Subsections

• Purpose
• Background
  • Exponential growth
  • Exponential Decay
  • Newton's law of cooling
  • Information Diffusion
  
  
• Exercises

Applications of exponential and logarithmic functions

Purpose

The purpose of this lab is to use Maple to study applications of exponential and logarithmic functions. These are used to model many types of growth and decay.

Background

Exponential growth

The simple model for growth is exponential growth, where it is assumed that $y'(t)$ is proportional to $y$. That is,

$$\frac{dy}{dt} = ky\quad\hbox{where $k$ is a positive constant.}$$

Separating the variables and integrating (see section 4.4 of the text), we have

$$\int\frac{dy}{y} = \int k dt$$

so that

$$\ln \mid y \mid = kt + C$$

In the case of exponential growth, we can drop the absolute value signs around $y$, because $y$ will always be a positive quantity. Solving for $y$, we obtain

$$\mid y \mid = e^{kt + C}$$

which we may write in the form $y = Ae^{kt}$, where $A$ is an arbitrary positive constant.

Exponential Decay

In a sample of a radioactive material, the rate at which atoms decay is proportional to the amount of material present. That is,

$$\frac{dA}{dt} = -kA$$

where $k>0$ is a constant. This is the same equation as in exponential growth, except that $-k$ replaces $k$. The solution is

$$A(t) = A_0 e^{-kt}$$

where $A_0$ is a positive constant. Physically, $A_0$ is the amount of material present at $t=0$.

Radioactivity is often expressed in terms of an element's half-life. For example, the half-life of carbon-14 is 5730 years. This statement means that for any given sample of ${}^{14}\hbox{C}$, after 5730 years, half of it will have undergone decay. So, if the half-life is of an element Z is $c$ years, it must be that $e^{-kc}=\frac{1}{2}$, so that $kc=\ln 2$ and $k=\frac{\ln 2}{c}$.

Newton's law of cooling

What is usually called Newton's law of cooling is a simple model for the change in temperature of an object that is in contact with an environment at a different temperature. It says that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the temperature of the environment. Mathematically, this can be expressed as the differential equation

$$\frac{dT}{dt} = -k \left( T-T_{\mathrm{out}} \right)$$

where $k$ is the constant of proportionality and $T_{\mathrm{out}}$ is the temperature of the environment. Using a technique called separation of variables it isn't hard to derive the solution

$$T(t) = T_{\mathrm{out}} + \left( T_0 - T_{\mathrm{out}} \right) e^{-kt}$$

where $T_0$ is the temperature of the object at $t=0$.

Information Diffusion

Information can be thought of as of a physical quantity which can be measured. According to the Gallup Institute, information news diffuses through a fixed adult population of size $P$ at a rate of time proportional to the number of people who have not heard the news.

If $N$ is the number of people who have heard the news after t days, then

$$\frac{dN}{dt}=k(P-N)$$

The initial condition $N(0)=0$ yields the solution

$$N(t) = P(1-exp(-kt))$$

Exercises

1. Suppose that the population of a certain bacteria can be modeled by an exponential function. In a particular experiment, the number of bacteria was $525$ at $t=0$. Suppose the population doubles every six hours. Find the value of the growth constant $k$ and use it to predict the number of bacteria that would have been present after a day and a half. Check your answer by repeated doubling of the initial number of bacteria.
2. Exponential growth can also be used to model the growth of investments. Suppose that the value $I$ of an investment satisfies the differential equation
   
   $$\frac{dI}{dt}=rI$$
   
   
   where $r$ is the interest rate. If the interest rate is $4.6 \%$ per year and you start with an investment of $8,500, what would the net gain be after 20 years? Approximately how many years would it take to save $50,000?
3. How long would it take for half of 75 grams of a radioactive element to decay if it decays at a yearly rate of 2.2%.
4. Suppose that a coffee was served extra hot ( $180^{\circ} \mathrm{F}$) in a $70^{\circ} \mathrm{F}$ room and was $145^{\circ} \mathrm{F}$ after 3 minutes. Approximately how many more minutes will it be too cold to drink ( $110^{\circ} \mathrm{F}$)?
5. Suppose that a group of 250 friends are on a social media website and 33% of them heard about a rumor 4 days after it happened. How many more days will it take for 90% of the people to have heard the rumor?