Question

The population of a community of foxes is observed to fluctuate on a 10-year cycle due to variations in the availability of prey. When population measurements began (t=0, the population was 35 foxes. The growth rate in units of foxes>>year was observed to be P′(t)=5+10sinπt/5​ 

a. What is the population 15 years later? 35 years later?
b. Find the population P(t) at any time t≥0.

Answer 1

Population 15 years later P(15) = 100 + 50/π

Population 35 years later P(35) = 200 + 50/π

Population any t ≥ 0 P(t) = 35 + 50 /π + 5*t + 10*cos(π*t/5)

Explanation:

P´(t) = 5 + 10*sinπt/5 ⇒ dP/dt = 5 + 10*sinπt/5

dP = ( 5 + 10*sinπt/5 ) *dt

P(t) = ∫ ( 5 + 10*sinπt/5 ) dt

P(t) = 5*t + 10 * ∫ sinπ*t/5* dt

P(t) = 5*t - 10*5/π *cos πt/5 + k

To determine k t = 0 P(t) = 35

P(0) = 5*0 - 50/π (1) + k

35 = - 50/π + k

k = 35 + 50/π and

P(t) = 5*t + 10*cos(π*t/5) + 35 + 50/π

b)P(t) = 35 + 50 /π + 5*t + 10*cos(π*t/5) (1)

a) Population 15 years later

P(15) = 35 + 50/π + 5*15 - 10

P(15) = 100 + 50/π

Again from equation (1)

P(35) = 35 + 50 /π + 5*35 + 10*cos(35*π/5)

P(35) = 35 + 50/π + 175 + 10*cos (7*π )

P(35) = 210 + 50/π - 10

P(35) = 200 + 50/π