지수적 증가와 감소

The amount of Radium-221 in a sample will be

$$\begin{aligned} Q(t)&=Ce^{-kt} \\ \text{where $C$ is the amount in the sample at time $t=0$, and $k$ is some positive constant. We know the half-life of the isotope, so we can find $e^{-k}$:} \\ \frac{C}{2}=Q(30)&=Ce^{-k\cdot 30}\\ \frac{1}{2}&=\left(e^{-k}\right)^{30}\\ 2^{-\tfrac{1}{30}}&=e^{-k} \\ \text{So,} \\ Q(t)&=C\left(e^{-k}\right)^t=C\cdot2^{-\tfrac{t}{30}} \\ \text{When only 0.01\% of the original sample is left, $Q(t)=0.0001C$:} \\ 0.0001C=Q(t)&=C\cdot 2^{-\tfrac{t}{30}}\\ 0.0001&=2^{-\tfrac{t}{30}}\\ \log(0.0001)&=\log\left(2^{-\tfrac{t}{30}}\right)\\ \log\left(10^{-4}\right)&=-\frac{t}{30}\log2\\ -4\log 10&=-\frac{t}{30}\log2\\ t&=120\cdot\frac{\log10}{\log2}\approx 398.6 \end{aligned}$$

It takes about 398.6 seconds (that is, roughly 6 and a half minutes) for all but $0.01$% of the sample to decay.

Remark: we can do another reality check here. The half-life is 30 seconds. 6 and a half minutes represents 13 half-lives. So, the sample is halved 13 times: $\left(\tfrac{1}{2}\right)^{13}\approx 0.00012=0.012\%$. So these 13 half-lives should reduce the sample to about 0.01% of its original amount, as desired.

The assumption that the animals grow according to the Malthusian model tells us that their population $t$ years after 2015 is given by $P(t)=121e^{bt}$ for some constant $b$, since $121=P(0)$, the population 0 years after 2015. The information about 2016 tells us

$$\begin{aligned} 136=P(1)&=121e^{b}\\ \frac{136}{121}&=e^b \\ \text{This gives us a better idea of $P(t)$:} \\ P(t)&=121e^{bt}=121\left(\frac{136}{121}\right)^t \\ \text{2020 is 5 years after 2015, so in 2020 (assuming the population keeps growing according to the Malthusian model) the population will be} \\ P(5)&=121\left(\frac{136}{121}\right)^5\approx217 \end{aligned}$$

In 2020, the Malthusian model predicts the herd will number 217 individuals.

Since the initial population of bacteria is 1000 individuals, the Malthusian model says that the population of bacteria $t$ hours after being placed in the dish will be $P(t)=1000e^{bt}$ for some constant $b$. Since $P(1)=2000$,

$$\begin{aligned} 2000=P(1)&=1000e^{b}\\ 2&=e^b \\ \text{So, the population at time $t$ is} \\ P(t)&=1000\cdot 2^t \\ \text{We want to know at what time the population triples, to 3,000 individuals.} \\ 3000&=1000\cdot 2^t\\ 3&=2^t\\ \log(3)&=\log\left(2^t\right)=t\log(2)\\ t&=\frac{\log(3)}{\log(2)}\approx 1.6 \end{aligned}$$

The population triples in about 1.6 hours. Note that the tripling time depends only on the constant $b$. In particular, it does not depend on the initial condition $P(0)$.

The Malthusian model suggests that, if we start with $P(0)$ cochineals, their population after 3 months will be

$$\begin{aligned} P(t)&=P(0)e^{bt} \\ \text{for some constant $b$. The constant $b$ is the net birthrate per population member per unit time. Assuming that the net birthrate for a larger population will be the same as the test population, we can use the data from the test to find $e^b$. Let $Q(t)$ be the number of individuals in the test population at time $t$.} \\ Q(t)&=Q(0)e^{bt}=200e^{bt}\\ 1000&=Q(3)=200e^{3b}\\ 5&=e^{3b}\\ e^b&=5^{1/3} \\ \text{Now that we have an idea of the birthrate, we predict} \\ P(t)&=P(0)\left(e^{b}\right)^t=P(0)\cdot5^{\tfrac{t}{3}} \\ \text{We want $P(12)=10^6+P(0)$.} \\ 10^6+P(0)=P(12)&=P(0)\cdot 5^{\tfrac{12}{3}}=P(0)\cdot 5^4\\ 10^6&=P(0)\cdot 5^4-P(0)=P(0)\left[5^4-1\right]\\ P(0)&=\frac{10^6}{5^4-1}\approx 1603 \end{aligned}$$

The farmer should use an initial population of (at least) about 1603 individuals.

Remark: if we hadn't specified that we need to save $P(0)$ individuals to start next year's population, the number of individual cochineals we would want to start with to get a million in a year would be 1600–almost the same!