Problem 31 - Thomas Calculus in SI Units

Question

Find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$f(\theta)=\sin \theta, \quad-\frac{\pi}{2} \leq \theta \leq \frac{5 \pi}{6}$$

Step-by-Step Solution

Answer

The absolute maximum is 1 at \(\theta = \frac{\pi}{2}\) and the absolute minimum is -1 at \(\theta = -\frac{\pi}{2}\).

Step 1: Understand the Problem

We need to find the absolute maximum and minimum values of the function \( f(\theta) = \sin \theta \) over the interval \( -\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6} \). This involves evaluating the function at critical points within the interval and at the endpoints of the interval.

Step 2: Determine Critical Points

A critical point occurs where the derivative is zero or undefined. The derivative of \( f(\theta) = \sin \theta \) is \( f'(\theta) = \cos \theta \). Set \( \cos \theta = 0 \) to find critical points: \[\theta = \frac{\pi}{2}\]This critical point \( \theta = \frac{\pi}{2} \) is within the given interval.

Step 3: Evaluate Function at Endpoints and Critical Points

Calculate \( f(\theta) \) at the endpoints and the critical point:- Endpoint \( \theta = -\frac{\pi}{2} \): \[ f\left(-\frac{\pi}{2}\right) = \sin\left(-\frac{\pi}{2}\right) = -1 \] - Critical point \( \theta = \frac{\pi}{2} \): \[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \] - Endpoint \( \theta = \frac{5\pi}{6} \): \[ f\left(\frac{5\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \]

Step 4: Identify Absolute Extrema

From the evaluations:- \( f\left(-\frac{\pi}{2}\right) = -1 \)- \( f\left(\frac{\pi}{2}\right) = 1 \)- \( f\left(\frac{5\pi}{6}\right) = \frac{1}{2} \)The absolute maximum value is 1 at \( \theta = \frac{\pi}{2} \).The absolute minimum value is -1 at \( \theta = -\frac{\pi}{2} \).

Step 5: Graph the Function

Graph \( f(\theta) = \sin \theta \) over the interval \( -\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6} \). Identify the points \( \left(\frac{\pi}{2}, 1\right) \) and \( \left(-\frac{\pi}{2}, -1\right) \) on the graph as the absolute extrema. The point \( \left(\frac{5\pi}{6}, \frac{1}{2}\right) \) is also marked.

Key Concepts

Absolute Extrema

Absolute extrema are the largest and smallest values that a function can achieve over a specified interval. When trying to find these values, it's important to evaluate not only the endpoints but also any critical points within the interval.
To identify the absolute extrema of a function like \( f(\theta) = \sin \theta \) over an interval \( -\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6} \), you follow these steps:

• First, analyze the critical points, which occur where the derivative of the function is zero or undefined.
• Next, calculate the function's value at those critical points and at the interval endpoints.
• Compare these values to find the absolute maximum and minimum.

In our example, the absolute maximum of \( f(\theta) = \sin \theta \) is \(1\) at \(\theta = \frac{\pi}{2}\) and the absolute minimum is \(-1\) at \(\theta = -\frac{\pi}{2}\). These represent the highest and lowest values of \( \sin \theta \) over the given interval.

Critical Points

Understanding critical points is crucial for finding extremas as they indicate where the graph of a function could potentially change direction. A critical point occurs where the derivative of the function equals zero or where the derivative is undefined.
For the function \( f(\theta) = \sin \theta \), the derivative is \( f'(\theta) = \cos \theta \). Finding the critical points involves solving \( \cos \theta = 0 \). This happens at \( \theta = \frac{\pi}{2} \), within the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6}\).
Once identified, these critical points are evaluated to determine if they are indeed local maxima, local minima, or saddle points. In this case, at \(\theta = \frac{\pi}{2}\), we get the maximum value for our function on the interval, making it a vital point for understanding the behavior of \( \sin \theta \). Remember, not every critical point is a point of extreme value, but they are always worth checking in the given range.

Graphing Trigonometric Functions

Graphing trigonometric functions is a powerful way to visualize how these functions behave over an interval. It shows the peaks, troughs, and periodic nature of trigonometric functions like sine and cosine.
For \( f(\theta) = \sin \theta \) within \(-\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6}\), the graph displays the smooth, oscillating wave pattern typical of sine functions. To accurately graph this:

• Plot key points derived from critical points and endpoints such as \((-\frac{\pi}{2}, -1)\), \((\frac{\pi}{2}, 1)\), and \((\frac{5\pi}{6}, \frac{1}{2})\).
• Use these points to sketch the wave, noting that the maximum is at \( \theta = \frac{\pi}{2} \), and the minimum at \( \theta = -\frac{\pi}{2} \).

When graphing trigonometric functions, understanding their periodic nature and symmetry can greatly assist in accurately depicting their behavior within a given interval. This visualization not only confirms analytic calculations but also provides a deeper comprehension of the function’s behavior across the interval.