Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the high temperature for the day is 81 degrees and the low temperature of 69 degrees occurs at 6 AM. Assuming it is the number of hours since midnight, find an equation for the temperature, D, in terms of t.

Therefore, we have:

At (t)  = 0 , D = - 6 sin 0 + 81 = 81  

At (t) = 6 AM, D = - 6 sin  ( π /2 ) + 81 = 79

t = 12 (noon), D = - 6 sin π + 81 = 81

t = 18 , (6:00 pm) , D = - 6 sin 3π / 2 + 81 = 79

Step-by-step explanation:

To start solving for this question, let us define or assume  a Sine Function for easy solving.

Defining the given parameters, we have:

Temperature =  81 - 69 = 12

Recalling from the question that

Temperature = number of hours before midnight,

Therefore Temperature (T) = 6

Also, we know that the give period of time is 24 hrs :

Therefore,

We have: 2π / k = 24 hours

Therefore,Making (K) the subject of formula, we have:

24 k = 2π

k = π / 12

Thus,

D = 6 sin ( π / 12 ) ( t ) + 81

With a given range of 81 and 69,

However, we know that the temperature after midnight would decrease, whereas our function has it increasing to 81 when t = 6 AM.

So therefore, we can attempt to swap the function, so we have:

D = -6 sin (π / 12 ) t + 81

Therefore, we have:

At (t)  = 0 , D = - 6 sin 0 + 81 = 81  

At (t) = 6 AM, D = - 6 sin  ( π /2 ) + 81 = 79

t = 12 (noon), D = - 6 sin π + 81 = 81

t = 18 , (6:00 pm) , D = - 6 sin 3π / 2 + 81 = 79

I hope this helps.