Problem 28 - University Calculus: Early Transcendentals

Question

Show that the functions have exactly one zero in the given interval. $$r(\theta)=\tan \theta-\cot \theta-\theta, \quad(0, \pi / 2)$$

Step-by-Step Solution

Answer

The function \( r(\theta) = \tan \theta - \cot \theta - \theta \) has exactly one zero in \((0, \pi/2)\).

Step 1: Define the Function

We are given the function \( r(\theta) = \tan \theta - \cot \theta - \theta \). We need to analyze this function over the interval \( (0, \frac{\pi}{2}) \).

Step 2: Investigate the Behavior at the Endpoints

As \( \theta \to 0^+ \), \( \tan \theta \to 0 \), and \( \cot \theta \to \infty \), thus \( r(\theta) \to -\infty \). As \( \theta \to \frac{\pi}{2}^- \), \( \tan \theta \to \infty \), and \( \cot \theta \to 0 \), making \( r(\theta) \to \infty \).

Step 3: Differentiate the Function

To find the critical points, we differentiate \( r(\theta) \). The derivative is \( r'(\theta) = \sec^2 \theta + \csc^2 \theta - 1 \).

Step 4: Analyze the Derivative

The derivative \( r'(\theta) = \sec^2 \theta + \csc^2 \theta - 1 \) is always positive in \( (0, \frac{\pi}{2}) \) because both \( \sec^2 \theta \, and \, \csc^2 \theta \) are greater than 1. Hence, \( r(\theta) \) is strictly increasing on \( (0, \frac{\pi}{2}) \).

Step 5: Apply the Intermediate Value Theorem

Since \( r(\theta) \) is continuous on \( (0, \frac{\pi}{2}) \), strictly increasing (as shown by positive derivative), and changes sign from negative (at \( \theta \to 0^+ \)) to positive (at \( \theta \to \frac{\pi}{2}^- \)), the Intermediate Value Theorem guarantees exactly one zero in the interval.

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus and is incredibly useful for proving the existence of zeros within a certain interval. The theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\), and \( N \) is any number between \( f(a) \) and \( f(b) \), then there is at least one number \( c \) in the interval \([a, b]\) such that \( f(c) = N \). In simpler terms, if you have a continuous curve and it crosses between two heights, then it must have touched every height between those two at least once.

In our problem, we applied the IVT to find that the function \( r(\theta) \) crosses the \( x \)-axis exactly once within \( (0, \frac{\pi}{2}) \). Because \( r(\theta) \) is continuous (as it is a combination of continuous functions) and changes from negative to positive value over the interval, the theorem assures us there must be a point \( \theta \) where \( r(\theta) = 0 \). This is because the function behavior moves from a negative range to a positive range, as was demonstrated at the endpoints with \( r(\theta) \) going from negative infinity to positive infinity.

Differentiation

Differentiation is a calculus technique used to determine the rate at which a function is changing at any given point, commonly known as the derivative. For any given function, the derivative helps identify critical points, which can tell you about the nature of the function, such as whether it is increasing or decreasing.

In the exercise, the derivative of \( r(\theta) \) is \( r'(\theta) = \sec^2 \theta + \csc^2 \theta - 1 \). This derivative was critical in analyzing the behavior of the function \( r(\theta) \). Since both \( \sec^2 \theta \) and \( \csc^2 \theta \) are always greater than 1 for \( \theta \) in \( (0, \frac{\pi}{2}) \), the derivative \( r'(\theta) \) is always positive. This tells us the original function is always increasing within the interval.

By knowing the function is strictly increasing, we reinforce the conclusion made using the Intermediate Value Theorem, allowing us to state confidently that \( r(\theta) \) can only have one zero in the interval \( (0, \frac{\pi}{2}) \).

Trigonometric Functions

Trigonometric functions, like tangents and cotangents, are an essential part of calculus, especially when dealing with oscillatory behavior and limits. They are periodic and have specific characteristics that differentiate them from other function types.

For our function \( r(\theta) = \tan \theta - \cot \theta - \theta \), understanding the behaviors of \( \tan \theta \) and \( \cot \theta \) is crucial. The \( \tan \theta \) rises from zero to infinity as \( \theta \) approaches \( \frac{\pi}{2} \) from the left, while \( \cot \theta \) decreases from infinity at a very small \( \theta \) to zero as \( \theta \) approaches \( \frac{\pi}{2} \). This relationship causes a sign change for \( r(\theta) \) and showcases the importance of these trigonometric components in defining the overall behavior of the function.

By utilizing trigonometric functions in calculus, you can effectively manage and predict oscillations and transitions within a defined interval, just like how they enable the understanding of \( r(\theta) \)'s tendency to transition from \(-\infty\) to \(\infty\) over the considered interval.