Problema Solution
The number of rabbits in a forest varies with time. During the period of 1980 to 1990 the number is given approximately by the formula 7000+3000 \sin(2 \pi t /5) where t is the number of years after 1980. (a) Sketch on graph paper the population during 1980-1990.
(b) What years was the population greatest (in increasing order)?
(c) What years was the population lowest (in increasing order)?
(d) How quickly was the population changing at the start of 1985?
Was it increasing or decreasing?
What are the units of your answer (write out all words)?
Answer provided by our tutors
(a) click here to see the graph of y=5000+5000sin(2πx/5), where x is the number of years after 1980
(b) y=5000+5000sin(2πx/5) has maximum when sin(2πx/5) = 1 that is
2πx/5 = π/2 + 2kπ, where k is integer
x = 5/4 + 5k, where k is integer
since 0<=x<=10 we have
for k = 0
x = 1.25 years
for k = 1
x = 6.25 years
in the years 1980 + 1.25 = 1980.25 and 1980 + 6.25 = 1986.25 the population is the greatest:
y=5000+5000*1 = 10,000 rabbits
(c) y=5000+5000sin(2πx/5) has minimum when sin(2πx/5) = -1 that is when
2πx/5 = 3π/2 + 2kπ, where k is integer
x = (15/4) + 5k
for k = 0
x = 3.75 years
for k = 1
x = 8.75 years
in the years 1980 + 3.75 = 1983.75 and 1980 + 8.75 = 1988.75 the population is the greatest:
y=5000+5000*(-1) = 0 rabbits
(d) Take derivative of the equation y=5000+5000sin(2πx/5) and put t =5 and find the answer.