Problem 63 - Thomas Calculus

Question

Show that the functions have local extreme values at the given values of \(\theta,\) and say which kind of local extreme the function has. $$ h(\theta)=3 \cos \frac{\theta}{2}, \quad 0 \leq \theta \leq 2 \pi, \quad \text { at } \theta=0 \text { and } \theta=2 \pi $$

Step-by-Step Solution

Answer

Local maximum at \( \theta = 0 \) and local minimum at \( \theta = 2\pi \).

Step 1: Understand the problem

We need to find out if the function \( h(\theta) = 3 \cos \frac{\theta}{2} \) has local extreme values at specific points: \( \theta = 0 \) and \( \theta = 2\pi \). We will determine the kind of extremum (if any) at these points within the interval \( 0 \leq \theta \leq 2\pi \).

Step 2: Find the derivative

To determine if there is a local extremum, we need to find the first derivative of the function. The derivative of \( h(\theta) = 3 \cos \frac{\theta}{2} \) is given by applying the chain rule:\[ h'(\theta) = 3 \cdot \left( -\sin \frac{\theta}{2} \right) \cdot \frac{1}{2} = -\frac{3}{2} \sin \frac{\theta}{2}. \]

Step 3: Evaluate the derivative at \( \theta = 0 \) and \( \theta = 2\pi \)

Evaluate \( h'(\theta) \) at the given points:- At \( \theta = 0 \), \( h'(0) = -\frac{3}{2} \sin(0) = 0 \).- At \( \theta = 2\pi \), \( h'(2\pi) = -\frac{3}{2} \sin(\pi) = 0 \).Since the derivative is 0 at both points, these are critical points where a local extremum can occur.

Step 4: Determine the second derivative

We use the second derivative test to determine the type of extremum. First, find the second derivative of \( h(\theta) \):\[ h''(\theta) = -\frac{3}{2} \cdot \cos \frac{\theta}{2} \cdot \frac{1}{2} = -\frac{3}{4} \cos \frac{\theta}{2}. \]

Step 5: Evaluate the second derivative at critical points

Now, evaluate \( h''(\theta) \) at \( \theta = 0 \) and \( \theta = 2\pi \):- At \( \theta = 0 \), \( h''(0) = -\frac{3}{4} \cos(0) = -\frac{3}{4} < 0 \). This indicates a local maximum.- At \( \theta = 2\pi \), \( h''(2\pi) = -\frac{3}{4} \cos(\pi) = \frac{3}{4} > 0 \). This indicates a local minimum.

Key Concepts

Critical Points

In the quest to find local extrema, critical points are the cornerstone. A critical point for a function occurs where the first derivative is either zero or undefined. It is at these points that the function might experience a local maximum or minimum.
To find critical points, you need to:

• Take the derivative of the function.
• Solve for values where the derivative equals zero or is undefined.

For the function \( h(\theta) = 3 \cos \frac{\theta}{2} \), the first derivative is \( h'(\theta) = -\frac{3}{2} \sin \frac{\theta}{2} \). Evaluating this derivative at \( \theta = 0 \) and \( \theta = 2\pi \), we find that the derivative equals zero at both points, confirming them as critical points. Understanding critical points is the first step in the journey to identify local extrema.

Chain Rule

The chain rule is an essential calculus tool, especially when differentiating composite functions. It states that if a function is composed of two functions, say \( f(g(x)) \), then its derivative is \( f'(g(x)) \cdot g'(x) \).
In our example, to differentiate \( h(\theta) = 3 \cos \frac{\theta}{2} \),

• First, identify the outer function as \( \cos(u) \) and take its derivative: \( -\sin(u) \).
• Next, differentiate the inner function \( u = \frac{\theta}{2} \), which gives \( \frac{1}{2} \).
• Combine them: \( h'(\theta) = -\frac{3}{2} \sin \frac{\theta}{2} \).

Using the chain rule correctly allows you to simplify the process of differentiation, especially when dealing with complex functions involving compositions, like trigonometric functions here.

Second Derivative Test

Once you've found the critical points, the second derivative test helps determine the nature of each point—whether it's a local maximum, local minimum, or a saddle point. The test involves computing the second derivative at the critical points:

• If \( h''(c) > 0 \), the function has a local minimum at \( c \).
• If \( h''(c) < 0 \), the function has a local maximum at \( c \).
• If \( h''(c) = 0 \), the test is inconclusive.

By evaluating the second derivative \( h''(\theta) = -\frac{3}{4} \cos \frac{\theta}{2} \) at \( \theta = 0 \) and \( \theta = 2\pi \), we find:

• \( h''(0) = -\frac{3}{4} \), indicating a local maximum.
• \( h''(2\pi) = \frac{3}{4} \), indicating a local minimum.

Thus, the second derivative test is a powerful tool for categorizing critical points effectively.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in understanding periodic behavior in mathematics. These functions often appear in calculus problems involving oscillations and waves.
In this exercise, the function \( h(\theta) = 3 \cos \frac{\theta}{2} \) is a scaled trigonometric function. Here, \( cos(\frac{\theta}{2}) \) means the angle input is halved, stretching the wave period, while the multiplier 3 scales the wave vertically.
Trigonometric functions exhibit unique properties:

• Periodicity: They repeat after a certain interval. Cosine and sine have periods of \( 2\pi \), but \( cos(\frac{\theta}{2}) \) extends this period to \( 4\pi \).
• Symmetry: Cosine functions are even, meaning \( \cos(-x) = \cos(x) \).
• Amplitude: Determines the maximum and minimum values the function can achieve.

Understanding these properties allows you to predict behavior, analyze oscillations, and effectively determine local extremum values in problems involving trigonometric functions.