# Population Growth with Immigration/Emmigration

A city has a population of 8,000,000 in 1900. The difference between the birth rate and death rate is 1.6%. 210,000 leave the city per year.

What is the population in 1910?

My Attempt: I manually did each year's population to get to the next year.

Ex:. P= Population

P in 1901 = 8000000(1.016)-210000 = 7918000

P in 1902 = (P in 1901)(1.016)-210000 = 7918000(1.016)-210000 = 7750043

After doing it few times, I get the population of 1910 is 7118369.

My question is, is there a way of doing this in one step using Calculus?

Yes, the calculus route is to use differential equations.

You could have done it algebraically with a summation:

$$P_0 = k$$ $$P_1 = 1.016*P-21*10^4$$ $$P_2 = 1.016*(1.016*P-21*10^4)-21*10^4$$ $$P_3 = 1.016*(1.016*(1.016*(1.016*P-21*10^4)-21*10^4)-21*10^4)-21*10^4$$ $$...$$ let $a=1.016$ and $c = -210,000$ $$P_1 = ax +c$$ $$P_2 = a(ax+c)+c=a^2x+ac+c$$ $$P_3=a(a(ax+c)+c)+c=a^3x+a^2c+ac+c$$ $$P_4=a(a(a(ax+c)x+c)+c)+c=a^4x+a^3c+a^2c+ac+c$$

See a pattern emerging? $$P_n=a^nx+c\sum_{i=0}^{n-1}a^i$$

Concentrate on the sum, you can eliminate it with some diligent manipulation $$S_n=\sum_{i=0}^{n-1}a^{i}$$ $$a S_n=\sum_{i=0}^{n-1}a^{i+1}$$ $$aS_n-S_n=S_n(a-1)=\sum_{i=0}^{n-1}a^{i+1} -\sum_{i=0}^{n-1}a^{i}=a^{n}-a^0$$

Therefore: $$S_n=\frac{a^{n}-1}{a-1}$$

Plug this result in $$P_n=a^nx+c\frac{a^{n}-1}{a-1}$$

$$P_n=(1.016)^n \times 8\times 10^6+(-210000)\frac{(1.016)^{n}-1}{0.016}$$

Since 1900 is year 0, the calculation collapses to $P_0=8\times 10^6$ which is the initial population. Substitute 10 for n to calculate the population for 1910. Running a simulation (the way you did it) in python3, yields the following results for comparison.

• Year 0: 8000000
• Year 1: 7918000.0
• Year 2: 7834688.0
• Year 3: 7750043.008
• Year 4: 7664043.6961280005
• Year 5: 7576668.395266049
• Year 6: 7487895.089590305
• Year 7: 7397701.411023751
• Year 8: 7306064.633600131
• Year 9: 7212961.667737733
• Year 10: 7118369.054421537
• I like your effort. – callculus May 9 '17 at 21:09
• Thank you for your feedback – DWD May 9 '17 at 21:17

If you look at the calculation you did the effect of each emigration grows by $1.6\%$ per year, so you have $P(1910)=8M1.016^{10}-210k1.016^9-210k1.016^8-\ldots 210k$ All the last terms form a geometric series, so it becomes $P(1910)=8M1.016^{10}-210k\frac {1.016^{10}-1}{1.016-1}$

• Thank you! That worked perfectly – Kaser May 9 '17 at 20:58

$P' = rP$ or $P = P_0e^{rt}$. Since you already have the rate in years this is just $r = 0.016$ and $t$ in years, so $P(1910) = P(1900)e^{0.016*10}$.

Edit To get emigration: $P' = rP - E$. $E$ is a constant so $$P(t) = e^{rt}P_0 - \int_0^t e^{r(t-s)}Eds = e^{rt} \left(P_0 + \frac{E}{r}(e^{-rt} - 1)\right).$$

Here $E$ is the emigration constant (the 210,000 leaving per year).

• This ignores the emigration. – Ross Millikan May 9 '17 at 20:23
• That will get me the growth, but how will I factor in the population decrease from 210000 people leaving each year? – Kaser May 9 '17 at 20:27
• @Kaser Are you familiar with differential equations ? If not, then you should calculate it like in Ross Millikan´s answer. – callculus May 9 '17 at 20:40
• @callculus the OP specifically asked for a "way of doing this in one step using Calculus", Ross's answer is interesting and probably correct, but doesn't really supply this... – Mortified Through Math May 9 '17 at 20:52
• @ALB You both have a closed formula at the end. That´s good. And in both cases it is necessary to derive the formula. I don´t see a difference in principle between both answers. The real question is if the OP gives a reply. – callculus May 9 '17 at 21:00