A population of rabbits oscillates 15 above and below average during the year

Question

A population of rabbits oscillates 15 above and below average during the year, hitting the lowest value in January (t = 0). The average population starts at 900 rabbits and increases by 190 each year. Find an equation for the population, P, in terms of the months since January, t.

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Answer 1

If you are still looking for help on this, it's definitely a real tricky problem because of that last bit about the average increasing each year. The first part isn't too bad if you break it into parts as a trigonometric function (oscillating is the keyword here). Knowing that it oscillates 15 above and below the average gives us an amplitude of 15. Since the lowest value is at t=0 we know that this is going to be a negative cosine function (positive cosine functions start at a high and sine starts at a midpoint). We can find the midpoint by looking at the average which is 900. Lastly, the period is found by seeing how often the function completes a cycle; since this is a yearly cycle the frequency is 12 months so the period is found by doing 2pi/12 which will give pi/6. With all of that, we can use the base formula: f(x) = acos(px) + m and fill in what we have to get

f(x) = -15cos(πx/6) + 900

To deal with the changing average population we need to add in a greatest common integer function that will span 12 months and change by 190 each time. This can be achieved by using g(x) = 190[[x/12]]

The final function will be: h(x) = -15cos(πx/6) + 900 + 190[[x/12]]