Quiz 2 Solution for Math 320, Lecture 2 - Spring 2009: Population Dynamics of Rabbits - Pr, Quizzes of Linear Algebra

The solution to quiz 2 problem of math 320, lecture 2 - spring 2009. The problem deals with the population dynamics of rabbits during an epidemic, where the death rate is proportional to the population. The solution involves writing the differential equation, solving the initial value problem, and determining the number of rabbits alive after 12 months.

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Pre 2010

Uploaded on 09/02/2009

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Quiz 2 Key
Math 320, Lecture 2 - Spring 2009
To get the most of this quiz, allow yourself no more than 20 minutes to completely answer it, and
do not use any notes or outside help. I will grade it if you hand it in during discussion on February
16 (for Monday sections), February 18 (for Wednesday sections).
1. Problem 1 (10 points)
An epidemic hits our favorite mammal to model: bunnies. As a result, the bunnies can no
longer reproduce. The death rate is proportional to the population of the remaining bunnies.
(a) Write the differential equation for the population P(t) of rabbits at time t.
(b) Assuming that P(0) = P0, solve the initial value problem.
(c) Suppose that when the epidemic hits at t=0, there are 1500 rabbits. Six months after
crisis strikes, there are only 500 rabbits remaining. How many rabbits will be alive in 12
months (t=12)?
Solution:
This problem was meant to use remind you of the logistic equation problems we did where
dP
dt = (βδ)P, where βand δare the rate of births and of deathes, respectively. Recall
homework problem # 17 from Section 2.2.
(a) The general equation is dP
dt = (βδ)P, where βand δare the birth and death rates,
respectively. In our problem, β= 0 and δ=kP for some positive constant k, because
there are no births and the death rate is propornal to P. So the differential equation is
dP
dt =kP 2.
(b) We can solve this by using the separation of variables technique.
dP
dt =kP 2
1
P2dP =k dt
Z1
P2dP =Z(k)dt
1
P=kt +C
1
P=kt +C0
P=1
kt +C0
At time t= 0, we have P(0) = P0. We must use this information in order to solve for our
constant C0.
P(t) = 1
kt +C0
P0=1
k(0) + C0
P0=1
C0
C0=1
P0
pf2

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Quiz 2 Key

Math 320, Lecture 2 - Spring 2009

To get the most of this quiz, allow yourself no more than 20 minutes to completely answer it, and do not use any notes or outside help. I will grade it if you hand it in during discussion on February 16 (for Monday sections), February 18 (for Wednesday sections).

  1. Problem 1 (10 points) An epidemic hits our favorite mammal to model: bunnies. As a result, the bunnies can no longer reproduce. The death rate is proportional to the population of the remaining bunnies. (a) Write the differential equation for the population P(t) of rabbits at time t. (b) Assuming that P (0) = P 0 , solve the initial value problem. (c) Suppose that when the epidemic hits at t=0, there are 1500 rabbits. Six months after crisis strikes, there are only 500 rabbits remaining. How many rabbits will be alive in 12 months (t=12)? Solution:

This problem was meant to use remind you of the logistic equation problems we did where dP dt = (β^ −^ δ)P^ , where^ β^ and^ δ^ are the rate of births and of deathes, respectively.^ Recall homework problem # 17 from Section 2.2.

(a) The general equation is dPdt = (β − δ)P , where β and δ are the birth and death rates, respectively. In our problem, β = 0 and δ = kP for some positive constant k, because there are no births and the death rate is propornal to P. So the differential equation is dP dt =^ −kP^

(b) We can solve this by using the separation of variables technique.

dP dt

= −kP 2 1 P 2

dP = −k dt ∫ 1 P 2

dP =

(−k) dt

P

= −kt + C 1 P

= kt + C′

P =

kt + C′

At time t = 0, we have P (0) = P 0. We must use this information in order to solve for our constant C′.

P (t) =

kt + C′ P 0 =

k(0) + C′

P 0 =

C′

C′^ =

P 0

Thus, we get

P (t) =

kt + (^) P^10

=

P 0

P 0 kt + 1

=

P 0

1 + P 0 kt

(c) We see from the given information that P 0 = 1500. Also, P (6) = 500. Using this piece of information, we can solve for the constant k.

P (t) =

1 + 1500kt 500 =

1 + 1500k(6) 1 + 9000k = 3 9000 k = 2

k =

Thus,

P (t) =

1 + 1500kt

4500

t

=

1 + 13 t

=

3 + t

We can now easily see that after 12 months, there will be P (12) = 300 bunnies remaining.

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