Question

. A population of rabbits oscillates 25 above and below an average of 129 during the year, hitting the lowest value in January (t = 0). a. Find an equation for the population, P, in terms of the months since January, t. b. What if the lowest value of the rabbit population occurred in April instead?

Answer 1

Answer:

Because we know that here we have an oscillation, we can model this with a sine or cosine function.

P = A*cos(k*t) + M

where:

k is the frequency

A is the amplitude

M is the midline

We know that at t = 0, we have the lowest population.

We know that the mean is 129, so this is the midline.

We know that the population oscillates 25 above and below this midline,

And we know that for t = 0 we have the lowest population, so:

P = A*cos(k*0) + 129 = 129 - 25

P = A + 129 = 129 - 25

A = -25

So, for now, our equation is

P = -25*cos(k*t) + 129

Because this is a yearly period, we should expect to see the same thing for t = 12 (because there are 12 months in one year).

And remember that the period of a cosine function is 2*pi

Then:

k*12 = 2*pi

k = (2*pi)/12 = pi/6

Finally, the equation is:

P = -25*cos(t*pi/6) + 129

Now we want to find the lowest population was in April instead:

if January is t = 0, then:

February is t = 2

March is t = 3

April is t = 4

Then we would have that the minimum is at t = 4

If we want to still use a cosine equation, we need to use a phase p, such that now our equation is:

P = -25*cos(k*t + p) + 129

Such that:

cos(k*4 + p) = 1

Then:

k*4 + p = 0

p =  -k*4

So our equation now is:

P = -25*cos(k*t  - 4*k) + 129

And for the periodicity, after 12 months, in t = 4 + 12 = 16, we should have the same population.

Then, also remembering that the period of the cosine function is 2*pi:

k*12 - 4*k = 2*pi

k*8 = 2*pi

k = 2*pi/8 = pi/4

And remember that we got:

p = -4*k = -4*(pi/4) = -pi

Then the equation for the population in this case is:

P = -25*cos( t*pi/4 - pi) + 129