Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In each of the Problems 1-21, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of c; if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ S(\theta)=\sin \theta ;[-\pi, \pi] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies, and \( c = -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).

Step 1: Determine the Criteria for the Mean Value Theorem

The Mean Value Theorem (MVT) states that for a function \( f \) which is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one point \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Step 2: Verify Continuity and Differentiability of S(θ)

The function \( S(\theta) = \sin \theta \) is a trigonometric function that is continuous and differentiable everywhere. Therefore, it is continuous on \([-\pi, \pi]\) and differentiable on \((-\pi, \pi)\).

Step 3: Compute the Average Rate of Change

First, we find the values of the function at the endpoints: \( S(-\pi) = \sin(-\pi) = 0 \) and \( S(\pi) = \sin(\pi) = 0 \). Now calculate \( \frac{S(\pi) - S(-\pi)}{\pi - (-\pi)} = \frac{0 - 0}{2\pi} = 0 \).

Step 4: Find the Derivative of S(θ)

The derivative of \( S(\theta) = \sin \theta \) is \( S'(\theta) = \cos \theta \).

Step 5: Solve for c Using the MVT Equation

Set \( S'(c) = 0 \). This gives \( \cos(c) = 0 \). Solving \( \cos(c) = 0 \) for \( c \) in the interval \((-\pi, \pi)\), we find \( c = -\frac{\pi}{2} \) and \( c = \frac{\pi}{2} \).

Step 6: Sketch the Graph

Sketch the graph of \( S(\theta) = \sin \theta \) from \(-\pi\) to \(\pi\) showing that it is a smooth wave-like shape passing through the x-axis at multiples of \(\pi\). It has peaks and troughs at \(\pm\frac{\pi}{2}\).

Key Concepts

Continuity in Mathematics

Continuity is a fundamental concept in calculus that is crucial for understanding the Mean Value Theorem. When we say a function is continuous on an interval, we mean that there are no holes, jumps, or breaks in its graph throughout that interval. This ensures that you can draw the function’s graph from start to finish without lifting your pencil.
For a function \( f(x) \) to be continuous on a closed interval \([a, b]\), it needs to satisfy the following conditions at each point \( x \) in the interval:

• The function \( f(x) \) is defined at \( x \).
• The limit of \( f(x) \) as \( x \) approaches any point \( c \) from both left and right exists.
• The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).

In this exercise, the function \( S(\theta) = \sin \theta \) is continuous on the interval \([-\pi, \pi]\), which means it does not have any breaks or gaps. This is a common characteristic of all standard trigonometric functions like sine and cosine. Understanding continuity in functions helps ensure they behave predictably across their specified intervals.

Differentiability and Its Importance

Differentiability indicates a function's smoothness and its capability to have a tangent line drawn at any given point. If a function is differentiable at a point, it means its graph is not broken, nor does it have any sharp corners or cusps there. A differentiable function can have a derivative defined at each point in its interval.
For the Mean Value Theorem to apply, a function must be differentiable on an open interval \((a, b)\). The function \( S(\theta) = \sin \theta \) is differentiable everywhere, which includes the open interval \((-\pi, \pi)\). This means we can calculate its derivative, \( S'(\theta) = \cos \theta \), wherever needed in this interval.
The derivation of the sine function results in the cosine function, which helps find points where the rate of change is zero (i.e., horizontal tangents). This is fundamental when solving Mean Value Theorem problems, as it helps identify specific points \( c \) where the average rate of change equals the instantaneous rate of change.

Understanding Trigonometric Functions

Trigonometric functions are some of the most important periodic functions in mathematics. They include sine, cosine, and tangent functions, all of which relate angles of a triangle to the ratios of its sides.
In this context, the function \( S(\theta) = \sin \theta \) is a fundamental trigonometric function. It outputs values between -1 and 1 as \( \theta \) varies. This periodic function repeats its values in cycles of \( 2\pi \).
Key features of the sine function include:

• Amplitude: the maximum distance the function moves from the horizontal axis (here, amplitude is 1).
• Period: the length of one complete cycle (for \( \sin \theta \), the period is \( 2\pi \)).
• Zeros: points where the function crosses the horizontal axis (like \( \pi \) and \(-\pi \)).

Understanding these features of the function helps visualize and sketch the graph, as needed for this exercise. Recognizing these traits makes analyzing changes in trigonometric functions more intuitive, especially when applying the Mean Value Theorem.