Subsections

# The exponential and logarithmic functions

## Exponential growth

The simple model for growth is exponential growth, where it is assumed that is proportional to . That is,

Separating the variables and integrating, we have

so that

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

which we may write in the form , where is an arbitrary positive constant. The same formula is used for exponential decay, except the decay constant is negative.

### 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

where is the constant of proportionality and is the temperature of the environment. Using a technique called separation of variables it isn't hard to derive the solution

where is the temperature of the object at .

## Inverse Functions

Consider the functions defined by

To be able to get an inverse the function must be one-to-one. You can plot the functions to get a hint as to whether they are invertible or not.
> f:=x->exp(x)+exp(-x);
> plot(f(x),x=-5..5);
> g:=x->exp(x)-exp(-x);
> plot(g(x),x=-5..5);
Both satisfy the vertical-line test but is not invertible since it does not satisfy the horizontal-line test. Indeed is not one-to-one, for instance . From the plot it seems that the function is one-to-one. In order to determine its inverse we solve for x.
> solve(g(x)=y,x);
We observe that one of the solutions is not defined since the arguement of the logarithm can only be positive. Thus:

> ginv:=y->ln(y/2+sqrt(y^2+4)/2);
Let's look at the plot along with the line to see if our functions seem to make sense.
> plot({x,g(x),ginv(x)},x=-20..20,y=-20..20,scaling=constrained);
Let's check that we have computed the right inverse. By definition the composot ion of the functions should be the line since an inverse is the reflectio n about this line.

> g(ginv(y));
> simplify(%);
> ginv(g(x));
> simplify(%);
We are having difficulty getting for the last composition. Think about what issue the computer has in dealing with this simplification as you will come across this in the exercises.

## Exercises

1. A pot of soup taken from the stovetop at and a jug of apple juice taken from a refrigerator at are put on a dinner table. The temperature in the dining room is held at . After 10 minutes, the soup temperature is and the juice temperature is
1. How much longer will it take the soup to lose its taste quality, i.e., to cool down to the temperature of ? What is the juice temperature at this moment?
2. Plot the graphs (temperature) versus (time) for both products. From the graphs, make an estimate when the soup should be eaten if it is supposed to be best at
http://www.math.wpi.edu/Course_Materials/MA1022B98/
Joseph Feribach

2. Copy every command from the background under the section titled Inverse Functions and paste them into Maple and execute each command. Be sure to shrink your graphs.

3. Given the functions and (Hint: Use Maple's surd function when entering fractional exponents.)
A)
Plot the function over the interval and plot the function over the same interval, but on a separate graph. Which function is not invertible and why?
B)
Find the inverse of the invertible function.
C)
Plot the function and its inverse along with the line on the domain of and .
D)
Show that you have the correct inverse by using the composite definition. (If you come across a simplifying problem type the command assume(x>0).)