 abreviatsjw

2022-01-03

The population of a certain country is known to increase at a rate proportional to the number of people presently living in the country. If after two years the population has doubled, and after three years the population is 20,000. Estimate the initial population of the nation and find an expression for the country's approximate population at any time. sirpsta3u

Given that the number of people living there now determines how quickly the population of the country grows,
, where P is the population
k is the proportionality constant

$\mathrm{ln}P=kt+C$
Let ${P}_{0}$ be the population initially
$t=0⇒P={P}_{0}$
$P={P}_{0}{e}^{kt}$
Given that at 2 years the population is dobuled
$2={e}^{2k}$
$k=\mathrm{ln}\left(\sqrt{2}\right)$
Given that at 3 years the population is 20000
$20000={P}_{i}{e}^{3\mathrm{ln}\sqrt{2}}$
Therefore the equation is $P={P}_{0}{e}^{\frac{\mathrm{ln}\left(2\right)}{2}t}$ Paul Mitchell

Rate of increase of population is proportional to Number of people living presently
After 2 years, Population has doubled
After 3 years, Population $=20000$
Let P be population in the country.
Thus, according to the question:
$\frac{dP}{dt}\mathrm{\infty }P$
$\frac{dP}{dt}=kP$, where k is any constant.
$\frac{dP}{P}=k.dt$
$\int \frac{dP}{P}=\int k.dt$
$\mathrm{ln}P=k.t+C$
When $t=0$ years, $P={P}_{0}$, where ${P}_{0}$ is the initial population.
Thus, ${\mathrm{ln}P}_{0}=0+C$
$C={\mathrm{ln}P}_{0}$
$\mathrm{ln}P=kt+{\mathrm{ln}P}_{0}$
$\mathrm{ln}P-{\mathrm{ln}P}_{0}=kt$
$\mathrm{ln}\left(\frac{P}{{P}_{0}}\right)=kt$
When $t=2$ years, $P=2{P}_{0}$
$\mathrm{ln}\left(\frac{2{P}_{0}}{{P}_{0}}\right)=kx2$
$k=\frac{\mathrm{ln}2}{2}$
$\mathrm{ln}\left(\frac{P}{{P}_{0}}\right)=\frac{\mathrm{ln}2}{2}.t$
When $t=3$ years, $P=20000$
$\mathrm{ln}\left(\frac{20000}{{P}_{0}}\right)=\frac{\mathrm{ln}2}{2}x3$
$\mathrm{ln}\left(\frac{20000}{{P}_{0}}\right)=1.0397$
$\frac{20000}{{P}_{0}}={e}^{1.0397}$
${P}_{0}=\frac{20000}{{e}^{1.0397}}=7071.2$
${P}_{0}=7072$
So, initial population, ${P}_{0}=7072$ karton

Let make N represent the population of the country at any given time t and ${N}_{o}$ as the initial population.
$\frac{dN}{dt}=kN$
where k is the constant of proportionality.
$\int \frac{1}{N}dN=k\int dt$
$\mathrm{ln}N=kt+C$
For $t=0,N={N}_{o}$
Therefore, $N={N}_{o}{e}^{kt}$ (1)
For $t=2$ yrs, $N=2No$
Substitute these values into (1)
$2{N}_{o}={N}_{o}{e}^{2t}$
$k=\frac{1}{2}\mathrm{ln}2$
$k=0.347$
Therefore (1) becomes
$N={N}_{o}{e}^{0.347t}$ (2)
For $t=3$ yrs, $N=20,000$
Substitute these values into (2)
$20000={N}_{o}{e}^{0.347\left(3\right)}$
${N}_{o}=\frac{20000}{{e}^{0.347\left(3\right)}}$
${N}_{o}=7062$
Therefore the number of people initially living in the country is 7062 Given that the population increases at a rate proportional to the number of people presently living in the country, we can express this relationship with the differential equation:
$\frac{dP}{dt}=kP$
Where $k$ is the proportionality constant. This is a separable differential equation, so we can solve it by separating variables and integrating:
$\int \frac{1}{P}dP=\int kdt$
Integrating both sides gives:
$\mathrm{ln}|P|=kt+C$
Where $C$ is the constant of integration. Applying the initial condition that after two years the population has doubled, we have:
$\mathrm{ln}|2{P}_{0}|=2k+C$
Simplifying, we get:
$C=\mathrm{ln}|2{P}_{0}|-2k$
Now, using the second condition that after three years the population is 20,000, we have:
$\mathrm{ln}|20|=3k+C$
Substituting the expression for $C$ from above, we get:
$\mathrm{ln}|20|=3k+\mathrm{ln}|2{P}_{0}|-2k$
Simplifying further:
$\mathrm{ln}|20|-\mathrm{ln}|2{P}_{0}|=k$
Applying the logarithmic identity $\mathrm{ln}\left(a\right)-\mathrm{ln}\left(b\right)=\mathrm{ln}\left(\frac{a}{b}\right)$, we have:
$\mathrm{ln}\left(\frac{20}{2{P}_{0}}\right)=k$
Now, to find the value of ${P}_{0}$, we substitute this value of $k$ into the first equation we obtained:
$\mathrm{ln}|P|=kt+C$
Using the initial condition $P\left(3\right)=20$, we have:
$\mathrm{ln}|20|=k\left(3\right)+C$
Substituting the value of $k$ we found earlier:
$\mathrm{ln}|20|=\mathrm{ln}\left(\frac{20}{2{P}_{0}}\right)\left(3\right)+C$
Simplifying:
$\mathrm{ln}|20|=3\mathrm{ln}\left(\frac{20}{2{P}_{0}}\right)+C$
Applying the logarithmic identity $\mathrm{ln}\left({a}^{b}\right)=b\mathrm{ln}\left(a\right)$, we have:
$\mathrm{ln}|20|=\mathrm{ln}\left(\frac{20}{\left(2{P}_{0}{\right)}^{3}}\right)+C$
Using the logarithmic identity $\mathrm{ln}\left(\frac{a}{b}\right)=\mathrm{ln}\left(a\right)-\mathrm{ln}\left(b\right)$, we get:
$\mathrm{ln}|20|=\mathrm{ln}\left(\frac{20}{8{P}_{0}^{3}}\right)+C$
Applying the logarithmic identity $\mathrm{ln}\left({e}^{a}\right)=a$, we have:
$\mathrm{ln}|20|=\mathrm{ln}\left(\frac{20}{8{P}_{0}^{3}}\right)+C$
Therefore, the expression for the country's approximate population at any time $t$ is:
$P\left(t\right)={e}^{kt+C}$ where $k$ and $C$ are determined as shown above.
To estimate the initial population of the nation, we substitute $t=0$ into the expression:
$P\left(0\right)={e}^{k\left(0\right)+C}={e}^{C}$
Given the value of $\mathrm{ln}|20|$ calculated earlier, we can write the expression for the initial population as:
${P}_{0}=\frac{20}{8{e}^{\mathrm{ln}|20|}}$
Therefore, the initial population of the nation is approximately $\frac{20}{8{e}^{\mathrm{ln}|20|}}$ thousand people. Eliza Beth13

Let's denote the initial population of the country as ${P}_{0}$. According to the given information, we know that after two years the population has doubled, so the population after two years is $2{P}_{0}$.
We are also told that after three years, the population is 20,000. Let's denote the population after three years as ${P}_{3}$.
We can set up a proportion to solve for ${P}_{0}$:
$\frac{2{P}_{0}}{{P}_{3}}=\frac{{P}_{0}}{20000}$
To solve this proportion, we can cross-multiply:
$2{P}_{0}×20000={P}_{0}×{P}_{3}$
Simplifying further, we have:
$40000{P}_{0}={P}_{0}{P}_{3}$
Dividing both sides by ${P}_{0}$, we get:
$40000={P}_{3}$
Therefore, the population after three years is 40,000.
To find the initial population of the nation, we substitute the value of ${P}_{3}$ back into one of the previous equations. Let's use the equation from after two years:
$2{P}_{0}=40000$
Dividing both sides by 2, we find:
${P}_{0}=20000$
Therefore, the initial population of the nation is 20,000.
Now, let's find an expression for the country's approximate population at any time, $t$. We can use the exponential growth formula:
$P\left(t\right)={P}_{0}{e}^{kt}$
Where:
- $P\left(t\right)$ is the population at time $t$
- ${P}_{0}$ is the initial population
- $k$ is the constant of proportionality
Since we know that the population doubles after two years, we can use this information to find the value of $k$. Let's substitute the values into the equation:
$2{P}_{0}={P}_{0}{e}^{2k}$
Dividing both sides by ${P}_{0}$, we get:
$2={e}^{2k}$
Taking the natural logarithm of both sides, we have:
$\mathrm{ln}\left(2\right)=2k$
Dividing both sides by 2, we find:
$k=\frac{\mathrm{ln}\left(2\right)}{2}$
Now we can substitute this value of $k$ into the equation for $P\left(t\right)$:
$P\left(t\right)={P}_{0}{e}^{\frac{\mathrm{ln}\left(2\right)}{2}t}$
Therefore, the expression for the country's approximate population at any time $t$ is $P\left(t\right)=20000{e}^{\frac{\mathrm{ln}\left(2\right)}{2}t}$.

Do you have a similar question?