full circle, this ensures the Ferris wheel completes one full rotation in \( 2\pi \) seconds.

Understanding angular velocity helps us determine the position of any point on the Ferris wheel at any given time, simply by multiplying the angular velocity \( \omega \) by time \( t \). This gives us the angle \( \theta = \omega t \), which can then be used with trigonometric functions to find the coordinates of the point on the Ferris wheel based on its radius.

Vertical Speed

Vertical speed refers to how fast something moves up or down. In the context of a Ferris wheel, it tells us how quickly a seat ascends or descends as the wheel turns. The vertical speed can be calculated using derivatives in calculus, specifically finding the rate of change of the height \( y \) of a point on the wheel as the wheel rotates.

In our example, the vertical speed is derived from the sine component of the Ferris wheel's position. This is because the height \( y \) at any point \( \theta \) is given by \( y = r\sin(\theta) \). To find how fast \( y \) is changing, we take the derivative with respect to time \( t \). This results in the formula \( \frac{dy}{dt} = r \cos(\theta) \cdot \omega \), where \( \omega \) is the angular velocity.

On the Ferris wheel, this provides a clear picture of how quickly a seat rises at any moment. For instance, at \( t = \pi/6 \), with \( \theta = \pi/6 \), computing the vertical speed helps us understand the dynamics of that motion.

Ferris Wheel Analysis

Analyzing a Ferris wheel's motion helps us apply trigonometric functions in a practical context. It combines aspects of geometry, rotational motion, and calculus to calculate specific movements, such as positions and speeds of seats on the wheel.

The analysis begins by understanding that each seat moves along a circular path. At time \( t = 0 \), a seat is positioned at \((20,0)\), meaning it is 20 feet from the center horizontally. Over time, using angular velocity, we calculate the changing angle \( \theta \) to determine the seat's new position via trigonometric functions:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

In our problem, the coordinates at \( t=\pi/6 \) were found to illustrate where the seat is on the wheel. We also delved into calculating how fast the seat moves upwards (vertical speed) at different points. This involved understanding the derivative \( \frac{dy}{dt} = r \cos(\theta) \cdot \omega \).

Analyzing these dynamics aids in predicting and understanding the real-life motion of Ferris wheels and similar rotating objects.