Zombies attack Oxford!!!

Zombie attack!

A zombie outbreak has struck Oxford!

Zombie attack!

The rate of transfer of the zombie virus is proportional to thenumber of students infected and the number not infected.

Let P(t) represent the number of students infected at time t (inhours), M = 17, 000 the total number of students, and k theinfection rate.

We can model this situation with the differential equation

dPdt = k P(M − P)

M .

The value M is known as the carrying capacity.

Zombie attack!

Initially, 1000 students are infected. Thus, P(0) = P0 = 1000.

The virus is spreading at a continuous rate of 10% an hour.

How long until half of the student body is infected?

Zombie attack!

Zombie attack!

We have a limited amount of time. Let’s change units so that thepopulation is out of thousands. Then M = 17 and P0 = 1.

Given the DEdPdt = .1P(17− P)

17 .

Separation givesdP

P(17− P) = .1 117 dt.

We now must integrate both sides.To integrate the left-hand side,we need to use Partial Fraction Decomposition.

Zombies hate Partial Fraction Decomposition.

Zombie attack!

∫ 117

( 1P + 1

17− P

)dP =

∫.1 1

17 dt

117 (ln |P| − ln |17− P|) = .1 t

17 + C

ln∣∣∣∣17− P

P

∣∣∣∣ = −.1t + C

17− PP = ±eCe−.1t .

Zombie attack!

Let A = ±eC , then17− P

P = Ae−.1t .

We have P = 1 when t = 0, so A = 16. Thus,

17P = 1 + 16e−.1t

and soP(t) = 17

1 + 16e−.1t .

Note thatlim

t→∞P(t) = 17.

Zombie attack!

How long until half the student body is infected? That is, at whattime t will P(t) = 8.5?

8.5 = P(t) = 171 + 16e−.1t

12 = 1

1 + 16e−.1t

1 + 16e−.1t = 2e−.1t = 1/16− .1t = ln(1/16)

t = ln(16)/.1 ≈ 27.73 hours.

Zombie attack!

Zombie attack!

The general solution to the logistic equation with population size(carrying capacity) M and initial value P(0) = P0 is

P(t) = M1 + Ae−kt where A = M − P0

P0.