Part I On a certain university campus there is an infestation of Norway rats. It

Question

Part I On a certain university campus there is an infestation of Norway rats. It is estimated that the number of rats on campus will follow a logistic model of the form P(t)= 5000/1+Be^kt

A) It is estimated that there were 500 rats on campus on January 1, 2010 and 750 on April 1, 2010. Using this information, find an explicit formula for P(t) where t is years since January 1, 2010. (Assume April 1, 2010 is t=.25.)
P(t)=

B) What was the rat population on October 1, 2010?

C) How fast was the rat population growing on April 1, 2010?

D) According to our logistic model, when will the rat population hit 2,500 rats?

_____________years after January 1, 2010.

E) Rats live in communal nests and the more rats there are, the closer they live together. Suppose the total volume of the rats' nests is F=(sqrt0.64P+4?2 )cubic meters when there are P rats on campus.
When there are 750 rats, what is the total volume of the rats' nests and how fast is the mass of nests growing with respect to time?
The total volume is_________cubic meters and the volume is increasing at

_____________cubic meters per year.

F) One of the reasons that the rats' population growth slows down is overcrowding. What is the population density of the rats' nests when there are 750 rats and how fast is the population density increasing at that time?
The population density is_______rats per cubic meter and the population density is increasing at

_________rats per cubic meter per year.

Explanation / Answer

A)
P(t) = 5000 / (1+B e^-kt)

Let t= 0 represent January 1, 2010
t = 4 represent April 1, 2010

P(0) = 500
500 = 5000 / (1+B)
500(1+B) = 5000
500 B = 4500
B = 9
P(t) = 5000 / ( 1+9e^-kt)

Substitute t=3
P(3) = 750
750 = 5000 / (1+9 e^-3k)
750 (1+ 9e^-3k) = 5000
1+9e^-3k = 6.6667
9e^-3k = 5.6667
e^-3k = 0.62963
-3k = ln(0.62963)
k = -ln(0.62963) / 3 = 0.154208

P(t) = 5000 / ( 1+ 9 e^-0.154208t)

B)
To find the October population, substitute t = 9 into P(t)
P(9) = 5000 / ( 1+ 9 e^[-0.154208 (9)) ] = 1540.14
P(9) = 1540

C)
dP/dt = 5000 B k e^(-kt) / ( 1+ B e^(-kt) )
k=0.154208
B=9
t=3
dP/dt = 1040.9 (when t=3)
655.4 rats per month

D)
Solve
2500 = 5000 / ( 1+ 9 e^-0.154208t ) for t
1+9 e^(-0.154208t) = 2
9 e^(-0.154208t) = 1
e^(-0.154208t) = 1/9
-0.154208t = ln(1/9)
t = -ln(1/9) / 0.154208 = 8.6468
t=14.25
January 1, 2010 is t=0
t=14.25 is second week of March 2011:
This is when the population would hit 2500.

E)
F = sqrt (0.64P +4) - 2
When P=750, total volume = sqrt( 0.64(750) +4) - 2
= 20 cubic metres. (total volume)

dF/dt = 0.6 / 2sqrt(0.6P+4)
substitute P=750
dF/dt = 6.39 cubic metres per year (increasing at)

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