Chapter 4: Problem 22
Show that the functions have exactly one zero in the given interval. \(r(\theta)=\tan \theta-\cot \theta-\theta, \quad(0, \pi / 2)\)
Short Answer
Expert verified
The function has exactly one zero in the interval \((0, \pi/2)\) because it is continuous and strictly increasing there.
Step by step solution
01
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 \(r(\theta) = \tan \theta - \cot \theta - \theta = 0\).
02
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 \(\theta\).
03
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 \(r(\theta) \to \infty\).
04
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 \((0, \pi/2)\).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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 \( r(\theta) \to \infty \).