Question

The temperature $T$ (in degree C) recorded in a city during a given day approximately followed the curve: $T = 0.002t^4 - 0.26t^2 + 21$, where $t$ is the number of hours from noon ($-12 \le t \le 12$ h). What was the average temperature during the day? Round 3 significant digits. Average temperature = Number $^\circ$C

          The temperature $T$ (in degree C) recorded in a city during a given day approximately followed the
curve: $T = 0.002t^4 - 0.26t^2 + 21$,
where $t$ is the number of hours from noon ($-12 \le t \le 12$ h).
What was the average temperature during the day?
Round 3 significant digits.
Average temperature = Number $^\circ$C

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The temperature T (in degree C) recorded in a city during a given day approximately followed the
curve: T = 0.002t^4 - 0.26t^2 + 21,
where t is the number of hours from noon (-12 ≤ t ≤ 12 h).
What was the average temperature during the day?
Round 3 significant digits.
Average temperature = Number ^∘C

Added by Benjamin C.

Calculus: Early Transcendentals

James Stewart 8th Edition

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The temperature in degrees Celsius recorded in a city during a given day approximately followed the curve T = 0.002t^4 + 0.26t^2 + 2t + 21, where t is the number of hours from noon (12h). What was the average temperature during the day? Round to 3 significant digits. Average temperature = Number

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Transcript

00:01 In this problem, the temperature t of t is given by 70 plus 10 times sine of pi t over 12, where t is the time in hours after 9 a .m.

00:25 In subpart a, we are asked to find out the temperature t at 9 a .m.

00:31 So this corresponds to t.

00:34 Equals to 0 since 0 hours have passed after 9 a .m.

00:39 So now substituting the value of t as 0, we get t of 0 equals to 70 plus 10 times sine of pi times 0 over 12 which is clearly equal to 0 and we know that sign of 0 equals to 0.

01:01 So we get t of 0 to be equal to 70.

01:06 And since the units is degrees fahrenheit, we get 70 degrees fahrenheit as the answer for subpart a.

01:18 Next, moving towards subpart b, in subpart b, we are asked to find the temperature at 3pm.

01:26 So this corresponds to t equals to 6 since 6 hours have passed after 9am.

01:34 So now substituting the value of t as 6, we get t of 6 equals to 70 plus 10 times sine of pi times 6 over 12.

01:48 Here 6 and 12 get cancelled 2 times.

01:53 So therefore we have t of 6 to be equal to 70 plus 10 times sine of pi over 2.

02:04 Since sine of pi over 2 equals to 1 we get 70 plus 10 which is equal to 80 so therefore t of 6 is equal to 80 degrees fahrenheit and this is our final answer for subpart b.

02:22 In subpart c we are asked to find out the average temperature which is denoted by t average average between 9am to 9 p .m.

02:38 So now we know that the value of t corresponding to 9am is 0 and the value of of t corresponding to 9 p m is 0 plus 12 which is 12.

02:51 So let us look at the formula for t average...

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