지수적 증가와 감소

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

$P'(t)=2.259{e}^{0.06407t}.$ The population is always increasing.

In the beginning of this section, the text says “A differential equation is an equation for an unknown function that involves the derivative of the unknown function." Our unknown function is $y$, so a differential equation is an equation that relates $y$ and $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$. This applies to (a) and (b), but not (c), (d), or (e).

Note that $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}x}=1$: this is the derivative of $x$ with respect to $x$.

What we're asked to find is when

$$\begin{aligned} Q(t)&=0 \\ \text{That is,} \\ Ce^{-kt}&=0 \end{aligned}$$

If $C=0$, then this is the case for all $t$. There was no isotope to begin with, and there will continue not being any undecayed isotope forever.

If $C>0$, then since $e^{-kt}>0$, also $Q(t)>0$: so $Q(t)$ is never 0 for any value of $t$. (But as $t$ gets bigger and bigger, $Q(t)$ gets closer and closer to 0.)

Remark: The last result is somewhat disturbing: surely at some point the last atom has decayed. The differential equation we use is a model that assumes $Q$ runs continuously. This is a good approximation only when there is a very large number of atoms. In practice, that is almost always the case.

The two pieces of information give us

$$\begin{aligned} f(0) &= A = 5 & f(7) &= A e^{7k}=\pi \end{aligned}$$

Thus we know that $A=5$ and so $\pi = f(7) = 5e^{7k}$. Hence

$$\begin{aligned} e^{7k} &= \frac{\pi}{5} \\ 7k &= \log(\pi/5) \\ k &= \frac{1}{7} \cdot \log(\pi/5). \end{aligned}$$

where we use $\log$ to mean natural logarithm, $\log_e$.

The amount of Carbon-14 in the sample $t$ years after the animal died will be $$Q(t)=5e^{-kt}$$ for some constant $k$ (where 5 is the amount of Carbon-14 in the sample at time $t=0$). So, the answer we're looking for is $Q(10000)$. We need to replace $k$ with an actual number to evaluate $Q(10000)$, and the key to doing this is the half-life. The text tells us that the half-life of Carbon-14 is 5730 years, so we know:

$$\begin{aligned} Q(5730)&=\frac{5}{2}\\ 5e^{-k\cdot5730}&=\frac{5}{2}\\ \left(e^{-k}\right)^{5730}&=\frac{1}{2}\\ e^{-k}&=\sqrt[5730]{\frac{1}{2}}=2^{-\tfrac{1}{5730}} \\ \text{So:} \\ Q(t)&=5\left(e^{-k}\right)^t\\ &=5\cdot2^{-\tfrac{t}{5730}} \\ \text{Now, we can evaluate:} \\ Q(10000)&=5\cdot 2^{-\tfrac{10000}{5730}}\approx 1.5~\mu g \end{aligned}$$

Remark: after $2(5730)=11,460$ years, the sample will have been sitting for two half-lives, so its remaining Carbon-14 will be a quarter of its original amount, or $1.25$ $\mu$g. It makes sense that at 10,000 years, the sample will contain slightly more Carbon-14 than at 11,460 years. Indeed, 1.5 is slightly larger than 1.25, so our answer seems plausible.

It's a good habit to look for ways to quickly check whether your answer seems plausible, since a small algebra error can easily turn into a big error in your solution.

Let 100 years ago be the time $t=0$. Then if $Q(t)$ is the amount of Radium-226 in the sample, $Q(0)=1$, and

$$\begin{aligned} Q(t)&=e^{-kt} \\ \text{for some positive constant $k$. When $t=100$, the amount of Radium-226 left is 0.9576 grams, so} \\ 0.9576=Q(100)&=e^{-k\cdot 100}=\left(e^{-k}\right)^{100}\\ e^{-k}&=0.9576^{\tfrac{1}{100}}\\ \text{This tells us} \\ Q(t)&=0.9576^{\tfrac{t}{100}} \\ \text{So, if half the original amount of Radium-226 is left,} \\ \frac{1}{2}&=0.9576^{\tfrac{t}{100}}\\ \log\left(\frac{1}{2}\right)&=\log\left(0.9576^{\tfrac{t}{100}}\right)\\ -\log 2&=\frac{t}{100}\log(0.9576)\\ t&=-100\frac{\log 2}{\log 0.9576}\approx 1600 \end{aligned}$$

So, the half life of Radium-226 is about 1600 years.

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 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.

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

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