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Course Notes for Calculus I

Section 3.2 Differential Equations

Subsection 3.2.1 Interpreting the Population Regression

In Section 3.1, I wondered about the connection between percentage growth to exponential functions. Percentage growth is a growth rate, but that’s not something I have a good mathematical understanding of — at least not yet. Algebraic techniques have a hard time understanding growth rates, or rates of change in general. Calculus includes a tool called the derivative of a function which measures its rate of change. For now, I am going to avoid the technical definition but work with the idea.
A derivative of a function is a measure of its rate of change or it growth rate.
I need some notation. For a function \(p(t)\text{,}\) there are two common ways of writing the derivative.
\begin{align*} \amp p^\prime(t) \amp \amp \frac{dp}{dt} \end{align*}
The first notation is called Newton’s notation, and the second is called Leibniz’s notation. I will use both in this course but I will rely on Leibniz’s notation most of the time. Leibniz’s notation has the advantage of including the independent variable (here \(t\)) in the notation; this is often very convenient, even though the notation is less succinct.
The derivative is the rate of change of the function, so for a population model, \(\frac{dp}{dt}\) is the growth rate of the population. Let me remind you of that data used in the Section 3.1.
Table 3.2.1.
Year Population Growth Rate
0 1032 Not Applicable
1 1214 17.6 %
2 1372 13.0 %
3 1629 18.7 %
4 2143 31.6 %
5 2520 17.6 %
6 2940 16.7 %
7 3292 12.0 %
8 3813 15.8 %
9 4757 24.8 %
10 5632 18.4 %
11 6842 21.5 %
12 8010 17.1 %
The average annual increase in the data is somewhere near \(20\)% growth. That means that the growth rate (the amount of population added in a year) is \(\frac{20}{100}\) (or \(\frac{1}{5}\)) times the current population. I can express this as an equation.
\begin{equation*} \frac{dp}{dt} = \frac{1}{5} p(t) \end{equation*}
This is a mathematical translation of the understanding of percentage growth. An equation of this form, involving a function and its derivative, is called a differential equation (a DE, in short). Very often in applied mathematics, it is easier to observe the rate of change of a function than the function itself. This leads to models expressed as differential equations. Many of the most important mathematical models are expressed as differential equations.
The regression provided the function \(p(t) = 2^{\frac{t}{4}}\text{.}\) I will show, later in the course, that the solution to this differential equation is a very similar function.
\begin{equation*} p(t) = e^{\frac{t}{5}} \end{equation*}

Subsection 3.2.2 The Concept of a Differential Equation

Solving differential equations is one of the major goals of calculus. The solution described above leads to the profound connection between percentage growth rate and exponential growth: if I observe a percentage growth rate, then the quantities I observe should be described by an exponential function. This is how I know that many populations grow exponentially, as do functions involving resource use, radioactive decay, heat dissipation, debt repayment and investment interest. The importance of the exponential function lies in its use as a solution to a differential equation that covers this wide variety of real world situations.
DEs provide a different approach to fiting a function to data. Instead of just using intuition to choose a class of functions and regression to get the specific function, I can look at the growth rates of the function in the data. If I can make a consistent observation about the growth rates, such as the observation that they grow by a consistent percentage, I can write that observation as a differential equation. I can then (hopefully) solve the differential equation to find the appropriate function.
Before worring about solving DEs, I simply want to understand what they say. The goal is essentially translation: a differential equation is a mathematical translation of a statement about the growth rate of a quantity. I want to be able to pass both ways between the equation and the associated statement. The example in this section already shows percentage growth with percentage \(c\)% translates into an equation.
\begin{equation*} \frac{dp}{dt} = \frac{c}{100} p \end{equation*}
I can use this example to translate similar sentences as well. Percentage decay says that a quantity looses a percentage \(c\)% of the current value per unit time. This translates into a DE with a negative growth rate.
\begin{equation*} \frac{dp}{dt} = -\frac{c}{100} p \end{equation*}
Not all differential equations are percentage growth. Perhaps the rate of change of some quantity is proportional to the square of the current value. That could likewise be translated into a DE.
\begin{equation*} \frac{dp}{dt} = c(p(t))^2 \end{equation*}