Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-10, 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 \(0 < t < \frac{\pi}{2}\) and \(\frac{3\pi}{2} < t < 2\pi\), decreasing on \(\frac{\pi}{2} < t < \frac{3\pi}{2}\).

Step 1: Find the derivative

First, we need to find the derivative of the function \(H(t) = \sin t\). The derivative of \(\sin t\) is \(\cos t\). So, \(H'(t) = \cos t\).

Step 2: Determine critical points

Next, we need to find the critical points of \(H(t)\). These occur where \(H'(t) = 0\). Set \(\cos t = 0\) and solve for \(t\). The solutions are \(t = \frac{\pi}{2}\) and \(t = \frac{3\pi}{2}\) within the interval \(0 \leq t \leq 2\pi\).

Step 3: Test intervals with the derivative

To determine where the function is increasing or decreasing, test the sign of \(H'(t) = \cos t\) in the intervals defined by the critical points: 1. \(0 < t < \frac{\pi}{2}\) 2. \(\frac{\pi}{2} < t < \frac{3\pi}{2}\) 3. \(\frac{3\pi}{2} < t < 2\pi\). - For \(0 < t < \frac{\pi}{2}\), \(\cos t > 0\), so \(H(t)\) is increasing.- For \(\frac{\pi}{2} < t < \frac{3\pi}{2}\), \(\cos t < 0\), so \(H(t)\) is decreasing.- For \(\frac{3\pi}{2} < t < 2\pi\), \(\cos t > 0\), so \(H(t)\) is increasing.

Step 4: State intervals of increase and decrease

Based on the sign of the derivative in each interval, \(H(t)\) is increasing on \(0 < t < \frac{\pi}{2}\) and \(\frac{3\pi}{2} < t < 2\pi\). It is decreasing on \(\frac{\pi}{2} < t < \frac{3\pi}{2}\).

Key Concepts

Monotonicity Theorem

The Monotonicity Theorem provides a way to understand how a function behaves in certain intervals. Specifically, it tells us where a function is increasing or decreasing.
To apply this theorem, we first need to find the derivative of the function. The derivative helps us determine the rate of change of the function.
Once we calculate the derivative, we look at its sign in different intervals:

• If the derivative is positive in an interval, the function is increasing there.
• If the derivative is negative, the function is decreasing.

Critical points are also found where the derivative equals zero. These points divide the number line into intervals that we test using the Monotonicity Theorem.

Derivative

In calculus, a derivative represents the rate of change of a function. For our problem, we deal with the derivative of a trigonometric function, specifically \(\sin t\).
For trigonometric functions, derivatives have specific rules that simplify the process. The derivative of \(\sin t\) is known to be \(\cos t\).
So, if our function is \(H(t) = \sin t\), then the derivative \(H'(t) = \cos t\). This derivative is key in applying the Monotonicity Theorem to determine where the function increases and decreases. By finding where this derivative equals zero, we can find critical points and test surrounding intervals.

Trigonometric Functions

Trigonometric functions such as \(\sin\), \(\cos\), and \(\tan\) have properties that make them unique. Their behaviors repeat over specific intervals, which makes them periodic.
In the context of our problem, we look at the function \(H(t) = \sin t\) in the interval \(0 \leq t \leq 2\pi\). This means we examine one full cycle of the sine function.
The derivative of \(\sin t\) within this interval gives us insight into where the function rises and falls. Understanding these behaviors helps in predicting the function's values between certain points, like critical points.

Critical Points

Critical points are crucial in calculus because they define boundaries in intervals where a function's behavior changes. A critical point occurs where the derivative is zero or undefined.
For \(H(t) = \sin t\) and \(H'(t) = \cos t\), critical points are where \(\cos t = 0\). Within \(0 \leq t \leq 2\pi\), this happens at \(t = \frac{\pi}{2}\) and \(t = \frac{3\pi}{2}\).
These points allow us to split the interval into smaller parts. We can then test each part to see if the function is increasing or decreasing by examining the sign of the derivative in these areas.