We are looking for a function of the form y = Asin(B(x - C)) + D , or y = Acos(B(x - C)) + D, and the given information will allow us to solve for those 4 parameters, A, B, C, and D:
D represents the average value of the periodic function, so we average the high and low temps to solve for D, which = 63.
A, which represents the amplitude, we can now determine easily by calculating the distance between D and the high or low temp. In this case, A = 9, which means the function varies between 9 below 63 and 9 above 63.
B controls the period of the function, and can be calculated by using B = 2π / period. So here, since we want to model a 24-hr cycle, B = 2π / 24 = π/12.
Lastly, since the temps start at a low at t = 0, we can use a negative cosine function to make C, the phase shift, = 0. (Negative cosine functions start at a low value.) Note that because we are using a negative periodic function, A will be = - 9 now.
Putting all this together gives the following:
Let t: time (in hrs after midnight). Let P(t): temperature (in °F)
P(t) = - 9cos(π/12·t) + 63
You can confirm that this function gives the correct values at the correct times. Then evaluate it t = 16 (4 pm) to answer the next part of the question. For the last part, you will need to set P(t) = 60 and use inverse trig (in this case cos-1) to solve for t. My approximation suggests that the temp = 60 sometime between 4 am and 6 am for the first time. Whatever t-value you get, you should look at the graph to recognize that the temp will again be = 60 that many hrs before midnight of the next day. (Most trig equations have two distinct solutions within one full period, the second of which we find by recognizing the symmetry of the periodic function.)
