Example 5.4.
Suppose the underlying mechanisms force \(f'\) to be proportional to \(f - c\) rather than \(f\text{.}\) Write down a guess as to what would the differential equation look like.
Unit 5.3 Exponentials revisited
Recall that \(e\) was defined to be the positive real number such that \(e^x\) has slope 1 at \(x=0\text{.}\) In other words, by definition,
\begin{equation*}
\lim_{h \to 0} \frac{e^h - 1}{h} = 1 \, .
\end{equation*}
From this we can compute the derivative of \(e^x\) at any point. Let \(f(x) := e^x\text{.}\) Then
\begin{equation*}
f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}
= \lim_{h \to 0} e^x \frac{e^h - 1}{h}
= e^x \lim_{h \to 0} \frac{e^h - 1}{h}
= e^x \, .
\end{equation*}
Next, for some \(a > 0\text{,}\) let \(f(x) := a^x = (e^{\ln a})^x = e^{x \ln a}\text{.}\) The chain gives
\begin{equation*}
f'(x) = e^{x \ln a} \frac{d}{dx} (x \ln a) = a^x \, \ln a \, .
\end{equation*}
At this point, we have derived parts 1, 3 and 4 of Proposition 4.8. For Part 5, letting \(f(x) := e^x\) so \(f^{-1} (x) = \ln x\text{,}\) we use the inverse function rule Proposition 5.1 and \((d/dx) e^x = e^x\) to obtain
\begin{equation*}
\frac{d}{dx} \ln x = \frac{1}{e^{\ln x}} = \frac{1}{x} \, .
\end{equation*}
Paying back a debt, this finishes off Part 2 of Proposition 4.8. For any real \(r\) and positive \(x\text{,}\) let \(f(x) := x^r = e^{r \ln x}\) and use the chain rule to obtain
\begin{equation*}
f' (x) = e^{r \ln x} \frac{d}{dx} (r \ln x) = x^r \frac{r}{x}
= r x^{r-1} \, .
\end{equation*}
Subsection 5.3.1 Differential equations
The course after this one studies differential equations. This semester we get only a tiny preview of this subject. A differential equation arises when you have a function that is unknown and your information about it includes something about the derivative. The simplest example is when you know the derivative outright, for example \(f' (t) = 5 + 4t\text{.}\) Integral calculus then produces a formula for \(f\text{.}\) In this case, because you are familiar with derivatives of polynomials, you can probably recognize the solution \(f(t) = 5 t + 2 t^2\text{.}\) There are other possible solutions, all differing by a constant, for example \(f(t) = 1 + 5t + 2t^2\text{.}\) The general solution is \(f(t) = c + 5t + 2t^2\) where \(c\) can be any constant. Further information is required to figure out \(c\text{;}\) if you know even a single value of \(f\text{,}\) such as \(f(7) = -2\text{,}\) you can solve for \(c\text{.}\)
The differential equation we will study here is the next simplest one: \(f' (t) = k f(t)\text{.}\) This is more subtle because the derivative is not given outright but rather is related to the function itself (of course \(f\) represents the same function on both sides of the equation). This method of solution of this equation is similar to the previous example. You can solve it because you can recall a function that behaves this way, namely the function \(f(t) := e^{kt}\text{.}\) That is the simplest looking solution but there are others. The most general solution is \(f(t) = A \, e^{kt}\) where \(A\) can be any constant. When you study methodical solutions to differential equations you will be able to prove that these are the only solutions. In the present course, we won’t discuss the problem at that level but you are free to assume this is true: if \(f'(t) = k \, f(t)\) for all \(t\) then \(f(t) = A \, e^{kt}\) for some real number \(A\text{.}\) Note that the constant \(k\) is not like the constant \(A\text{:}\) the constant \(k\) is part of the equation you were given altering it will make the function no longer a solution to the equation.
Because \(f' = k f\) is such a basic equation, it occurs pretty commonly in applications. For this reason it pays to study the functions \(a e^{kt}\) in detail.
When \(k > 0\) this represents exponential growth. Some things that behave this way under the right circumstances are populations, money (both assets and debt), epidemics, adoption of new technology, and pyramid schemes. In all of these cases, it’s easy to argue that the rate of increase should, to a first approximation, be proportional to the present size; in other words, \(f' = k f\text{.}\)
Aside
When \(k = -\ell < 0\text{,}\) this is called exponential decay. The classic example of exponential decay is a radioactive material breaking down through alpha or beta decay. Other things that decay exponentially under the right circumstances are temperature difference, correlations in time series data and valuations of future goods. These examples were mentioned briefly in Unit 1.4. Calculus gives a a reason to believe why exponential growth and decay are plausible models for these physical phenomena. It is because the underlying mechanisms force \(f'\) to be proportional to \(f\text{.}\)
Subsection 5.3.2 Time constants
Suppose \(f(t) := A e^{kt}\) where \(t\) is in units of time and \(f\) is a quantity in some units we will just refer to as ``units of \(f\)’’. Recall from the introduction to units early in the course that the exponent \(kt\) is required to be unitless if the expression is to make physical sense. That means the constant \(k\) has to have units of inverse time. Such constants are called time constants.
At first these can be difficult to make physical sense of. We understand the quantity 0.02 days, but what is the physical significance of the quantity 0.02 inverse days? Most directly it means that if \(t\) is the reciprocal, namely 50 days, then \(kt = 1\) (unitless) and the quantity \(A e^{kt}\) is \(A \cdot e\text{,}\) a factor of \(e\) greater than it was at the start (because at the start, \(A e^{k \cdot 0} = A\text{.}\)
Example 5.5.
In March, 2020, the U.S. COVID-19 epidemic was increasing exponentially with a time constant of 1.4 inverse weeks. By roughly what factor did the number of total cases increase each week in March?
If \(k\) is negative, then \(f\) represents exponential decay. For example if \(k = -0.02\) inverse days, then after 50 days, the function will have decreased by a factor of \(e\text{.}\)
Which is bigger, an inverse second or an inverse minute? Minutes of course are much longer than second: one minute equals 60 seconds. On paper one can convert between inverse time units as well. For example,
\begin{equation*}
1 {\rm sec}^{-1} = \frac{1}{\rm sec} \cdot \frac{60 {\rm sec}}{1 {\rm minute}}
= \frac{60}{\rm minute} = 60 {\rm min}^{-1}
\end{equation*}
so one inverse second is 60 inverse minutes. To make this a little more intuitive, think of one inverse second as \(1 / {\rm sec}\) which we might write say aloud as ``one per second’’. The phrases ``one per second’’ and ``sixty per minute’’ should sound believably the same.
Consider a quantity that is decaying exponentially. As a function of time, the quantity is represented as a function \(f(t) := A e^{-kt}\text{.}\) Such a quantity is said to have a half-life. Regardless of how much of the quantity there is originally, the time until half remains is always the same. It’s too bad the concept wasn’t first conceived as \(e\)th-life, the time it takes to reduce by a factor of \(e\text{,}\) because that is clearly the time for \(kt\) to become \(-1\text{,}\) in other words the reciprocal of \(k\) (it’s a good thing that \(k\) has inverse time units so its reciprocal is a time). No matter, if instead of \(kt = -1\) we say \(kt = -\ln 2 \approx 0.7\text{,}\) then \(e^{-kt}\) will be \(1/2\text{.}\) So the half-life is just \((\ln 2)\) times the \(e\)th-life, that is, \((\ln 2) / k\text{.}\)
Checkpoint 94.
Polonium-210 is a radioactive substance and decays to lead with a half-life of about 138 days. What is the present rate of decay of a sample of 5 micrograms of Polonimum-210? Give units.
