Exponential Growth and Decay — a First Look at Differential Equations

Section 3.3 Exponential Growth and Decay — a First Look at Differential Equations

A differential equation is an equation for an unknown function that involves the derivative of the unknown function. For example, Newton’s law of cooling says:

The rate of change of temperature of an object is proportional to the difference in temperature between the object and its surroundings.

We can write this more mathematically using a differential equation — an equation for the unknown function $T(t)$ that also involves its derivative $\diff{T}{t}(t)\text{.}$ If we denote by $T(t)$ the temperature of the object at time $t$ and by $A$ the temperature of its surroundings, Newton’s law of cooling says that there is some constant of proportionality, $K\text{,}$ such that

\begin{align*} \diff{T}{t}(t) &= K\big[T(t)-A\big] \end{align*}

Differential equations play a central role in modelling a huge number of different phenomena, including the motion of particles, electromagnetic radiation, financial options, ecosystem populations and nerve action potentials. Most universities offer half a dozen different undergraduate courses on various aspects of differential equations. We are barely going to scratch the surface of the subject. At this point we are going to restrict ourselves to a few very simple differential equations for which we can just guess the solution. In particular, we shall learn how to solve systems obeying Newton’s law of cooling in Section 3.3.2, below. But first, here is another slightly simpler example.

Subsection 3.3.1 Carbon Dating

Scientists can determine the age of objects containing organic material by a method called carbon dating or radiocarbon dating

Willard Libby, of Chicago University was awarded the Nobel Prize in Chemistry in 1960, for developing radiocarbon dating.

. Cosmic rays hitting the atmosphere convert nitrogen into a radioactive isotope of carbon, ${}^{14}C\text{,}$ with a half–life of about 5730 years

A good question to ask yourself is “How can a scientist (who presumably doesn’t live 60 centuries) measure this quantity?” One way exploits the little piece of calculus we are about to discuss.

. Vegetation absorbs carbon dioxide from the atmosphere through photosynthesis and animals acquire ${}^{14}C$ by eating plants. When a plant or animal dies, it stops replacing its carbon and the amount of ${}^{14}C$ begins to decrease through radioactive decay. More precisely, let $Q(t)$ denote the amount of ${}^{14}C$ in the plant or animal $t$ years after it dies. The number of radioactive decays per unit time, at time $t\text{,}$ is proportional to the amount of ${}^{14}C$ present at time $t\text{,}$ which is $Q(t)\text{.}$ Thus

Equation 3.3.1. Radioactive decay.

\begin{align*} \diff{Q}{t}(t) &=-k Q(t) \end{align*}

Here $k$ is a constant of proportionality that is determined by the half–life. We shall explain what half-life is and also determine the value of $k$ in Example 3.3.3, below. Before we do so, let’s think about the sign in equation 3.3.1.

Recall that $Q(t)$ denotes a quantity, namely the amount of ${}^{14}C$ present at time $t\text{.}$ There cannot be a negative amount of ${}^{14}C\text{,}$ nor can this quantity be zero (otherwise we wouldn’t use carbon dating, so we must have $Q(t) \gt 0\text{.}$
As the time $t$ increases, $Q(t)$ decreases, because ${}^{14}C$ is being continuously converted into ${}^{14}N$ by radioactive decay
³
The precise transition is ${}^{14}C\rightarrow {}^{14}N+ e^- + \bar{\nu}_e$ where $e^-$ is an electron and $\bar{\nu}_e $ is an electron neutrino.
. Thus $\diff{Q}{t}(t) \lt 0\text{.}$
The signs $Q(t) \gt 0$ and $\diff{Q}{t}(t) \lt 0$ are consistent with equation 3.3.1 provided the constant of proportionality $k \gt 0\text{.}$
In equation 3.3.1, we chose to call the constant of proportionality “$-k$”. We did so in order to make $k \gt 0\text{.}$ We could just as well have chosen to call the constant of proportionality “$K$”. That is, we could have replaced equation 3.3.1 by $\diff{Q}{t}(t)=K Q(t)\text{.}$ The constant of proportionality $K$ would have to be negative, (and $K$ and $k$ would be related by $K=-k$).

Now, let’s guess some solutions to equation 3.3.1. We wish to guess a function $Q(t)$ whose derivative is just a constant times itself. Here is a short table of derivatives. It is certainly not complete, but it contains the most important derivatives that we know.

$F(t)$	$1$	$t^a$	$\sin t$	$\cos t$	$\tan t$	$e^t$	$\log t$	$\arcsin t$	$\arctan t$
$\diff{}{t}F(t)$	$0$	$at^{a-1}$	$\cos t$	$-\sin t$	$\sec^2 t$	$e^t$	$\frac{1}{t}$	$\frac{1}{\sqrt{1-t^2}}$	$\frac{1}{1+t^2}$

There is exactly one function in this table whose derivative is just a (nonzero) constant times itself. Namely, the derivative of $e^t$ is exactly $e^t = 1\times e^t\text{.}$ This is almost, but not quite what we want. We want the derivative of $Q(t)$ to be the constant $-k$ (rather than the constant $1$) times $Q(t)\text{.}$ We want the derivative to “pull a constant” out of our guess. That is exactly what happens when we differentiate $e^{at}\text{,}$ where $a$ is a constant. Differentiating gives

\begin{gather*} \diff{}{t}e^{at} = a e^{at} \end{gather*}

i.e. “pulls the constant $a$ out of $e^{at}$”.

We have succeeded in guessing a single function, namely $e^{-kt}\text{,}$ that obeys equation 3.3.1. Can we guess any other solutions? Yes. If $C$ is any constant, $Ce^{-kt}$ also obeys equation 3.3.1:

\begin{gather*} \diff{}{t}(Ce^{-kt}) = C\diff{}{t}e^{-kt} = Ce^{-kt}(-k) = -k (Ce^{-kt}) \end{gather*}

You can try guessing some more solutions, but you won’t find any, because with a little trickery we can prove that a function $Q(t)$ obeys equation 3.3.1 if and only if $Q(t)$ is of the form $Ce^{-kt}\text{,}$ where $C$ is some constant.

The trick

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Notice that is very similar to what we needed in Example 3.1.2, except that here the constant is multiplicative rather than additive. That is $const \times f(t)$ rather than $const + f(t)\text{.}$

is to imagine that $Q(t)$ is any (at this stage, unknown) solution to equation 3.3.1 and to compare $Q(t)$ and our known solution $e^{-kt}$ by studying the ratio $Q(t)/e^{-kt}\text{.}$ We will show that $Q(t)$ obeys equation 3.3.1 if and only if the ratio $Q(t)/e^{-kt}$ is a constant, i.e. if and only if the derivative of the ratio is zero. By the product rule

\begin{gather*} \diff{}{t}\big[Q(t)/e^{-kt}\big]= \diff{}{t}\big[e^{kt}Q(t)\big] =ke^{kt} Q(t)+e^{kt}Q'(t) \end{gather*}

Since $e^{kt}$ is never $0\text{,}$ the right hand side is zero if and only if $k Q(t)+Q'(t)=0\text{;}$ that is $Q'(t)=-kQ(t)\text{.}$ Thus

\begin{gather*} \diff{}{t}Q(t) = -k Q(t) \iff \diff{}{t}\big[Q(t)/e^{-kt}\big] =0 \end{gather*}

as required.

We have succeed in finding all functions that obey 3.3.1. That is we have found the general solution to 3.3.1. This is worth stating as a theorem.

Theorem 3.3.2.

A differentiable function $Q(t)$ obeys the differential equation

\begin{gather*} \diff{Q}{t}(t)=-k Q(t) \end{gather*}

if and only if there is a constant $C$ such that

\begin{gather*} Q(t)= C e^{-kt} \end{gather*}

Before we start to apply the above theorem, we take this opportunity to remind the reader that in this text we will use $\log x$ with no base to indicate the natural logarithm. That is

\begin{gather*} \log x = \log_e x = \ln x \end{gather*}

Both of the notations $\log(x)$ and $\ln(x)$ are used widely and the reader should be comfortable with both.

Example 3.3.3. Carbon dating and half–life.

In this example, we determine the value of the constant of proportionality $k$ in equation 3.3.1 that corresponds to the half–life of ${}^{14}C\text{,}$ which is 5730 years.

Imagine that some plant or animal contains a quantity $Q_0$ of ${}^{14}C$ at its time of death. Let’s choose the zero point of time $t=0$ to be the instant that the plant or animal died.
Denote by $Q(t)$ the amount of ${}^{14}C$ in the plant or animal $t$ years after it died. Then $Q(t)$ must obey both equation 3.3.1 and $Q(0)=Q_0\text{.}$
Since $Q(t)$ must obey equation 3.3.1, Theorem 3.3.2 tells us that there must be a constant $C$ such that $Q(t)= C e^{-kt}\text{.}$ To also have $Q_0=Q(0) =Ce^{-k\times 0}\text{,}$ the constant $C$ must be $Q_0\text{.}$ That is, $Q(t) = Q_0 e^{-kt}$ for all $t\ge 0\text{.}$
By definition, the half–life of ${}^{14}C$ is the length of time that it takes for half of the ${}^{14}C$ to decay. That is, the half–life $t_{1/2}$ is determined by
\begin{align*} Q(t_{1/2})=\half Q(0)&=\half Q_0 & \text{but we know }Q(t) = Q_0 e^{-kt}\\ Q_0 e^{-kt_{1/2}}&=\half Q_0 & \text{now cancel } Q_0\\ e^{-kt_{1/2}}&=\half\\ \end{align*}
Taking the logarithm of both sides gives
\begin{align*} -k t_{1/2} &=\log \frac{1}{2} = -\log 2 & \text{ and so}\\ k &=\frac{\log 2}{t_{1/2}}. \end{align*}
We are told that, for ${}^{14}C\text{,}$ the half–life $t_{1/2}=5730\text{,}$ so
\begin{align*} k&=\frac{\log 2}{5730} = 0.000121 &\text{ to 6 digits} \end{align*}

From the work in the above example we have accumulated enough new facts to make a corollary to Theorem 3.3.2.

Corollary 3.3.4.

The function $Q(t)$ satisfies the equation

\begin{align*} \diff{Q}{t} &= -k Q(t) \end{align*}

if and only if

\begin{align*} Q(t) &= Q(0) \cdot e^{-kt}. \end{align*}

The half-life is defined to be the time $t_{1/2}$ which obeys

\begin{align*} Q(t_{1/2}) &= \frac{1}{2} \cdot Q(0). \end{align*}

The half-life is related to the constant $k$ by

\begin{align*} t_{1/2} &= \frac{\log 2}{k} \end{align*}

Now here is a typical problem that is solved using Corollary 3.3.4.

Example 3.3.5. Determining the age of an artefact.

A particular piece of parchment contains about 64$\%$ as much ${}^{14}C$ as plants do today. Estimate the age of the parchment.

Solution Let $Q(t)$ denote the amount of ${}^{14}C$ in the parchment $t$ years after it was first created.

By equation 3.3.1 and Example 3.3.3,

\begin{gather*} \diff{Q}{t}=-k Q(t)\qquad\text{with }k = \frac{\log 2}{5730} = 0.000121. \end{gather*}

By Corollary 3.3.4

\begin{align*} Q(t) &= Q(0) \cdot e^{-kt} \end{align*}

The time at which $Q(t)$ reaches $0.64 Q(0)$ is determined by

\begin{align*} Q(t) &=0.64 Q(0) & \text{ but } Q(t) = Q(0) e^{-kt}\\ Q(0)e^{-kt} &=0.64 Q(0) & \text{cancel $Q(0)$}\\ e^{-kt} &=0.64 & \text{take logarithms}\\ -kt &=\log 0.64\\ t &=\frac{\log 0.64}{-k} =\frac{\log 0.64}{-0.000121} = 3700 & \text{to 2} \text{significant digits.} \end{align*}

That is, the parchment

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The British Museum has an Egyptian mathematical text from the seventeenth century B.C.

is about 37 centuries old.

We have stated that the half-life of ${}^{14}C$ is 5730 years. How can this be determined? We can explain this using the following example.

Example 3.3.6. Computing a half-life.

A scientist in a B-grade science fiction film is studying a sample of the rare and fictitious element, implausium

⁶

Implausium leads to even weaker plots than unobtainium.

. With great effort he has produced a sample of pure implausium. The next day — 17 hours later — he comes back to his lab and discovers that his sample is now only 37% pure. What is the half-life of the element?

Solution We can again set up our problem using Corollary 3.3.4. Let $Q(t)$ denote the quantity of implausium at time $t\text{,}$ measured in hours. Then we know

\begin{align*} Q(t)&= Q(0) \cdot e^{-kt} \end{align*}

We also know that

\begin{align*} Q(17) &= 0.37 Q(0). \end{align*}

That enables us to determine $k$ via

\begin{align*} Q(17) = 0.37 Q(0) &= Q(0) e^{-17k} & \text{ divide both sides by $Q(0)$}\\ 0.37 &= e^{-17k}\\ \end{align*}

and so

\begin{align*} k &= -\frac{\log 0.37}{17} = 0.05849 \end{align*}

We can then convert this to the half life using Corollary 3.3.4:

\begin{align*} t_{1/2} &= \frac{\log 2}{k} \approx 11.85 \text{ hours} \end{align*}

While this example is entirely fictitious, one really can use this approach to measure the half-life of materials.

Subsection 3.3.2 Newton’s Law of Cooling

Recall Newton’s law of cooling from the start of this section:

The rate of change of temperature of an object is proportional to the difference in temperature between the object and its surroundings.

The temperature of the surroundings is sometimes called the ambient temperature. We then translated this statement into the following differential equation

Equation 3.3.7. Newton’s law of cooling.

\begin{gather*} \diff{T}{t}(t) = K\big[T(t)-A\big] \end{gather*}

where $T(t)$ is the temperature of the object at time $t\text{,}$ $A$ is the temperature of its surroundings, and $K$ is a constant of proportionality. This mathematical model of temperature change works well when studying a small object in a large, fixed temperature, environment. For example, a hot cup of coffee in a large room

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It does not work so well when the object is of a similar size to its surroundings since the temperature of the surroundings will rise as the object cools. It also fails when there are phase transitions involved — for example, an ice-cube melting in a warm room does not obey Newton’s law of cooling.

Before we worry about solving this equation, let’s think a little about the sign of the constant of proportionality. At any time $t\text{,}$ there are three possibilities.

If $T(t) \gt A\text{,}$ that is, if the body is warmer than its surroundings, we would expect heat to flow from the body into its surroundings and so we would expect the body to cool off so that $\diff{T}{t}(t) \lt 0\text{.}$ For this expectation to be consistent with equation 3.3.7, we need $K \lt 0\text{.}$
If $T(t) \lt A\text{,}$ that is the body is cooler than its surroundings, we would expect heat to flow from the surroundings into the body and so we would expect the body to warm up so that $\diff{T}{t}(t) \gt 0\text{.}$ For this expectation to be consistent with equation 3.3.7, we again need $K \lt 0\text{.}$
Finally if $T(t)=A\text{,}$ that is the body and its environment have the same temperature, we would not expect any heat to flow between the two and so we would expect that $\diff{T}{t}(t)=0\text{.}$ This does not impose any condition on $K\text{.}$

In conclusion, we would expect $K \lt 0\text{.}$ Of course, we could have chosen to call the constant of proportionality $-k\text{,}$ rather than $K\text{.}$ Then the differential equation would be $\diff{T}{t} = -k\big(T-A\big)$ and we would expect $k \gt 0\text{.}$

Now to find the general solution to equation 3.3.7. Since this equation is so similar in form to equation 3.3.1, we might expect a similar solution. Start by trying $T(t) = Ce^{Kt}$ and let’s see what goes wrong. Substitute it into the equation:

\begin{align*} \diff{T}{t} &= K( T(t)- A)\\ K C e^{Kt} &= KCe^{KT} - KA\\ ?0 & = -KA? & \text{the constant $A$ causes problems!} \end{align*}

Let’s try something a little different — recall that the derivative of a constant is zero. So we can add or subtract a constant from $T(t)$ without changing its derivative. Set $Q(t) = T(t)+B\text{,}$ then

\begin{align*} \diff{Q}{t}(t) &= \diff{T}{t}(t) & \text{by Newton's law of cooling}\\ & = K(T(t)-A) = K(Q(t)-B-A) \end{align*}

So if we choose $B=-A$ then we will have

\begin{align*} \diff{Q}{t}(t) &= K Q(t) \end{align*}

which is exactly the same form as equation 3.3.1, but with $K=-k\text{.}$ So by Theorem 3.3.2

\begin{align*} Q(t) &= Q(0) e^{Kt} \end{align*}

We can translate back to $T(t)\text{,}$ since $Q(t)=T(t)-A$ and $Q(0)=T(0)-A\text{.}$ This gives us the solution.

Corollary 3.3.8.

A differentiable function $T(t)$ obeys the differential equation

\begin{gather*} \diff{T}{t}(t) = K\big[T(t)-A\big] \end{gather*}

if and only if

\begin{gather*} T(t) = [T(0)-A]\,e^{Kt} + A \end{gather*}

Just before we put this into action, we remind the reader that $\log x = \log_e x = \ln x\text{.}$

Example 3.3.9. Warming iced tea.

The temperature of a glass of iced tea is initially $5^\circ\text{.}$ After 5 minutes, the tea has heated to $10^\circ$ in a room where the air temperature is $30^\circ\text{.}$

Determine the temperature as a function of time.
What is the temperature after 10 minutes?
Determine when the tea will reach a temperature of $20^\circ\text{.}$

Solution Part (a)

Denote by $T(t)$ the temperature of the tea $t$ minutes after it was removed from the fridge, and let $A=30$ be the ambient temperature.
By Newton’s law of cooling,
\begin{gather*} \diff{T}{t}=K(T-A) = K(T-30) \end{gather*}
for some, as yet unknown, constant of proportionality $K\text{.}$
By Corollary 3.3.8,
\begin{gather*} T(t) = [T(0)-30]\,e^{Kt} + 30 =30-25 e^{Kt} \end{gather*}
since the initial temperature $T(0)=5\text{.}$
This solution is not complete because it still contains an unknown constant, namely $K\text{.}$ We have not yet used the given data that $T(5)=10\text{.}$ We can use it to determine $K\text{.}$ At $t=5\text{,}$
\begin{align*} T(5) &=30-25 e^{5K}=10 & \text{rearrange}\\ e^{5K} &=\frac{20}{25}\\ 5K &=\log\frac{20}{25} & \text{and so}\\ K &=\frac{1}{5}\log\frac{4}{5}=-0.044629 & \text{ to 6 digits} \end{align*}

Part (b)

To find the temperature at 10 minutes we can just use the solution we have determined above.
\begin{align*} T(10)&=30-25 e^{10K}\\ &=30-25 e^{10\times\frac{1}{5}\log\frac{4}{5}}\\ &=30-25 e^{2\log\frac{4}{5}} = 30-25 e^{\log\frac{16}{25}}\\ &=30-16=\text{$14^\circ$} \end{align*}

Part (c)

We can find when the temperature is $20^\circ$ by solving $T(t)=20\text{:}$
\begin{align*} 20 &= 30-25 e^{Kt} & \text{rearrange}\\ e^{Kt} &=\frac{10}{25} = \frac{2}{5}\\ K t &= \log \frac{2}{5}\\ t &= \frac{\log \frac{2}{5}}{K}\\ &= \text{20.5 minutes} & \text{ to 1 decimal place} \end{align*}

A slightly more gruesome example.

Example 3.3.10. Determining a time from temperatures.

A dead body is discovered at 3:45pm in a room where the temperature is 20$^\circ$C. At that time the temperature of the body is 27$^\circ$C. Two hours later, at 5:45pm, the temperature of the body is 25.3 $^\circ$C. What was the time of death? Note that the normal (adult human) body temperature is $37^\circ\text{.}$

Solution We will assume

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We don’t know any other method!

that the body’s temperature obeys Newton’s law of cooling.

Denote by $T(t)$ the temperature of the body at time $t\text{,}$ with $t=0$ corresponding to 3:45pm. We wish to find the time of death — call it $t_d\text{.}$
There is a lot of data in the statement of the problem; we are told that
- the ambient temperature: $A=20$
- the temperature of the body when discovered: $T(0)=27$
- the temperature of the body 2 hours later: $T(2)=25.3$
- assuming the person was a healthy adult right up until he died, the temperature at the time of death: $T(t_d)=37\text{.}$
Since we assume the temperature of the body obeys Newton’s law of cooling, we use Corollary 3.3.8 to find,
\begin{gather*} T(t) = [T(0)-A]\,e^{Kt} + A =20+7 e^{Kt} \end{gather*}
Two unknowns remain, $K$ and $t_d\text{.}$
We can find the constant $K$ by using $T(2)=25.3\text{:}$
\begin{align*} 25.3=T(2)&= 20+7 e^{2K} & \text{rearrange}\\ 7 e^{2K}&=5.3 & \text{rearrange a bit more}\\ 2K &= \log\big(\tfrac{5.3}{7}\big)\\ K &= \tfrac{1}{2} \log\big(\tfrac{5.3}{7}\big) = -0.139 & \text{to 3 decimal places} \end{align*}
Since we know
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Actually, we are assuming again.
that $t_d$ is determined by $T(t_d)=37\text{,}$ we have
\begin{align*} 37 = T(t_d) &= 20+7 e^{-0.139 t_d} & \text{rearrange}\\ e^{-0.139 t_d} &= \tfrac{17}{7}\\ -0.139 t_d &=\log\big(\tfrac{17}{7}\big)\\ t_d &= -\tfrac{1}{0.139}\log\big(\tfrac{17}{7}\big)\\ & = - 6.38 &\text{to 2 decimal places} \end{align*}
Now $6.38$ hours is $6$ hours and $0.38\times 60 = 23$ minutes. So the time of death was $6$ hours and $23$ minutes before 3:45pm, which is 9:22am.

A slightly tricky example — we need to determine the ambient temperature from three measurements at different times.

Example 3.3.11. Finding the temperature outside.

A glass of room-temperature water is carried out onto a balcony from an apartment where the temperature is $22^\circ$C. After one minute the water has temperature $26^\circ$C and after two minutes it has temperature $28^\circ$C. What is the outdoor temperature?

Solution We will assume that the temperature of the thermometer obeys Newton’s law of cooling.

Let $A$ be the outdoor temperature and $T(t)$ be the temperature of the water $t$ minutes after it is taken outside.
By Newton’s law of cooling,
\begin{gather*} T(t)=A+\big(T(0)-A\big)e^{Kt} \end{gather*}
by Corollary 3.3.8. Notice there are 3 unknowns here — $A\text{,}$ $T(0)$ and $K$ — so we need three pieces of information to find them all.
We are told $T(0)=22\text{,}$ so
\begin{align*} T(t) &=A+\big(22-A\big)e^{Kt}. \end{align*}
We are also told $T(1)=26\text{,}$ which gives
\begin{align*} 26 &=A+\big(22-A\big)e^{K} & \text{rearrange things}\\ e^K&=\frac{26-A}{22-A} \end{align*}
Finally, $T(2)=28\text{,}$ so
\begin{align*} 28&=A+\big(22-A\big)e^{2K} & \text{rearrange}\\ e^{2K} &= \frac{28-A}{22-A} & \text{but $e^K=\frac{26-A}{22-A}$, so}\\ \left(\frac{26-A}{22-A}\right)^2 &=\frac{28-A}{22-A} & \text{multiply through by $(22-A)^2$}\\ (26-A)^2 &= (28-A)(22-A) \end{align*}
We can expand out both sides and collect up terms to get
\begin{align*} \underbrace{26^2}_{=676}-52A+A^2 &= \underbrace{28\times22}_{=616}-50A+A^2\\ 60 &= 2A\\ 30 &= A \end{align*}
So the temperature outside is $30^\circ\text{.}$

Subsection 3.3.3 Population Growth

Suppose that we wish to predict the size $P(t)$ of a population as a function of the time $t\text{.}$ In the most naive model of population growth, each couple produces $\beta$ offspring (for some constant $\beta$) and then dies. Thus over the course of one generation $\beta\tfrac{P(t)}{2}$ children are produced and $P(t)$ parents die so that the size of the population grows from $P(t)$ to

\begin{gather*} P(t+t_g)= \underbrace{P(t)+\beta\frac{P(t)}{2}}_{\text{parents+offspring}} -\underbrace{P(t)}_{\text{parents die}}=\frac{\beta}{ 2 } P(t) \end{gather*}

where $t_g$ denotes the lifespan of one generation. The rate of change of the size of the population per unit time is

\begin{gather*} \frac{P(t+t_g)-P(t)}{t_g} =\frac{1}{t_g}\Big[\frac{\beta}{2}P(t) -P(t)\Big] = b P(t) \end{gather*}

where $b=\tfrac{\beta-2}{2t_g}$ is the net birthrate per member of the population per unit time. If we approximate

\begin{gather*} \tfrac{P(t+t_g)-P(t)}{t_g}\approx\diff{P}{t}(t) \end{gather*}

we get the differential equation

Equation 3.3.12. Simple population model.

\begin{gather*} \diff{P}{t} = bP(t) \end{gather*}

By Corollary 3.3.4, with $-k$ replaced by $b\text{,}$

\begin{align*} P(t) &= P(0)\cdot e^{bt} \end{align*}

This is called the Malthusian

¹⁰

This is named after Rev. Thomas Robert Malthus. He described this model in a 1798 paper called “An essay on the principle of population”.

growth model. It is, of course, very simplistic. One of its main characteristics is that, since $P(t+T) = P(0)\cdot e^{b(t+T)} = P(t)\cdot e^{bT}\text{,}$ every time you add $T$ to the time, the population size is multiplied by $e^{bT}\text{.}$ In particular, the population size doubles every $\frac{\log 2}{b}$ units of time. The Malthusian growth model can be a reasonably good model only when the population size is very small compared to its environment

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That is, the population has plenty of food and space to grow.

. A more sophisticated model of population growth, that takes into account the “carrying capacity of the environment” is considered in the optional subsection below.

Example 3.3.13. A simple prediction of future population.

In 1927 the population of the world was about 2 billion. In 1974 it was about 4 billion. Estimate when it reached 6 billion. What will the population of the world be in 2100, assuming the Malthusian growth model?

Solution We follow our usual pattern for dealing with such problems.

Let $P(t)$ be the world’s population $t$ years after 1927. Note that 1974 corresponds to $t=1974-1927 = 47\text{.}$
We are assuming that $P(t)$ obeys equation 3.3.12. So, by Corollary 3.3.4 with $-k$ replaced by $b\text{,}$
\begin{gather*} P(t)=P(0)\cdot e^{bt} \end{gather*}
Notice that there are 2 unknowns here — $b$ and $P(0)$ — so we need two pieces of information to find them.
We are told $P(0)=2\text{,}$ so
\begin{gather*} P(t)=2\cdot e^{bt} \end{gather*}
We are also told $P(47)=4\text{,}$ which gives
\begin{align*} 4 &=2\cdot e^{47b} & \text{clean up}\\ e^{47b}&=2 & \text{take the log and clean up}\\ b&=\frac{\log 2}{47} = 0.0147 & \text{to 3 significant digits} \end{align*}
We now know $P(t)$ completely, so we can easily determine the predicted population
¹²
The 2015 Revision of World Population, a publication of the United Nations, predicts that the world’s population in 2100 will be about 11 billion. They are predicting a reduction in the world population growth rate due to lower fertility rates, which the Malthusian growth model does not take into account.
in 2100, i.e. at $t=2100-1927 = 173\text{.}$
\begin{gather*} P(173) = 2 e^{173 b} = 2 e^{173\times 0.0147} = 25.4\text{ billion} \end{gather*}
Finally, our crude model predicts that the population is 6 billion at the time $t$ that obeys
\begin{align*} P(t) &= 2 e^{b t} = 6 & \text{clean up}\\ e^{b t}&=3 & \text{take the log and clean up}\\ t&=\frac{\log 3}{b} = 47\frac{\log 3}{\log 2} = 74.5 \end{align*}
which corresponds
¹³
The world population really reached 6 billion in about 1999.
to the middle of 2001.

Subsubsection 3.3.3.1 (Optional) — Logistic Population Growth

Logistic growth adds one more wrinkle to the simple population model. It assumes that the population only has access to limited resources. As the size of the population grows the amount of food available to each member decreases. This in turn causes the net birth rate $b$ to decrease. In the logistic growth model $b=b_0\left(1-\tfrac{P}{K}\right)\text{,}$ where $K$ is called the carrying capacity of the environment, so that

\begin{gather*} P'(t) =b_0\left(1-\frac{P(t)}{K}\right)P(t) \end{gather*}

We can learn quite a bit about the behaviour of solutions to differential equations like this, without ever finding formulae for the solutions, just by watching the sign of $P'(t)\text{.}$ For concreteness, we’ll look at solutions of the differential equation

\begin{gather*} \diff{P}{t}(t)=\big(\,6000-3P(t)\,\big)\,P(t) \end{gather*}

We’ll sketch the graphs of four functions $P(t)$ that obey this equation.

For the first function, $P(0)=0\text{.}$
For the second function, $P(0)=1000\text{.}$
For the third function, $P(0)=2000\text{.}$
For the fourth function, $P(0)=3000\text{.}$

The sketches will be based on the observation that $(6000-3P)\,P=3(2000-P)\,P$

is zero for $P=0,\ 2000\text{,}$
is strictly positive for $0 \lt P \lt 2000$ and
is strictly negative for $P \gt 2000\text{.}$

Consequently

\begin{align*} \diff{P}{t}(t)\ \begin{cases} =0 & \text{if }P(t)=0\\ \gt 0 & \text{if }0 \lt P(t) \lt 2000\\ =0 & \text{if }P(t)=2000 \\ \lt 0 & \text{if }P(t) \gt 2000 \end{cases} \end{align*}

Thus if $P(t)$ is some function that obeys $\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)\text{,}$ then as the graph of $P(t)$ passes through $\big(t,P(t)\big)$

\begin{align*} \text{the graph has } \begin{cases} \text{slope zero,}& \text{i.e. is horizontal, if }P(t)=0 \\ \text{positive slope,}& \text{i.e. is increasing, if } 0 \lt P(t) \lt 2000 \\ \text{slope zero,}& \text{i.e. is horizontal, if }P(t)=2000 \\ \text{negative slope,}& \text{i.e. is decreasing, if }P(t) \gt 2000 \end{cases} \end{align*}

as illustrated in the figure

As a result,

if $P(0)=0\text{,}$ the graph starts out horizontally. In other words, as $t$ starts to increase, $P(t)$ remains at zero, so the slope of the graph remains at zero. The population size remains zero for all time. As a check, observe that the function $P(t)=0$ obeys $\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)$ for all $t\text{.}$
Similarly, if $P(0)=2000\text{,}$ the graph again starts out horizontally. So $P(t)$ remains at $2000$ and the slope remains at zero. The population size remains 2000 for all time. Again, the function $P(t)=2000$ obeys $\diff{P}{t}(t)=\big(6000-3P(t)\big)P(t)$ for all $t\text{.}$
If $P(0)=1000\text{,}$ the graph starts out with positive slope. So $P(t)$ increases with $t\text{.}$ As $P(t)$ increases towards 2000, the slope $(6000-3P(t)\big)P(t)\text{,}$ while remaining positive, gets closer and closer to zero. As the graph approaches height 2000, it becomes more and more horizontal. The graph cannot actually cross from below 2000 to above 2000, because to do so, it would have to have strictly positive slope for some value of $P$ above 2000, which is not allowed.
If $P(0)=3000\text{,}$ the graph starts out with negative slope. So $P(t)$ decreases with $t\text{.}$ As $P(t)$ decreases towards 2000, the slope $(6000-3P(t)\big)P(t)\text{,}$ while remaining negative, gets closer and closer to zero. As the graph approaches height 2000, it becomes more and more horizontal. The graph cannot actually cross from above 2000 to below 2000, because to do so, it would have to have negative slope for some value of $P$ below 2000. which is not allowed.

These curves are sketched in the figure below. We conclude that for any initial population size $P(0)\text{,}$ except $P(0)=0\text{,}$ the population size approaches $2000$ as $t\rightarrow\infty\text{.}$

Subsection 3.3.4 Exercises

Exercises Exercises for § 3.3.1

Exercises — Stage 1 .

1.

Which of the following is a differential equation for an unknown function $y$ of $x\text{?}$

\begin{align*} \amp\mbox{(a) } y=\diff{y}{x} \amp \amp\mbox{(b) } \diff{y}{x}=3\left[y-5\right] \amp \amp\mbox{(c) } y=3\left[y-\diff{x}{x}\right]\\ \amp\mbox{(d) } e^x=e^y+1 \amp \amp\mbox{(e) } y=10e^x \end{align*}

2.

Which of the following functions $Q(t)$ satisfy the differential equation $Q(t)=5\ds\diff{Q}{t}\text{?}$

\begin{align*} \amp\mbox{(a) } Q(t)=0\amp \amp\mbox{(b) } Q(t)=5e^t\amp \amp\mbox{(c) } Q(t)=e^{5t}\\ \amp\mbox{(d) } Q(t)=e^{t/5}\amp \amp\mbox{(e) } Q(t)=e^{t/5}+1 \end{align*}

3.

Suppose a sample starts out with $C$ grams of a radioactive isotope, and the amount of the radioactive isotope left in the sample at time $t$ is given by

\begin{equation*} Q(t)=Ce^{-kt} \end{equation*}

for some positive constant $k\text{.}$ When will $Q(t)=0\text{?}$

Exercises — Stage 2 .

4. (✳).

Consider a function of the form $f(x) = A e^{kx}$ where $A$ and $k$ are constants. If $f(0)=5$ and $f(7)=\pi\text{,}$ find the constants $A$ and $k\text{.}$

5. (✳).

Find the function $y(t)$ if $\ds\diff{y}{t} +3y = 0\text{,}$ $y(1) = 2\text{.}$

6.

A sample of bone belongs to an animal that died 10,000 years ago. If the bone contained 5 $\mu$g of Carbon-14 when the animal died, how much Carbon-14 do you expect it to have now?

7.

A sample containing one gram of Radium-226 was stored in a lab 100 years ago; now the sample only contains 0.9576 grams of Radium-226. What is the half-life of Radium-226?

8. (✳).

The mass of a sample of Polonium--210, initially 6 grams, decreases at a rate proportional to the mass. After one year, 1 gram remains. What is the half--life (the time it takes for the sample to decay to half its original mass)?

9.

Radium-221 has a half-life of 30 seconds. How long does it take for only 0.01% of an original sample to be left?

Exercises — Stage 3 .

10.

Polonium-210 has a half life of 138 days. What percentage of a sample of Polonium-210 decays in a day?

11.

A sample of ore is found to contain $7.2 \pm 0.3\;\mu$g of Uranium-232, the half-life of which is between 68.8 and 70 years. How much Uranium-232 will remain undecayed in the sample in 10 years?

Exercises Exercises for § 3.3.2

Exercises — Stage 1 .

1.

Which of the following functions $T(t)$ satisfy the differential equation $\ds\diff{T}{t}=5\left[T-20\right]\text{?}$

\begin{align*} \amp\mbox{(a) } T(t)=20 \amp \amp\mbox{(b) } T(t)=20e^{5t}-20 \amp \mbox{(c) } T(t)=e^{5t}+20\\ \amp\mbox{(d) } T(t)=20e^{5t}+20 \end{align*}

2.

At time $t=0\text{,}$ an object is placed in a room, of temperature $A\text{.}$ After $t$ seconds, Newton’s Law of Cooling gives the temperature of the object is as

\begin{equation*} T(t)=35e^{Kt}-10 \end{equation*}

What is the temperature of the room? Is the room warmer or colder than the object?

3.

A warm object is placed in a cold room. The temperature of the object, over time, approaches the temperature of the room it is in. The temperature of the object at time $t$ is given by

\begin{equation*} T(t)=[T(0)-A]e^{Kt}+A. \end{equation*}

Can $K$ be a positive number? Can $K$ be a negative number? Can $K$ be zero?

4.

Suppose an object obeys Newton’s Law of Cooling, and its temperature is given by

\begin{equation*} T(t)=[T(0)-A]e^{kt}+A \end{equation*}

for some constant $k\text{.}$ At what time is $T(t)=A\text{?}$

Exercises — Stage 2 .

5.

A piece of copper at room temperature (25$^\circ$) is placed in a boiling pot of water. After 10 seconds, it has heated to 90$^\circ\text{.}$ When will it be 99.9$^\circ\text{?}$

6.

Today is a chilly day. We heated up a stone to 500$^\circ$ C in a bonfire, then took it out and left it outside, where the temperature is 0$^\circ$ C. After 10 minutes outside of the bonfire, the stone had cooled to a still-untouchable 100$^\circ$ C. Now the stone is at a cozy 50$^\circ$ C. How long ago was the stone taken out of the fire?

Exercises — Stage 3 .

7. (✳).

Isaac Newton drinks his coffee with cream. To be exact, 9 parts coffee to 1 part cream. His landlady pours him a cup of coffee at $95^\circ$ C into which Newton stirs cream taken from the icebox at $5^\circ$ C. When he drinks the mixture ten minutes later, he notes that it has cooled to $54^\circ$ C. Newton wonders if his coffee would be hotter (and by how much) if he waited until just before drinking it to add the cream. Analyze this question, assuming that:

The temperature of the dining room is constant at $22^\circ$ C.
When a volume $V_1$ of liquid at temperature $T_1$ is mixed with a volume $V_2$ at temperature $T_2\text{,}$ the temperature of the mixture is $\dfrac{V_1T_1+V_2T_2}{V_1+V_2}\text{.}$
Newton’s Law of Cooling: The temperature of an object cools at a rate proportional to the difference in temperature between the object and its surroundings.
The constant of proportionality is the same for the cup of coffee with cream as for the cup of pure coffee.

8. (✳).

The temperature of a glass of iced tea is initially $5^\circ\text{.}$ After 5 minutes, the tea has heated to $10^\circ$ in a room where the air temperature is $30^\circ\text{.}$

Use Newton’s law of cooling to obtain a differential equation for the temperature $T(t)$ at time $t\text{.}$
Determine when the tea will reach a temperature of $20^\circ\text{.}$

9.

Suppose an object is changing temperature according to Newton’s Law of Cooling, and its temperature at time $t$ is given by

\begin{equation*} T(t)=0.8^{kt}+15 \end{equation*}

Is $k$ positive or negative?

Exercises Exercises for § 3.3.3

Exercises — Stage 1 .

1.

Let a population at time $t$ be given by the Malthusian model,

\begin{equation*} P(t)=P(0)e^{bt}\mbox{ for some positive constant } b. \end{equation*}

Evaluate $\ds\lim_{t \to \infty}P(t)\text{.}$ Does this model make sense for large values of $t\text{?}$

Exercises — Stage 2 .

2.

In the 1950s, pure-bred wood bison were thought to be extinct. However, a small population was found in Canada. For decades, a captive breeding program has been working to increase their numbers, and from time to time wood bison are released to the wild. Suppose in 2015, a released herd numbered 121 animals, and a year later, there were 136

¹⁴

These numbers are loosely based on animals actually released near Shageluk, Alaska in 2015. Watch the first batch being released here.

. If the wood bison adhere to the Malthusian model (a big assumption!), and if there are no more releases of captive animals, how many animals will the herd have in 2020?

3.

A founding colony of 1,000 bacteria is placed in a petri dish of yummy bacteria food. After an hour, the population has doubled. Assuming the Malthusian model, how long will it take for the colony to triple its original population?

4.

A single pair of rats comes to an island after a shipwreck. They multiply according to the Malthusian model. In 1928, there were 1,000 rats on the island, and the next year there were 1500. When was the shipwreck?

5.

A farmer wants to farm cochineals, which are insects used to make red dye. The farmer raises a small number of cochineals as a test. In three months, a test population of cochineals will increase from 200 individuals to 1000, given ample space and food.

The farmer’s plan is to start with an initial population of $P(0)$ cochineals, and after a year have $1\,000\,000+P(0)$ cochineals, so that one million can be harvested, and $P(0)$ saved to start breeding again. What initial population $P(0)$ does the Malthusian model suggest?

Exercises — Stage 3 .

6.

Let $f(t)=100e^{kt}\text{,}$ for some constant $k\text{.}$

If $f(t)$ is the amount of a decaying radioactive isotope in a sample at time $t\text{,}$ what is the amount of the isotope in the sample when $t=0\text{?}$ What is the sign of $k\text{?}$
If $f(t)$ is the number of individuals in a population that is growing according to the Malthusian model, how many individuals are there when $t=0\text{?}$ What is the sign of $k\text{?}$
If $f(t)$ is the temperature of an object at time $t\text{,}$ given by Newton’s Law of Cooling, what is the ambient temperature surrounding the object? What is the sign of $k\text{?}$

Exercises Further problems for § 3.3

1. (✳).

Find $f(2)$ if $f'(x) = \pi f(x)$ for all $x\text{,}$ and $f(0) = 2\text{.}$

2.

Which functions $T(t)$ satisfy the differential equation $\ds\diff{T}{t}=7T+9\text{?}$

3. (✳).

It takes 8 days for 20% of a particular radioactive material to decay. How long does it take for 100 grams of the material to decay to 40 grams?

4.

A glass of boiling water is left in a room. After 15 minutes, it has cooled to 85$^\circ$ C, and after 30 minutes it is 73$^\circ$ C. What temperature is the room?

5. (✳).

A 25-year-old graduate of UBC is given $50,000 which is invested at 5% per year compounded continuously. The graduate also intends to deposit money continuously at the rate of $2000 per year. Assuming that the interest rate remains 5%, the amount $A(t)$ of money at time $t$ satisfies the equation

\begin{equation*} \diff{A}{t}= 0.05 A+2000 \end{equation*}

Solve this equation and determine the amount of money in the account when the graduate is 65.
At age 65, the graduate will withdraw money continuously at the rate of $W$ dollars per year. If the money must last until the person is 85, what is the largest possible value of $W\text{?}$

6. (✳).

An investor puts $120,000 which into a bank account which pays 6% annual interest, compounded continuously. She plans to withdraw money continuously from the account at the rate of $9000 per year. If $A(t)$ is the amount of money at time $t\text{,}$ then

\begin{equation*} \diff{A}{t}= 0.06 A-9000 \end{equation*}

Solve this equation for $A(t)\text{.}$
When will the money run out?

7. (✳).

A particular bacterial culture grows at a rate proportional to the number of bacteria present. If the size of the culture triples every nine hours, how long does it take the culture to double?

8. (✳).

An object falls under gravity near the surface of the earth and its motion is impeded by air resistance proportional to its speed. Its velocity $v$ satisfies the differential equation

\begin{equation*} \dfrac{dv}{dt}=-g-kv \end{equation*}

where $g$ and $k$ are positive constants.

Find the velocity of the object as a function of time $t\text{,}$ given that it was $v_0$ at $t=0\text{.}$
Find $\lim\limits_{t\rightarrow\infty} v(t)\text{.}$

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\(F(t)\)	\(1\)	\(t^a\)	\(\sin t\)	\(\cos t\)	\(\tan t\)	\(e^t\)	\(\log t\)	\(\arcsin t\)	\(\arctan t\)
\(\diff{}{t}F(t)\)	\(0\)	\(at^{a-1}\)	\(\cos t\)	\(-\sin t\)	\(\sec^2 t\)	\(e^t\)	\(\frac{1}{t}\)	\(\frac{1}{\sqrt{1-t^2}}\)	\(\frac{1}{1+t^2}\)