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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

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  • $\begingroup$ 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
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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.

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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).

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  • $\begingroup$ 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

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