Problem 31 - Thomas Calculus

Question

In Exercises \(21-36,\) 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

Max value is 1 at \(\theta=\frac{\pi}{2}\); Min value is -1 at \(\theta=-\frac{\pi}{2}\).

Step 1: Determine the Critical Points

The critical points occur where the derivative of the function is zero or undefined. For the function \(f(\theta) = \sin \theta\), the derivative is \(f'(\theta) = \cos \theta\). Setting \(f'(\theta) = 0\) gives \(\cos \theta = 0\). The solutions to this equation within the given interval are \(\theta = \frac{\pi}{2}\).

Step 2: Evaluate the Function at Critical Points and Endpoints

Evaluate the function \(f(\theta) = \sin \theta\) at the critical point \(\theta = \frac{\pi}{2}\) and at the endpoints \(\theta = -\frac{\pi}{2}\) and \(\theta = \frac{5\pi}{6}\). We find: \[f\left(-\frac{\pi}{2}\right) = \sin\left(-\frac{\pi}{2}\right) = -1,\] \[f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1,\] \[f\left(\frac{5\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}.\]

Step 3: Identify Absolute Extrema

Based on the calculated values, the absolute maximum value of \(f(\theta)\) on the interval occurs at \(\theta = \frac{\pi}{2}\) with \(f(\theta) = 1\). The absolute minimum value occurs at \(\theta = -\frac{\pi}{2}\) with \(f(\theta) = -1\).

Step 4: Graph the Function and Indicate Extrema

Graph the function \(f(\theta) = \sin \theta\) over the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{5 \pi}{6}\). On the graph, mark the points \(\left(-\frac{\pi}{2}, -1\right)\) for the absolute minimum, and \(\left(\frac{\pi}{2}, 1\right)\) for the absolute maximum, along with \(\left(\frac{5\pi}{6}, \frac{1}{2}\right)\).

Key Concepts

Critical Points

Critical points of a function are where its derivative is either zero or does not exist. These points are essential for identifying where functions change direction, which helps us find extrema, such as maximum or minimum values. For the function \(f(\theta) = \sin \theta\), the derivative is \(f'(\theta) = \cos \theta\). To find the critical points, you set this derivative equal to zero and solve for \(\theta\).
In this case, \(\cos \theta = 0\) yields a critical point at \(\theta = \frac{\pi}{2}\), which lies within the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6}\). There are no points where the derivative is undefined, making \(\theta = \frac{\pi}{2}\) the only critical point within this range.

Derivative

The derivative of a function measures how the function's output value changes as the input changes. For trigonometric functions like \(f(\theta) = \sin \theta\), derivatives provide crucial information about the function's behavior.
The derivative \(f'(\theta) = \cos \theta\) tells us how steep the function is at any given point. When \(f'(\theta) = 0\), the function has a horizontal tangent line, indicating a potential critical point where extrema might occur. In this exercise, solving \(\cos \theta = 0\) informs us of a pivotal point at \(\theta = \frac{\pi}{2}\), which is important for analyzing the function's absolute extrema on the given interval.

Function Evaluation

Evaluating the function at specific points allows us to compare these values and determine the absolute extrema. After identifying the critical point and the endpoints of the interval, we substitute them into the function.
For \(f(\theta) = \sin \theta\), we evaluate:

• At \(\theta = -\frac{\pi}{2}\), \(f(\theta) = -1\).
• At \(\theta = \frac{\pi}{2}\), \(f(\theta) = 1\).
• At \(\theta = \frac{5\pi}{6}\), \(f(\theta) = \frac{1}{2}\).

By comparing these results, we discovered the absolute maximum value occurs at \(\theta = \frac{\pi}{2}\), and the absolute minimum occurs at \(\theta = -\frac{\pi}{2}\), with "1" and "-1," respectively, as the function values at those points.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are periodic and have specific characteristics that make them unique compared to other functions. In the exercise, \(f(\theta) = \sin \theta\) is a sine function. Its period is \(2\pi\), meaning it repeats its values every \(2\pi\) units.
Within the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6}\), the sine function behaves predictably, swinging between -1 and 1. Understanding the nature of these trigonometric functions is vital. Using the identities and properties of sine, we can easily find values such as \(\sin \left( \frac{5\pi}{6} \right)\), which equals \(\frac{1}{2}\). Their predictable patterns simplify finding and verifying extrema within specified ranges.