Question

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.

Answer 1

Answer:

(a)$N(t)=Noe^{kt}$

(b)5,832 Mosquitoes

(c)5 days

Step-by-step explanation:

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

$N(t)=Noe^{kt}$

(b)If $N_o=1000$ and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

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

Therefore, our model is:

$N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t$

In 3 days time

$N(3)=1000\cdot1.8^3=5832$

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

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

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

Answer 2

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