Question

1. a population of 265 swans was introduced to circle lake. the population’s birth rate is 0.341 swans/year per capita, and the death rate is 0.296 swans/year per capita. what is the rate of population growth per capita, and is it increasing or decreasing? round to the nearest thousandth.
2. there are 190 grey tree frogs in a swamp. the population is under carrying capacity. if r = (-0.093) frogs/year, predict the population size next year. round to the nearest whole number.
3. a population of 1492 baltimore orioles was introduced to an area of nerstrand woods. over the next year, the orioles show a death rate of 0.395 while the population drops to 1134. what’s the birth rate for this population? round to the nearest hundredth.
4. there are 780 turkeys living in merriam township, which is 92 acres in size. the birth rate is 0.472 turkeys/year per capita. the death rate is 0.331 turkeys/year per capita.

a. what is the population density? round to the nearest tenth.
b. what is dn/dt? round to the nearest whole number.
c. predict n after one year, assuming dn/dt stays constant. round to the nearest whole number.

1. one dandelion plant can produce many seeds leading to a high growth rate for dandelion populations. if a population of dandelions is currently 40 individuals and r_max = 0.2 dandelions/month per capita, predict how many dandelions would be in this population after 4 months. round to the nearest whole number.
2. imagine the dandelion population of 40 (in #18) cannot continue to grow exponentially due to lack of space. the carrying capacity for their patch of lawn is 70 dandelions. what is their dn/dt in this logistic growth situation? round to the nearest tenth.
3. a hypothetical population has a carrying capacity of 1,500 individuals and r_max is 1.0. fill out the following table. round all answers to the nearest whole number and record your answers here (no grids provided).

population size	show work here	population growth
1600
1750
2000

Explanation:

Step1: Solve problem 14

The formula for per - capita population growth rate $r$ is $r = b - d$, where $b$ is the birth rate per capita and $d$ is the death rate per capita. Given $b = 0.341$ swans/year per capita and $d=0.296$ swans/year per capita.
$r=0.341 - 0.296=0.045$ swans/year per capita. Since $r>0$, the population is increasing.

Step2: Solve problem 15

The formula for population size next year $N_{t + 1}$ is $N_{t+1}=N_t+rN_t$, where $N_t = 190$ and $r=- 0.093$.
$N_{t + 1}=190+( - 0.093\times190)=190(1 - 0.093)=190\times0.907 = 172.33\approx172$

Step3: Solve problem 16

First, find the number of deaths. The number of deaths $D=(1492 - 1134)=358$. The death - rate $d = 0.395$. Let the number of births be $B$. We know that $d=\frac{D}{N_0}$, so $N_0 = 1492$. Let the birth rate be $b$. The change in population $\Delta N=B - D$. We know $N_1 = 1134$ and $N_0 = 1492$, so $\Delta N=1134 - 1492=-358$. Also, $b=\frac{B}{N_0}$. First, find $B$: $B=\Delta N + D$. Since $\Delta N=-358$ and $D = 358$, $B = 0$. So $b = 0$

Step4: Solve problem 17a

The population density $D=\frac{N}{A}$, where $N = 780$ turkeys and $A = 92$ acres.
$D=\frac{780}{92}\approx8.5$ turkeys/acre

Step5: Solve problem 17b

The per - capita growth rate $r=b - d$, where $b = 0.472$ turkeys/year per capita and $d = 0.331$ turkeys/year per capita. So $r=0.472-0.331 = 0.141$ turkeys/year per capita. The formula for $\frac{dN}{dt}=rN$. Substituting $N = 780$ and $r = 0.141$, we get $\frac{dN}{dt}=0.141\times780 = 109.98\approx110$

Step6: Solve problem 17c

If $\frac{dN}{dt}$ is constant, the population after one year $N_{t+1}=N_t+\frac{dN}{dt}$. Substituting $N_t = 780$ and $\frac{dN}{dt}=110$, we get $N_{t + 1}=780 + 110=890$

Step7: Solve problem 18

The formula for exponential growth is $N_t=N_0e^{rt}$, where $N_0 = 40$, $r = 0.2$ dandelions/month per capita and $t = 4$ months. First, since it's per - capita growth, we assume a single - individual growth factor and multiply by the initial population. The growth factor for one individual in 4 months is $e^{0.2\times4}=e^{0.8}$. $N = 40\times e^{0.8}=40\times2.22554\approx89$

Step8: Solve problem 19

The formula for logistic growth is $\frac{dN}{dt}=r_{max}N(1-\frac{N}{K})$, where $N = 40$, $K = 70$ and $r_{max}=0.2$ dandelions/month per capita.
$\frac{dN}{dt}=0.2\times40\times(1-\frac{40}{70})=8\times\frac{30}{70}\approx3.4$

Step9: Solve problem 20

The formula for logistic growth is $\frac{dN}{dt}=r_{max}N(1-\frac{N}{K})$, where $K = 1500$ and $r_{max}=1.0$

• For $N = 1600$: $\frac{dN}{dt}=1\times1600\times(1-\frac{1600}{1500})=1600\times(-\frac{1}{15})\approx - 107$
• For $N = 1750$: $\frac{dN}{dt}=1\times1750\times(1-\frac{1750}{1500})=1750\times(-\frac{1}{6})\approx - 292$
• For $N = 2000$: $\frac{dN}{dt}=1\times2000\times(1-\frac{2000}{1500})=2000\times(-\frac{1}{3})\approx - 667$

Answer:

1. The rate of population growth per capita is $0.045$ and it is increasing.
2. 172
3. 0

17a. 8.5
17b. 110
17c. 890

1. 89
2. 3.4

20.

Population Size	Show Work Here	Population Growth
1750	$\frac{dN}{dt}=1\times1750\times(1-\frac{1750}{1500})$	- 292
2000	$\frac{dN}{dt}=1\times2000\times(1-\frac{2000}{1500})$	- 667