The right answer will get 5 stars right away. Chapter 2, Section 2.3, Additional

Question

The right answer will get 5 stars right away.

Chapter 2, Section 2.3, Additional Question 01 Your answer is incorrect. Try again. The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of other factors, the population doubles each week. There are 600,000 mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 20,000 mosquitoes/day. Determine the population of mosquitoes in the area at any time. (Note that the variable t represents days.) Enter an exact answer. Do not use decimal approximations. P (t)(600000"ln(2)-140000)

Explanation / Answer

In Absence of other factors

P'(t)=kP(t)

ln(P(t))=kt + ln(C)

P(t)=Cexp(kt)

Every 7 days the population doubles, implies that exp(7k)=2 or k=ln(2)/7=0.1

In presence of predators

P'(t)=0.1P(t)-20000

ln(P(t)-200000)=0.1t +ln(C)

P(t)-200000=Cexp(0.1t)

P(t)=Cexp(0.1t)+200000

P(0)=600000 implies that C=400000

Therefore P(t)=400000exp(0.1t)+200000

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.