Question

Watching a Ferris wheel An observer stands \(20 \mathrm{m}\) from the bottom of a Ferris wheel on a line that is perpendicular to the face of the wheel, with her eyes at the level of the bottom of the wheel. The wheel revolves at a rate of \(\pi\) rad/min, and the observer's line of sight to a specific seat on the Ferris wheel makes an angle \(\theta\) with the horizontal (see figure). At what time during a full revolution is \(\theta\) changing most rapidly?

Short answer

Answer: To find the time during a full revolution when the angle 𝜃 is changing most rapidly, we need to analyze the critical points of the second derivative of 𝜃 with respect to time, \(\frac{d^2\theta}{dt^2}\), and determine when the maximum rate of change occurs.

Step-by-step solution

Analyzing the given information

We know that the observer's distance to the bottom of the Ferris wheel is 20m, and the rotation rate is \(\pi\) rad/min. Let the radius of the Ferris wheel be \(r\) meters. Let's consider a point on the Ferris wheel which represents the seat that the observer is watching, and let's say it is at point P(x,y) coordinates. We can start by modeling the rotation of the Ferris wheel, and then find the coordinates of the seat as it rotates with respect to the angle of rotation \(\omega\). Using the equation of a circle: (radius*cosine(angle), radius*sine(angle)). As the Ferris wheel rotates clockwise, the angle of rotation \(\omega= -\pi t\), where t represents time in minutes. Hence, the coordinates of point P are given by \((x, y) = (r \cos(-\pi t), r \sin(-\pi t))\). Now, let's find the relationship between the angle 𝜃 and the angle of rotation \(\omega\).

Finding the relationship between angles

The observer is standing at a distance of 20m, so the horizontal distance between the observer and point P is \(20 + r\cos(-\pi t)\). Since the observer's eyes are at the level of the bottom of the wheel, the vertical distance from point P to the ground is \(r\sin(-\pi t)\). Using the tangent function, we can relate the angle 𝜃 with the distances: \(\tan(\theta) = \frac{r \sin(-\pi t)}{20 + r\cos(-\pi t)}\)

Finding the rate of change of the angle 𝜃

Now we need to differentiate both sides of the equation in Step 2 with respect to time t: \(\frac{d}{dt}\tan(\theta) = \frac{d}{dt} \left(\frac{r \sin(-\pi t)}{20 + r\cos(-\pi t)}\right)\) Using the quotient rule, we have: \(\frac{d}{dt}\tan(\theta) = \frac{(r\cos(-\pi t) \cdot (-\pi)\sin(-\pi t)) - (r\sin(-\pi t) \cdot (-\pi)\cos(-\pi t))}{(20 + r\cos(-\pi t))^2}\) Now, since \(\frac{d}{dt}\tan(\theta) = \sec^2(\theta) \cdot \frac{d\theta}{dt}\), we need to find the value of \(\sec^2(\theta)\) to isolate the rate of change of angle 𝜃, \(\frac{d\theta}{dt}\). \(\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left(\frac{r\sin(-\pi t)}{20 + r\cos(-\pi t)}\right)^2\) We will now substitute the value of sec^2(𝜃) in the equation for the rate of change of the angle 𝜃: \(\frac{d\theta}{dt} = \frac{\frac{(r\cos(-\pi t) \cdot (-\pi)\sin(-\pi t)) - (r\sin(-\pi t) \cdot (-\pi)\cos(-\pi t))}{(20 + r\cos(-\pi t))^2}}{1 + \left(\frac{r\sin(-\pi t)}{20 + r\cos(-\pi t)}\right)^2}\)

Finding the maximum rate of change

To find the maximum rate of change, we will differentiate \(\frac{d\theta}{dt}\) with respect to time t: \(\frac{d^2\theta}{dt^2} = \frac{d}{dt}\left(\frac{d\theta}{dt}\right)\) We will find the critical points by setting \(\frac{d^2\theta}{dt^2} = 0\) and analyzing their values. The critical points will give us the maximum, minimum, or inflection points of the rate of change of the angle 𝜃. To find out whether we have a maximum rate, we will use the second derivative test: If \(\frac{d^2\theta}{dt^2} > 0\) at a critical point, it is a relative minimum. If \(\frac{d^2\theta}{dt^2} < 0\) at a critical point, it is a relative maximum. If \(\frac{d^2\theta}{dt^2} = 0\) at a critical point, the test is inconclusive. Once we find the time t at which we have the maximum rate of change, we can conclude at what time during a full revolution, \(\theta\) is changing most rapidly.

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how things change. It allows us to calculate the rate of change of one quantity concerning another.
In the context of this problem, we want to know how fast the angle \( \theta \) is changing as the Ferris wheel rotates. By differentiating the expression for \( \tan(\theta) \) with respect to time \( t \), we can find \( \frac{d\theta}{dt} \), the rate at which the angle \( \theta \) changes.

This involves using the chain rule and the quotient rule:

• The chain rule helps us differentiate composite functions. If a function is inside another, this rule is handy.
• The quotient rule helps when dividing two functions. It tells us how to differentiate a ratio.

Differentiation gives us the ability to pinpoint exactly how the angle \( \theta \) changes over time, which is crucial for understanding behavior in rotational motions and other applied contexts.

Trigonometry

Trigonometry deals with the relationships between the sides and angles of triangles. It's particularly useful in this problem where angles and rotations are involved.
When the observer views the Ferris wheel, the position of the seat relative to her forms a right triangle:

• The observer is \( 20m \) away horizontally.
• The vertical position changes as the wheel turns.

By using the tangent function, \( \tan(\theta) = \frac{r \sin(-\pi t)}{20 + r\cos(-\pi t)} \), we relate the angle \( \theta \) with the continuous rotating position of a point on the Ferris wheel.

This relationship is essential because it helps convert between linear movement (like how far the seat has moved) and angular movement (like how much the observer's angle of sight has changed). Understanding these trigonometric relationships allows us to effectively model and solve real-world problems involving rotations and angles.

Critical Points

In mathematics, critical points are where the derivative equals zero or does not exist. They are essential in understanding where a function's rates of change may reach a maximum or minimum.
For this exercise, we differentiated \( \frac{d\theta}{dt} \) to analyze how \( \theta \) changes most rapidly. By setting \( \frac{d^2\theta}{dt^2} = 0 \), we find these critical points.

• A positive second derivative at these points means a local minimum for rate of change.
• A negative second derivative indicates a maximum rate of change.

The critical points are especially important here because they tell us when during the Ferris wheel's rotation the observer sees the fastest change in angle. This helps in comprehending the overall dynamics of the system and provides insights into optimizing viewing experiences or mechanical functions in rotating structures.