Periodic Functions

Question

A man claimed that the inside temperature of his shop should be less than the temperature of the outside for 75 % of the day (24 hours). He immediately found his recording instruments and measured the temperatures for 24 hours starting at 4am the next morning. The instruments indicated the following functions:

Inside of house temperature: T=21 - 3cos(πt/12) for 0 ≤ t ≤ 24

Outside temperature: T=22 - 5cos(πt/12) for 0 ≤ t ≤ 24

Does he have reasons from complaint? (Justify your answer)

THIS IS THE ANSWER THAT I GOT:

The temperatures in and out are:

Tin= 21 - 3cos(πt/12) for 0 ≤ t ≤ 24
Tout= 22 - 5cos(πt/12)

He wants
Tin < Tout for 75% of the time, or:

Tin - Tout < 0 for 75% of the time.

Putting in the expressions for the temps we get:

(21 - 3cos(*)) - (22-5cos(*)) < 0 75% of the time
where * = πt/12
This simplifies to:

2 cos(πt/12) < 1 75% of the time, 0 ≤ t ≤ 24

The function on the left will start at 2 at t=0 and decrease down to -2 at t=12 and back to 2 at t=24 in a usual cosine way.
The place where it first gets to/below 1 is when:
cos(πt/12) = 1/2
taking the inverse cosine of 1/2:
πt/12 = π/3 (fyi, this is 60 degrees)
and so this happens when:
t = 4 hours

So for t=0 to 4 the condition is not satisfied,
AND again for t=20 to 24 it will not be satisfied.
All together that's 8 hours out of 24, or 33% not satisfied.
So that's 67% of the time his condition is satisfied, less than the 75% he requires, and so he does have reason to complain.


IF YOU FIND THAT THERE IS A PROBLEM WITH THE INSIDE TEMPERATURE OF THE SHOP, PROVIDE A FULLY JUSTIFIED ALTERNATIVE PERIODIC FUNCTION WHICH WILL MATCH THE REQUIRED CONDITIONS. YOUR FUNCTION MUST ABOVE THAT THE INSIDE TEMPERATURE OF THE SHOP IS LESS THAN THE OUTSIDE TEMPERATURE FOR BETWEEN 79 % AND 90 % OF THE 24 HOUR DAY.

Answer 1

Good work so far!

 

Now you just need to find another periodic function that does better, without doing too much better.

 

 

You could either adjust the average indoor temperature (the vertical offset), or the swing (the amplitude of the periodic indoor temperature), or both.  Which choice would you prefer? Regardless of which you choose, you must now solve a new pair of inequalities.

 

I would suggest trying to write your inequality to calculate the length of time the interior temp is less than the exterior temp. You are doing the right things when you take the inverse cosine, noting that the both positive and negative 60 degrees are solutions (this corresponds to your first four hours of the day and the last four hours of the previous day).

 

Realize that what you have done so far establishes a statement explaining what the duration for which Tin < Tout is.

That statement can be used in a second separate inequality because that duration is what must be between 79% and 90%.

 

You will be able to rearrange your second inequality statement to solve for constraints on your selected parameter.  It is possible that just changing the swing won't work.