Question

In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function T(t) = 54 + 11 sin πt 12 . Find the average temperature Tave during the period from 9 AM to 9 PM. (Round your answer to the nearest whole number.)

Answer 1

`T_(ave) \approx 55^(\circ)F`

Explanation:

Given :
`T(t) = 54 + 11 sin ((\pi t)/(12))`

To Find : Find the average temperature Tave during the period from 9 AM to 9 PM.

Solution:

`T(t) = 54 + 11 sin ((\pi t)/(12))`

Where t is hours after 9 AM

Now we are supposed to find the average temperature Tave during the period from 9 AM to 9 PM

Period : 9 a.m. to 9 p.m.

Hours between 9 am to 9 pm = 12

Interval of function: [0,12]

So, we will use mean value theorem

`T_(ave)=(1)/(b-a)\int\limits^b_a {T(t)} \, dt`

`T_(ave)=(1)/(12-0)\int\limits^(12)_0{ 54 + 11 sin ((\pi t)/(12))} \, dt`

`T_(ave)=(1)/(12)\int\limits^(12)_0{54}\, dt+(1)/(12)\int\limits^(12)_0{ 11 sin ((\pi t)/(12))} \, dt`

`T_(ave)=(1)/(12)\int\limits^(12)_0{54}\, dt+(1)/(12)\int\limits^(12)_0{ 11 sin ((\pi t)/(12))} \, dt`

let
`(\pi t)/(12)=u`

`du =(\pi)/(12)dt\\(12)/(\pi)du=dt`

u(0)=0

u(12)=π

Now apply this substitution on right integral

`T_(ave)=(1)/(12)[54t]^(12) _0+(1)/(12)  *(12)/(\pi)* 11 \int\limits^(\pi)_0{ sin u} \, du`

`T_(ave)=(1)/(12)(54 * 12-0)+(11)/(\pi)(- cos u)^(\pi)_0`

`T_(ave)=54+(11)/(\pi)(- cos {\pi}-(-cos 0))^(\pi)_0`

`T_(ave)=54+(11)/(\pi)(- (-1)-(-1))`

`T_(ave)=54+(11)/(\pi)(2)`

`T_(ave)=54+(22)/(\pi)`

So,
`T_(ave)=54+(22)/(180) ^(\circ)F`

`T_(ave) \approx 55^(\circ)F`

Hence the average temperature Tave during the period from 9 AM to 9 PM is
`55^(\circ)F`