Problem 107 - Thomas Calculus in SI Units

Question

You are sitting in a classroom next to the wall looking at the blackboard at the front of the room. The blackboard is \(4 \mathrm{m}\) long and starts \(1 \mathrm{m}\) from the wall you are sitting next to. a. Show that your viewing angle is \(\alpha=\cot ^{-1} \frac{x}{5}-\cot ^{-1} x\) if you are \(x\) m from the front wall. b. Find \(x\) so that \(\alpha\) is as large as possible.

Step-by-Step Solution

Answer

The viewing angle is \(\alpha = \cot^{-1} \frac{x}{5} - \cot^{-1} x\) and is maximized when \(x = \sqrt{5}\).

Step 1: Understand the Geometry

Visualize the classroom as a right triangle where your position next to the wall and looking at the blackboard determines the viewing angle \(\alpha\). The position from the blackboard to the wall is not directly in front of you, starting 1 m from the wall, not at the corner.

Step 2: Setup the Tangent Functions

The blackboard starts 1 m from the wall and is 4 m long. Therefore, one side of the right triangle has a length of \(x + 1\) m and the other side (covering the blackboard’s length) is \(x + 5\) m. The tangent of the angle at these points is given by \(\tan^{-1}\left(\frac{1}{x}\right)\) and \(\tan^{-1}\left(\frac{5}{x}\right)\).

Step 3: Derive the Viewing Angle Equation

To find the viewing angle \(\alpha\), we recognize that it is the difference between these two inverse tangent functions: \(\alpha = \tan^{-1}\left(\frac{5}{x}\right) - \tan^{-1}\left(\frac{1}{x}\right)\). Simplifying this using the cotangent function results in \(\alpha = \cot^{-1}\left(\frac{x}{5}\right) - \cot^{-1}(x)\).

Step 4: Maximize the Viewing Angle

The viewing angle \(\alpha\) is maximized when the derivative of \(\alpha(x) = \tan^{-1}\left(\frac{5}{x}\right) - \tan^{-1}\left(\frac{1}{x}\right)\) is zero. Differentiating with respect to \(x\) and setting it to zero involves using the chain rule on the tangent inverse functions.

Step 5: Solve for Maximum Condition

Calculating the derivatives, set \(\frac{d}{dx}[\tan^{-1}\left(\frac{5}{x}\right) - \tan^{-1}\left(\frac{1}{x}\right)] = \frac{-5}{x^2 + 25} + \frac{1}{x^2 + 1} = 0\). Solving this equation involves finding a common denominator and simplifying. The result gives the optimal \(x\) when \(5(x^2 + 1) = x^2 + 25\).

Step 6: Solve for Optimal x

Solving \(5x^2 + 5 = x^2 + 25\) simplifies to \(4x^2 = 20\), hence \(x^2 = 5\). Therefore, \(x = \sqrt{5}\).

Key Concepts

Trigonometric Functions

Trigonometric functions are essential tools in understanding angles and relationships in triangles. In this problem, we are working with a right triangle formed by the position in the classroom and the blackboard. Here, trigonometric functions help us express relationships in terms of angles and distances.

These functions include sine, cosine, and tangent, as well as their inverses, such as arcsine, arccosine, and arctangent. They help us relate different sides of a triangle to its angles. In our case, the tangent and inverse tangent functions are primarily used. This is because dealing with these aspects allows one to measure the width across the angle at which the blackboard is viewed.

Derivatives

Derivatives in calculus are used to understand how a function changes. It's like finding the slope of a function at any given point. In optimization, finding maximum or minimum values often involves taking derivatives.

In this case, to maximize the viewing angle, we used derivatives to find the point where the angle at which the blackboard is viewed is largest. By differentiating the expression for the viewing angle (\( \tan^{-1}\left(\frac{5}{x}\right) - \tan^{-1}\left(\frac{1}{x}\right) \)), and setting it equal to zero, we can find those critical points that might give us the maximum angle. The chain rule is particularly useful here because we are dealing with composite functions.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find the angles when the sides of the triangle are known. This helps when we want to express angles explicitly. In our classroom problem, the inverse tangent function, (\( \tan^{-1} \)), is used since we know the opposite side lengths relative to the angle.

• \( \tan^{-1}\left(x\right) \) yields an angle whose tangent is \( x \).
• By using inverse functions like \( \tan^{-1}\left(\frac{5}{x}\right) \) and \( \tan^{-1}\left(\frac{1}{x}\right) \), we get the angles \( \alpha \) necessary to describe the viewing angle more succinctly.

This gives rise to the derived expression for the viewing angle, illustrating why inverse functions are crucial for solving such geometry problems.

Maximization Problem

A maximization problem in calculus involves finding the point or points where a function reaches its highest value, known as the maximum. This is key in many real-world applications, from maximizing efficiency to optimizing viewing angles as in our scenario here.

By setting the derivative of the angle equation to zero, we locate critical points. Solving the resulting equation, (\( \frac{-5}{x^2 + 25} + \frac{1}{x^2 + 1} = 0 \)), gives us the value of \( x \) that provides the largest viewing angle. After simplification, we find that \( x = \sqrt{5} \) is the optimum position to sit for the best view of the blackboard. Understanding how derivatives and inverse trigonometric functions interact is key to solving such problems.