Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Monotonicity Theorem to find where the given function is increasing and where it is decreasing. $$ H(t)=\sin t, 0 \leq t \leq 2 \pi $$

Step-by-Step Solution

Answer

\( H(t) \) is increasing on \([0, \frac{\pi}{2}) \) and \((\frac{3\pi}{2}, 2\pi] \); decreasing on \((\frac{\pi}{2}, \frac{3\pi}{2}) \).

Step 1: Find the Derivative

The first step is to find the derivative of the function \( H(t) = \sin t \). The derivative of \( \sin t \) with respect to \( t \) is \( \cos t \). So we have \( H'(t) = \cos t \).

Step 2: Determine Where the Derivative is Positive, Negative, or Zero

To find where the function is increasing or decreasing, we need to determine where the derivative \( \cos t \) is positive, negative, or zero. We know that \( \cos t = 0 \) at \( t = \frac{\pi}{2} \) and \( t = \frac{3\pi}{2} \) within the interval \( [0, 2\pi] \).

Step 3: Analyze the Intervals Using the Derivative

-**Interval 1: \( 0 < t < \frac{\pi}{2} \)**: In this interval, \( \cos t > 0 \), so \( H(t) \) is increasing. -**Interval 2: \( \frac{\pi}{2} < t < \frac{3\pi}{2} \)**: Here, \( \cos t < 0 \), meaning \( H(t) \) is decreasing. -**Interval 3: \( \frac{3\pi}{2} < t < 2\pi \)**: In this interval, \( \cos t > 0 \), so \( H(t) \) is again increasing.

Step 4: Conclusion about Increasing and Decreasing Intervals

The function \( H(t) = \sin t \) is increasing on the intervals \( [0, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, 2\pi] \), and decreasing on the interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \).

Key Concepts

Derivative

A derivative represents the rate at which a function changes at any given point in its domain. In simpler terms, it tells us how fast a function is increasing or decreasing with respect to its independent variable. For any function, the derivative is often denoted as \( f'(x) \) or \( \frac{df}{dx} \).

In this exercise, we look at the function \( H(t) = \sin t \). To find where the function is increasing or decreasing, we start by determining its derivative. The derivative of the sine function, \( \sin t \), with respect to \( t \) is \( \cos t \). Therefore, the derivative \( H'(t) = \cos t \).

The derivative holds valuable information about the function's behavior. Where the derivative is positive, the function is increasing. Where the derivative is negative, the function is decreasing. These insights are crucial for understanding the nature of the given function across its domain, especially within specified intervals.

Trigonometric functions

Trigonometric functions such as sine, cosine, and tangent play a pivotal role in understanding periodic phenomena. These functions have unique properties and are defined using the unit circle concept in trigonometry.

For instance, the sine function, \( \sin t \), represents the y-coordinate of a point on the unit circle. Its derivative, \( \cos t \), gives us the slope of the tangent to the sine curve at any point \( t \).

Within the interval \( [0, 2\pi] \), the behavior of trigonometric functions showcases their periodic nature, exhibiting specific patterns:

• \( \sin t = 0 \) at \( t = 0, \pi, \text{and } 2\pi \).
• \( \cos t = 0 \) at \( t = \frac{\pi}{2} \text{ and } \frac{3\pi}{2} \).

These roots help determine intervals where these functions are increasing or decreasing, which is crucial for understanding their overall behavior.

Increasing and decreasing intervals

Increasing and decreasing intervals of a function tell us where the function grows or shrinks as we move through its domain. By leveraging the derivative, we can identify these intervals. According to the Monotonicity Theorem:

• A function is increasing on intervals where its derivative is positive.
• A function is decreasing on intervals where its derivative is negative.

For the function \( H(t) = \sin t \), the derivative \( H'(t) = \cos t \) was used to determine where \( H(t) \) is increasing or decreasing.

- **Interval 1:** From \( 0 < t < \frac{\pi}{2} \), \( \cos t > 0 \), so \( H(t) \) is increasing.
- **Interval 2:** From \( \frac{\pi}{2} < t < \frac{3\pi}{2} \), \( \cos t < 0 \), so \( H(t) \) is decreasing.
- **Interval 3:** From \( \frac{3\pi}{2} < t < 2\pi \), \( \cos t > 0 \), so \( H(t) \) is again increasing.

Understanding these intervals helps paint a full picture of how and where the function changes, which can be applied to various problems involving trigonometric functions or other types of functions with well-defined derivatives.