# h(t)= cot(t) [(3.14)/(4), ((3)(3.14)/(4))] Find the average rate of change of the function over the giveninterval or intervals.

Bernard Boyer 2022-07-27 Answered
$h\left(t\right)=\mathrm{cot}\left(t\right)$
[(3.14)/(4), ((3)(3.14)/(4))]
Find the average rate of change of the function over the given interval or intervals.
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## Answers (2)

thenurssoullu
Answered 2022-07-28 Author has 13 answers
The formula the poster listed above is used to find the averagefunction value...not the average rate of change.
the average rate change of f(x) from a to b is $\frac{f\left(b\right)-f\left(a\right)}{b-a}$
so in this case then.
$\frac{h\left(\frac{3\pi }{4}\right)-h\left(\frac{\pi }{4}\right)}{\frac{3\pi }{4}-\frac{\pi }{4}}=\frac{\mathrm{cot}\frac{3\pi }{4}-\mathrm{cot}\frac{\pi }{4}}{\frac{\pi }{2}}=\frac{2}{\pi }\left(\frac{\mathrm{cos}\frac{3\pi }{4}}{\mathrm{sin}\frac{3\pi }{4}}-\frac{\mathrm{cos}\frac{\pi }{4}}{\mathrm{sin}\frac{\pi }{4}}\right)=\frac{2}{\pi }\left(\frac{\frac{-\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}-\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\right)=\frac{2}{\pi }\left(-1-1\right)$
$=\frac{-4}{\pi }$
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posader86
Answered 2022-07-29 Author has 3 answers
average rate of change means:
$\frac{1}{t2-t1}{\int }_{t1}^{t2}\frac{d}{dt}\mathrm{cot}\left(t\right)dt$
Which is just:
$\left(\mathrm{cot}\left(t2\right)-\mathrm{cot}\left(t1\right)\right)/\left(t2-t1\right)$
Here we have
$t1=\pi /4$
$t2=3\pi /4$
So the average is:
$\left(\mathrm{cot}\left(3\pi /4\right)-\mathrm{cot}\left(\pi /4\right)\right)/\left(\pi /2\right)$
$=\left(-1-1\right)/\left(\pi /2\right)=-4/\pi$
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