Are there direct practical applications of differentiating natural logarithms?

Question

The textbook I am using to teach Calculus I includes in the exercises of most chapters a number of interesting real-world applications of the concepts from that chapter. However, the chapter on the derivative of the natural logarithm is remarkably abstract in its exercises.

Are there not scenarios in which it would be useful to differentiate a logarithm to answer a real-world problem? Something to do with determining the stimuli needed to accomplish a particular exponential rate of growth?

Or is differentiating natural logarithms primarily motivated by its usefulness for simplifying differentiation using logarithmic differentiation?

Since a number of practical exponential growth and logarithm exercises problems revolve around population growth, I've been trying to contrive an example in those terms. But it feels very odd and abstract:

Exponential function: What will be the population after x years? Derivative of exponential: How quickly will the population be growing x years from now? Logarithmic function: How many years will it take to reach a particular target population? Derivative of logarithm: How much would increasing or decreasing a given target population affect the time it takes to reach it?

Is that even really a practical question? How do I make it interesting?

I've tried searching this question here and on Google, but haven't found anything. Thanks in advance for your help!

Answer 1

Have you thought about the fact that you’re asking this in the middle of a pandemic for which log plots are being used all over the place to visualize the growth of COVID cases?

At any rate, $${d \over dt} \ln f(t) = {f’(t) \over f(t)} = \text{relative growth rate}\,.$$

Thus where the graph is roughly straight with slope $m$, we have the number of cases growing exponentially, proportional to $e^{mt}$.

And when the graph is concave up, the growth is greater than exponential (and worrisome).

When it’s concave down, it can be much harder to judge by eye. The slope of the tangent when $f$ is a power function $A t^k$ is $k/t$ — that is, they all look like log graphs (obviously, since $\ln A t^k = \ln A + k \ln t$). That’s why, as the pandemic went on and the growth of cases became subexponential, log graphs of daily new cases began to appear along with total cases. For $t$ measured in days, $f'(t)$ is the daily rate, and the slope of $\ln f’(t)$, which is $f’’(t)/f’(t)$, when the log graph is an upward-trending line still indicates exponential growth, but linear growth ($f’’(t)$ constant) yields a flat log graph of new cases (since $f’’(t)=0$).

I think some of this can be understood without derivatives. For instance, increasing and decreasing are pretty elementary notions and easy to see. But understanding the slope of a log plot beyond just when it is going up or down takes some analysis. Caveat: Beware the students measuring the slope. If they take you down this route, you might have to explain to them what to do with the powers of ten on the vertical axis, since that is what is usually shown (i.e., $\log_{10} f(t)$ instead of $\ln f(t)$).

Answer 2

Whenever we measure a quantity on a log scale (such as Richter, decibels, musical pitch, or a log-plot axis), we are focusing attention on relative variation in that quantity. If $y = \ln x$, we have $$\frac{dy}{dx} = \frac{1}{x}$$ and thus, for small finite changes, $$\Delta y \approx \frac{\Delta x}{x}.$$ That is, an absolute change in the logarithm equates to a relative change in the original magnitude. For example, a 1% change in $x$ will change $y$ additively by about 0.01.

This property is what makes log scales useful. People are naturally attuned to relative changes (often using percentages to express them). The logarithm helps express the "impact" of a phenomenon, because to "move the needle" on the logarithm ($y$), the quantity ($x$) must change by a noticeable amount compared to its own size. It is also very useful that the change in the logarithm does not depend on the physical unit in which $x$ is measured, because this unit cancels between $\Delta x$ and $x$.

Example 1: A chorus with 10 equal singers produces a sound level measured at 70.0 dB. What will be the approximate sound level after an 11th equal singer joins the chorus?

Answer: Sound level is defined as $L = 10\,\mathrm{dB} \times \log_{10} P$, where $P$ is a measure of sound energy. The derivative of this logarithmic function gives $$\Delta L \approx \frac{10\,\mathrm{dB}}{\ln 10}\, \frac{\Delta P}{P}.$$ Adding one more singer to a group of 10 means $\Delta P/P = 1/10$, so $\Delta L \approx 0.4\,\mathrm{dB}$. Thus, the new sound level is about 70.4 dB. This illustrates that there is very little difference in perceived loudness between 10 and 11 singers.

Example 2: How many semitones is the musical interval between 1000 Hz and 1100 Hz?

Answer: The pitch in semitones is defined as $S = 12 \log_2 f$, where $f$ is the frequency. The derivative of this logarithmic function gives $$\Delta S \approx \frac{12}{\ln 2}\, \frac{\Delta f}{f}.$$ With $\Delta f/f = 100/1000$, we have $\Delta S \approx 1.7$. The interval is about 1.7 semitones.

Answer 3

I couldn't find a lot either. Suggest playing with some logarithmic properties and constructing problems based on that.

E.g. pH is log10 of the hydronium ion concentration. Could ask how the pH changes with hydronium concentration addition (assume strong acid addition, to an unbuffered solution). Of course this brings in chemistry, which weirds the kids out more than logs!

Maybe play with decibels or Richter or NPV (discount rate). [Sorry, I can't give you the fish, just a direction to some lakes.]

All that said, I think applied problems are a bit of a double edged sword. And I say this as one of the knuckledraggers usually fighting the theorists. The issue with applied problems is they're usually harder. "Word problems are hard."

Answer 4

Boltzmann's equation for entropy is $S=k\ln W$, and the second law of thermodynamics is all about change in entropy. Maybe this is a place to start with your quest for a practical application of the derivative of a logarithmic function.

Answer 5

A couple of direct applications:

Showing that a power law appears on a log-log graph as a straight line with gradient equal to the exponent of the power law (although that can be done by other, probably easier, means too).

Thinking about a wind turbine that's well within the turbulent regime of the atmospheric boundary layer, the average wind speed incident on that turbine will depend on its height according to the logarithmic law of the wall, so the derivative of a logarithm is needed to tell you how much additional wind speed you can get by increasing the height of the turbine slightly.

In addition, the inverse operation, i.e. integrating $1/x$ with respect to $x$ to get $\ln\left(x\right)$, turns up in several places in thermodynamics and fluid mechanics, e.g.

Deriving the Poisson adiabat, starting from the ideal gas law, the non-flow energy equation, and the definition of "specific heat capacity at constant volume" (or "specific internal energy capacity"). (Batchelor, 2000, An introduction to fluid dynamics, Cambridge University Press, pp. 43-45)

Deriving an algebraic expression for the entropy of an ideal gas in terms of pressure and temperature, also starting from the ideal gas law, the non-flow energy equation, and the definition of "specific heat capacity at constant volume" (or "specific internal energy capacity"). (Adkins, 1983, Equilibrium thermodynamics, Cambridge University Press, p. 119)

Deriving the relationship between pressure gradient and flow rate, in isothermal compressible flow of an ideal gas in a pipe of uniform cross-section, starting from the continuity equation, the Euler momentum equation, the ideal gas law, and a dimensional analysis of the shear traction at the pipe walls. (Douglas et al., 2005, Fluid mechanics, Pearson Prentice Hall, section 17.9)

But the trouble is, the typical order in which concepts are presented in education systems is such that students are expected to know about the derivative of $\ln\left(x\right)$ before they know about any of the physical principles that are the starting points for those derivations, so these applications may not be particularly helpful if you're just introducing calculus for the first time.

Answer 6

In physics, in particular statistical mechanics, it is very common to take derivatives of logarithms. The basic idea is this:

Assume we have some system that has many possible states, labeled $s$, and the probability of any given state is given by the Boltzmann distribution, so that $P_s \propto e^{-E(s)/T}$ where $E(s)$ is the energy of the state and $T$ is the temperature of the system (expressed in units of energy, i.e. my $T$ here is really $k_B T$ where $k_B$ is Boltzmann's constant with units of energy / temperature). The normalized probability is then given by $$P_s = \frac{e^{-E(s)/T}}{\sum_{s'} e^{-E(s')/T}}$$ where in the denominator we sum over every state. We define the denominator to be the parititon function, and for convenience define the inverse temperature $\beta = 1/T$, $$Z(\beta) = \sum_{s} e^{-\beta E(s)}$$ Now consider we want to calculate some quantity of interest, for example the average energy of the system, $\langle E \rangle$. Well that is just given by $$\langle E \rangle = \sum_{s} E(s) P_s = \frac{\sum_s E(s) e^{-\beta E(s)}}{Z(\beta)}$$ Note that we always have to divide by $Z$ to get the right normalized expectation value. With this in mind, we can write the above with a moment's thought as $$\langle E \rangle = -\frac{\partial}{\partial \beta} \ln Z(\beta) = -\frac{1}{Z} \frac{\partial Z}{\partial \beta}$$ Moreover, we can calculate the heat capacity of the system (at constant volume to be precise), $$C_v = \frac{\partial \langle E\rangle}{\partial T} = - \frac{\partial}{\partial T} \frac{\partial}{\partial \beta} \ln Z(\beta) = \frac{1}{T^2} \frac{\partial^2}{\partial \beta^2}\ln Z(\beta)$$ if we work this out we will find $T^2 C_v = \langle E^2 \rangle - \langle E \rangle^2 = \mathrm{var}(E)$.

It is generally of more use to work with the Helmholtz free energy $F$ rather than the energy $E$, which is defined by $$Z = e^{-\beta F} \leftrightarrow F = -T \ln Z$$ The entropy is defined as $$S = \langle -\ln P_s \rangle = -\sum_s P_s \ln P_s$$ and again we can easily show that this is $$S = -\frac{\partial F}{\partial T}$$

All of this is much more general than just thermodynamics. From a statistician's point of view, $Z$ is the generating function of moments of the distribution, while $\ln Z$ is the generating function of connected moments (or cumulants) of the distribution. For a general case, consider some probability distribution for a random variable $x$, $P(x)$. Then consider the quantity (let's just assume this converges, e.g. if $P$ is gaussian) $$Z(t) = \int_{-\infty}^\infty dx\, e^{tx} P(x)$$ Expanding the exponential, we get $$Z(t) = \sum_{n} \frac{t^n}{n!} \int_{-\infty}^\infty dx\, x^n P(x)=\sum_n \frac{t^n}{n!} \langle x^n \rangle$$ then clearly $$\left.\frac{d^n Z}{dt^n}\right|_{t=0} = \langle x^n \rangle$$ For the connected moments, akin to $F$ above, we have $$W(t) = \ln Z(t) = \sum_n \frac{t^n}{n!}\langle x^n \rangle_c$$ The first connected moment is the mean, the second is the variance, the third is the skewness, the fourth is the kurtosis, etc. (Note there is a nice insight here about why the normal distribution is special: it is fully characterized by only the first two connected moments, the mean and variance, and all of its higher connected moments are exactly zero). Once again, $$\langle x^n \rangle_c = \left. \frac{d^n W}{dt^n}\right|_{t=0}$$

A more complicated but exactly parallel method is used in Quantum Field Theory for the computation of scattering amplitudes, where $Z$ is the Feynman Path integral (note this looks very similar to the definition of the partition function), $$Z = \int \mathcal{D}\phi\, e^{i S[\phi]}$$ Using essentially the same derivative tricks as above leads to a series expansion for any expectation we want to compute where each term in the expansion can be represented by a Feynman diagram. Then the quantity $W = \ln(Z)$ is the generator of the fully-connected, 1-particle-irreducible diagrams, where again the logarithm takes care of proper normalization in the calculation of probability amplitudes.

In all of these cases, the importance of the logarithm is to ensure that on taking the derivative we get a properly normalized result, i.e. we divide by $Z$, which is the sum of all the un-normalized probabilities.