Problem 299 - Calculus Volume 1

Question

A local movie theater with a 30 -foot-high screen that is 10 feet above a person's eye level when seated has a viewing angle \(\theta\) (in radians) given by \(\theta=\cot ^{-1} \frac{x}{40}-\cot ^{-1} \frac{x}{10}, \quad\) where \(x\) is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure.

Step-by-Step Solution

Answer

The viewing angle \(\theta = \tan^{-1}\left(\frac{120x}{400 + x^2}\right)\).

Step 1: Understand the Problem

The problem involves a movie theater screen and a person's viewing angle from their seat. The theater screen is 30 feet high and starts 10 feet above the person's eye level. We need to understand how the viewing angle \(\theta\) changes as the distance \(x\) changes.

Step 2: Identify the Function Components

The viewing angle \(\theta\) is given by the formula \(\theta = \cot^{-1} \frac{x}{40} - \cot^{-1} \frac{x}{10}\). This involves inverse cotangent functions, which will need simplification.

Step 3: Recall Trigonometric Identities

Recall that \(\cot^{-1}(A) - \cot^{-1}(B) = \tan^{-1}\left(\frac{B-A}{1+AB}\right)\). This identity will help simplify the expression for \(\theta\).

Step 4: Apply the Trigonometric Identity

Substitute \(A = \frac{x}{40}\) and \(B = \frac{x}{10}\) into the identity: \(\theta = \tan^{-1}\left(\frac{\frac{x}{10} - \frac{x}{40}}{1 + \frac{x}{40} \cdot \frac{x}{10}}\right)\). Simplify the numerator and denominator.

Step 5: Simplify the Expressions

The numerator becomes \(\frac{x}{10} - \frac{x}{40} = \frac{4x}{40} - \frac{x}{40} = \frac{3x}{40}\). The denominator is \(1 + \frac{x^2}{400} = \frac{400 + x^2}{400}\).

Step 6: Simplify \(\theta\) Further

Substitute simplified expressions into the tangent inverse formula: \[ \theta = \tan^{-1}\left(\frac{\frac{3x}{40}}{\frac{400 + x^2}{400}}\right) = \tan^{-1}\left(\frac{1200x}{40(400 + x^2)}\right) \]. Further simplify: \(\theta = \tan^{-1}\left(\frac{3x}{400 + x^2}\right)\).

Step 7: Conclude the Simplification

The simplified expression for the viewing angle \(\theta\) is now \(\theta = \tan^{-1}\left(\frac{120x}{400 + x^2}\right)\), capturing how the viewing angle changes as the distance from the screen changes.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the standard trigonometric functions. They are often used to find angles when the ratios of a right-angled triangle are known. In this case, we're working with the inverse cotangent function, denoted as \(\cot^{-1}(x)\). The cotangent function, \(\cot(\theta)\), is the reciprocal of the tangent function, therefore \(\cot(\theta) = \frac{1}{\tan(\theta)}\).
Inverse functions specify the angle which corresponds to a given ratio.
This means that if \(\cot(\theta) = x\), then \(\theta = \cot^{-1}(x)\). While challenging to directly visualize inverse cotangent, simplifying it using known identities and transformations can make problem-solving easier.
In the provided exercise, the angle \(\theta\) is expressed as the difference of two inverse cotangents, \(\cot^{-1} \frac{x}{40} - \cot^{-1} \frac{x}{10}\). By using trigonometric identities, complex expressions involving these functions become more manageable.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for every value within their domain. They play a crucial role in simplifying trigonometric expressions. One useful identity utilized in this problem is:

• \(\cot^{-1}(A) - \cot^{-1}(B) = \tan^{-1}\left(\frac{B-A}{1+AB}\right)\)

In the problem, we have set \(A = \frac{x}{40}\) and \(B = \frac{x}{10}\). By substituting these values into the identity, the expression simplifies, transforming it into a single inverse tangent function. This is a powerful simplification tool that turns a potentially cumbersome problem into a more straightforward calculation.
Understanding and memorizing common identities can dramatically streamline the solving of trigonometric problems, making it possible to approach the exercise efficiently.

Angle of Elevation

The angle of elevation refers to the angle between the horizontal line and the line of sight when an observer looks upwards towards an object.
In the context of the exercise, the angle of elevation, \(\theta\), relates to the viewing angle of a person looking at a movie screen.
The screen starts 10 feet above the person's eye level and is 30 feet tall.
This creates the potential for a dynamic angle of elevation depending on how far the person sits from the screen.
The angle changes as the distance \(x\) varies, illustrating how observers further from the screen have a different visual perspective compared to those seated closer. By calculating \(\theta\), we are effectively determining how sharp the viewer must tilt their head up to see the entire display.
This concept is crucial in understanding real-world applications such as properly setting seat heights and viewing arrangements in audiences.

Calculus Problem Solving

Calculus comes into play when we need to understand how the viewing angle \(\theta\) changes as one variable, the distance \(x\), changes.
This can be explored through functions involving derivatives and integrals.
During the simplification of \(\theta\), taking the derivative with respect to \(x\) would reveal how the angle of elevation intensifies as one moves closer or further from the screen.
The final simplified expression for \(\theta\), \(\tan^{-1}(\frac{3x}{400 + x^2})\), can allow us to examine its behavior over different values of \(x\). This behavior is influenced by calculus principles such as critical points and points of inflection.
These techniques help solve problems by assessing rates of change, particularly useful when planning layouts for optimal viewing experiences or when designing items sensitive to angles of vision.