Bernard Boyer

2022-07-27

$h\left(t\right)=\mathrm{cot}\left(t\right)$
[(3.14)/(4), ((3)(3.14)/(4))]
Find the average rate of change of the function over the given interval or intervals.

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.
$\frac{h\left(\frac{3\pi }{4}\right)-h\left(\frac{\pi }{4}\right)}{\frac{3\pi }{4}-\frac{\pi }{4}}=\frac{\mathrm{cot}\frac{3\pi }{4}-\mathrm{cot}\frac{\pi }{4}}{\frac{\pi }{2}}=\frac{2}{\pi }\left(\frac{\mathrm{cos}\frac{3\pi }{4}}{\mathrm{sin}\frac{3\pi }{4}}-\frac{\mathrm{cos}\frac{\pi }{4}}{\mathrm{sin}\frac{\pi }{4}}\right)=\frac{2}{\pi }\left(\frac{\frac{-\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}-\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\right)=\frac{2}{\pi }\left(-1-1\right)$
$=\frac{-4}{\pi }$

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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