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QuestionFind the average rate of change of h(t)=cot⁡th(t)=\cot t over the intervals: a. [3π4,5π4]\left[\frac{3 \pi}{4}, \frac{5 \pi}{4}\right] b. [5π6,3π2]\left[\frac{5 \pi}{6}, \frac{3 \pi}{2}\right]

Studdy Solution

STEP 1

Assumptions1. The function is h(t)=cot⁡th(t)=\cot t . We are looking for the average rate of change over two intervals [3π4,5π4]\left[\frac{3 \pi}{4}, \frac{5 \pi}{4}\right] and [5π6,3π]\left[\frac{5 \pi}{6}, \frac{3 \pi}{}\right]

STEP 2

The average rate of change of a function over an interval [a,b][a, b] is given by the formulaAverage rate of change=h(b)−h(a)b−aAverage\, rate\, of\, change = \frac{h(b) - h(a)}{b - a}

STEP 3

First, we will find the average rate of change over the interval [3π,5π]\left[\frac{3 \pi}{}, \frac{5 \pi}{}\right]. Plug in the values for aa and bb into the formula.
Average rate of change=h(5π)−h(3π)5π−3πAverage\, rate\, of\, change = \frac{h\left(\frac{5 \pi}{}\right) - h\left(\frac{3 \pi}{}\right)}{\frac{5 \pi}{} - \frac{3 \pi}{}}

STEP 4

Evaluate h(π4)h\left(\frac{ \pi}{4}\right) and h(3π4)h\left(\frac{3 \pi}{4}\right).
h(π4)=cot⁡(π4)=−1h\left(\frac{ \pi}{4}\right) = \cot\left(\frac{ \pi}{4}\right) = -1h(3π4)=cot⁡(3π4)=−1h\left(\frac{3 \pi}{4}\right) = \cot\left(\frac{3 \pi}{4}\right) = -1

STEP 5

Substitute the values of h(5π4)h\left(\frac{5 \pi}{4}\right) and h(3π4)h\left(\frac{3 \pi}{4}\right) into the formula.
Average rate of change=−1−(−1)5π4−3π4=0π2=0Average\, rate\, of\, change = \frac{-1 - (-1)}{\frac{5 \pi}{4} - \frac{3 \pi}{4}} = \frac{0}{\frac{\pi}{2}} =0

STEP 6

Now, we will find the average rate of change over the interval [5π6,3π2]\left[\frac{5 \pi}{6}, \frac{3 \pi}{2}\right]. Plug in the values for aa and bb into the formula.
Average rate of change=h(3π2)−h(5π6)3π2−5π6Average\, rate\, of\, change = \frac{h\left(\frac{3 \pi}{2}\right) - h\left(\frac{5 \pi}{6}\right)}{\frac{3 \pi}{2} - \frac{5 \pi}{6}}

STEP 7

Evaluate h(3π2)h\left(\frac{3 \pi}{2}\right) and h(5π6)h\left(\frac{5 \pi}{6}\right).
h(3π2)=cot⁡(3π2)=0h\left(\frac{3 \pi}{2}\right) = \cot\left(\frac{3 \pi}{2}\right) =0h(5π6)=cot⁡(5π6)=−3h\left(\frac{5 \pi}{6}\right) = \cot\left(\frac{5 \pi}{6}\right) = -\sqrt{3}

STEP 8

Substitute the values of h(3π2)h\left(\frac{3 \pi}{2}\right) and h(5π6)h\left(\frac{5 \pi}{6}\right) into the formula.
Average rate of change=0−(−3)3π2−5π6=3π6=63Average\, rate\, of\, change = \frac{0 - (-\sqrt{3})}{\frac{3 \pi}{2} - \frac{5 \pi}{6}} = \frac{\sqrt{3}}{\frac{\pi}{6}} =6\sqrt{3}The average rate of change of the function h(t)=cot⁡th(t)=\cot t over the interval [3π4,5π4]\left[\frac{3 \pi}{4}, \frac{5 \pi}{4}\right] is 00 and over the interval [5π6,3π2]\left[\frac{5 \pi}{6}, \frac{3 \pi}{2}\right] is 636\sqrt{3}.

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