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 for complaint? (Justify your answer)

Answer 1

The temperatures in and out are:

 

  T_in= 21 - 3cos(πt/12)      for 0 ≤ t ≤ 24 

T_out= 22 - 5cos(πt/12)

 

He wants

  T_in < T_out  for 75% of the time, or:

 

 T_in - T_out < 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.