Skip to main content

Exercises 3.7 Exponential Growth and Decay

Recall that the solution of the initial-value problem \(y^\prime=ky\text{,}\) \(y(0)=A\text{,}\) is given by \(\ds y=Ae^{kx}\text{.}\)

Solve the following problems.

  1. An amount of \(A_0\) CAD is invested against yearly interest of \(p\%\text{.}\) Give the expression for \(A(t)\text{,}\) the value of the investment in CAD after \(t\) years if the interest is compounded continuously by writing down the differential equation that \(A\) satisfies and solving it.

  2. Jane invests 10,000 CAD against a yearly interest \(p\%\text{,}\) compounded continuously. After 4 years the value of her investment is 15,000 CAD. What is \(p\text{?}\)

  1. \(A=A_0e^{pt}\text{.}\)

  2. \(\displaystyle p \approx 0.101. \)

  1. We notice that the value of the investment satisfies the differential equation,

    \begin{equation*} \frac{dA}{dt}=pA\text{,} \end{equation*}

    with initial condition \(A(0)=A_0\text{.}\) Hence, we must have that

    \begin{equation*} A(t) = A_0e^{pt}. \end{equation*}
  2. We solve \(15,000=10,000\cdot e^{4p}\) to get \(p = \frac{1}{4} \log \left(\frac{3}{2}\right). \)


The rate at which a student learns new material is proportional to the difference between a maximum, \(M\text{,}\) and the amount she already knows at time \(t\text{,}\) \(A(t)\text{.}\) This is called a learning curve.

  1. Write a differential equation to model the learning curve described.

  2. Solve the differential equation you created in part (a).

  3. If took a student 100 hours to learn \(50\%\) of the material in Math 151 and she would like to know \(75\%\) in order to get a \(B\text{,}\) how much longer she should study? You may assume that the student began knowing none of the material and that the maximum she might achieve is \(100\%\text{.}\)

  1. \(\ds \frac{dA}{dt}=k(M-A(t))\text{.}\)

  2. \(A(t) = M- \left(M-A_0\right)e^{-kt}.\text{.}\)

  3. 100 hours.

  1. Let \(A(t) \) be the amount of material a student knows. We are told that the rate \(\frac{dA}{dt}\) is proportional to \(M - A(t) \text{.}\) That is, \(\ds \frac{dA}{dt}=k(M-A(t))\text{,}\) for some constant \(k \text{.}\)

  2. We separate variables:

    \begin{equation*} \int \frac{dA}{M-A} = k \int dt \implies -\ln|M-A| = kt + C, \end{equation*}

    for some constant \(C \text{.}\) Therefore, we get

    \begin{equation*} A(t) = M - Ce^{-kt} \text{.} \end{equation*}

    Now let \(A(0) = A_0\text{.}\) Then we have \(C = M - A_0, \) and so

    \begin{equation*} A(t) = M- \left(M-A_0\right)e^{-kt}. \end{equation*}
  3. It is given that \(M=100\text{,}\) \(A(0)=0\text{.}\) Hence,

    \begin{equation*} \ds A(t)=100(1-e^{-kt})\text{.} \end{equation*}

    We first solve for \(k \text{:}\)

    \begin{equation*} A(100)=50 \implies 100(1-e^{-k(100)}) = 50 \implies k = -\frac{\ln 2}{100}. \end{equation*}

    We now wish to solve

    \begin{equation*} \ds 75=100(1-e^{-\frac{t\ln 2}{100}}) \end{equation*}

    for \(t\text{.}\) It follows that the student needs to study another 100 hours.


The concentration of alcohol (in \(\%\)) in the blood, \(C(t)\text{,}\) obeys the decay differential equation:

\begin{equation*} \frac{dC}{dt}=-\frac{1}{k}C\text{,} \end{equation*}

where \(k=2.5\) hours is called the elimination time. It is estimated that a male weighing 70 kg who drinks 3 pints of beer over a period of one hour has a concentration of \(1\%\) of alcohol in his blood. The allowed legal concentration for driving is a maximum of \(0.5\%\text{.}\)

  1. If a person has a blood alcohol concentration of \(1\%\text{,}\) how long should she/he wait before driving in order not to disobey the law. You may need the value \(\ln 2\approx 0.7\text{.}\)

  2. What is the initial (\(t=0\)) rate of change in the concentration?

Note: The permissible BAC limit in the Criminal Code of Canada is .08 (80 milligrams of alcohol in 100 millilitres of blood). Some advocate a lower criminal limit of .05 (50 milligrams of alcohol in 100 millilitres of blood).

  1. \(t=2.5\ln2\approx 1.75\) hours.

  2. \(-\frac{2}{7}\) hours.

  1. The model is \(\ds C(t)=C_0e^{-\frac{t}{2.5}}\) where \(C_0=1\) and the question is to solve \(\ds 0.5=e^{-\frac{t}{2.5}}\) for \(t\text{.}\) Hence \(t=2.5\ln2\approx 1.75\) hours.

  2. \(\ds C'(0)=-\frac{1}{2.5}=-\frac{2}{7}\) hours.


The concentration of alcohol (in \(\%\)) in the blood, obeys the decay equation \(\displaystyle \frac{dC}{dt}=-0.4C\text{.}\) If a person has a blood alcohol concentration of \(2\%\text{,}\) how long would it take for blood alcohol concentration to drop to \(1\%\text{?}\) Take that the elimination time is given in hours.

\(\approx 1.73\) hours.

The percentage of alcohol in the blood at time \(t\) can be modelled as \(\ds c(t)=c_0 e^{-0.4t}\text{.}\) If \(c(t)=1\) and \(c_0=2\text{,}\) it follows that \(\ds t=\frac{5\ln 2}{2}\approx 1.73\) hours.


Carbon dating is used to estimate the age of an ancient human skull. Let \(f(t)\) be the proportion of original \(^{14}C\) atoms remaining in the scull after \(t\) years of radioactive decay. Since \(^{14}C\) has a half life of 5700 years we have \(f(0)=1\) and \(f(5700)=0.5\text{.}\)

  1. Sketch the graph of \(f(t)\) versus \(t\) in the domain \(0\leq t\leq 20000\text{.}\) Label at least two points of your plot and be sure to label the axes.

  2. Write an expression for \(f(t)\) in terms of \(t\) and other numerical constants such as \(\ln 2\text{,}\) \(\sin 5\text{,}\) \(e^3\text{,}\) and \(1/5700\text{.}\) (Note: Not all of these constants need appear in your answer!)

  3. Suppose that only \(15\%\) of the original \(^{14}C\) is found to remain in the skull. Derive from your previous answer, an expression for the estimated age of the skull.

  1. \(\ds f(t)=e^{-\frac{t\ln 2}{5700}}\text{.}\)

  2. The question is to solve \(\ds 0.15=e^{-\frac{t\ln 2}{5700}}\) for \(t\text{.}\) Hence the age of the skull is \(\ds t=-\frac{5700\ln 0.15}{\ln 2}\approx 15600\) years.


The mass of a sample of a radioactive particle decays according to the rule \(\displaystyle \frac{dm}{dt}=-5m\text{.}\) Determine the half-life of this particle.

\(\approx 0.138\) units of time.

The proportion of radioactive sample remaining after time \(t\) can be modelled as \(\ds m(t)=m_0 e^{-5t}\text{.}\) If \(m(t)=0.5\text{,}\) \(m_0=1\text{,}\) it follows that the half-life of the particle is \(t=\frac{\ln 2}{5}\approx 0.138\) units of time.


Plutonium-239 is part of the highly radioactive waste that nuclear power plans produce. The half-life of plutonium-239 is 24,110 years. Suppose that 10 kg of plutonium-239 has leaked into and contaminated a lake. Let \(m(t)\) denote the mass of plutonium-239 that remains in the lake after \(t\) years.

  1. Find an expression for \(m(t)\) based on the information given.

  2. How much mass remains in the lake after 1000 years.

  3. Suppose the lake is considered safe for use after only 1 kg of the plutonium-239 remains. How many years will this take?

  1. \(m(t)=10e^{-\frac{t\ln 2}{24110}}\text{.}\)

  2. \(m(1000)=10e^{-\frac{\ln 2}{24.11}}\approx 9.716\) kilograms.

  3. About \(80091.68\) years.

  1. The model is \(\ds m(t)=10e^{-kt}\) where \(t\) is in years, \(m(t)\) is in kilograms, and \(k\) is a constant that should be determined from the fact that \(m(24110)=5\text{.}\) Hence \(\ds k=-\frac{\ln 2}{24110}\) and \(m(t)=10e^{-\frac{t\ln 2}{24110}}\text{.}\)

  2. \(m(1000)=10e^{-\frac{\ln 2}{24.11}}\approx 9.716\) kilograms.

  3. We solve \(\ds 1=10e^{-\frac{t\ln 2}{24110}}\) to get \(\ds t=24110\frac{\ln 10}{\ln 2}\approx 80091.68\) years.


On a certain day, a scientist had 1 kg of a radioactive substance \(X\) at 1:00 pm. After six hours, only 27 g of the substance remained. How much substance \(X\) was there at 3:00 pm that same day?

\(\approx 0.1392476650\) kg.

The model is \(\ds A=A(0)e^{-kt}\text{.}\) It is given that \(A(0)=1\) kg and \(A(6)=0.027\) kg. Hence \(\ds A(t)= e^{\frac{t\ln 0.0027}{6}}\text{.}\) It follows that at 3:00 there are \(A(2)=e^{\frac{\ln 0.0027}{3}} \approx 0.1392476650\) kg of substance \(X\text{.}\)


In a certain culture of bacteria, the number of bacteria increased tenfold in 10 hours. Assuming natural growth, how long did it take for their number to double?

\(\approx 3.01\) hours.

The model is \(\ds P=P(t)=P_0e^{kt}\) where \(k\) is a constant, \(P_0\) is the initial population and \(t\) is the time elapsed. It is given that \(\ds 10P_0=P_0e^{10k}\) which implies that \(\ds k=\frac{\ln 10}{10}\text{.}\) The question is to solve \(\ds 2=e^{\frac{t\ln 10}{10}}\) for \(t\text{.}\) Hence \(\ds t=\frac{10\ln 2}{\ln 10}\approx 3.01\) hours.


A bacterial culture starts with 500 bacteria and after three hours there are 8000. Assume that the culture grows at a rate proportional to its size.

  1. Find an expression in \(t\) for the number of bacteria after \(t\) hours.

  2. Find the number of bacteria after six hours.

  3. Find an expression of the form \(\ds m\frac{\ln a}{\ln b}\) with \(m\text{,}\) \(a\text{,}\) and \(b\) positive integers for the number of hours it takes the number of bacteria to reach a million.

  1. \(\ds P=500e^{\frac{4t\ln 2}{3}}\text{.}\)

  2. \(128000\) bacteria.

  3. \(\approx 4.7414\) hours.

  1. The model is \(\ds P=500e^{kt}\text{.}\) From \(\ds 8000=500e^{3k}\) it follows that \(\ds k=\frac{4\ln 2}{3}\text{.}\) Thus the model is \(\ds P=500e^{\frac{4t\ln 2}{3}}\text{.}\)

  2. \(128000\) bacteria.

  3. Solve \(\ds 10^6=500e^{\frac{4t\ln 2}{3}}\) for \(t\text{.}\) It follows that \(\ds t=\frac{3(4\ln 2+3\ln 5)}{4\ln 2}\approx 4.7414\) hours.


A bacteria culture starts with 500 bacteria and grows at a rate proportional to its size. After three hours there are 8000 bacteria. Find the number of bacteria after four hours.


The bacteria population \(P(t)\) quadruples every 15 minutes. The initial bacteria population is \(P(0)=10\text{.}\) You might need the following values: \(\ln 6\approx 1.6\text{,}\) \(\ln 8\approx 2.08\text{,}\) \((\ln 10)/(\ln 2)\approx 3.32\text{,}\) \(\ln 2\approx 0.69\text{,}\) \(4^3=64\text{,}\) \(4^6=4096\text{,}\) \(4^9=262144\text{,}\) \(4^{12}=16777216\text{.}\)

  1. What is the population after three hours?

  2. How much time does it take for the population to grow to 1 billion?

  1. \(167,772,160\) bacteria.

  2. \(3.32\) hours.

  1. The model is \(\ds P(t)=10e^{kt}\) where \(t\) is time in hours. From \(40=10e^{\frac{k}{4}}\) it follows that \(k=4\ln 4\text{.}\) Hence \(P(3)=10e^{12\ln 4}=167,772,160\) bacteria.

  2. \(3.32\) hours.


The population of a bacteria culture grows at a rate that is proportional to the size of the population.

  1. Let \(P\) denote the population of the culture at time \(t\text{.}\) Express \(dP/dt\) in terms of the proportional constant \(k\) and \(P\text{.}\)

  2. If the population is 240 at time \(t=1\) and is 360 at time \(t=2\text{,}\) find a formula for the number of bacteria at time \(t\text{.}\) (\(t\) in hours.)

  3. How many bacteria were there at time \(t=0\text{.}\) Your answer should be a positive integer.

  4. What is the value of \(dP/dt\) when \(t=0\text{.}\)

  1. \(\displaystyle \frac{dP}{dt}=kP\text{.}\)

  2. \(\displaystyle P=160e^{t\ln \frac{3}{2}}\text{.}\)

  3. \(P(0)=160\) bacteria.

  4. \(\displaystyle \frac{dP}{dt}(0)=160\ln \frac{3}{2}\text{.}\)


Assume that Math 151 in fall of 2000 had an enrolment of 500 students and in fall 2002 had an enrolment of 750 students. Assume also that if \(P(t)\) is the enrolment at time \(t\) (let \(t\) be in years, with \(t=0\) corresponding to year 2000), then \(P'(t)=kP(t)\) for some constant \(k\text{.}\) Calculate \(P(500)\) (the enrolment in Math 151 in fall of 2500). Simplify your answer as much as possible. The answer will be quite large.

5.27\cdot 10^{46}.

The model is \(\ds P(t)=500e^{kt}\) where \(t\) is time in years. From \(P(2)=750\) it follows that \(\ds k=\frac{\ln 3-\ln 2}{2}\text{.}\) Thus \(\ds P(500)=500e^{250(\ln 3-\ln 2)}=5.27\cdot 10^{46}\text{.}\)


A freshly brewed cup of coffee has temperature \(95^\circ\)C and is in a \(20^\circ\)C room. When its temperature is \(70^\circ\)C, it is cooling at the rate of \(1^\circ\)C per minute. When does this occur?

Use Newton's Law of Cooling.
\(t\approx 20.2\) minutes.

A cup of coffee, cooling off in a room at temperature \(20^0\)C, has cooling constant \(k=0.09\)min\(^{-1}\text{.}\)

  1. How fast is the coffee cooling (in degrees per minute) when its temperature is \(T=80^\circ\)C?

  2. Use linear approximation to estimate the change in temperature over the next 6 seconds when \(T=80^\circ\)C.

  3. The coffee is served at a temperature of \(90^\circ\text{.}\) How long should you wait before drinking it if the optimal temperature is \(65^\circ\)C?

  1. \(\ds \left. \frac{dT}{dt}\right| _{T=80^0}=-0.09\cdot(80-20)=-5.4\ ^0\)C/min.

  2. \(\ds T-80\approx -0.54^0\)C.

  3. About \(6.4\) minutes.

  1. \(\ds \left. \frac{dT}{dt}\right| _{T=80^0}=-0.09\cdot(80-20)=-5.4\ ^0\)C/min.

  2. Note that 6 seconds should be used as 0.1 minutes. From \(T\approx 80- 5.4\Delta T=80-5.4\cdot 0.1\) it follows that the change of temperature will be \(\ds T-80\approx -0.54^0\)C.

  3. \(\ds t= -\frac{100}{9}\ln \frac{9}{16}\approx 6.4\) minutes. Find the function \(T=T(t)\) that is the solution of the initial value problem \(\ds \frac{dT}{dt}=-0.09(T-20)\text{,}\) \(T(0)=90\text{,}\) and then solve the equation \(T(t)=65\) for \(t\text{.}\)


A cold drink is taken from a refrigerator and placed outside where the temperature is \(32^\circ\)C. After 25 minutes outside its temperature is \(14^\circ\)C, and after 50 minutes outside its temperature is \(20^\circ\)C. assuming the temperature of drink obeys Newton's Law of Heating, what was the initial temperature of the drink?


The model is \(\ds \frac{dT}{dt}=k(T-32)\) where \(T=T(t)\) is the temperature after \(t\) minutes and \(k\) is a constant. Hence \(\ds T=32+(T_0-32)e^kt\) where \(T_0\) is the initial temperature of the drink. From \(14=32+(T_0-32)e^{25k}\) and \(20=32+(T_0-32)e^{50k}\) it follows \(\ds \frac{3}{2}=e^{-25k}\) and \(\ds k=-\frac{1}{25}\ln \frac{3}{2}\text{.}\) Therefore, the initial temperature was approximately \(T_0=5^0\)C.


On Hallowe'en night you go outside to sit on the porch to hand out candy. It is a cold night and the temperature is only \(10^\circ\)C so you have made a cup of hot chocolate to drink. If the hot chocolate is \(90^\circ\)C when you first go out, how long does it take until it is a drinkable \(60^\circ\)C given that \(k=0.03s^{-1}\text{?}\)

\(t\approx 4.5\) minutes.

In a murder investigation the temperature of the corpse was \(32.5^\circ\)C at 1:30PM and \(30.3^\circ\)C an hour later. Normal body temperature is \(37^\circ\)C and the temperature of the surroundings was \(20^\circ\)C. When did murder take place?

\(t\approx 2.63\) hours.