A country currently has a population of $N_0$ and growth rate of $a_0$. The country introduces, at $t = 0$, a birth control scheme which hopes to gradually reduced the growth rate to $a_1 < a_0$ over a period of time $T$.

Using the formula for birth control which I have already worked out, derive the ratio of the population size with the birth control policy to that without the policy at time T.

$N_{\text{birth control}}(t) = N_0\exp\left[a_0 t - (a_0 - a_1)\frac{t^2}{2T}\right]$

Attempted solution:

Since there are no restrictions or boundaries to the original growth rate, then I assumed, the formula for no birth control at time T, would just be:

$N_{\text{no birth control}}(t) = N_0\text{exp}[a_0t]$

Then inputting $T$ into the birth control equation, I would get:

$N_{\text{birth control}}(T) = N_0\exp\left[(a_0 + a_1)\frac{T}{2}\right]$

So then I would simply have to find the ratio between the two, resulting in:

$$\frac{N_{\text{birth control}}(T)}{N_{\text{no birth control}}(T)} = \frac{\text{exp}[(a_0 + a_1)T/2]}{\text{exp}[a_0T]}$$

However upon finding the solution, to be:

$$\frac{N_{\text{birth control}}(T)}{N_{\text{no birth control}}(T)} =\exp\left[-\frac{T}{2}(a_0 - a_1)\right]$$

I realise that my formula for no birth control is probably wrong, so could you please explain to me where I went wrong and why it is so?

Thanks in advance

  • $\begingroup$ The two equations for the ratio agree, because $\exp(a)/\exp(b)=\exp(a-b)$. $\endgroup$ – Tom-Tom Apr 30 '15 at 8:35
  • $\begingroup$ Thanks for the help, for some reason I completely forgot exponential division was simply subtraction :) $\endgroup$ – Tarius Apr 30 '15 at 8:40

You are right. Note that for the exponential funcition, we have $$ \exp(x) \cdot \exp(y) = \exp(x+y), \quad \frac 1{\exp(x)} = \exp(-x) $$ Hence, your solution simplifies \begin{align*} \frac{\exp\left(\frac T2 (a_0 + a_1)\right)}{\exp(a_0 T)} &= \exp \left(\frac T2(a_0 + a_1)\right) \cdot \exp(-a_0 T)\\ &= \exp\left(\frac T2(a_0 + a_1)-a_0 T\right)\\ &= \exp\left(-\frac T2(a_0 - a_1)\right) \end{align*} as in the given solution.


This seems fine. A tool you could use for checking your formulas is to compute the rate of growth of the population $N$: namely, this rate $r$ is $$ r = \frac{dN/dt}{N} = \frac{d (\log N)}{dt}.$$ For a population with constant rate of growth $a_0$, this gives $d(\log N)/dt = a_0$, which we integrate as $\log N(t) - \log N(0) = a_0 t$, or $$N(t) = N(0) \exp (a_0 t).$$ For the population with birth control, the rate of growth (during the period $[0, T]$ is an affine function starting at $a_0$ and ending at $a_1$, so it is indeed $a(t) = a_0 + (t/T) (a_1 - a_0)$. We may again integrate the formula $d(\log N)/dt = a_0 + t (a_1-a_0)/T$ as $$ \log N(t) - \log N(0) = a_0 t + \frac{t^2}{2} \frac{a_1-a_0}{T},$$ which gives your formula for the population with birth control.

However, this formula is only true for $t \in [0,T]$. Once $t > T$, the growth rate is stable at $a_1$, which gives a new formula for $N(t)$. I leave the computation of this formula to you as an exercise! (You should integrate $\log N(t) - \log N(T)$ using the same methods as above).

  • $\begingroup$ Please note that, as is standard in at least some fields of math, I used $\log$ for the natural logarithm (what the physicists call $\ln$). (In various other fields of math, $\log$ might be the logarithm in base $2$, or any prime number - the decimal logarithm is actually quite useless in math). $\endgroup$ – Circonflexe Apr 30 '15 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.