Introducing a Differential Equation
Newton's Law of Cooling
UBC Calculus Online Course Notes

Initial Value Problems for Growth and Decay


Example 1: Unlimited Population Growth
The number of bacteria in a liquid culture is observed to grow at a rate proportional to the number of cells present. At the begining of the experiment there are 10,000 cells and after three hours there are 500,000. How many will there be after one day of growth if this unlimited growth continues? What is the doubling time of the bacteria?

Solution: We might start by summarising the information briefly and defining the variables in the problem.
Let

$ t $ = time elapsed from the begining of the experiment (in hours).
$ y(t) $ = the number of cells at time t.

Given: The starting number of cells at the begining of the experiment (which we may as well call $ t=0 $ is:

$  y(0) = y_0 = 10,000 $

The rate of change of the number of cells, i.e. the growth rate is simply $ dy/dt $ , so that the statement that the growth rate is proportional to the number of cells just means that:


\[ 
\frac{dy}{dt} = k y. 
\]

We know from previous work that this differential equation has the solution


\[ 
y(t) = y_o e^{kt} 
\]

and now our task is to put in values for the constants $ y_o, k $ .

It is straightforward to put in $ y_o $ , since this value is known. Thus


\[ 
y(t) = 10,000 e^{kt} 
\]

To find $ k $ we must do a little more work. We can use the fact that after three hours, $ t=3, $ there are 500,000 cells so, plugging $ t=3 $ in for t, we have


\[ 
y(3) = 10,000 e^{k3}= 500,000 
\]

Cancelling a factor of 10,000 and writing the exponent in a tidier way, this implies that


\[ 
 e^{3k}= 50 
\]

We can use this relationship to find the value of $ k $ with the following standard little "trick". Remembering that the natural logarithm is the inverse of the exponential function, we take the natural log of both sides, getting:


\[ 
\ln(e^{3k})= \ln (50) \\ 
~~{\rm ~so~}~~~ 3k = \ln(50),~~~~k =\ln(50)/3 \approx 3.91/3 = 1.3~ 
 {\rm per~hour} 
\]

Thus, having found the rate constant, $ k=1.3 $ , we find that the solution to the differential equation that also statisfies the initial value is the function.


\[ 
y(t) = 10,000 e^{1.3 t} 
\]

The graph of this function is shown above. The units on the y axis correspond to multiples of 1,000.

We can now predict how many bacteria there will be after 1 day. (This is not shown on the graph, because it would be way off scale!) Remembering to convert to consistent time units, namely hours, we compute that after 1 day (24 hours) of this unlimited exponential growth the total number of cells should be:


\[ 
y(24) = 10,000 e^{1.3 (24)} = 10,000 e^{31.2} = 3.5 \times 10^{17} 
\]

Doubling Time: The amount of time it takes for a given population to double in size.
(Note: In the Andromeda strain, the bacteria doubled every twenty minutes. We used that fact to determine how the bacteria would grow. Here we are not given the doubling time directly, but we can calculate it from the other information given in the problem, as shown below.)
Let's refer to the doubling time by the symbol $ \tau $ . (This is the Greek letter tau. Mathematicians have a special affection for the Greek alphabet, which comes in handy whenever the Roman alphabet runs out of convenient letters to use.)
Since $ \tau $ is the time it takes for a population to double, the size of the population at $ t=\tau $ is twice its size at $ t=0 $ , i.e. $ y(\tau)=2y_o $ . So,


\[ 
y(\tau) = 10,000 e^{k \tau } = 2 y_o = 20,000 
\]

We can now find the value of $ \tau $ by solving


\[ 
10,000 e^{k \tau } = 20,000 
\]

Cancelling a factor of 10,000, taking natural logs as before, and simplifying the result leads to:


\[ 
 e^{k \tau } = 2, ~~~\ln(e^{k \tau}) = \ln(2), ~~~~k \tau = \ln(2), 
 ~~~~\tau = \frac{\ln(2)}{k} 
\]

In the problem we are looking at, $ k=1.3 $ so that


\[ 
\tau = \frac{\ln(2)}{k} = \frac{0.69}{1.3}=0.533 
\]

Thus, the doubling time in this problem is 0.533 hours, which is roughly 32 minutes. These bacteria grow a little more slowly than the Andromeda strain.



Example 2: Radioactive Decay
Radioactivity is a property characteristic of substances whose atoms undergo spontaneous decomposition. Such substances can exist in the unstable "radioactive" isotope , or in the stable form to which this decays. The decay usually happens at some constant rate, releasing "bursts" that can be detected by a geiger counter. The more radioactive the sample, the more frequent the bursts, and the more intense the measured level of bursts. As the atoms decay, the rate of change of the mass of the radioactive isotope in the sample is proportional to the mass present. (Think of it this way: if there is twice as much radioactive material, then twice as many atoms would break apart in a given period of time).

We can use the same kind of mathematics to investigate the phenomenon of radioactive decay as we used to study growth of a population. We proceed by defining the variables of interest:


$ t $ = time.
$ m(t) $ = mass of radioactive isotope.

The mass of the radioactive isotope is always a positive number, but, with time, it will get smaller and smaller as more and more of the substance is converted into the stable, nonradioactive material.

Using the fact that the rate of change of the mass $ dm/dt $ of the radioactive isotope is proportional to its mass $ m $ at the given time, we write


\[ 
\frac{dm}{dt} {\rm~is~proportional~to~} m. 
\]

However, here we encounter a slight difference: the mass is decreasing. This must mean that the derivative, $ dm/dt $ is negative, which tells us that the constant of proportionality must be negative. We can represent this by writing


\[ 
\frac{dm}{dt} = - k m. 
\]

Comment: The independent variable is still time, but now the mass $ m(t) $ is the function solved for. In this equation, the constant $ k $ is positive, the mass $ m $ is positive, so the derivative must be negative, signifying a decreasing mass !

By the previous work, we know that the solution to this differential equation is


\[ 
y(t) = y_o e^{-kt} 
\]

Note that when $  t > 0 $ , the exponent in this function will be negative. Thus, we need to acquaint ourselves with functions of the above form for negative exponents. A few members of this family of functions (for various values of the constant $ y_o $ are shown below:


For your consideration:

  • Why do these functions decrease rather than increase with time? (Hint: remember that the value of the base is $ e=2.17.. $ , and that $ e^{-kt}=1/(e^{kt}) $ . What can you say about the behaviour of this expression?

With this preparation, we are ready to turn attention to solving the folowing problem:

Problem:
Carbon-14 is a radioactive isotope of carbon that has a half life of 5600 years. It is used extensively in dating organic material that is tens of thousands of years old. What fraction of the original amount of Carbon-14 in a sample would be present after 10,000 years?


Definition: Problem:
The half life of a substance or a decaying material (or population) is the amount of time it takes for 50% of the original amount of substance (or material or population) to decay.


If we use the symbol $ \tau $ to denote the half life of a process, and $ m_o $ to represent the amount of a substance initially present then


\[ 
m(\tau) = \frac{1}{2}m_o 
\]

Thus


\[ 
\frac{1}{2}m_o =m_o e^{-k \tau} 
\]

Cancelling a factor of $ m_o $ , we get


\[ 
\frac{1}{2}=e^{-k \tau} 
\]

We can take the reciprocals of both sides of this equation to get it in a more familiar form:


\[ 
 2=e^{k \tau} 
\]

We are back to a relation that we already solved when we studying doubling time for a process of growth (see Example 1 on this page). Taking the natural logarithm of both sides, we find that


\[ 
 \ln(2)=\ln(e^{k \tau}) = k \tau 
\]

So we conclude that the half life and the rate constant $ k $ are related in the same way as before:


\[ 
\tau = \frac{\ln(2)}{k}~~~~{\rm or} ~~~~  k= \frac{\ln(2)}{\tau} 
\]

Solution:

In this example we are given the half life, $  \tau=5600 $ years, and we should find the constant k from it:


\[ 
 k= \frac{\ln(2)}{\tau} = \frac{0.69}{5600} = 1.2 \times 10^{-4} {\rm ~per~ 
 year} 
\]

Thus the solution we are looking for is:


\[ 
m(t) = m_o e^{-1.2 \times 10^{-4} t} 
\]

After 10,000 years, the fraction of the original amount that will be left is found by substituting $ t=10,000 $ into this expression, and noting that


\[ 
\frac{m(10,000)}{m_o} =  e^{-1.2 \times 10^{-4} \times 10,000} = e^{-1.2} 
=0.3 
\]

Thus 30% of the original sample will be left.