미분방정식 전체 문제

We know that the amount of Polonium-210 in a sample after $t$ days is given by

$$\begin{aligned} Q(t)&=Ce^{-kt} \\ \text{where $C$ is the original amount of the sample, and $k$ is some positive constant.} \\ \\ \text{The question asks us what percentage of the sample decays in a day. Since $t$ is measured in days, the amount that decays in a day is $Q(t)-Q(t+1)$. The percentage of $Q(t)$ that this represents is $100\dfrac{Q(t)-Q(t+1)}{Q(t)}$. (For example, if there were two grams at time $t$, and one gram at time $t+1$, then $100\dfrac{2-1}{1}=50$: 50\% of the sample decayed in a day.)} \\ \\ \text{In order to simplify, we should figure out a better expression for $Q(t)$. As usual, we make use of the half-life.} \\ Q(138)&=\frac{C}{2}\\ Ce^{-k\cdot138}&=\frac{C}{2}\\ \left(e^{-k}\right)^{138}&=\frac{1}{2}=2^{-1}\\ e^{-k}&=2^{-\tfrac{1}{138}} \\ \text{Now, we have a better formula for $Q(t)$:} \\ Q(t)&=C\left(e^{-k}\right)^t\\ Q(t)&=C\cdot2^{-\tfrac{t}{138}} \\ \text{Finally, we can evaluate what percentage of the sample decays in a day.} \\ 100\frac{Q(t)-Q(t+1)}{Q(t)}&=100\frac{C\cdot2^{-\tfrac{t}{138}}-C\cdot2^{-\tfrac{t+1}{138}}}{C\cdot2^{-\tfrac{t}{138}}}\left(\frac{\frac{1}{C}}{\frac{1}{C}}\right)\\ &=100\frac{2^{-\tfrac{t}{138}}-2^{-\tfrac{t+1}{138}}}{2^{-\tfrac{t}{138}}}\\ &=100\left(2^{-\tfrac{t}{138}}-2^{-\tfrac{t+1}{138}}\right)2^{\tfrac{t}{138}}\\ &=100\left(1-2^{-\tfrac{1}{138}}\right)\approx 0.5 \end{aligned}$$

About 0.5% of the sample decays in a day.

Remark: when we say that half a percent of the sample decays in a day, we don't mean half a percent of the original sample. If a day starts out with, say, 1 microgram, then what decays in the next 24 hours is about half a percent of that 1 microgram, regardless of what the “original" sample (at some time $t=0$) held.

In particular, since the sample is getting smaller and smaller, that half of a percent that decays every day represents fewer and fewer actual atoms decaying. That's why we can't say that half of the sample (50%) will decay after about 100 days, even though 0.5% decays every day and $100\times 0.5 = 50$.

The amount of Uranium-232 in the sample of ore at time $t$ will be

$$\begin{aligned} Q(t)&=Q(0)e^{-kt} \\ \text{where $6.9 \leq Q(0) \leq 7.5$. We don't exactly know $Q(0)$, and we don't exactly know the half-life, so we also won't exactly know $Q(10)$: we can only say that is it between two numbers. Our strategy is to find the highest and lowest possible values of $Q(10)$, given the information in the problem.} \\ \\ \text{In order for the most possible Uranium-232 to be in the sample after 10 years, we should start with the most and have the longest half-life (since this represents the slowest decay). So, we take $Q(0)=7.5$ and $Q(70)=\frac{1}{2}(7.5)$.} \\ Q(t)&=7.5e^{-kt}\\ \frac{1}{2}(7.5)=Q(70)&=7.5e^{-k(70)}\\ \frac{1}{2}&=\left(e^{-k}\right)^{70}\\ 2^{-\tfrac{1}{70}}&=e^{-k} \\ \text{So, in this secenario,} \\ Q(t)&=7.5\cdot 2^{-\tfrac{t}{70}} \\ \text{After ten years,} \\ Q(10)&=7.5\cdot 2^{-\tfrac{10}{70}}\approx 6.79 \\ \text{So after ten years, the sample contains \textit{at most} 6.8 $\mu$g.} \\ \\ \text{Now, let's think about the least possible amount of Uranium-232 that could be left after 10 years. We should start with as little as possible, so take $Q(0)=6.9$, and the sample should decay quickly, so take the half-life to be 68.8 years.} \\ Q(t)&=6.9e^{-kt}\\ \frac{1}{2}6.9=Q(68.8)&=6.9e^{-k(68.8)}\\ \frac{1}{2}&=\left(e^{-k}\right)^{68.8}\\ 2^{-\tfrac{1}{68.8}}&=e^{-k} \\ \text{In this scenario,} \\ Q(t)&=6.9\cdot 2^{-\tfrac{t}{68.8}} \\ \text{After ten years,} \\ Q(10)&=6.9\cdot 2^{-\tfrac{10}{68.8}}\approx 6.24 \\ \text{So after ten years, the sample contains \textit{at least} 6.2 $\mu$g.} \\ \end{aligned}$$

After ten years, the sample contains between 6.2 and 6.8 $\mu$g of Uranium-232.

• 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

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

What we're given is an equation relating $y$ to the integral of a function of $y$. What we know how to solve is an equation relating the derivative of $y$ to a function of $y$. We can create this by differentiating the given integral equation. By the Fundamental Theorem of Calculus, part 1:

$$y'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\int_0^x\big(y(t)^2-3y(t)+2\big)\sin t\ \mathrm{d}t\right\} =\big(y(x)^2-3y(x)+2\big)\sin x$$

So $y(x)$ satisfies the differential equation $y'=\big(y^2-3y+2\big)\sin x =(y-2)(y-1)\sin x$ and the initial equation $y(0)=3$ (just substitute $x=0$ into $(*)$). For $y\ne 1,2$:

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=(y-2)(y-1)\sin x\\ \frac{\mathrm{d}y}{(y-2)(y-1)}&=\sin x\ \mathrm{d}x \\\int \frac{\mathrm{d}y}{(y-2)(y-1)}&=\int \sin x\ \mathrm{d}x \\ \text{Using the method of partial fractions, we see $\frac{1}{(y-2)(y-1)}=\frac{1}{y-2}-\frac{1}{y-1}$.} \\ \int\Big[\frac{1}{y-2}-\frac{1}{y-1}\Big]\mathrm{d}y&=\int\sin x\ \mathrm{d}x\\ \log|y-2|-\log|y-1|&=-\cos x+c\\ \log\left|\frac{y-2}{y-1}\right|&=-\cos x+c\\ \left|\frac{y-2}{y-1}\right|&= e^{c-\cos x} \\ \text{The condition $y(0)=3$ forces $\big|\frac{3-2}{3-1}\big|= e^{c-1}$ or $e^c=\frac{1}{2} e$, hence} \\ \left|\frac{y-2}{y-1}\right|&=\frac{1}{2} e^{1-\cos x} \end{aligned}$$

Observe that, when $x=0$, $\frac{y-2}{y-1}=\frac{1}{2}>0$. Furthermore $\frac{1}{2} e^{1-\cos x}$, and hence $\big|\frac{y-2}{y-1}\big|$, can never take the value zero. As $y(x)$ varies continuously with $x$, $y(x)$ must remain larger than 2. Consquently, $\frac{y-2}{y-1}$ remains positive and we may drop the absolute value signs. Hence

$$\begin{aligned} \frac{y-2}{y-1}=\frac{1}{2} e^{1-\cos x} \\ \text{ Solving for $y$,} \\ \frac{y-2}{y-1}&=\frac{1}{2} e^{1-\cos x}\\ 2(y-2)&=e^{1-\cos x}(y-1)\\ 2y-4&=ye^{1-\cos x}-e^{1-\cos x}\\ y\big(2-e^{1-\cos x}\big)&=4-e^{1-\cos x} \\ y&=\frac{4-e^{1-\cos x}}{2-e^{1-\cos x}} \end{aligned}$$

To avoid division by zero in the last step, we need

$$\begin{aligned} e^{1-\cos x}&\neq2\\ 1-\cos x&\neq \log 2\\ \cos x&\neq 1-\log 2 \end{aligned}$$

Let $L=1-\log 2$, for brevity, and note that $L>0$. (This can be seen by observing $2<e$, so, $\log 2<\log e = 1$, hence $1-\log 2>0$.)

We know $x=0$ is in the domain of our function, but the points $x=\pm \arccos(L) = \pm\arccos(1-\log 2)$ are not.

Therefore, the largest interval for which our answer makes sense is $$-\arccos(1-\log 2))>x>\arccos(1-\log 2)$$ or approximately $- 1.259 < x < 1.259$.

Suppose that in a very short time interval $\mathrm{d}t$, the height of water in the tank changes by $\mathrm{d}h$ (which is negative). Then in this time interval the amount of the water in the tank decreases by $\mathrm{d}V=-\pi(3)^2\mathrm{d}h$. This must be the same as the amount of water that flows through the hole in this time interval. The water flowing through the hole makes a cylinder of radius 1 cm (that is, 0.01 m) with length $v(t)\mathrm{d}t$, the distance the water moves out of the hole in $\mathrm{d}t$ seconds. So, the amount of water leaving the hole over the time interval $\mathrm{d}t$ is $\pi(0.01)^2 v(t)\,\mathrm{d}t =\pi(0.01)^2 \sqrt{2gh(t)}\,\mathrm{d}t$.

This gives us a separable differential equation. Recall $g$ is a constant.

$$\begin{aligned} -\pi(3)^2\mathrm{d}h&=\pi(0.01)^2 \sqrt{2gh(t)}\,\mathrm{d}t\\ \frac{\mathrm{d}h}{\sqrt{h}}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,\mathrm{d}t\\ \int\frac{\mathrm{d}h}{\sqrt{h}}&=\int-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,\mathrm{d}t\\ 2\sqrt{h}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,t+C \\ \text{At time $0$, the height is $6$, so $C=2\sqrt{6}$ and} \\ 2\sqrt{h}&=-{\Big(\frac{0.01}{3}\Big)}^{\!2} \sqrt{2g}\,t+2\sqrt{6} \\ \text{We want to know when the height of the water in the tank is 0.} \\ 0&=-\Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}\,t+2\sqrt{6}\\ \Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}\,t&=2\sqrt{6}\\ t&=\frac{2\sqrt6}{\Big(\frac{0.01}{3}\Big)^{\!2} \sqrt{2g}}\\ &=2{\Big(\frac{3}{0.01}\Big)}^{\!2} \sqrt{\frac{3}{g}}\\ &=180,000 \sqrt{\frac{3}{g}}\approx 99,591\text{ sec} \approx 27.66\text{ hr} \end{aligned}$$