If N is the population of the colony and t is the time in days, express N as a function of t. Consider Upper N 0 is the original amount at tequals0 and knot equals0 is a constant that represents the growth rate. N(t)equals eN0et e Superscript t(Type an expression using t as the variable and in terms of e.) (b) The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there are 1800 after 1 day, what is the size of the colony after 3 days? Approximately nothing mosquitoes. (Do not round until the final answer. Then round to the nearest whole number as needed.) (c) How long is it until there are 20 comma 000 mosquitoes? About nothing days.

(a)Given an original amount [tex]N_o[/tex] at t=0. The population of the colony with a growth rate [tex]k \neq 0[/tex], where k is a constant is given as:

[tex]N(t)=Noe^{kt}[/tex]

(b)If [tex]N_o=1000[/tex] and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

[tex]1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)[/tex]

Therefore, our model is:

[tex]N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t[/tex]

In 3 days time

[tex]N(3)=1000\cdot1.8^3=5832[/tex]

The population of mosquitoes in 3 days time will be approximately 5832.

(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.

From our model

[tex]N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days[/tex]

In approximately 5 days, the population of mosquitoes will be 20,000.

abidemiokin abidemiokin · Answer 2 · 2020-05-06T11:28:02+02:00

Answer:

a) N(t) = Noe^kt

b) 5832mosquitoes

c) Approximately 5years

Step-by-step explanation:

If he rate of growth is equal the number of population present in a given time, we have:

dN/dt ∝ N

N is the number of population.

dN/dt = kN

k is the constant of proportionality

dN = kNdt

On separating the variables

dN/N = kdt

Integrating both sides of the equation, we have

∫dN/N = k∫dt

ln(N/No) = kt

Taking exp of both sides

e^ln(N/No) = e^kt

N/No = e^kt

N = Noe^kt

N(t) = Noe^kt

No is the original amount of population

t is the time in days

b) If there are 1000 mosquitoes initially and there are 1800 after 1 day,

N = 1800, No = 1000 at t = 1

1800 = 1000e^k(1)

1800/1000 = e^k

e^k = 1.8

Taking ln of both sides

lne^k = ln1.8

k = ln 1.8

k = 0.5878

To get the size of the colony after 3 days,

N(t) = 1000e^0.5878(3)

N(3) = 1000e^1.7634

N(3) = 1000×5.832

N(3) = 5832

The size if the colony after 3years is 5,832mosquitoes

c) To determine how long is it until there are 20,000 mosquitoes?

N(t) = Noe^kt

20,000 = 1000e^(0.5878)t

20,000/1,000 = e^0.5878t

20 = e^0.59t

ln20 = lne^0.5878t

ln 20 = 0.5878t

t = ln20/0.5878

t = 5.07years

It will take 5years until there are 20,000mosquitoes