Question

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

Short answer

The function is increasing on \([0, \frac{\pi}{2})\) and \((\frac{3\pi}{2}, 2\pi)\), and decreasing on \((\frac{\pi}{2}, \frac{3\pi}{2})\).

Step-by-step solution

Find the Derivative

To start, we'll find the derivative of the function \(H(t) = \sin t\). The derivative of \(\sin t\) is \(\cos t\). Thus, \(H'(t) = \cos t\).

Determine Critical Points

Set the derivative \(H'(t) = \cos t\) equal to zero to find the critical points. \(\cos t = 0\) at \(t = \frac{\pi}{2}\) and \(t = \frac{3\pi}{2}\) within the interval \([0, 2\pi]\).

Test Intervals

Divide the interval \([0, 2\pi]\) into subintervals: \([0, \frac{\pi}{2}])\), \((\frac{\pi}{2}, \frac{3\pi}{2})\), and \((\frac{3\pi}{2}, 2\pi]).\) Test a point from each interval in \(H'(t)\).

Analyze Each Subinterval

1. For \(t \in (0, \frac{\pi}{2})\), choose \(t = \frac{\pi}{4}\). \(\cos(\frac{\pi}{4}) > 0\), so \(H(t)\) is increasing.2. For \(t \in (\frac{\pi}{2}, \frac{3\pi}{2})\), choose \(t = \pi\). \(\cos(\pi) < 0\), so \(H(t)\) is decreasing.3. For \(t \in (\frac{3\pi}{2}, 2\pi)\), choose \(t = \frac{7\pi}{4}\). \(\cos(\frac{7\pi}{4}) > 0\), so \(H(t)\) is increasing.

Conclusion

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

Key concepts

Derivative

The derivative is a fundamental mathematical tool used to describe how a function changes at any given point. In simple terms, it tells us the slope of the tangent line to the curve of a function at a specific point. For trigonometric functions like the sine function, the derivative becomes particularly interesting.
To compute the derivative of the function \(H(t) = \sin t\), we use a basic trigonometric derivative rule: the derivative of \(\sin t\) is \(\cos t\). This means, for any value of \(t\), the rate of change of \(\sin t\) is \(\cos t\).

• \(H'(t) = \cos t\) tells us how steep the sine wave is at any point \(t\).
• This derivative helps us find out where the function \(H(t)\) is increasing or decreasing.

It's essential to recognize that the derivative forms the basis for determining critical points and analyzing intervals.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These are points on the graph where the function's rate of change switches direction, meaning it can change from increasing to decreasing or vice versa.
For the function \(H(t) = \sin t\), we set \(H'(t) = \cos t = 0\) to find these critical points. Solving \(\cos t = 0\), we get:

• \(t = \frac{\pi}{2}\)
• \(t = \frac{3\pi}{2}\)

Both values are within the interval \([0, 2\pi]\). This tells us that at \(t = \frac{\pi}{2}\) and \(t = \frac{3\pi}{2}\), the direction of \(H(t)\) might change. Understanding critical points helps us segment the function into smaller parts for further analysis in interval testing.

Interval Testing

Interval testing involves dividing the domain of the function into smaller segments around critical points, then testing each segment to see if the function is increasing or decreasing in those intervals.
For \(H(t) = \sin t\), the interval \([0, 2\pi]\) is segmented into:

• \([0, \frac{\pi}{2})\)
• \((\frac{\pi}{2}, \frac{3\pi}{2})\)
• \((\frac{3\pi}{2}, 2\pi])\)

We test points within each interval to determine the sign of \(H'(t) = \cos t\):

• In \((0, \frac{\pi}{2})\), test \(t = \frac{\pi}{4}\) and find \(\cos(\frac{\pi}{4}) > 0\); thus, \(H(t)\) is increasing.
• In \((\frac{\pi}{2}, \frac{3\pi}{2})\), test \(t = \pi\) and find \(\cos(\pi) < 0\); thus, \(H(t)\) is decreasing.
• In \((\frac{3\pi}{2}, 2\pi)\), test \(t = \frac{7\pi}{4}\) and find \(\cos(\frac{7\pi}{4}) > 0\); thus, \(H(t)\) is increasing.

Through interval testing, you gain a clear understanding of where the function's behavior shifts.

Increasing and Decreasing Functions

Increasing functions are those where the output (or y-value) grows as the input (or x-value) increases over an interval. Conversely, decreasing functions show a decline in y-values as x-values rise.
For the function \(H(t) = \sin t\):

• The function is increasing on the intervals \([0, \frac{\pi}{2})\) and \((\frac{3\pi}{2}, 2\pi)\).
• It is decreasing on the interval \((\frac{\pi}{2}, \frac{3\pi}{2})\).

This behavior corresponds to the positive and negative signs of \(H'(t) = \cos t\) in those intervals, as revealed by interval testing.
Understanding where a function increases or decreases is essential for sketching graphs, optimizing problems, and many applications in calculus.