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Applications of exponential and logarithmic functions


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, as well as in many scales, such as the Richter and decibel scales.


Exponential growth

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

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

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

\int\frac{dy}{y} = \int k\,dt \end{displaymath}

so that

\ln y = kt + C\end{displaymath}

(If y is a positive quantity, we may drop the absolute value signs around y.) Solving for y

y = ekt + C

which we may write in the form y = Aekt, 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\end{displaymath}

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) = A0 e-kt

where A0 is a positive constant. Physically, A0 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) \end{displaymath}

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} \end{displaymath}

where T0 is the temperature of the object at t=0.

Effective medicine dosage

If a drug is adminstered to a patient intravenously, the concentration jumps to its highest level almost immediately. The concentration subsequently decays exponentially. If we use C(t) to represent the concentration at time t, and C0 to represent the concentration just after the dose is administered then our exponential decay model would be given by

C(t) = C0 e-kt

A problem facing physicians is the fact that for most drugs, there is a concentration, m, below which the drug is ineffective and a concentration, M, above which the drug is dangerous. Thus the physician would like the have the concentration C(t) satisfy

m < C(t) < M

This means that the initial dose must not produce a concentration larger than M and that another dose will have to be administered before the concentration reaches m.

Maple commands

The main functions you need are the natural exponential and natural logarithm. The Maple commands for these functions are exp and ln. Here are a few examples.
  > f := x -> exp(-2*x);

{f} := {x} \rightarrow {\rm e}^{(\, - 2\,{x}\,)}\end{displaymath}\end{maplelatex}
  > simplify(ln(3)+ln(9));

3\,{\rm ln}(\,3\,)\end{displaymath}\end{maplelatex}
  > ln(exp(x));

{\rm ln}(\,{\rm e}^{{x}}\,)\end{displaymath}\end{maplelatex}
  > simplify(ln(exp(x)));

  > solve(exp(-3*x)=0.5,x);

  > plot(log[10](x),x=0..100);

Sometimes you need to use experimental data to determine the value of constants in models. For example, suppose that for a particular drug, the following data were obtained. Just after the drug is injected, the concentration is 1.5 mg/ml (milligrams per milliliter). After four hours the concentration has dropped to 0.25 mg/ml. From this data we can determine values of C0 and k as follows. The value of C0 is the initial concentration, so we have

C0 = 1.5

To find the value of k we need to solve the equation

0.25 = 1.5 e-4k

which we get by plugging in t=4 and using the data C(4)=0.25. Maple commands for solving for k and defining and plotting the function C(t) are shown below.
  > k1 := solve(0.25=1.5*exp(-4*k),k);

{\it k1} := .4479398673\end{displaymath}\end{maplelatex}
  > C1 := t -> 1.5*exp(-k1*t);

{\it C1} := {t} \rightarrow 1.5\,{\rm e}^{(\, - {\it k1}\,{t}\,)}\end{displaymath}\end{maplelatex}
  > plot(C1(t),t=0..6);


Suppose that b satisfies b > 1. Explain the relationship between the graphs of

f(x) = bx


g(x) = \left( \frac{1}{b}\right) ^x \end{displaymath}

Suppose that the population of a certain bacteria can be modeled by an exponential function. In a particular experiment, the number of bacteria was 10,000 at t=0. Four hours later, the number of bacteria was 250,000. Suppose a second experiment is performed under the same conditions, but the number of bacteria at t=0 is only 2000. Show, using the equation for exponential growth, that the predicted number of bacteria after four hours in this second experiment is 50,000.

Suppose that for a certain drug, the following results were obtained. Immediately after the drug was administered, the concentration was 3.3 mg/ml. Six hours later, the concentration had dropped to 1.55 mg/ml. Determine the value of k for this drug.

Suppose that for the drug in the previous exercise, the maximum safe level is $M=8 \mbox{ mg/ml}$ and the minimum effective level is $m=1.8 \mbox{ mg/ml}$. What is the maximum possible time between doses for this drug? (Hint - the initial dose should give an initial concentration of C0 = M = 8.)

A thermometer is taken from a room at $72 \, ^{\circ} \mathrm{F}$to the outdoors where the temperature is $20 \, ^{\circ}
\mathrm{F}$. Determine the reading on the thermometer after 5 minutes, if the reading drops to $48 \, ^{\circ} \mathrm{F}$ after 1 minute.

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Jane E Bouchard