Problem 25 - Thomas Calculus

Question

In Exercises \(15-30\) , 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

Absolute maximum is 1 at \( \theta = \frac{\pi}{2} \); absolute minimum is -1 at \( \theta = -\frac{\pi}{2} \).

Step 1: Identify Critical Points

To find the absolute extrema, we first need to identify any critical points within the interval. For the function \( f(\theta) = \sin \theta \), we take the derivative and set it to zero to find critical points: \( f'(\theta) = \cos \theta = 0 \). Solving \( \cos \theta = 0 \) gives critical points at \( \theta = \frac{\pi}{2} \). However, we must check if this lies within the interval \( \left[-\frac{\pi}{2}, \frac{5\pi}{6}\right]\).

Step 2: Evaluate Endpoints and Critical Points

Evaluate the function \( f(\theta) \) at the endpoints and the critical point within the interval.- 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 3: Identify Absolute Extrema

Compare the function values from Step 2:- \( f\left(-\frac{\pi}{2}\right) = -1 \)- \( f\left(\frac{\pi}{2}\right) = 1 \)- \( f\left(\frac{5\pi}{6}\right) = \frac{1}{2} \)The absolute maximum is \( 1 \) at \( \theta = \frac{\pi}{2} \) and the absolute minimum is \(-1\) at \( \theta = -\frac{\pi}{2} \).

Step 4: Graph the Function

Draw the graph of \( f(\theta) = \sin \theta \) over the interval \( -\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6} \). Identify and mark the extrema on the graph, noting:- Point with absolute maximum: \( (\frac{\pi}{2}, 1) \)- Point with absolute minimum: \( (-\frac{\pi}{2}, -1) \)

Key Concepts

Understanding Critical Points

In calculus, critical points are where a function's derivative equals zero or does not exist. These points are important because they help us determine where the function might have local maxima or minima. For the function \( f(\theta) = \sin \theta \), we find critical points by setting its derivative equal to zero.
To compute the derivative of \( \sin \theta \), we use the basic derivative rule of trigonometric functions: the derivative of \( \sin \theta \) is \( \cos \theta \). When setting \( \cos \theta = 0 \), we solve for \( \theta \) to find possible critical points:

• The equation \( \cos \theta = 0 \) is satisfied at \( \theta = \frac{\pi}{2} \) and repeating at increments of \( \pi \).

Since we are focusing on the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6} \), we verify that \( \theta = \frac{\pi}{2} \) is indeed within this range. This makes it a critical point for our specific problem, needing further evaluation to determine its significance in finding extrema.

Exploring Trigonometric Functions

Trigonometric functions, like sine and cosine, capture the periodic nature of angles and circles. They are fundamental in the study of calculus and appear frequently in problems involving oscillations and waves.
The function \( f(\theta) = \sin \theta \) represents the y-coordinate of a point on the unit circle as \( \theta \) changes. This periodic function repeats its values every \( 2\pi \) radians and provides a smooth wave-like graph.
In this exercise, the function's behavior is crucial in determining maximum and minimum values within specified intervals. For \( \sin \theta \), we know:

• It reaches maximum value of 1 when \( \theta = \frac{\pi}{2} + 2k\pi \) where \( k \) is an integer.
• It reaches minimum value of -1 when \( \theta = \frac{3\pi}{2} + 2k\pi \).

Understanding these properties helps us evaluate the function at critical points and endpoints effectively to find absolute extrema.

Calm Calculus Problem Solving

Solving calculus problems like finding absolute extrema involves a series of logical steps. It's essential to consider critical points, evaluate endpoints, and compare values. Let's break this into simpler steps:
First, identify and calculate the derivative of the function \( f(\theta) = \sin \theta \), which leads us to \( f'(\theta) = \cos \theta \). We determine the critical points by setting \( \cos \theta = 0 \).
Next, evaluate the function at the endpoints of the interval and any critical points. Here, the critical point is \( \theta = \frac{\pi}{2} \) and the endpoints are \(-\frac{\pi}{2}\) and \(\frac{5\pi}{6}\):

• Computing \( \sin(-\frac{\pi}{2}) = -1 \).
• Computing \( \sin(\frac{\pi}{2}) = 1 \).
• Computing \( \sin(\frac{5\pi}{6}) = \frac{1}{2} \).

Finally, compare these values to identify the absolute maximum and minimum within the given interval.
This structured approach ensures that you thoroughly understand the problem-solving process in calculus, making it easier to tackle similar problems in the future. Remember, practice and familiarity with these steps will build confidence and intuition in calculus problem solving.