1. Facts: N₀ = 400

N_t = 200

r = -0.14

t = (lnN_t-lnN₀)/r

2. Facts: r = -0.15

N₀ = 100 birds

N_t = 1 bird

N_t = N₀e^rt

ln N_t = ln(N₀e^rt)

ln N_t = ln N₀ (rt)

ln N_t – ln N₀ = rt

t = (ln N_t – ln N₀)/r

t = [ln(1) – ln(100)]/ -0.15

t = 30.7 Birds will go extinct in 31 years.

3. Facts: r = (ln N_t – ln N₀)/t

N_t =5

N₀ =1

T = 7

r = (ln5-ln1)/7

r = (1.6-0)/7

r = 0.23 daily growth rate

4. Facts: see equation in Problem 2

t = 30

N_t = N₀e^rt

N_t = 1e^(0.23)(30)

N_t = 992.28

 

5. use N_t = N₀e^rt to generate N at each Saturday in June and the first two in July

This population is increasing geometrically. The slope at any given point is dN/dT = rN. The rate of increase is dependent on population size.

6.

7.

8. Facts: N₀ = 6

K = 35

r = 0.6

t = 1 and 2

b = (35-6)/6

b = 4.83

N_t = K/1+be^-rt

N_1year = 35/(1+(4.83)e^-(0.6)(1))

N_1year = 9.59 wolves

N_2years = 35/(1+(4.83)e^-(0.6)(2))

N_2years = 14.26 wolves

9. The initially high population of algae will drop after the introduction and increase in grass carp population. The grass carp population will rise sharply then decrease sharply once they eliminate most of their food supply. This is one example. There are multiple correct answers.

5 0.9 0.1

10 0.8 0.2

15 0.7 0.3

20 0.6 0.4

25 0.5 0.5

30 0.4 0.6

35 0.3 0.7

40 0.2 0.8

10. Natality rate (b) is density dependent; as density increases, natality rate decreases. Mortality rate (d) is density dependent; as density increases, mortality rate increases.

11. Carrying capacity (K) is met when birth rate (b) = death rate (d). b = d at density 25; thus, k = 25 moose. See graph.

13. N_t = N₀e^rt

ln N_t = ln(N₀e^rt)

ln N_t = ln N₀+(rt)

ln N_t – ln N₀ = rt

and because log_e and ln are both notations to indicate "natural log"

r = (log_eN_t - log_eN₀)/t