|
CEE 370 |
Fall 2010 |
Homework #5
1. Do Problem 5-1 in the D&M text.
A
population of purple rabbits lives on the island of Zulatop. The rabbits have a net growth rate of 0.09 yr-1. At the present time there are 176 rabbits on
the island. What is the predicted
number of rabbits 5, 10, and 20 years from now?
Use the simple exponential growth equation to calculate the number of
rabbits. In addition, make a plot of purple rabbit population versus time for a
period from 0 to 25 years.
5-1 Population projection of rabbits
Given: Net growth rate = 0.09 yr-1, initial population = 176.
Solution:
a. Using Eqn 5-11
Nt = Noe(rt)
Nt =
176e(0.09)t
b. At 5, 10 and 20 years,
t = 5
Nt = 176e(0.09)(5)
= 276 rabbits
t = 10
Nt = 176e(0.09)(10)
= 432.89 or 433 rabbits
t = 20
Nt = 176e(0.09)(20) = 1064.74 or 1065 rabbits
2. Do Problem 5-2 in the D&M text.
Recalculate the number of purple rabbits if the carrying capacity is 386 and you use the logistic equation. Assume the number of rabbits at the present time and use the same intervals. In addition, make a plot of purple rabbit population versus time for a period from 0 to 25 years.
5-2 Population projection using logistic equation
Given: Problem 5-1, carrying capacity = 386.
Solution:
a. Using Eqn 4-14
K(No)
N(t) = ------------------------
(No + (K – No)e-rt)
386(176)
N(t) = -------------------------------
(176 + (386 – 176)e-(0.09)t)
67936
N(t) = ------------------------
(176 + (210)e-(0.09)t)
b. At 5, 10, and 20 years
t = 5
67936
N(t) =
-------------------------- = 219.23
or 219 rabbits
(176 +
(210)e-(0.09)(5))
t = 10
67936
N(t) =
--------------------------- = 259.92
or 260 rabbits
(176 +
(210)e-(0.09)(10))
t = 20
67936
N(t) =
---------------------------- = 322.42
or 322 rabbits
(176 +
(210)e-(0.09)(20))
3. Do Problem 5-12 in the D&M text.
The
concentration of diazinon has been measured to be 23.3 µg/L in Lake Pekko. The biocentration factor for diazinon is 337
L/kg. What is the expected concentration
of diazinon in fish living in Lake Pekko?
Given: Diazinon = 23.3 μg/L, bioconcentration factor = 337 L/kg.
Solution:
Conc.
= (23.3 μg/L)(337
L/kg) = 7850 μg/kg
4. Do Problem 5-14 in the D&M text.
The
bioconcentration factor for bis(2-ethylhexyl)phthalate, a commonly used
plasticizer, in the organism Daphnia
is 5200 L/kg. If the concentration of
bis(2-ethylhexyl)phthalate in a lake is 3.6 µg/L, estimate the concentration of
bis(2-ethylhexyl)phthalate in Daphnia
in units of µg/kg.
Given: bis(2-ethylhexyl)phthalate = 3.6 μg/L, bioconcentration factor = 5200 L/kg.
Solution:
Conc.
= (3.6 μg/L)(5200
L/kg) = 18,720 μg/kg