지수적 증가와 감소

$20$ hours

$5$ days $6$ hours $27$ minutes

$9$ hours $13$ minutes

The population reaches $10$ billion people in $2032.$

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.

We want to know when

$$\begin{aligned} [T(0)-A]e^{kt}+A&=A \\ \text{That is, when} \\ [T(0)-A]e^{kt}&=0 \\ \text{Since $e^{kt}>0$ for all values of $k$ and $t$, this happens exactly when} \\ T(0)-A=0 \end{aligned}$$

So: if the initial temperature of the object is not the same as the ambient temperature, then according to this model, it never will be! (However, as $t$ gets larger and larger, $T(t)$ gets closer and closer to $A$–it just never exactly reaches there.)

If the initial temperature of the object starts out the same as the ambient temperature, then $T(t)=A$ for all values of $t$.

• First scenario: At time $0$, Newton mixes 9 parts coffee at temperature $95^\circ$ C with 1 part cream at temperature $5^\circ$ C. The resulting mixture has temperature $$\frac{9\times 95+1\times 5}{9+1}=86^\circ$$ The mixture cools according to Newton's Law of Cooling, with initial temperature 86$^\circ$ and ambient temperature 22$^\circ$:

$$\begin{aligned} T(t)&=[86-22]e^{-kt}+22\\ T(t)&=64e^{-kt}+22 \\ \text{After 10 minutes,} \\ 54=T(10)&=22+64e^{-10 k}\\ e^{-10 k}&=\frac{54-22}{64}=\frac{1}{2} \end{aligned}$$

We could compute $k$ from this, but we don't need it.

• Second scenario: At time $0$, Newton gets hot coffee at temperature $95^\circ$ C. It cools according to Newton's Law of Cooling

$$\begin{aligned} T(t)=[T(0)-22]e^{-kt}+22 \end{aligned}$$

In this second scenario, $T(0)=95$, so $$T(t)=[95-22]e^{-kt}+22=73e^{-kt}+22$$ The value of $k$ is the same as in the first scenario, so after 10 minutes $$T(10)=22+73e^{-10k}=22+73\frac{1}{2}=58.5$$ This cooled coffee is mixed with cold cream to yield a mixture of temperature $$\frac{9\times 58.5+1\times 5}{9+1}=53.15$$

Under the second (add cream just before drinking) scenario, the coffee ends up cooler by $0.85^\circ$ C$\,$.

(a) By Newton's law of cooling, the rate of change of temperature is proportional to the difference between $T(t)$ and the ambient temperature, which in this case is $30^\circ$. Thus $$\frac{\mathrm{d}T}{\mathrm{d}t}=k[T(t)-30]$$ for some constant of proportionality $k$. To answer part (a), all we have to do is find $k$.

Under Newton's Law of Cooling, the temperature at time $t$ will be given by

$$\begin{aligned} T(t)&=[T(0)-A]e^{kt}+A\\ &=[5-30]e^{kt}+30\\ &=-25e^{kt}+30 \\ \text{We are told $T(5)=10$:} \\ 10&=-25e^{5k}+30\\ 25e^{5k}&=20\\ e^{5k}&=\frac{4}{5}\\ 5k&=\log(4/5)\\ k&=\tfrac{1}{5}\log(4/5) \\ \text{So, the differential equation is} \\ \frac{\mathrm{d}T}{\mathrm{d}t}(t)&=\frac{1}{5}\log(4/5)[T(t)-30] \end{aligned}$$ (b) Since $T(t)=30-25e^{kt}$, the temperature of the tea is $20^\circ$ when

$$\begin{aligned} 30-25 e^{kt}&=20\\ e^{kt}&=\frac{10}{25}\\ kt&=\log\left(\frac{10}{25}\right)\\ t&=\frac{1}{k}\log\frac{2}{5}\\ &=\frac{5\log(2/5)}{\log(4/5)}\\ &\approx{20.53 ~\mathrm{ min}} \end{aligned}$$

Since $b$ is a positive constant, $\displaystyle \lim_{t \to \infty}e^{bt}=\infty$. Therefore:

$$\begin{aligned} \lim_{t \to \infty}P(t)&=\lim_{t \to \infty} P(0)e^{bt} =\left\{\begin{array}{cc} 0&\text{ if }P(0)=0\\ \infty&\text{ if }P(0)>0 \end{array}\right. \end{aligned}$$

If $P(0)=0$, then the model simply says “a population that starts with no individuals continues to have no individuals indefinitely," which certainly makes sense. If $P(0) \neq 0$, then (since we can't have a negative population) $P(0)>0$, and the model says “a population that starts out with some individuals will end up with any gigantically huge number you can think of, given enough time." This one doesn't make so much sense. Populations only grow to a certain finite amount, due to scarcity of resources and such. In the derivation of the Malthusian model, we assume a constant net birth rate–that the birth and death rates (per individual) don't depend on the population, which is not a reasonable assumption long-term.

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!

Let $t=0$ be the time the boiling water is left in the room, and let $T(t)$ be the temperature of the water $t$ minutes later, so $T(0)=100$. Using Newton's Law of Cooling, we model the temperature of the water at time $t$ as

$$\begin{aligned} T(t)&=[100-A]e^{Kt}+A \\ \text{where $A$ is the room temperature, and $K$ is some constant. We are told that $T(15)=85$ and $T(30)=73$, so:} \\ 85=T(15)&=[100-A]e^{15K}+A\\ 73=T(30)&=[100-A]e^{30K}+A \\ \text{Rearranging both equations:} \\ \frac{85-A}{100-A}&=e^{15K}\\ \frac{73-A}{100-A}&=e^{30K}=\left(e^{15K}\right)^2 \\ \text{Using these equations:} \\ \left(\frac{85-A}{100-A}\right)^2=\left(e^{15K}\right)^2&=e^{30K}=\frac{73-A}{100-A}\\ \frac{(85-A)^2}{100-A}&=73-A\\ (85-A)^2&=(73-A)(100-A)\\ 85^2-170A+A^2&=7300-173A+A^2\\ 173A-170A&=7300-85^2\\ 3A&=75\\ A&=25 \end{aligned}$$

The room temperature is 25$^\circ$ C.