Please help Im so confused!!!

Question

Suppose that the monthly average precipitation at a particular location varies sinusoidally between a maximum of 6 inches in March and a minimum of 3 inches in September. Let t=1 correspond to January and t=12 correspond to December.

​(a) Use f(t)=acos(b(t−c))+d to model these conditions.

​(b) Evaluate ​f(3​) and f(9​).

Answer 1

We need to know what each of the 4 parameters (a,b,c, and d) does to the cosine function. It is often useful to think of these in terms of transformations, 2 vertical, 2 horizontal.

Let's begin with d, which causes a shift up or down, which we call the average value of the cosine function. It is the average of the highest y-value and the lowest y-value, which are 6 and 3. So d = 4.5.

The other parameter that transforms the function vertically, a stretch or shrink, is a, which we call the amplitude. It controls how far away up or down the y-values get from the average value, d. Because we know already d = 4.5, it is easy to see that a = 1.5. a can be calculated by (highest y-value - lowest y-value) / 2.

c controls a horizontal shift left or right, called a phase shift for sinusoidal functions. An untransformed cosine function starts at its peak when t = 0. This one has a peak when t = 3, so c = 3 (it will look like - 3 when we write the function, because we want to shift the graph 3 to the right).

Lastly, b is the horizontal stretch or shrink, and it controls the period of the sinusoid in the following way period = 2π / b or, since we know period and want b, b = 2π / period. The period here is 12 months, so b = 2π/12 = π/6

Let t: month of the year (with Jan = 1). f(t): avg. monthly precipitation (in inches)

f(t) = 1.5cos(π/6(t - 3)) + 4.5

You should graph this to confirm that it gives us a graph that fits the description given.

Answer 2

As the peak is 6 and the minimum is 3, a = (6 - 3) / 2 = 3/2

The period is 12. b = 2 π/ 12 = π/ 6

The peak is in March at t = 3, so c = 3

As the average is (6 + 3)/2 = 9/2, d = 9/2

3/2cos(π/6(t-3)) + 9/2

f(3) = 3/2cos(π/6(t-3)) + 9/2 = 3/2cos(π/6(3-3)) + 9/2 = 3/2cos0 + 9/2 = 3/2*1 + 9/2 = 12/2 = 6

f(9) = 3/2cos(π/6(t-3)) + 9/2 = 3/2cos(π/6(9-3)) + 9/2 = 3/2cosπ/ + 9/2 = 3/2*-1 + 9/2 = 6/2 = 3