# 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, , 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 and then evaluate at x = 2.

```  > f:=exp(x);
```

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

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

```  > g:=ln(x);
```

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

Plotting the two functions and 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 and which curve is .

```  > 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);
```

## 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

where 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 grams. After two hours, 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);
```

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

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

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

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

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

## 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

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.

2. Anti-differentiate both sides.

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

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 into a non-insulated cup in a room at . After a minute, the coffee has cooled to . How much time is required before the coffee reaches a drinkable temperature of ?

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

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

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

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

```  > k:=solve(eq3,k);
```

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

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

```  > solve(eq4,t);
```

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

## Exercises

1. A bacteria colony initially has mass g. After two hours, the colony has mass 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 to in 20 minutes when surrounded by air at a temperature of . What will the temperature of the plate be be after one hour of cooling? When will the temperature of the plate reach ?
3. Just before midday the body of an apparent homicide victim is found in a room that is kept at a constant temperature of . At 12 noon the temperature of the body is and at 2 P.M. it is . Assume that the temperature of the body at the time of death was . 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.