Answers to exercises

Exercise 1

Since (as calculated in the example)

\[ e^{-0.000121 \times 5730} \approx \dfrac{1}{2}, \]

we may rewrite the equation \(m = 100 \, e^{-0.000121 \, t}\) as

\begin{align*} m &= 100 \, e^{-0.000121 \, t} \\ &= 100 \, \Bigl( e^{-0.000121 \times 5730} \Bigr)^{\tfrac{t}{5730}} \\ &\approx 100 \, \Bigl( \dfrac{1}{2} \Bigr)^{\tfrac{t}{5730}}. \end{align*}

From this equation we can see that, when \(t\) increases by 5730, the exponent \(\dfrac{t}{5730}\) increases by 1 and so \(m\) is multiplied by \(\dfrac{1}{2}\).

Exercise 2

Since the population is given by \(x(t) = 1000 \, e^{0.1 \, t}\), the population is double its initial value when

\[ 2000 = 1000 \, e^{0.1 \, t} \quad \implies \quad t = \dfrac{\log_e 2}{0.1} \approx 6.93 \text{ years.} \]

Exercise 3

If the growth rate is \(k\), then the population \(P\) will be given by \(P(t) = C e^{kt}\) after \(t\) years. The population will double over a time period of length \(T\) if

\begin{align*} P(t + T) = 2P(t) \quad &\iff \quad C e^{k(t + T)} = 2Ce^{kt} \\ &\iff \quad e^{kT} = 2 \\ &\iff \quad T = \dfrac{1}{k} \log_e 2. \end{align*}

Exercise 4

Let \(P\) be the number of pheasants \(t\) years after introduction. With continuous growth rate \(k=1\), we have \(P(t) = C e^{t}\) for a constant \(C\). Since \(P(0) = 8\), we have \(C=8\) and so \(P(t) = 8 e^t\). After 60 years, the pheasant population is \(P(60) = 8 e^{60} \approx 9.14 \times 10^{26}\).
At 1.5 kg per pheasant, we have \(9.14 \times 10^{26} \times 1.5 \approx 1.37 \times 10^{27}\) kg of pheasants. This is much greater than the mass of the earth.
A continuous growth rate as rapid as 100% cannot persist for long. The pheasants will rapidly run out of food, land and other resources. The population's growth will slow down as these limits are reached. A more realistic model is discussed in the section Links forward (Logistic growth).

Exercise 5

Let \(P(t)\) denote the population after \(t\) years.

In this case we effectively have a constant continuous growth rate of \(k=0.1\), starting from a population of 1700. We obtain \(P(t) = 1700 \, e^{0.1 \, t}\). After seven years, the number of birds is \(P(7) = 1700 \, e^{0.7} \approx 3420\) (to three significant figures).
This is the situation in the example, which gives 3030 birds after seven years.
This is effectively the situation in the earlier example, but with an extra 700 birds added at the end. Thus, to three significant figures, there are \(2010 + 700 = 2710\) birds after seven years.

We see that the earlier the birds migrate, the larger the final population. The earlier they arrive, the more time they have to reproduce.

Exercise 6

With the patch on, we have \(\dfrac{dA}{dt} = 1 - 0.12 A\), which has general solution \[ A(t) = C e^{-0.12 \, t} + \dfrac{1}{0.12}. \] Since \(A(0) = 0\), we obtain \(C = - \dfrac{1}{0.12}\). Thus \begin{align*} A(t) &= - \dfrac{1}{0.12} e^{-0.12 \, t} + \dfrac{1}{0.12}\\ &= \dfrac{1}{0.12} \bigl( 1 - e^{-0.12 \, t} \bigr), \end{align*} so \begin{align*} A(16) &= \dfrac{1}{0.12} \bigl(1 - e^{-0.12 \times 16} \bigr) \\ &\approx 7.11\ \mu\text{g/L (to three significant figures)}. \end{align*}
With the patch off, \(\dfrac{dA}{dt} = -0.12 A\), which has general solution \(A(t) = C e^{-0.12 \, t}\). Since \(A(16) \approx 7.11\), we obtain \(C \approx 48.5\), so \begin{align*} A(24) &\approx 48.5 \, e^{-0.12 \times 24} \\ &\approx 2.72\ \mu\text{g/L (to three significant figures)}. \end{align*}

Exercise 7

The population reaches 2 million when \[ 2\,000\,000 = 500\,000 \, e^{0.01 \, t} + 500\,000. \] Rearranging this gives \[ e^{0.01 \, t} = 3 \quad \iff \quad t = 100 \, \log_e 3. \] So \(t \approx 109.86\) hours (to two decimal places).
If there were no extraction, then we would have the differential equation \[ \dfrac{dp}{dt} = 0.015 \, p - 0.005 \, p = 0.01 \, p. \] Thus the growth rate is \(k=0.01\) and hence the doubling time is \begin{align*} T &= \dfrac{1}{k} \log_e 2 \\ &= 100 \, \log_e 2 \\ &\approx 69.31 \text{ hours (to two decimal places).} \end{align*}

Exercise 8

\(M(t) = 100 \, (1+r)^t\)
The amount in the account doubles when \((1+r)^t = 2\). Taking the natural logarithm of both sides gives \(t \, \log_e (1+r) = \log_e 2\), so
\[ t = \dfrac{ \log_e 2}{ \log_e (1+r) }. \]
Rewriting \((1+r)^t\) as \(e^{t \, \log_e (1+r)}\) gives \(M = 100 \, e^{t \, \log_e (1+r)}\). Therefore \begin{align*} \dfrac{dM}{dt} &= 100 \, \log_e (1+r) \, e^{t \, \log_e (1+r)} \\ &= M \, \log_e (1+r), \end{align*} and hence the continuous growth rate is \(\log_e (1+r)\).

Exercise 9

The discrete growth rate \(r\) and the continuous growth rate \(k\) are related by \(k = \log_e (1 + r)\). So \(k = 1\) gives \(1 = \log_e (1+r)\), and therefore

\[ r = e - 1 \approx 1.72 \ \ \text{(to two decimal places)}. \]

This discrete growth rate of 1.72 per year is analogous to an interest rate of 172% per annum; the number of pheasants was multiplied by 2.72 each year.

Exercise 10

From

\[ \dfrac{dx}{dt} = kx \Bigl( 1 - \dfrac{x}{R} \Bigr) = \dfrac{kx(R-x)}{R}, \]

we obtain

\[ \dfrac{dt}{dx} = \dfrac{R}{kx(R-x)}. \]

Writing this expression in terms of partial fractions we obtain

\[ \dfrac{dt}{dx} = \dfrac{1}{kx} + \dfrac{1}{k(R-x)}. \]

Integrating gives

\begin{align*} t &= \dfrac{1}{k} \log_e x - \dfrac{1}{k} \log_e (R-x) + c \\ &= \dfrac{1}{k} \log_e \Bigl(\dfrac{x}{R-x}\Bigr) + c, \end{align*}

where \(c\) is a constant of integration. Rearranging, we obtain

\[ x = \dfrac{R}{e^{-k(t-c)} + 1}, \]

as desired.

Exercise 11

The equilibrium value of \(n\) is given by \(-\dfrac{m}{k}\), which is assumed to be the existing number of HIV viruses before the experiment.

Exercise 12

After administering the agent so that \(m=0\), the number of viruses \(n\) obeys the differential equation \(\dfrac{dn}{dt} = kn\), giving exponential decay. The half-life \(T\) is given by \(T = - \dfrac{1}{k} \log_e 2\). Since \(T=0.24\) days, we have

\[ 0.24 = - \dfrac{1}{k} \log_e 2 \quad \iff \quad k = - \dfrac{1}{0.24} \log_e 2 \approx -2.88811. \]

Exercise 13

Since \(-\dfrac{m}{k} \approx 100\,000\) and \(k \approx -2.88811\), we have \[ m \approx 100\,000 \times 2.88811 = 288\,811. \] That is, approximately \(288\ 811\) viruses are produced per day per millilitre of extracellular fluid.
Multiplying \(m\) by the \(15\ 000\) millilitres of extracellular fluid gives a production rate of \(4.3 \times 10^9\) viruses per day. That's a lot!

Exercise 14

Let \(T\) be the temperature of the coffee (in \(^\circ\)C) after \(t\) minutes. By Newton's law we have

\[ \dfrac{dT}{dt} = -k(T - 20) = -kT + 20k, \]

so the general solution is \(T(t) = C e^{-kt} + 20\), where \(C\) is a constant. We have \(T(0) = 100\), so \(100 = C + 20\), and therefore \(C=80\). From \(T(2) = 80\), we have \(80 = 80 \, e^{-2k} + 20\), and hence \(k = - \dfrac{1}{2} \, \log_e \dfrac{3}{4}\). Thus

\[ T(t) = 80 \, e^{ \dfrac{1}{2} (\log_e \dfrac{3}{4}) \, t} + 20 = 80 \, \Bigl( \dfrac{3}{4} \Bigr)^{\tfrac{t}{2}} + 20. \]

After five minutes, the temperature is

\[ T = 80 \, \Bigl( \dfrac{3}{4} \Bigr)^{\tfrac{5}{2}} + 20 \approx 58.97\ ^\circ\text{C}. \]

The temperature is 40 \(^\circ\)C when

\begin{align*} 40 &= 80 \, \Bigl( \dfrac{3}{4} \Bigr)^{\tfrac{t}{2}} + 20 \\ \Bigl( \dfrac{3}{4} \Bigr)^{\tfrac{t}{2}} &= \dfrac{1}{4} \\ \dfrac{t}{2} \, \log_e \dfrac{3}{4} &= \log_e \dfrac{1}{4} \\ t &= \dfrac{2 \, \log_e \dfrac{1}{4}}{ \log_e \dfrac{3}{4}} \approx 9.64 \text{ minutes.} \end{align*}

Exercise 15

Let \(T\) be the temperature of the metal (in \(^\circ\)C) after \(t\) minutes. From Newton's law of cooling we have

\[ \dfrac{dT}{dt} = -k(T-20) = -kT + 20 k, \]

which has general solution \(T(t) = Ce^{-kt} + 20\). From \(T(6) = 80\) and \(T(8) = 50\) we obtain

\[ C e^{-6k} + 20 = 80 \qquad \text{and} \qquad C e^{-8k} + 20 = 50, \]

giving \(Ce^{-6k} = 60\) and \(Ce^{-8k} = 30\). Dividing these two equations gives \(e^{2k} = 2\), so that \(k = \dfrac{1}{2} \log_e 2\). From \(Ce^{-6k} = 60\) we then have

\[ 60 = C e^{- \dfrac{6}{2} \log_e 2} = C \times 2^{-3} = \dfrac{C}{8}, \]

so that \(C = 480\). We have thus found the temperature \(T(t)\) to be

\begin{align*} T(t) &= 480 \, e^{- \dfrac{1}{2} (\log_e 2) \, t} + 20 \\ &= 480 \, \Bigl(\dfrac{1}{\sqrt 2}\Bigr)^t + 20, \end{align*}

and so the initial temperature was \(T(0) = 500\) \(^\circ\)C.

Next page - References