Applications of the Natural Logarithm and
Exponential Function

As you have seen in class, the exponential and logarithmic functions have many real life applications. In this lab, we will gain experience working with two of the applications: exponential growth and decay and Newton's Law of Cooling. First, a little background on the exponential and natural logarithmic functions.

Background

In order to enter the exponential function, tex2html_wrap_inline201 , in Maple, you must use the exp command. The syntax of this command is similar to that of sin and cos. For example, here we enter the function tex2html_wrap_inline203 and then evaluate at x = 2.

  > f:=exp(x);

displaymath191

  > evalf(subs(x=2,f));

displaymath192

The natural log function, ln, is also used in this manner.

  > g:=ln(x);

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  > evalf(subs(x=2,g));

displaymath194

Plotting the two functions tex2html_wrap_inline203 and tex2html_wrap_inline209 on the same coordinate system gives an idea of the symmetry around the line y=x. Make sure that you can identify which curve is tex2html_wrap_inline201 and which curve is tex2html_wrap_inline215 .

  > plot(f,g,x,x=-5..5,y=-5..5);

Looking at the graph of the two functions, you can get a pretty good idea of what happens to f(x) as x approaches negative infinity; however,in order to be sure, you must take a limit.

  > limit(f,x=-infinity);

displaymath195

Exponential Growth and Decay

The growth (decay) of bacteria colonies can be modeled using exponential functions. For example, one type of bacteria commonly used in research is Escherichia coli (also known as E. coli), a bacteria residing in the human digestive system. During the growth phase, approximately 12 hours, of E. coli the colony grows exponentially. For practicality, bacteria colonies are measured by mass rather than the number of cells present. The equation used to represent growth is

displaymath221

where tex2html_wrap_inline223 is the initial mass of the bacteria population, r is the growth rate and M is the mass of the population after time t.

Example

At the start of an experiment, a bacteria colony has mass tex2html_wrap_inline241 grams. After two hours, tex2html_wrap_inline243 grams. Find the growth rate r of the colony. Find the mass of the colony after 10 hours.

  > solve(M=M0*exp(r*t),r);

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  > M0:=2*10^(-9);

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  > M2:= 1.5*10^(-8);

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  > r:=solve(M2 = M0 * exp(r*2), r);

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  > mass_10 := M0 * exp(r*10);

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The growth rate of the colony is r = 1.00745. Using this information, the mass of the colony after 10 hours is tex2html_wrap_inline249 .

Newton's Law of Cooling

Newton's Law of Cooling is used to model the temperature change of an object of some temperature placed in an environment of a different temperature. The law states that

displaymath251

where T is the temperature of the object at time t, R is the temperature of the surrounding environment (constant) and k is a constant of proportionality. What this law says is that the rate of change of temperature is proportional to the difference between the temperature of the object and that of the surrounding environment.

In order to get the previous equation to something that we can use, we must solve the differential equation. The steps are given below.

  1. Separate the variables. Get all the T's on one side and all the t's on the other side. The constants can be on either side.

    displaymath265

  2. Anti-differentiate both sides.

    displaymath267

  3. Leave in the previous form or solve for T.

    displaymath271

We now have a useful equation. When you are working with Newton's Law of Cooling, remember that t is the variable. The other letters, R, k, C, are all constants. In order to find the temperature of the object at a given time, all of the constants must first have numerical values.

Example

A cup of coffee is poured from a potwhose contents are tex2html_wrap_inline293 into a non-insulated cup in a room at tex2html_wrap_inline295 . After a minute, the coffee has cooled to tex2html_wrap_inline297 . How much time is required before the coffee reaches a drinkable temperature of tex2html_wrap_inline299 ?

  1. Enter the equation.
      > eq1:= T = exp(k*t+C) +20;
    

    displaymath277

  2. Solve for C by using the initial temperature.
      > eq2:= subs(t=0,T=95,eq1);
    

    displaymath278

      > C:=evalf(solve(eq2,C));
    

    displaymath279

  3. Solve for k by using the temperature at t=1.
      > eq3:=subs(t=1,T=90,eq1);
    

    displaymath280

      > k:=solve(eq3,k);
    

    displaymath281

  4. Look at the original equation and notice that the only variable is t. Proceed to enter desired temperature and solve for t.
      > eq1;
    

    displaymath282

      > eq4:=subs(T=65,eq1);
    

    displaymath283

      > solve(eq4,t);
    

    displaymath284

The time it takes for the coffee to reach tex2html_wrap_inline299 is 7.4 minutes.

Exercises

  1. A bacteria colony initially has mass tex2html_wrap_inline313 g. After two hours, the colony has mass tex2html_wrap_inline315 g. Find the mass after 6 hours. Find the time it takes for the original mass of the colony to triple.
  2. A metal plate that has been heated cools from tex2html_wrap_inline317 to tex2html_wrap_inline319 in 20 minutes when surrounded by air at a temperature of tex2html_wrap_inline321 . What will the temperature of the plate be be after one hour of cooling? When will the temperature of the plate reach tex2html_wrap_inline323 ?
  3. Just before midday the body of an apparent homicide victim is found in a room that is kept at a constant temperature of tex2html_wrap_inline325 . At 12 noon the temperature of the body is tex2html_wrap_inline327 and at 2 P.M. it is tex2html_wrap_inline329 . Assume that the temperature of the body at the time of death was tex2html_wrap_inline331 . What is the time of death?
    1. Radioactive radium has a half-life of approximately 1600 years. What percentage of a present amount remains after 200 years?
    2. If radioactive material decays continuously at a rate proportional to the amount present, find the half-life of the material if after one year, 99.25 % of an initial amount still remains.

About this document ...

This document was generated using the LaTeX2HTML translator Version 96.1-e (April 9, 1996) Copyright © 1993, 1994, 1995, 1996, Nikos Drakos, Computer Based Learning Unit, University of Leeds.

The command line arguments were:
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The translation was initiated by William W. Farr on Tue Sep 16 11:35:57 EDT 1997


William W. Farr
Tue Sep 16 11:35:57 EDT 1997