Does Euler number $e$ have a role in kinematics?

Question

Euler number $e$ is often explained with the example of compound continuous interest.

I was wondering if it could also be illustrated with an example about the displacement of a body (although not an oscillating one, with which I have less difficulty in finding e, but extending endlessly in a straight line).

My attempts so far to find an answer:

• One of the ways to calculate Euler’s number is with the series of the inverse factorials:

$e = \sum\limits_{n = 0}^\infty {(\frac{1}{{n!}})} = \frac{1}{{0!}} + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + \frac{1}{{5!}} + ... = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{{24}} + \frac{1}{{120}} + ... % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyzaiabg2 % da9maaqahabaGaaiikamaalaaabaGaaGymaaqaaiaad6gacaGGHaaa % aiaacMcaaSqaaiaad6gacqGH9aqpcaaIWaaabaGaeyOhIukaniabgg % HiLdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIWaGaaiyiaaaacqGH % RaWkdaWcaaqaaiaaigdaaeaacaaIXaGaaiyiaaaacqGHRaWkdaWcaa % qaaiaaigdaaeaacaaIYaGaaiyiaaaacqGHRaWkdaWcaaqaaiaaigda % aeaacaaIZaGaaiyiaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI0a % GaaiyiaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI1aGaaiyiaaaa % cqGHRaWkcaGGUaGaaiOlaiaac6cacqGH9aqpcaaIXaGaey4kaSIaaG % ymaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdaaaGaey4kaSYaaSaa % aeaacaaIXaaabaGaaGOnaaaacqGHRaWkdaWcaaqaaiaaigdaaeaaca % aIYaGaaGinaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaIXaGaaGOm % aiaaicdaaaGaey4kaSIaaiOlaiaac6cacaGGUaaaaa!69B7! $

• You get a similar series if you consider the displacement, after 1 second, of an object starting at position = 1 m, with v = 1 m/s and subjected to acceleration = 1 m/s^2 and to successive derivatives of position with respect to time (jerk, snap, crackle, pop... and so on ad infinitum), all of value = 1:

$\Delta r = {r_o} + {v_o}\Delta t + a\frac{{\Delta {t^2}}}{2} + j\frac{{\Delta {t^3}}}{6} + c\frac{{\Delta {t^4}}}{{24}} + p\frac{{\Delta {t^5}}}{{120}}... = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{{24}} + \frac{1}{{120}} + ... % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % OCaiabg2da9iaadkhadaWgaaWcbaGaam4BaaqabaGccqGHRaWkcaWG % 2bWaaSbaaSqaaiaad+gaaeqaaOGaeyiLdqKaamiDaiabgUcaRiaadg % gadaWcaaqaaiabgs5aejaadshadaahaaWcbeqaaiaaikdaaaaakeaa % caaIYaaaaiabgUcaRiaadQgadaWcaaqaaiabgs5aejaadshadaahaa % WcbeqaaiaaiodaaaaakeaacaaI2aaaaiabgUcaRiaadogadaWcaaqa % aiabgs5aejaadshadaahaaWcbeqaaiaaisdaaaaakeaacaaIYaGaaG % inaaaacqGHRaWkcaWGWbWaaSaaaeaacqGHuoarcaWG0bWaaWbaaSqa % beaacaaI1aaaaaGcbaGaaGymaiaaikdacaaIWaaaaiaac6cacaGGUa % GaaiOlaiabg2da9iaaigdacqGHRaWkcaaIXaGaey4kaSYaaSaaaeaa % caaIXaaabaGaaGOmaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI2a % aaaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdacaaI0aaaaiabgUca % RmaalaaabaGaaGymaaqaaiaaigdacaaIYaGaaGimaaaacqGHRaWkca % GGUaGaaiOlaiaac6caaaa!6F1D! $

But this situation looks very unreal.

However, there should be a real-life example, because you can easily get displacement = e (if I am not mistaken) when the acceleration affects only direction, not modulus. Take for example a satellite that is 1 unit away from a gravitation source and that moves tangentially by inertia at v = 1 of whatever units. If gravity acts perpendicularly to the tangential motion and the acceleration caused by gravity is $\frac{v^2}{r}$ (that is to say, with these numbers, 1), then the displacement after 1 time unit will be circular and of modulus 1 radian = $e^i$, wouldn’t it? This situation looks to me very much like compound interest, because you start with a capital (radius) and an interest rate (velocity) and then you make the interest compound and continuous, although only in terms of direction (centripetal acceleration).

But what if the satellite were heading towards the earth and what changed were the modulus of its velocity, do we lose then the possibility to involve number e?

Answer 1

A simple example is just an object starting at rest falling with a drag force proportional to the velocity of the object, $F_D=-bv$. Then the acceleration is given by

$$a=\frac{dv}{dt}=g-\frac bmv$$

Therefore, the velocity over time is given by $$v(t)=\frac{mg}{b}(1-e^{-bt/m})$$

Typically you get $e$ popping up when the rate of change of something is proportional to itself.

Answer 2

Of course, $e$ is ubiquitous in kinematics. For example, consider a repulsive force proportional to $x$, $$F = kx.$$ Then the acceleration is $$a = \frac{k}{m} x = \omega^2 x, \quad \omega = \sqrt{\frac{k}{m}}.$$ This differential equation has solutions of the form $e^{\omega t}$ and $e^{- \omega t}$. In particular, suppose that $x(0) = 1$ and $v(0) = \omega$. In that case the solution is exactly $$x(t) = e^{\omega t}.$$ In general, for any linear force law, the solutions will be exponentials or complex exponentials, so it's honestly hard to avoid using $e$.

Answer 3

Euler number ${\rm e}$ shows up in plenty of situations in dynamics, but not so many in kinematics. Any example given that involves some kind of force definition, or integration of acceleration is an example in dynamics. But you are asking about kinematics. Kinematics is the study of allowable motions.

I personally cannot think of any involving ${\rm e}$, except maybe particles moving along paths that already have ${\rm e}$ in their definitions. For example a catenary shape. And those seem contrived to me.

The real answer would be one where ${\rm e}$ arises naturally, out the process if differentiating the path to get velocity or acceleration. But if the derivative involves ${\rm e}$ then must also the things are being differentiated, and then we are back to the previous examples.