미분방정식 전체 문제

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!

The given differential equation is separable and we solve it accordingly.

$$\begin{aligned} \frac{\mathrm{d}P}{\mathrm{d}t}&=-k\sqrt{P}\\ \frac{\mathrm{d}P}{\sqrt{P}}&=-k\,\mathrm{d}t\\ \int\frac{\mathrm{d}P}{\sqrt{P}}&=\int-k\,\mathrm{d}t\\ 2\sqrt{P}&=-kt+C \\ \text{At $t=0$, $P=90,000$ so } \\ 2\sqrt{90,000}&=-k\times 0+C\\ C&=2\times 300=600 \\ \text{Therefore,} \\ 2\sqrt{P}&=-kt+600\tag{$*$} \\ \text{Now, we find $k$. Let $t$ be measured in weeks. Then when $t=6$, $P=40,000$.} \\ 2\sqrt{40,000}&=-6k+600\\ 2\cdot 200&=-6k+600\\ k&=\frac{200}{6} = \frac{100}{3} \\ \text{Substituting our value of $k$ into ($*$):} \\ 2\sqrt{P}&=-\frac{100}{3}t+600 \\ \text{To find when the population will be 10,000, we set $P=10,000$ and solve for $t$.} \\ 2\sqrt{10,000}&=-\frac{100}{3}t+600\\ 2\cdot 100&=-\frac{100}{3}t+600\\ \frac{100}{3}t&=400\\ t&=12 \end{aligned}$$

Since we measured $t$ in weeks when we found $k$, we see that in 12 weeks the population will decrease to 10,000 individuals.

The given differential equation is separable and we solve it accordingly.

$$\begin{aligned} m\frac{\mathrm{d}v}{\mathrm{d}t}&=-(mg+kv^2)\\ \frac{m}{mg+kv^2}\,\mathrm{d}v&=-\mathrm{d}t\\ \int\frac{m}{mg+kv^2}\,\mathrm{d}v&=\int-\mathrm{d}t \\ \text{The left integral looks something like the antiderivative of arctangent. Let's factor out that $mg$ from the denominator.} \\ \frac{1}{mg}\int\frac{m}{1+\frac{k}{mg}v^2}\,\mathrm{d}v&=-t+C\\ \frac{1}{g}\int\frac{1}{1+\left(\sqrt{\frac{k}{mg}}v\right)^2}\,\mathrm{d}v&=-t+C \\ \text{Now it looks even more like the derivative of arctangent. We can guess the antiderivative from here, or use the substitution $u=\sqrt{\frac{k}{mg}}v$, $\mathrm{d}u=\sqrt{\frac{k}{mg}}\,\mathrm{d}v$.} \\ \frac{1}{g}\sqrt{\frac{mg}{k}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=-t+C\\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=-t+C\tag{$*$} \\ \text{At $t=0$, $v=v_0$, so: } \\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)&=C \\ \text{Plug $C$ into $(*)$.} \\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v\right)&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t \\ \text{ At its highest point, the object has velocity $v=0$. This happens when $t$ obeys:} \\ \sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}0\right)&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t\\ 0&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right)-t\\ t&=\sqrt{\frac{m}{gk}}\arctan\left(\sqrt{\frac{k}{mg}}v_0\right) \end{aligned}$$

(a) The given differential equation is separable and we solve it accordingly.

$$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}t}&=-k\,v^2\\ -\frac{\mathrm{d}v}{v^2}&=k\,\mathrm{d}t\\ \int- \frac{\mathrm{d}v}{v^2}&=\int k\,\mathrm{d}t\\ \frac{1}{v}&=kt+C \\ \text{At $t=0$, $v=40$ so } \\ \frac{1}{40}&=k\times 0+ C\\ C&=\frac{1}{40} \\ \text{Therefore,} \\ v(t)&=\frac{1}{kt+C}=\frac{1}{kt+1/40}=\frac{40}{40kt+1}\tag{$*$} \\ \text{The constant of proportionality $k$ is determined by} \\ v(10)&=20\\ 20&=\frac{40}{40k\times 10+1}\\ \frac{1}{2}&=\frac{1}{400k+1}\\ 400k+1&=2 \\ k&=\frac{1}{400} \end{aligned}$$

(b) Subbing in the value of $k$ to $(*)$,

$$\begin{aligned} v(t)&=\frac{40}{40kt+1}=\frac{40}{t/10+1}\\ \\ \text{We want to know the value of $t$ that gives $v(t)=5$.} \\ 5&=\frac{40}{t/10+1}\\ \frac{t}{10}+1&=8 \\ t&=70\text{ sec} \end{aligned}$$

(a) The rate of change of speed at time $t$ is $-kv(t)^2$ for some constant of proportionality $k$ (to be determined–but we assume it is positive, since the speed is decreasing). So $v(t)$ obeys the differential equation $\frac{\mathrm{d}v}{\mathrm{d}t}=-kv^2$ . (b) The equation $\frac{\mathrm{d}v}{\mathrm{d}t}=-kv^2$ is a separable differential equation, which we can solve in the usual way.

$$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}t}&=-kv^2\\ \frac{\mathrm{d}v}{-v^2}&=k\mathrm{d}t\\ \int -\frac{\mathrm{d}v}{v^2} &=\int k\mathrm{d}t\\ \frac{1}{v}&=kt+C \\ \text{At time $t=0$, $v=400$, so $C=\frac{1}{400}$. Then:} \\ \frac{1}{v}&=kt+\frac{1}{400}\tag{$*$} \\ \text{At time $t=1$, $v=200$, so } \\ \frac{1}{200}&=k+\frac{1}{400}\\ k&=\frac{1}{400} \\ \text{Therefore, from ($*$),} \\ \frac{1}{v}&=\frac{t}{400}+\frac{1}{400} = \frac{t+1}{400}\\ v&=\frac{400}{t+1} \end{aligned}$$ (c) To find when the speed is 50, we set $v=50$ in the equation from (b) and solve for $t$.

$$\begin{aligned} 50&=\frac{400}{t+1}\\ 50(t+1)&=400\\ t+1&=8\\ t&=7 \end{aligned}$$