미분방정식 전체 문제

(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.

Let $Q(t)$ be the quantity of morphine in a patient's bloodstream at time $t$, where $t$ is measured in minutes.

Using the definition of a derivative, $$\frac{\mathrm{d}Q}{\mathrm{d}t}=\lim_{h \to 0}\frac{Q(t+h)-Q(t)}{h}\approx \frac{Q(t+1)-Q(t)}{1}$$ So, $\frac{\mathrm{d}Q}{\mathrm{d}t}$ is roughly the change in the amount of morphine in one minute, from $t$ to $t+1$.

The sentence tells us that the change in the amount of morphine in one minute is about $-0.003Q$, where $Q$ is the quantity in the bloodstream. That is: $$\frac{\mathrm{d}Q}{\mathrm{d}t}=-0.003Q(t)$$

If $p(t)$ is the proportion of times speakers use the new form, measured between 0 and 1, then $1-p(t)$ is the proportion of times speakers use the old form.

The law, then, states that $\frac{\mathrm{d}p}{\mathrm{d}t}$ is proportional to $p(t)\times \big(1-p(t)\big)$. When we say two quantities are proportional, we mean that one is a constant multiple of the other. So, the law says $$\frac{\mathrm{d}p}{\mathrm{d}t}=\alpha p(t)\big(1-p(t)\big)$$ for some constant $\alpha$.

Remark: it follows from this model that, when a new form is either very rare or entirely ubiquitous, the rate of change of its adoption is small. This makes sense: if the new form is used all the time ($p(t)\approx 1$), there's nobody left to convert; if the new form is almost never used ($p(t)\approx 0$) then people don't know about it, so they won't pick it up.

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 $(*)$,

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