|UBC Calculus Online Course Notes|
Initial Value Problems for Growth and Decay
Solution: We might start by summarising the information
briefly and defining the variables in the problem.
Given: The starting number of cells at the begining of the experiment (which we may as well call is:
The rate of change of the number of cells, i.e. the growth rate is simply , so that the statement that the growth rate is proportional to the number of cells just means that:
We know from previous work that this differential equation has the solution
and now our task is to put in values for the constants .
It is straightforward to put in , since this value is known. Thus
To find we must do a little more work. We can use the fact that after three hours, there are 500,000 cells so, plugging in for t, we have
Cancelling a factor of 10,000 and writing the exponent in a tidier way, this implies that
We can use this relationship to find the value of 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:
Thus, having found the rate constant, , we find that the solution to the differential equation that also statisfies the initial value is the function.
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:
Doubling Time: The amount of time it takes for a given
population to double in size.
We can now find the value of by solving
Cancelling a factor of 10,000, taking natural logs as before, and simplifying the result leads to:
In the problem we are looking at, so that
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
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:
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 of the radioactive isotope is proportional to its mass at the given time, we write
However, here we encounter a slight difference: the mass is decreasing. This must mean that the derivative, is negative, which tells us that the constant of proportionality must be negative. We can represent this by writing
Comment: The independent variable is still time, but now the mass is the function solved for. In this equation, the constant is positive, the mass 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
Note that when , 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 are shown below:
For your consideration:
With this preparation, we are ready to turn attention to solving the
If we use the symbol to denote the half life of a process, and to represent the amount of a substance initially present then
Cancelling a factor of , we get
We can take the reciprocals of both sides of this equation to get it in a more familiar form:
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
So we conclude that the half life and the rate constant are related in the same way as before:
| In this example we are given the half life,
years, and we should find the
constant k from it:
Thus the solution we are looking for is:
After 10,000 years, the fraction of the original amount that will be left is found by substituting into this expression, and noting that
Thus 30% of the original sample will be left.