Question

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

Short answer

Answer: The rate of change of the viewing angle when the observer is 30 ft away from the screen is \(-\frac{1}{28}\) radians per second.

Step-by-step solution

Draw a diagram and label it with given values

We have a triangle formed by an vertical line with the heights 3 ft and 7 ft and a horizontal line with 30 ft. The angle between the horizontal line and the line connecting the bottom and top of the screen is \(\theta\). For better illustration, split the screen into two parts: the bottom part whose top end is at the observer's eye level (3 ft high) and the top part (7 ft high).

Set up a function using tangent of \(\theta\)

Observe that: \(\tan(\theta) = \frac{7 \mathrm{ft}}{x} - \frac{3 \mathrm{ft}}{x}\), where \(x\) is the horizontal distance from the observer to the screen.

Implicit differentiation with respect to time

Differentiate both sides with respect to time. Use the chain rule for tangent. \(\sec^2(\theta) \frac{d\theta}{dt} = \frac{3-7}{x^2} \frac{dx}{dt}\) We are given \(\frac{dx}{dt} = 3 \mathrm{ft}/\mathrm{s}\), and we want to find \(\frac{d\theta}{dt}\) when \(x = 30 \mathrm{ft}\).

Find the value of \(\sec^2(\theta)\)

When \(x = 30 \mathrm{ft}\), we have \(\tan(\theta) = \frac{7\,{\rm ft}/30{\rm \,ft} - 3\,{\rm ft}/30{\rm \,ft}}{1} = \frac{4}{30} = \frac{2}{15}\). Because \(\sec^2(\theta) = 1 + \tan^2(\theta)\), we can substitute the value of \(\tan(\theta)\) and compute \(\sec^2(\theta)\): \(\sec^2(\theta) = 1 + \left(\frac{2}{15}\right)^2 = 1 + \frac{4}{225} = \frac{229}{225}\)

Calculate the rate of change of \(\theta\)

Substituting \({\rm sec}^2(\theta)\), \(\frac{dx}{dt}\), and \(x\) values into the equation: \(\frac{d\theta}{dt} \frac{229}{225} = \frac{-4}{(30)^2}\cdot 3\). Now, solve for \(\frac{d\theta}{dt}\): \(\frac{d\theta}{dt} = \frac{-4\cdot 3}{30^2\cdot \frac{229}{225}} = \frac{-36}{30600/225}\) \(\frac{d\theta}{dt} = -\frac{36\cdot 225}{30600} = -\frac{1}{28}\) When the observer is 30 ft away from the screen, the viewing angle \(\theta\) is decreasing at a rate of \(-\frac{1}{28}\) radians per second.

Key concepts

Related Rates

The concept of related rates deals with the idea that we can determine the rate at which one quantity changes with respect to time by understanding its relationship to another quantity that also changes with respect to time.

In the context of our problem, we are looking at the rate of change of the viewing angle, \( \theta \), as you walk away from a screen. This is a classic related rates problem because the viewing angle's rate of change is directly related to the rate at which you are increasing your distance from the screen, \( \frac{dx}{dt} \).

To solve such problems, you first need to find a relation between the quantities (here, \( \theta \) and \( x \)) and then differentiate this relation with respect to time. This process usually involves chain rule because we are dealing with a composition of functions - the function that relates \( \theta \) and \( x \) and the function of \( x \) with respect to time.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where one variable is not explicitly solved for in terms of another. In related rates problems, we often need to use this method because the relationship between the variables and time is implicit.

In our example, we have the tangent function involving \( \theta \) and \( x \) implicitly defined. By differentiating both sides with respect to time, we can find \( \frac{d\theta}{dt} \) without having to solve for \( \theta \) explicitly in terms of \( x \) beforehand. We use the chain rule to differentiate the trigonometric function of \( \theta \) and apply the quotient rule or the properties of division to differentiate the expressions involving \( x \).

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are pivotal in understanding angles and their relationships with the sides of triangles. These functions are often used in calculus problems, particularly those involving right triangles and angles, like the viewing angle in our problem.

The tangent function, used in our example, relates the angle \( \theta \) to the ratio of the opposite side over the adjacent side. By using these trigonometric functions, we can express complex geometric relationships in algebraic terms, which we can then differentiate or integrate depending on the problem's requirements.

In related rates problems, we frequently encounter \( sec^2(\theta) \) because it comes from differentiating \( tan(\theta) \) as a result of the chain rule. Understanding the identities between these functions, such as \( sec^2(\theta) = 1 + tan^2(\theta) \) allows us to convert and manipulate our equations as needed to find the rates of change.

Calculus Applications

Calculus applications involve using the principles and techniques of calculus - differentiation and integration - to solve problems in real-life situations. This includes computing rates of change, optimizing functions, and analyzing the behavior of physical systems.

Our related rates problem is an application of differential calculus. Differential calculus allows us to study the rates at which quantities change, which is fundamental in physics, engineering, biology, economics, and more.

In the theater screen problem, we applied calculus to relate the geometric situation to a rate of change of \( \theta \) as one moves away from the screen. These applications showcase the incredible utility of calculus in interpreting and solving dynamic problems where variables depend on one another in complex ways.