Find the average rate of change of the function over the given intervals.
​h(t)equalscotangent t
a. left bracket StartFraction 5 pi Over 4 EndFraction comma StartFraction 7 pi Over 4 EndFraction right bracket
b. left bracket StartFraction pi Over 3 EndFraction comma StartFraction pi Over 2 EndFraction right bracket

The average rate of change of a function \( h(t) \) over an interval \([a, b]\) is given by the formula:

\[
\text{Average Rate of Change} = \frac{h(b) - h(a)}{b - a}
\]

For the given function \( h(t) = \cot(t) \), we will compute the average rate of change for both intervals.

### Part a: Interval \( \left[ \frac{5\pi}{4}, \frac{7\pi}{4} \right] \)

1. Compute \( h\left(\frac{5\pi}{4}\right) \) and \( h\left(\frac{7\pi}{4}\right) \):
   \[
   h\left(\frac{5\pi}{4}\right) = \cot\left(\frac{5\pi}{4}\right)
   \]
   Since \( \cot\left(\frac{5\pi}{4}\right) = \cot(225^\circ) = 1 \),
   \[
   h\left(\frac{7\pi}{4}\right) = \cot\left(\frac{7\pi}{4}\right)
   \]
   Since \( \cot\left(\frac{7\pi}{4}\right) = \cot(315^\circ) = -1 \).

2. Compute the average rate of change:
   \[
   \text{Average Rate of Change} = \frac{h\left(\frac{7\pi}{4}\right) - h\left(\frac{5\pi}{4}\right)}{\frac{7\pi}{4} - \frac{5\pi}{4}} = \frac{-1 - 1}{\frac{7\pi}{4} - \frac{5\pi}{4}} = \frac{-2}{\frac{2\pi}{4}} = \frac{-2}{\frac{\pi}{2}} = -\frac{4}{\pi}
   \]

### Part b: Interval \( \left[ \frac{\pi}{3}, \frac{\pi}{2} \right] \)

1. Compute \( h\left(\frac{\pi}{3}\right) \) and \( h\left(\frac{\pi}{2}\right) \):
   \[
   h\left(\frac{\pi}{3}\right) = \cot\left(\frac{\pi}{3}\right)
   \]
   Since \( \cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}} \),
   \[
   h\left(\frac{\pi}{2}\right) = \cot\left(\frac{\pi}{2}\right) = 0
   \]

2. Compute the average rate of change:
   \[
   \text{Average Rate of Change} = \frac{h\left(\frac{\pi}{2}\right) - h\left(\frac{\pi}{3}\right)}{\frac{\pi}{2} - \frac{\pi}{3}} = \frac{0 - \frac{1}{\sqrt{3}}}{\frac{\pi}{2} - \frac{\pi}{3}} = \frac{-\frac{1}{\sqrt{3}}}{\frac{\pi}{6}} = -\frac{6}{\pi \sqrt{3}} = -\frac{2\sqrt{3}}{\pi}
   \]

### Summary:

- **Part a**: The average rate of change is \( -\frac{4}{\pi} \).
- **Part b**: The average rate of change is \( -\frac{2\sqrt{3}}{\pi} \).

Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. How is the cotangent function related to the tangent function?
2. What does the average rate of change represent geometrically for trigonometric functions?
3. Can the average rate of change be positive for the cotangent function?
4. How would the average rate of change change if the interval were shifted?
5. How is the cotangent function derived from the sine and cosine functions?

**Tip**: When dealing with trigonometric functions, it's helpful to know key values at standard angles to simplify your calculations.

Find the average rate of change of the function over the given intervals. h(t) = cotangent(t) a. [5π/4, 7π/4] b. [π/3, π/2]

Learn how to calculate the average rate of change for the cotangent function over intervals [5π/4, 7π/4] and [π/3, π/2]. This problem involves using the formula for the average rate of change and key trigonometric values for the cotangent function at standard angles. Suitable for students in Grades 10-12.

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