Problem 26 - Thomas Calculus in SI Units

Question

Show that the functions that 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 has exactly one zero in the interval \((0, \pi/2)\).

Step 1: Define the Function

First, we define the function given in the problem: \( r(\theta) = \tan \theta - \cot \theta - \theta \). This function needs to be investigated in the interval \((0, \frac{\pi}{2})\).

Step 2: Analyze Endpoints

Check the behavior of \( r(\theta) \) as \( \theta \) approaches the boundary of the interval. At \( \theta = 0 \), \( \tan \theta = 0 \) and \( \cot \theta \) is undefined, but approaches infinity. At \( \theta = \frac{\pi}{2} \), \( \tan \theta \) approaches infinity, and \( \cot \theta = 0 \).

Step 3: Derivative and Monotonicity

Calculate the derivative \( r'(\theta) \) to assess monotonicity:\[ r'(\theta) = \sec^2 \theta + \csc^2 \theta - 1. \]Both \( \sec^2 \theta \) and \( \csc^2 \theta \) are positive within the interval, hence \( r'(\theta) > 0 \). This implies \( r(\theta) \) is strictly increasing over \( (0, \frac{\pi}{2}) \).

Step 4: Evaluate Function for a Zero Point

Now evaluate \( r(\theta) \) at a few points to determine its sign change. As \( \theta \to 0^+ \), \( r(\theta) \to -\infty \). For \( \theta = \frac{\pi}{4} \), \( r(\frac{\pi}{4}) = 0 - \frac{\pi}{4} - \theta \), since \( \tan \frac{\pi}{4} = \cot \frac{\pi}{4} = 1 \), results in a negative value (since \( \pi/4 \) is positive and greater than 1). As \( \theta \to \frac{\pi}{2}^- \), \( r(\theta) \to \infty \).

Step 5: Apply Intermediate Value Theorem

Since \( r(\theta) \) is continuous and strictly increasing, and it goes from negative to positive, there exists exactly one zero of the function \( r(\theta) \) in the interval \((0, \frac{\pi}{2})\). By the Intermediate Value Theorem, since \( r(\theta) \) changes sign, a solution exists.

Key Concepts

Intermediate Value Theorem

When investigating whether a function has a zero within a specific interval, the Intermediate Value Theorem (IVT) is a powerful tool. It states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), and \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one \( c \in (a, b) \) such that \( f(c) = 0 \). This is because a continuous function without a break must hit every value between \( f(a) \) and \( f(b) \). In our case, we applied IVT to the function \( r(\theta) = \tan \theta - \cot \theta - \theta \) in the interval \((0, \frac{\pi}{2})\). As we evaluated the behavior of the function at the interval's boundaries and noticed a sign change—from negative to positive—the theorem assured us that exactly one zero must exist within the interval.

Monotonicity

Monotonicity refers to the consistent behavior of a function in terms of either increasing or decreasing within an interval. If a function is continuously increasing, it cannot repeat any of its output values or have a local maximum. This property aids in analyzing zeros and the behavior of functions. To check monotonicity of \( r(\theta) = \tan \theta - \cot \theta - \theta \), we compute its derivative: \( r'(\theta) = \sec^2 \theta + \csc^2 \theta - 1 \). Since \( \sec^2 \theta \) and \( \csc^2 \theta \) are both positive over \( (0, \frac{\pi}{2}) \), the derivative is positive too, indicating that \( r(\theta) \) is strictly increasing. This strict increase supports the existence of a single zero in the given interval once a sign change is identified.

Trigonometric Functions

Trigonometric functions, such as tangent and cotangent, play a crucial role in the behavior of periodic phenomena. The function \( \tan \theta \) has a range from \(-\infty\) to \(+\infty\) within its period, which is \( (0, \pi) \). It is undefined at \( \frac{\pi}{2} \). Similarly, \( \cot \theta \), which is \( 1/\tan \theta \), is undefined at \( 0 \) and also ranges from \(+\infty\) to \(-\infty\) in the same interval.These characteristics of trigonometric functions are vital in our problem, where \( r(\theta) = \tan \theta - \cot \theta - \theta \) mixes these elements. Understanding how each function behaves near the endpoints of \((0, \frac{\pi}{2})\) helps us analyze \( r(\theta) \) as we observe its transition from large negative to large positive values, which indicates a zero exists.

Derivative Analysis

Derivative analysis provides insights into a function's rate of change, telling us if and where it is increasing or decreasing, and helps in identifying critical points. Calculating the derivative \( r'(\theta) = \sec^2 \theta + \csc^2 \theta - 1 \) entails recognizing how different terms contribute to the growth or decline of the original function \( r(\theta) \).In the case of \( r \) given in the exercise, both \( \sec^2 \theta \) and \( \csc^2 \theta \) are greater than 1 in \( (0, \frac{\pi}{2}) \), ensuring that \( r'(\theta) > 0 \), thus guaranteeing that \( r(\theta) \) is increasing. This assurance of an increasing function when paired with IVT, supplies a robust conclusion about the existence and uniqueness of a zero within the specified interval.