Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ r(\theta)=\sin \theta ; I=\left[-\frac{\pi}{4}, \frac{\pi}{6}\right] $$

Step-by-Step Solution

Answer

Maximum is 0.5 at \(\theta = \frac{\pi}{6}\), minimum is \(-\frac{\sqrt{2}}{2}\) at \(\theta = -\frac{\pi}{4}\).

Step 1: Identify the function and interval

The function we are analyzing is \( r(\theta) = \sin \theta \). The interval given for the analysis is \( \left[-\frac{\pi}{4}, \frac{\pi}{6}\right] \). We will evaluate the function within this interval.

Step 2: Find the derivative

To find the critical points, we first need the derivative of \( r(\theta) \). The derivative of \( r(\theta) = \sin \theta \) is \( r'(\theta) = \cos \theta \).

Step 3: Determine critical points

Critical points occur where the derivative is zero or undefined. Set \( r'(\theta) = \cos \theta = 0 \). Since \( \cos \theta = 0 \) at \( \theta = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \), we check within the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{6}\right] \). However, there are no values where \( \cos \theta = 0 \) within this range, so only the endpoints need consideration.

Step 4: Evaluate endpoints

Evaluate \( r(\theta) \) at the endpoints of the interval: \( r\left(-\frac{\pi}{4}\right) = \sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \) and \( r\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \).

Step 5: Compare values for extrema

Since there are no critical points within the interval, compare the values at the endpoints: \(-\frac{\sqrt{2}}{2} \approx -0.707\) and \(\frac{1}{2} = 0.5\). The maximum value is \(0.5\) and the minimum value is \(-\frac{\sqrt{2}}{2}\).

Key Concepts

Understanding Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These are significant in calculus because they mark where a function might change from increasing to decreasing or vice versa. To find them, you take the derivative of your function and set it equal to zero. For trigonometric functions like \( \sin \theta \), it's all about finding where its derivative (\( \cos \theta \)) becomes zero.

It's essential to evaluate within the given interval, which is \([-\frac{\pi}{4}, \frac{\pi}{6}]\). Sometimes, within such intervals, the derivative doesn't become zero, which was the case here. Thus, the critical points were only at the interval's endpoints. Endpoints are crucial as they can often turn out to be the maximum or minimum values for a function, especially when there are no internal critical points.

Evaluating Intervals for Extremes

Interval evaluation involves checking function values at critical points and endpoints within a specified interval. In this problem, since there were no critical points inside the interval \([-\frac{\pi}{4}, \frac{\pi}{6}]\), it was necessary to evaluate the function at the endpoints.

• The endpoints here are \( \theta = -\frac{\pi}{4} \) and \( \theta = \frac{\pi}{6} \).
• Calculating \( r(-\frac{\pi}{4}) \) gave \( -\frac{\sqrt{2}}{2} \) and \( r(\frac{\pi}{6}) \) yielded \( \frac{1}{2} \).

By comparing these endpoint values, we could determine which was the maximum and which was the minimum, which is critical for understanding how the function behaves within the interval. Always consider endpoints in problems where internal critical points do not exist.

Trigonometric Functions in Calculus

Trigonometric functions like \( \sin \theta \) and their derivatives are prevalent in calculus because they model periodic behavior. When dealing with these functions, understanding their fundamental derivatives is crucial. For instance, the derivative of \( \sin \theta \) is \( \cos \theta \), which is pivotal in finding critical points and analyzing intervals.

Practically, we know that \( \cos \theta \) equals zero at specific values like \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \), but these did not appear within our interval \([-\frac{\pi}{4}, \frac{\pi}{6}]\), hence the critical points are absent.

When facing such functions:

• First, calculate the derivative.
• Find where it equals zero within your interval.
• If absent, focus on evaluating endpoints.

Understanding trigonometric functions and their behaviors at various intervals is essential for solving problems on maxima and minima in calculus.