## Exercises3.7Exponential 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.
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. $\displaystyle p \approx 0.101.$

Solution
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 $p = \frac{1}{4} \log \left(\frac{3}{2}\right).$

###### 2.

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.

Solution
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.

###### 3.

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.

Solution
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.

###### 4.

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.
Solution

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.

###### 5.

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.

###### 6.

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.
Solution

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.

###### 7.

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.

Solution
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.

###### 8.

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.
Solution

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 $X\text{.}$

###### 9.

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.
Solution

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.

###### 10.

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.

Solution
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.

###### 11.

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.

$20158\text{.}$
###### 12.

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.

Solution
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.

###### 13.

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

###### 14.

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}.
Solution

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 $\ds P(500)=500e^{250(\ln 3-\ln 2)}=5.27\cdot 10^{46}\text{.}$

###### 15.

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?

Hint
Use Newton's Law of Cooling.
$t\approx 20.2$ minutes.
###### 16.

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.

Solution
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 $t\text{.}$

###### 17.

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?

$T_0=5^0$C.
Solution

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.

###### 18.

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.