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QuestionFind the average rate of change of h(t)=cot⁡th(t)=\cot t over [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 considering the interval [5π6,3π]\left[\frac{5 \pi}{6}, \frac{3 \pi}{}\right]
3. We are calculating the average rate of change of the function over this interval, which is given by ΔyΔt=cot⁡(3π)−cot⁡(5π6)3π−5π6\frac{\Delta y}{\Delta t}=\frac{\cot \left(\frac{3 \pi}{}\right)-\cot \left(\frac{5 \pi}{6}\right)}{\frac{3 \pi}{}-\frac{5 \pi}{6}}

STEP 2

First, we need to calculate the values of the cotangent function at the endpoints of the interval.Let's start with cot⁡(π2)\cot \left(\frac{ \pi}{2}\right).

STEP 3

We know that the cotangent function is the reciprocal of the tangent function, and that the tangent of 3π2\frac{3 \pi}{2} is undefined. Therefore, the cotangent of 3π2\frac{3 \pi}{2} is also undefined.
cot⁡(3π2)=1tan⁡(3π2)=1undefined=undefined\cot \left(\frac{3 \pi}{2}\right) = \frac{1}{\tan \left(\frac{3 \pi}{2}\right)} = \frac{1}{\text{undefined}} = \text{undefined}

STEP 4

Next, let's calculate cot⁡(π6)\cot \left(\frac{ \pi}{6}\right).

STEP 5

We know that the cotangent function is the reciprocal of the tangent function, and that the tangent of 5π\frac{5 \pi}{} is −1-1. Therefore, the cotangent of 5π\frac{5 \pi}{} is also −1-1.
cot⁡(5π)=1tan⁡(5π)=1−1=−1\cot \left(\frac{5 \pi}{}\right) = \frac{1}{\tan \left(\frac{5 \pi}{}\right)} = \frac{1}{-1} = -1

STEP 6

Now that we have the values of the function at the endpoints of the interval, we can substitute these values into the formula for the average rate of change.
ΔyΔt=cot⁡(3π2)−cot⁡(5π6)3π2−5π6=undefined−(−1)3π2−5π6\frac{\Delta y}{\Delta t}=\frac{\cot \left(\frac{3 \pi}{2}\right)-\cot \left(\frac{5 \pi}{6}\right)}{\frac{3 \pi}{2}-\frac{5 \pi}{6}} = \frac{\text{undefined} - (-1)}{\frac{3 \pi}{2}-\frac{5 \pi}{6}}

STEP 7

Since the numerator of the fraction is undefined, the whole fraction is undefined.
Therefore, the average rate of change of the function h(t)=cot⁡th(t)=\cot t over the interval [5π6,3π2]\left[\frac{5 \pi}{6}, \frac{3 \pi}{2}\right] is undefined.

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