The population of City A increases by 8% every 10 years. The population of City B triples every 120 years. The two cities had equal populations of 10,000 residents each in the year 2000. In what year will city B have twice as many residents as city A?
[Math] Exponential growth precalc population
algebra-precalculusexponential function
Related Solutions
We find an expression for $\frac{dP}{dt}$, the rate of change of the population at time $t$. There are three items that contribute: births, deaths, and people leaving. births at time $t$ are occurring at the rate $\beta P$. Deaths are occurring at the rate $\alpha P$. We conclude that $$\frac{dP}{dt}=\beta P -\alpha P-m=(\beta-\alpha)P-m.$$ We are asked to find the general solution of this equation. I don't know the style you use for such things.
There are several possible approaches. One is to note that our DE is separable, and can be rewritten as $$\frac{dP}{(\beta-\alpha)P-m}=dt.$$ Integrate. We get $$\frac{1}{\beta-\alpha}\ln\left|(\beta-\alpha)P-m\right|=t+C.$$ Multiply both sides by $\beta-\alpha$, and take the exponential. We get $$(\beta-\alpha)P-m=D\ast e^{(\beta-\alpha)t} $$ for some constant $D$. Now we can solve explicitly for $P$. The general solution is $$P(t)=\frac{m}{\beta-\alpha}+Ee^{(\beta-\alpha)t},$$ where $E$ is an arbitrary constant.
The above is the general solution of the DE that you were asked for.
I would slightly prefer to make instead the change of variable $y=(\beta-\alpha)P-m$. Then $\frac{dy}{dt}=(\beta-\alpha)\frac{dP}{dt}=(\beta-\alpha)y$.
The equation $\frac{dy}{dt}=(\beta-\alpha)y$ is the familiar differential equation of exponential growth (if $\beta\gt \alpha$). We can write down the general solution from memory, and then use that to get an expression for $P(t)$.
The above is what you explicitly asked about. Now take over. If you have difficulties, please leave a message.
So, we have that after $x$ years, there are 3000 people where we started from 2000. Thus
$$3000=2000 e^{rx}$$
which can be rewritten as
$$e^{rx}=1.5$$
We also know that after 1 year, there is a 11.5% increase. Thus
$$e^r=1.115$$
From the conjugation of those two formulas, we have that
$$(1.115)^x=1.5$$
So all we have to do is see how many times we have to multiply 1.115 by itself to obtain 1.5. Since this will likely involve a non-integer number, we just take the integer solution such that we exceed 1.5. Which is 4. So somewhere in the fourth year, population exceeds 3000.
Best Answer
Here are the relevant formulas you need. $t$ is measured in years, and $t=0$ corresponds to the year 2000. $A(t)$ represents the population of city $A$, and $B(t)$ likewise for the other city.
$$A(t)=10000e^{k_1t}$$ $$B(t)=10000e^{k_2t}$$
Now, find the constants $k_1, k_2$ using the data you were given. Then, solve the equation $$B(t)=2A(t)$$ This is an equation entirely in $t$, after you've plugged in $k_1, k_2$. You may solve it by taking logs of both sides and rearranging. Find the value(s) of $t$ that make this hold. Then, translate to a year by adding 2000.