Problem 22 - Thomas Calculus

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 has exactly one zero in the interval \((0, \pi/2)\) because it is continuous and strictly increasing there.

Step 1: Understand the definition of zero of the function

A function has a zero at a point if the function evaluates to zero at that point. Thus, we need to find when \(r(\theta) = \tan \theta - \cot \theta - \theta = 0\).

Step 2: Investigate continuity and behavior of the function

On the interval \((0, \pi/2)\), both \(\tan \theta\) and \(\cot \theta\) are defined and continuous, therefore \(r(\theta)\) is continuous as a combination of these functions and \(\theta\).

Step 3: Analyze function behavior at boundaries

Calculate the behavior of \(r(\theta)\) as \(\theta\to 0^+\): \(\tan \theta \to 0\) and \(\cot \theta \to \infty\), thus \(r(\theta) \to -\infty\). As \(\theta \to \pi/2^-\), \(\tan \theta \to \infty\) and \(\cot \theta \to 0\), thus \(r(\theta) \to \infty\).

Step 4: Apply Intermediate Value Theorem

Since \(r(\theta)\) is continuous on the interval \((0, \pi/2)\) and changes from \(-\infty\) to \(\infty\), by the Intermediate Value Theorem there must be at least one zero in \((0, \pi/2)\).

Step 5: Verify uniqueness of zero by evaluating derivative

Find the derivative \(r'(\theta) = \sec^2 \theta + \csc^2 \theta - 1\). Since both \(\sec^2 \theta\) and \(\csc^2 \theta\) are positive, \(r'(\theta) > 0\) for all \(\theta\) in \((0, \pi/2)\). This indicates \(r(\theta)\) is strictly increasing, ensuring only one zero in the interval.

Key Concepts

Continuity of Functions

Continuity is a foundational concept in calculus and plays a significant role when analyzing functions, particularly when discussing zeros within an interval. A function is said to be continuous on an interval if, intuitively, it can be drawn without lifting the pen from the paper across that range. For the function \( r(\theta) = \tan \theta - \cot \theta - \theta \) on the interval \((0, \pi/2)\), continuity is assured since both the tangent and cotangent functions, as well as the linear function \(\theta\), are continuous in this domain.

• \(\tan \theta\) and \(\cot \theta\) are defined and continuous between \(0\) and \(\pi/2\), excluding the endpoints.
• The combination of continuous functions results in \(r(\theta)\) being continuous.
• This continuity is crucial for applying the Intermediate Value Theorem (IVT) later.

Being able to assert that a function is continuous over an interval allows us to make powerful conclusions about its behavior and access useful theorems in calculus, such as the IVT.

Derivative Analysis

Analyzing the derivative of a function gives insights into its behavior, such as whether it is increasing or decreasing over a particular interval. For \( r(\theta) = \tan \theta - \cot \theta - \theta \), the derivative is computed as \[ r'(\theta) = \sec^2 \theta + \csc^2 \theta - 1. \]This derivative helps in understanding the function's monotonicity.

• Both \(\sec^2 \theta\) and \(\csc^2 \theta\) are positive for \(\theta\) in \((0, \pi/2)\).
• Thus, \( r'(\theta) > 0 \) over the interval, showing that the function is strictly increasing.

When a function's derivative is positive like this over interval \((0, \pi/2)\), it implies that the function is moving upwards, with no local minima or maxima, confirming that it cannot loop back and have more than one zero in the interval.

Behavior at Boundaries

Understanding a function's behavior as it approaches the boundaries of its domain is essential for analyzing its properties over an interval. For \( r(\theta) = \tan \theta - \cot \theta - \theta \), examining its limits as \(\theta\) approaches 0 and \(\pi/2\) provides crucial insights.

• As \( \theta \to 0^+ \), \( \tan \theta \to 0 \), \( \cot \theta \to \infty \), thus \( r(\theta) \to -\infty \).
• As \( \theta \to \pi/2^- \), \( \tan \theta \to \infty \), \( \cot \theta \to 0 \), leading \( r(\theta) \to \infty \).

These boundary behaviors are critical because they set up the conditions for using the Intermediate Value Theorem. Seeing \( r(\theta) \) transition from \(-\infty\) to \(\infty\) indicates that at some point in \((0, \pi/2)\), the function must cross the x-axis, meaning there is at least one zero in the interval. This information complements our derivative analysis by confirming that if there’s one zero, it’s unique due to the function's strict increase.