The fox population in a certain region has an annual growth rate of 6 percent per year. It is estimated that the population in the year 2000 was 26600

Question

The fox population in a certain region has an annual growth rate of 6 percent per year. It is estimated that the population in the year 2000 was 26600

(a) Find a function that models the population t years after 2000 (t=0 for 2000).Your answer is P(t)=

(b) Use the function from part (a) to estimate the fox population in the year 2008.Your answer is (the answer should be an integer)

Answer 1

P(t) = 26600e^.06t

P(8)= 26600e^.06(8)

= 26600e^.48

=about 26600(1.6160744)

= about 42,988 foxes in 2008

that's for continuous compounding, which best models biological growth

if for some reason the fox population compounded only yearly

such as a once a year breeding season? once a year between December and March, but the male is with each female for only 3 weeks then moves to a new mate in the next 3 weeks, so it's kind of more continuous for that 3 month period.

then

P(t) = 26600(1.06)^t

and

P(8) = 26600(1.06)^8 = about 42,396 foxes in 2008

since foxes tend to procreate continually, the continous compounding formula seems more realistic

but just annual compounding isn't that much less 42,396 instead of 42,988. about 592 less foxes

about 1% difference

Answer 2

How the equation comes:

t =0, at the beginning which is 2000

P(0) = 13600

for t=1

P(1) = 13600 + 6%(13600)

if we say r = rate then

P(1) = 13600 + r (13600) = 13600 (1 + r)

for t=2

P(2) =(previous year pop.) + r(previous year pop.)

P(2) = [13600(1 +r)] + r[(13600(1 + r)]. Factor out 13600(1 + r):

P(2) = 13600(1+r)(1+r) =13600(1+r)2.

for t=3, you'll see that:

P(3) = 13600(1+r)3

(a) P(t) = 13600(1 + r)t, where r=6% or 0.06

2008 is 8 years from year 2000, hence t=8

(b) P(8) = 13600 (1.06)8 = 21676