CIV 220F ENGINEERING ECOLOGY
Practice Problems #2: Population Growth
The solutions to these problems will be posted on the class website.
1. 10 pairs of cougars are reintroduced into a region that had historically contained cougars.
Growth of the cougar population was exponential for four years after introduction with b = 4
yr–1 and d = 3.5 yr–1. In the fifth year, after a new provincial government weakened habitat
protection of the region, a road was constructed through the region. Ecologists observed that
the death rate, d, increased for 5 years after construction of the road before stabilizing in year
10. The expression d = 3.5 + 0.08·t, described the death rate during that 5-year period,
where t = time after road construction.
a. What was the cougar population after 3 years?
b. What was the cougar population growth rate (cougars/yr) 3 years after road
construction if the population at that time was 575 cougars?
c. When would the cougar population first begin to decline if the death rate did not
stabilize in year 10?
2. The current population of beluga whales in the St. Lawrence river system is approximately
500. Assume that the instantaneous specific reproduction rate, b, is equal to 0.2 yr–1 and that
the instantaneous specific death rate, d, is 0.1 yr–1. Also assume that the exponential growth
model adequately describes the population.
a. What is the initial growth rate for beluga whales in the St. Lawrence system?
b. What is the growth rate when the beluga whale population is 750 whales?
c. When will the beluga whale population reach 1,250 whales?
d. Autopsies of dead beluga whales indicate that the cause of death in many cases is
pollution. Compounds such as PCBs, DDT, and other pesticides accumulate in the
whale blubber to toxic levels. If the population of beluga whales in the St. Lawrence
river system 20 years ago was 1,500, what was the instantaneous specific death rate
over the last 20 years? Assume the instantaneous specific reproduction rate has
remained constant at 0.2 yr–1.
3. The squirrel population in a large urban park has been studied after the introduction of 3
pairs of squirrels 10 years ago. Migration into and out of the park has been determined to be
negligible. The population data are as follows:
1
Year
0 (squirrels introduced)
2
4
7
10 (present)
Squirrel Population
6
12
24
65
167
a. Estimate the net specific growth rate, r, for the squirrels in the park. Assume that
initially growth is exponential.
b. If the rate of population change, dN/dt, for the squirrels in the park at year 10 is 50
squirrels/year, estimate the carrying capacity, K, for the park.
c. Predict the squirrel populations at years 15 and 40.
Competition
4. (previous exam question)
The populations of two species in competition for resources may be described by the LotkaVolterra equations. At steady-state, these equations reduce to two linear equations:
N1 = K1 – αN2
N2 = K2 – βN1
Two species of grass have the following parameters:
K1 = 100 plants/m2
K2 = 120 plants/m2
r1 = 0.5 yr–1
r2 = 1.5 yr–1
α = 0.8 and β = 0.6
a) Plot the steady-state equations for species populations N1 and N2
Your diagram should be a neat sketch, drawn approximately to scale.
b) If the populations start away from steady-state at values N1 = 25 and N2 = 25, plot on your
graph the starting point and the approximate path to the steady-state position. What are the two
populations at steady-state?
c) Consider a different starting point where the populations begin at N1 =100 and N2 = 25.
2
Use the full Lotka-Volterra equations to determine the rates of change of the two species at the
starting point, i.e., determine
dN 1 and dN 2 at the specified values of N and N
1
2.
dt
dt
What will be the approximate change in species populations over the next one year? Show this
change as a vector on your diagram.
What will be the populations at steady-state?
5. Competing species usually have traits that improve their competitiveness under certain
environmental conditions. For example, one grass may be more competitive under dry
conditions and another more competitive under wet conditions. Consider two populations of
prairie grass. The first, a native grass, has r1 = 0.2 yr–1, K1 = 50 + 3(P – 15) plants/m2 (P =
precipitation, cm/yr), and β = 0.75. The second grass, an imported species, has r2 = 0.25 yr–
1, K = 100 + 1.5(P – 10) plants/m2, and α = 0.8.
2
a.
b.
c.
d.
At what precipitation rate are the carrying capacities equal?
If P is 30 cm/yr, what are the steady-state populations of the two species?
If P is 60 cm/yr, what are the steady-state populations of the two species?
Because of varying precipitation rates from year to year, the steady-state populations
may never be reached. At the start of the year of interest, the populations of both grass
species are 20 plants/m2. What are the growth rates for each species initially if the
precipitation rate is expected to be 50 cm/yr? Does the initial growth rate affect the
final steady-state population?
Predator / Prey
6. A very simplistic predator-prey model was considered in class. A more realistic model
accounts for the carrying capacity of the environment for the prey and interactions of the
predators with each other. For a system with no migration in or out, assume:
⎛K − N⎞
Prey Net Reproduction Rate = rN⎜
⎟
⎝ K ⎠
⎛ C ⎞
Prey Predation Rate = aN⎜
⎟
⎝ K1 + C ⎠
⎛ C ⎞
Predator Production Rate = faN⎜
⎟
⎝ K1 + C ⎠
Predator Death Rate = qC
3
where r = 0.75 year–1, K = 10,000 prey organisms, a = 1.0 year–1, K1 = 1,000 predator
organisms, f = 0.1, q = 0.3 year–1, N = prey population at any time, organisms, and C =
predator population at anytime, organisms.
a. Write the differential equations describing the time-varying organism populations for
both the predators and the prey.
b. Determine the steady-state predator and prey populations.
7. The following information is available to model polar bear populations in a 70 km2 area:
Polar bear growth rate on seals:
fS ⋅ a′S ⋅ B⋅ S
Polar bear growth rate on fish:
fF ⋅ a′F ⋅ B⋅ F
Polar bear death rate:
q·B
Polar bear migration rate out of the area:
⎛ b
c ⎞
a ⋅ B⎜
+
⎟
⎝ b +S c + F ⎠
Polar bear migration rate into the area:
0
Seal growth rate:
Seal death rate:
where:
⎛ K − S⎞
r ⋅ S⎜ S
⎟
⎝ KS ⎠
a′S ⋅ B⋅ S
B = polar bear population
S = seal population
F = fish population
fS , fF , a ′S , a ′F , q, r, KS, a, b, and c are constants.
a. Write organism balances for both polar bears and seals.
b. Given r = 2 seals·seal–1·yr–1, KS = 700,000 seals, and a S′ = 0.001 bear–1·yr–1, determine
the polar bear population at steady-state if the seal population at steady-state is
300,000.
c. Assume the seal population begins to decrease due to external influences such as
hunting. Will the polar bear population increase, decrease, or stay the same? Justify
your answer.
4
EDV 220 – Practice Problem Set #2: Population Growth
Question 1 a) For exponential growth:
dN
= rN = (b - d) N
dt
Integrate with respect to t for b and d = constants
N = Noexp [(b-d)t]
b = 4 yr -1
d = 3.5 yr -1
t = 3 years
No = 10 pairs x 2 cougars / pair = 20 cougars
N = 20 exp [(4-3.5) x 3]
N = 89.6 cougars
Question 1 b) Population growth rate is:
dN
= population growth rate = rN = (b - d) N
dt
After road construction,
d = 3.5 + 0.08t, so
dN
= (b - 3.5 - 0.08t) N
dt
b = 4 yr -1
t = 3 yr (after road construction)
N at t = 3 yr = 575 cougars
dN
at t = 3 year (after road construction = (4 – 3.5 – 0.08(3))575 = 149.5 cougars / yr
dt
Note, because this is a rate, rounding to an integer is not necessarily required.
Question 1 c) Decline begins when b – d < 0 or b < d
4 − 3.5
= 6.25 yrs
0.08
Cougar population begins to decline 6.25 years after the road construction (11.25 years
after cougar reintroduction).
Set b < 3.5 + 0.08 t ; t =
Page 1 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 2 a)
dN
= initial growth rate = r No = ( b – d ) No ; b = 0.2 yr -1, d = 0.1 yr -1, No = 500
dt t =0
dN
= ( 0.2 – 0.1) 500 = + 50 whales / yr
dt t =0
Question 2 b)
dN
= (b − d ) N 750
dt N =750
Assuming b and d remain constant;
dN
= (0.2 − 0.1)750 = + 75 whales / yr
dt N =750
Question 2 c)
dN
= (b − d ) N
dt
; Integrate to get
N
No
N
ln
= (b − d )t ; so t =
b−d
No
ln
N = 1250 whales, No = 500 whales
1250
ln(
)
500
= 9.16 yr
t=
0.2 − 0.1yr −1
Question 2 d)
Use integrated exponential equation again; but integrate from t = -20 yr to t = 0
N
ln( 0 ) = (b − d )(t 0 − t − 20 ) ; solve for d
N − 20
N
N
ln( 0 )
ln( 0 )
N −20
N − 20
= b − d ; so it follows that d = b −
(t 0 − t − 20 )
(t 0 − t − 20 )
Page 2 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Given that b = 0.2 yr -1, No = 500 whales, N-20 = 1500 whales
500
ln(
)
1500
d = 0.2 −
= 0.255 yr -1 (> b which accounts for the decline in whale pop’n)
(0 − (−20))
Question 3 a)
Exponential growth is
Integrate to get ln(
ln(
So: r =
dN
= rN
dt
N
) = rt
No
N
)
No
t
Use t = 0 and t = 2 results in N0 = 6, N2 = 12
12
ln( )
6 = 0.347 yr -1
r=
2
Question 3 b)
Must use logistic expression for carrying capacity
dN
⎛K −N⎞
-1
= rN ⎜
⎟ ; r = 0.347 yr
dt
⎝ K ⎠
N10 = 167 squirrels
dN
= 50 squirrels / yr ; so
dt
⎛ K − 167 ⎞
50 sq / yr = 0.347 yr -1 (167 sq) ⎜
⎟
⎝ K ⎠
⎛ K − 167 ⎞
0.8628 = ⎜
⎟ ; 167 = K – 0.8628k ; so K = 1217.4 ; K = 1217 squirrels
⎝ K ⎠
Page 3 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 3 c)
Must use the integrated term of the logistic equation
KN o exp(rt )
K − N o + N o exp(rt )
N=
r = 0.347 yr -1, No = 6, K = 1217
;
for t = 15 yrs N =
1217(6) exp(0.347 × 15)
= 577 squirrels
1217 − 6 + 6 exp(0.347 × 15)
for t = 40 yrs N =
1217(6) exp(0.347 × 40)
= 1216.8 squirrels = 1217 squirrels = K
1217 − 6 + 6 exp(0.347 × 40)
Question 4 a)
K1/α = 100/0.8 = 125; K2 = 120 ;
∴ K1/ α > K2
K2/β = 120/0.6 = 200; K2 = 100 ;
∴ K2/β > K1
150
(K1/α)
125
Steady State
(K2)
100
75
50
25
0
0
25
50
75
100
(K1)
125
150
175
200
225
(K2/β)
Page 4 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 4 b)
The steady state equations are:
N1 = K1 – α N2 Æ N1 = 100 – 0.8 N2
(1)
N2 = K2 – β N1
N2 / β = K2 / β – N1 Æ N1 = 200 – N2 / 0.6
(2)
Solving (1) and (2):
100 – 0.8 N2 = 200 – N2 / 0.6 Æ 0.866 N2 = 100
N2 = 115.4 (i.e. 115)
∴ N1 = 100 – 0.8 x 115.4 = 7.7 (i.e. 8)
∴ The populations at steady state are N1 = 8 and N2 = 115
Page 5 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 4 c)
⎡ ( K − αN 2 ) − N 1 ⎤
dN 1
= r1 N 1 ⎢ 1
⎥
dt
K1
⎣
⎦
dN1
⎡ (100 − 0.5 ×100) − 25⎤
= 0.5 ×100⎢
⎥⎦ = -10
100
dt
⎣
⎡ ( K − βN1 ) − N 2 ⎤
dN 2
= r2 N 2 ⎢ 2
⎥
dt
K2
⎣
⎦
dN 2
⎡ (120 − 0.6 × 100) − 25 ⎤
= 1.5 × 25⎢
⎥⎦ = +11
120
dt
⎣
∴ Over one year the approximate changes in populations are -10 for N1 and +11 for N2.
Page 6 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 5 a)
Set K1 = K2
50 + 3 ( P – 15 ) = 100 + 1.5 ( P – 10 )
50 – 45 + 3 P = 100 – 15 + 1.5 P
1.5 P = 80; P = 53.3 cm / year
Question 5 b)
Calculate K1, K1/α,K2, K2/β
K1 = 50 + 3 ( 30 – 15 ) = 95 plants / m2
K1 / α = 95 / 0.8 = 118.75
K2 = 100 + 1.5 ( 30 – 10 ) = 130 plants / m2
K2/β = 130 / 0.75 = 173.3
At steady state, N1 = 0, N2 = 130 plants / m2
150
(K2)
Steady State
125
(K1/α)
100
75
50
25
0
0
25
50
75
(K1)
100
125
150
175
(K2/β)
200
Page 7 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 5 c)
Calculate K1, K1/α,K2, K2/β
K1 = 50 + 3 ( 60 – 15 ) = 185 plants / m2
K1 / α = 185 / 0.8 = 231.25
K2 = 100 + 1.5 ( 60 – 10 ) = 175 plants / m2
K2/β = 175 / 0.75 = 233.33
250
(K1/α)
225
200
(K2)
175
150
125
100
Steady State
75
50
25
0
0
25
50
75
100
125
150
Exact solution comes from the intersection point:
N1 =
K 1 − αK 2 185 − 0.8(175)
=
= 112.5 plants / m3
1 − αβ
1 − 0.8(0.75)
N2 =
K 2 − αK1 175 − 0.8(185)
=
= 90.6 plants / m3
1 − αβ
1 − 0.8(0.75)
175
(K1)
200
225
250
(K2/β)
At steady state, N1 = 113 plants / m3 and N2 = 91 plants / m3.
Because the carrying capacities are functions of P, there is a shift in the steady-state
solution.
Page 8 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 5 d)
⎡ ( K − αN 2 ) − N 1 ⎤
dN 1
= r1 N 1 ⎢ 1
⎥
dt
K1
⎣
⎦
⎡ ( K − βN1 ) − N 2 ⎤
dN 2
= r2 N 2 ⎢ 2
⎥
K2
dt
⎣
⎦
K1 = 50 + 3 (50 – 15) = 155 plants / m2
K2 = 100 + 1.5 (50 – 10) = 160 plants / m2
Initial growth rate when N1 = N2 = 20
dN 1
⎡ (155 − (0.8)(20) − 20 ⎤
= (0.2 yr −1 )(20 plants / m 2 ) ⎢
= 3.1 plants /(m 2 • yr )
⎥
dt
155
⎣
⎦
dN 2
⎡ (160 − (0.75)(20) − 20 ⎤
= (0.25 yr −1 )(20 plants / m 2 ) ⎢
= 3.9 plants /(m 2 • yr )
⎥
dt
160
⎣
⎦
The initial growth rates are:
For N1 : 3.1 plants/(m2-yr)
For N2 : 3.9 plants/(m2-yr)
In theory, the initial growth would not affect the final steady-state population. That is
determined as per parts b) and c). In practice, ecological systems are more complex and
if there is a lot of variability in rainfall from year to year, initial rates may matter.
Question 6 a)
⎛ C ⎞
dN
⎛K−N⎞
⎟⎟
= rN ⎜
⎟ − aN ⎜⎜
dt
⎝ K ⎠
⎝ K1 + C ⎠
⎛ C ⎞
dC
⎟⎟ − qC
= faN ⎜⎜
dt
⎝ K1 + C ⎠
Page 9 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 6 b)
At steady-state, d/dt = 0
⎛ C ⎞
⎛K −N⎞
⎟⎟ Å from dN/dt equation
0 = rN ⎜
⎟ − aN ⎜⎜
⎝ K ⎠
⎝ K1 + C ⎠
⎛ C ⎞
⎛K −N⎞
⎟⎟
rN ⎜
⎟ = aN ⎜⎜
⎝ K ⎠
⎝ K1 + C ⎠
r
(K − N ) = C
aK
K1 + C
r r
(N ) = C
−
a ak
K1 + C
r
(N ) = r − C
aK
a K1 + C
N=
Ka ⎛ r
C ⎞
⎜⎜ −
⎟
r ⎝ a K 1 + C ⎟⎠
⎛ a ⎛ C ⎞⎞
⎟⎟ ⎟ Å dN/dt = 0 curve
N = K ⎜⎜1 − ⎜⎜
⎟
r
K
+
C
⎠⎠
⎝ 1
⎝
⎛ C ⎞
⎟⎟ − qC ÅFrom dC/dt equation
0 = faN ⎜⎜
⎝ K1 + C ⎠
⎛ C ⎞
⎟⎟ = qC
faN ⎜⎜
⎝ K1 + C ⎠
⎛ C ⎞ qC
⎟⎟ =
N ⎜⎜
⎝ K 1 + C ⎠ fa
q
(K1 ) + q (C ) Å dC/dt = 0 curve
N=
fa
fa
Now solve simultaneously:
⎛ a ⎛ C ⎞⎞ q
⎟⎟ ⎟ =
(K1 ) + q (C )
K ⎜⎜1 − ⎜⎜
⎟
fa
⎝ r ⎝ K 1 + C ⎠ ⎠ fa
⎛ qK
aK
q ⎞
KK 1 + KC −
C = ⎜⎜ 1 + C ⎟⎟(K 1 + C )
r
fa ⎠
⎝ fa
Page 10 of 12
EDV 220 – Practice Problem Set #2: Population Growth
2
qK
qK
q
aK ⎞ qK 1
⎛
KK 1 + C ⎜ K −
+ 1 C + 1 C + C2
⎟=
fa
fa
r ⎠
fa
fa
⎝
2
⎞
⎞ ⎛ qK
⎛ 2qK 1 aK
q 2
+
− K ⎟⎟ + ⎜⎜ 1 − KK 1 ⎟⎟ = 0
C + C ⎜⎜
r
fa
⎠ ⎝ fa
⎝ fa
⎠
q
,b=
Let a =
fa
2
⎞
⎛ 2qK1 aK
qK1
⎜⎜
− KK1
+
− K ⎟⎟ , c =
fa
r
⎠
⎝ fa
Use the quadratic equation to solve for C
C=
− b ± b 2 − 4ac
2a
a=
0.3
q
=
=3
fa 0.1 × 1.0
⎞ 2(0.3)(1000) (1.0)(10,000)
⎛ 2qK1 aK
+
− 10,000 = 9,333.33
b = ⎜⎜
+
− K ⎟⎟ =
(0.1)(1.0)
0.75
r
⎠
⎝ fa
2
(0.3)(1000) 2
qK1
−) − (10,000)(1,000) = -7.0 x 106
− KK1 =
c=
(0.1)(1.0)
fa
− 9,333.33 + (9,333.33) 2 − 4(3)(−7.0 × 10 6 )
C=
2(3)
= 624.6 or ~ 625 predators
Substitute into
dC
= 0 curve (because it’s easier)
dt
N=
q
(K1 ) + q (C ) = 0.3 (1,000) + 0.3 (625) = 4,875 prey
fa
fa
(0.1)(1.0)
(0.1)(1.0)
Page 11 of 12
EDV 220 – Practice Problem Set #2: Population Growth
Question 7 a)
For Polar Bears (B):
dB
c ⎞
⎛ b
= f s a s ' BS + f f a f ' BF − qB − aB⎜
+
⎟
dt
⎝b+ S c+ F ⎠
For Seals (S):
K −S
dS
= rS s
− a s ' BS
dt
Ks
Question 7 b)
At steady state, d/dt = 0 ; so ds/dt = 0
0 = rS
Ks − S
− a s ' BS
Ks
Solve for B:
a s ' BS = rS
B=
Ks − S
Ks
2 ⎛ 700,000 − 300,000 ⎞
r Ks − S
=
⎜
⎟ = 1,143 bears
0.001 ⎝
700,000
as ' K s
⎠
Question 7 c)
b
to increase. So the polar bear growth
b+S
rate on seals deceases and the migration rate out of the area increases. This leads to a
decrease in dB/dt.
Decreasing S causes fsas’BS to decrease and
Æ If dB/dt ≤ 0 before the decrease in S, B will decrease.
Æ If dB/dt > 0 before the decrease in S, B may increase but at a slower rate if the new
dB/dt is still > 0.
Æ If the new dB/dt = 0 then B stays the same and if the new dB/dt < 0, B decreases
Page 12 of 12