지수적 증가와 감소

The two pieces of information give us

$$\begin{aligned} f(0) &= A = 5 & f(7) &= A e^{7k}=\pi \end{aligned}$$

Thus we know that $A=5$ and so $\pi = f(7) = 5e^{7k}$. Hence

$$\begin{aligned} e^{7k} &= \frac{\pi}{5} \\ 7k &= \log(\pi/5) \\ k &= \frac{1}{7} \cdot \log(\pi/5). \end{aligned}$$

where we use $\log$ to mean natural logarithm, $\log_e$.

The amount of Carbon-14 in the sample $t$ years after the animal died will be $$Q(t)=5e^{-kt}$$ for some constant $k$ (where 5 is the amount of Carbon-14 in the sample at time $t=0$). So, the answer we're looking for is $Q(10000)$. We need to replace $k$ with an actual number to evaluate $Q(10000)$, and the key to doing this is the half-life. The text tells us that the half-life of Carbon-14 is 5730 years, so we know:

$$\begin{aligned} Q(5730)&=\frac{5}{2}\\ 5e^{-k\cdot5730}&=\frac{5}{2}\\ \left(e^{-k}\right)^{5730}&=\frac{1}{2}\\ e^{-k}&=\sqrt[5730]{\frac{1}{2}}=2^{-\tfrac{1}{5730}} \\ \text{So:} \\ Q(t)&=5\left(e^{-k}\right)^t\\ &=5\cdot2^{-\tfrac{t}{5730}} \\ \text{Now, we can evaluate:} \\ Q(10000)&=5\cdot 2^{-\tfrac{10000}{5730}}\approx 1.5~\mu g \end{aligned}$$

Remark: after $2(5730)=11,460$ years, the sample will have been sitting for two half-lives, so its remaining Carbon-14 will be a quarter of its original amount, or $1.25$ $\mu$g. It makes sense that at 10,000 years, the sample will contain slightly more Carbon-14 than at 11,460 years. Indeed, 1.5 is slightly larger than 1.25, so our answer seems plausible.

It's a good habit to look for ways to quickly check whether your answer seems plausible, since a small algebra error can easily turn into a big error in your solution.

Let 100 years ago be the time $t=0$. Then if $Q(t)$ is the amount of Radium-226 in the sample, $Q(0)=1$, and

$$\begin{aligned} Q(t)&=e^{-kt} \\ \text{for some positive constant $k$. When $t=100$, the amount of Radium-226 left is 0.9576 grams, so} \\ 0.9576=Q(100)&=e^{-k\cdot 100}=\left(e^{-k}\right)^{100}\\ e^{-k}&=0.9576^{\tfrac{1}{100}}\\ \text{This tells us} \\ Q(t)&=0.9576^{\tfrac{t}{100}} \\ \text{So, if half the original amount of Radium-226 is left,} \\ \frac{1}{2}&=0.9576^{\tfrac{t}{100}}\\ \log\left(\frac{1}{2}\right)&=\log\left(0.9576^{\tfrac{t}{100}}\right)\\ -\log 2&=\frac{t}{100}\log(0.9576)\\ t&=-100\frac{\log 2}{\log 0.9576}\approx 1600 \end{aligned}$$

So, the half life of Radium-226 is about 1600 years.

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 $P(0)$.

According to the Malthusian Model, if the ship wrecked at year $t=0$ and 2 rats washed up on the island, then $t$ years after the wreck, the population of rats will be

$$\begin{aligned} P(t)&=2e^{bt} \\ \text{for some constant $b$. We want to get rid of this extraneous variable $b$, so we use the given information. If 1928 is $a$ years after the wreck:} \\ 1000=P(a)&=2e^{ba}\\ 1500=P(a+1)&=2e^{b(a+1)}=2e^{ba}e^b \\ \text{So,} \\ (1000)\left(e^b\right)&=\left(2e^{ba}\right)\left(e^b\right)=1500 \\ \text{Which tells us} \\ e^b&=\frac{1500}{1000}=1.5 \\ \text{Now, our model is complete:} \\ P(t)&=2\left(e^b\right)^t=2\cdot1.5^t \\ \text{Since $P(a)=1000$, we can find $a$:} \\ 1000=P(a)&=2\cdot1.5^a\\ 500&=1.5^a\\ \log(500)&=\log\left(1.5^a\right)=a\log(1.5)\\ a&=\frac{\log(500)}{\log(1.5)}\approx 15.3 \end{aligned}$$

So, the year 1928 was about 15.3 years after the shipwreck. Since we aren't given a month when the rats reached exactly 1000 in number, that puts the shipwreck at the year 1912 or 1913.

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!