Bernard Boyer

2022-07-27

$h(t)=\mathrm{cot}(t)$

[(3.14)/(4), ((3)(3.14)/(4))]

Find the average rate of change of the function over the given interval or intervals.

[(3.14)/(4), ((3)(3.14)/(4))]

Find the average rate of change of the function over the given interval or intervals.

thenurssoullu

Beginner2022-07-28Added 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 $\frac{f(b)-f(a)}{b-a}$

so in this case $a=\pi /4\text{}and\text{}b=3\pi /4$ then.

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

$=\frac{-4}{\pi}$

the average rate change of f(x) from a to b is $\frac{f(b)-f(a)}{b-a}$

so in this case $a=\pi /4\text{}and\text{}b=3\pi /4$ then.

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

$=\frac{-4}{\pi}$

posader86

Beginner2022-07-29Added 3 answers

average rate of change means:

$\frac{1}{t2-t1}{\int}_{t1}^{t2}\frac{d}{dt}\mathrm{cot}(t)dt$

Which is just:

$(\mathrm{cot}(t2)-\mathrm{cot}(t1))/(t2-t1)$

Here we have

$t1=\pi /4$

$t2=3\pi /4$

So the average is:

$(\mathrm{cot}(3\pi /4)-\mathrm{cot}(\pi /4))/(\pi /2)$

$=(-1-1)/(\pi /2)=-4/\pi $

$\frac{1}{t2-t1}{\int}_{t1}^{t2}\frac{d}{dt}\mathrm{cot}(t)dt$

Which is just:

$(\mathrm{cot}(t2)-\mathrm{cot}(t1))/(t2-t1)$

Here we have

$t1=\pi /4$

$t2=3\pi /4$

So the average is:

$(\mathrm{cot}(3\pi /4)-\mathrm{cot}(\pi /4))/(\pi /2)$

$=(-1-1)/(\pi /2)=-4/\pi $

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