Problem 113 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(109-114\) \(\begin{array}{l}{\text { You are sitting in a classroom next to the wall looking at the }} \\ {\text { blackboard at the front of the room. The blackboard is } 12 \text { ft long }} \\ {\text { and starts } 3 \mathrm{ft} \text { from the wall you are sitting next to. }} \\ {\text { a. Show that your viewing angle is }}\end{array}\) $$ \alpha=\cot ^{-1} \frac{x}{15}-\cot ^{-1} \frac{x}{3} $$ if you are \(x\) ft from the front wall. b. Find \(x\) so that \(\alpha\) is as large as possible.

Step-by-Step Solution

Answer

The angle \(\alpha\) is maximized when \(x = 3\sqrt{5}\).

Step 1: Understanding the Scenario

Visualize the classroom. You are sitting next to a wall, looking at a blackboard that is 12 feet long and starts 3 feet from the wall that you are against. Your distance from the front wall is denoted by \(x\).

Step 2: Determine the Viewing Angles

You want to find the angle you view the blackboard. Remember, \(\alpha = \cot^{-1} \frac{x}{15} - \cot^{-1} \frac{x}{3}\). This represents the angle subtended by the blackboard at your eye level.

Step 3: Differentiate Viewing Angle to Maximize

To maximize \(\alpha\), differentiate: \(\alpha = \cot^{-1} \frac{x}{15} - \cot^{-1} \frac{x}{3}\). Use the derivative formula for \(\cot^{-1} u\), which is \(-\frac{1}{1+u^2}\).

Step 4: Set the Derivative to Zero

Calculate the derivative: \(-\frac{1}{1+(\frac{x}{15})^2} \cdot \frac{1}{15} + \frac{1}{1+(\frac{x}{3})^2} \cdot \frac{1}{3}\), then set this expression equal to zero to find x.

Step 5: Solve for Critical Points

Solving \(-\frac{1}{1+(\frac{x}{15})^2} \cdot \frac{1}{15} + \frac{1}{1+(\frac{x}{3})^2} \cdot \frac{1}{3} = 0\) gives the crucial x for maximizing \(\alpha\). Find common denominators and solve the resulting equation.

Step 6: Determine Value of x

Simplify the terms of the equation: \(\frac{1}{3+\frac{x^2}{15^2}} = \frac{1}{5+\frac{x^2}{3^2}}\). Cross-multiply and solve for \(x\). This should result in a specific \(x\) value.

Step 7: Confirm Maximization Conditions

Verify that the critical point is a maximum using the second derivative test or checking the values around the found \(x\). If the second derivative is negative, it indicates a maximum.

Key Concepts

Initial Value Problem

An initial value problem in calculus involves finding a function that satisfies a specific differential equation and meets a specified initial condition. In simpler terms, this is about determining the exact behavior of a system given an initial starting point.

• First, you set up a differential equation based on the given context. This is often related to how a system changes over time.
• Next, you'll be given an 'initial condition', which tells you the starting point of the system at a certain time.
• Your task is to solve the differential equation so that the solution satisfies the initial condition.

For instance, in the given problem about maximizing the viewing angle in a classroom, you need to find the value of \(x\) that maximizes the angle \(\alpha\). The initial condition here is related to the values \(x\) based on the classroom setup. This prepares you to tailor the function to fit the particular scenario described.

Maximization

Maximization problems are about finding the point at which a function reaches its highest value. In the context of this problem, you're looking for the specific distance \(x\) that makes the viewing angle \(\alpha\) as large as possible.

• To solve a maximization problem, first express the quantity you want to maximize—in this case, the angle \(\alpha\)—in terms of a variable, here \(x\).
• Next, determine the derivative of this function. Derivatives give you the rate of change and help identify where these rates change signs, showing potential maxima or minima.
• Set the derivative equal to zero and solve for \(x\). This identifies potential points where the function could have a maximum value.

In this exercise, the difference of two inverse cotangent functions defines \(\alpha\). The goal is to use calculus to compute where this expression, when finally differentiated with respect to \(x\), hits zero, indicating critical points of potential maximization.

Differentiation

Differentiation is a key concept in calculus that involves finding a function's derivative, which represents the rate of change of a function with respect to a variable. In practical terms, it tells you how a quantity changes as another quantity changes.

• The process begins with identifying the function needing differentiation. Here, \(\alpha = \cot^{-1} \frac{x}{15} - \cot^{-1} \frac{x}{3}\) needs to be differentiated with respect to \(x\).
• Using the differentiation formula for inverse trigonometric functions, you'll apply the derivative of \(\cot^{-1} u\), which is \(-\frac{1}{1+u^2}\).
• After finding the derivatives, combining and simplifying them helps in identifying the points where the rate of change is zero, aiding in solving max-min problems.

Differentiation in this problem allows us to analyze the change of the viewing angle \(\alpha\) in relation to the distance \(x\). Setting the derivative to zero pinpoints where \(\alpha\) may be maximized.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to solve for angles when the trigonometric ratios are known. They are essentially the "undo" buttons for trigonometric functions and are pivotal in many calculus problems.

• The problem uses inverse cotangent, represented as \(\cot^{-1}\), which helps solve scenarios where the tangent value is known, and you need the angle.
• Understanding inverse trigonometric functions is essential as they frequently appear in real-world problems where angles need to be found from given positions or lengths.
• In particular, this problem calculates the viewing angle \(\alpha\) using inverse cotangent based on the classroom's layout to determine the limits of the blackboard visibility angle.

These functions hold central importance in trigonometry and calculus because they extend the capability to evaluate and manipulate angles in various geometric and physical problems, much like the one described in this exercise involving classroom geometry.