지수적 증가와 감소

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 $Q(t)$ denote the amount of radioactive material after $t$ days. Then $Q(t)=Q(0)e^{kt}$. We are told

$$\begin{aligned} Q(8)&=0.8\,Q(0) \\ \text{So,} \\ Q(0)e^{8k}&=0.8\,Q(0)\\ e^{8k}&=0.8\\ e^k&=0.8^{\tfrac{1}{8}} \\ \text{If $Q(0)=100$, the time $t$ at which $Q(t)=40$ is determined by} \\ 40=Q(t)&=Q(0)e^{kt}=100e^{kt}=100\left(0.8^{\tfrac{1}{8}}\right)^t=100\cdot 0.8^{\tfrac{t}{8}} \\ \text{Solving for $t$:} \\ \frac{40}{100}&=0.8^{\tfrac{t}{8}}\\ \log\left(0.4\right)&=\log\left(0.8^{\tfrac{t}{8}}\right)=\frac{t}{8}\log(0.8)\\ t&=\frac{8\log(0.4)}{\log(0.8)}\approx 32.85\text{ days} \end{aligned}$$

100 grams will decay to 40 grams in about 32.85 days.

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.