[begin transmission]

Dear You,

I'm sorry.

Love,
-▒▒▒▒▒▒▒

[end transmission]

 

[begin transmission]

Every so often a conversation haunts me long after the exchange has concluded.
It's typically relegated to the background, w/ demands of daily life taking precedence.
But when there is a break in the action...
A quiet lull when lights are out, the door is locked, and contemplation takes hold...

My mind revisits the exchange w/ none other than the Professor, re: Marxism.

If I recall correctly, it was your position that critical race theory (CRT) cannot be considered meaningfully Marxist.
I hear this exact argument all the time; particularly from those on the Left that are trying to condescend to me, gaslight me, or make me out as some caricature of a paranoid Right-winger that believes in conspiracy theories (to any conservatives out there reading this, don't let anyone tell you that cultural Marxism or Jordan Peterson's notion of "Post Modern Neo-Marxists" aren't a thing. They are, and all it takes is some careful tracing of philosophical genealogy to confirm this for yourself). It typically doesn't work since their arguments end up ringing hollow, and I've actually bothered to read up on a fair amount of Marxism to know what I am talking about. But you, you came at me w/ that argument in good faith, and you were ready to fight it out on a doctrinal level; I have to say I really appreciated that back-and-forth--w/ good reason I still think about it to this very day.

At the halfway mark of our original conversation, we came to the consensus that CRT cannot be considered part of orthodox Marxism, but as I laid out the case for, it can be considered part of Neo-Marxism. I think we're both fine w/ that; I still certainly am.

I recall a specific statement that you made, however. You said that b/c CRT doesn't make any overtures towards class in it's dialectic, it cannot be considered Marxist. It should be considered more Hegelian if anything. I agreed w/ you on that point, and I still do. Though--and this is largely why I'm even bothering to write this, to clear up and sharpen our understanding--I think we may have not appreciated enough of the distinction between the the Hegelian dialectic and Marxist dialectic; the two of us seemed to have presumed all that Marx did was take Hegel's homework, and, in naughty schoolboy fashion, copy it and swap around some words to submit to the instructor as his own.

In reality, the Marxist dialectic is NOT merely the same as the Hegelian dialectic, just narrowed in scope to focus on material, economic conditions and lasered-in on class-capital relations. What we failed to appreciate--and this is a gap in our knowledge that I hope will henceforth be filled, is that the Hegelian dialectic presupposes what he calls the "objective spirit". The objective spirit is what we would consider the social and cultural norms that people live and abide by. Without getting too much into it, I'd like to illustrate that this is different from Hegel's "subjective spirit", which is akin to individual consciousness--something for your own edification. From the objective spirit, we get acceptance and understanding of the existing social order--the thesis, which eventually forms it's own negation--the antithesis, which is eventually combined w/ the thesis to form a new and improved version of the thesis: the synthesis.

Okay, we knew that; we're familiar w/ the Hegelian dialectic. Working under the assumption that the Marxist dialectic is just the Hegelian dialectic w/ a materialist spin, the process should go something like this: the working class accepts working for the bourgeois, this eventually gives rise through material privation and resentment to a negation that leads to class conflict. Out of class conflict will rise a new and improved material reality. Sounds about right, doesn't it? Well, it turns out that this is wrong:

My dialectic method is not only different from the Hegelian, but is its direct opposite...the ideal is nothing else than the material world reflected by the human mind, and translated into forms of thought...With him it is standing on its head. It must be turned right side up again, if you would discover the rational kernel within the mystical shell.

Karl Marx. Das Kapital. 1873

Where Hegel starts w/ the objective spirit, where people's understanding and embracing of sociocultural norms gives rise to a material reality (and thus it is within the purview of society and culture--ultimately people--to change material reality), Marx starts w/ material reality giving rise to sociocultural norms. From there, change in material reality cannot come from people rejecting/embracing sociocultural norms, but from the state. This is a VERY important distinction between Hegelian and Marxist dialectics. The former supposes that the objective spirit--the culture--comes first and shapes reality through the dialectic. The latter supposes that the reality comes first and eventually shapes the culture. They're exactly the opposite of one another.

...and I will admit that's all that I have for you. Maybe you still think that the Marxist dialectic is the Sanic to Hegel's Sonic, but I'm coming to realize that they're very different beasts.

What's interesting is that the type of Marxism (i.e. cultural Marxism) espoused by Antonio Gramsci is a departure from the Marxist dialectic and a return to the Hegelian dialectic. Gramsci's cultural Marxism went on to influence the Frankfurt school, and from there CRT. So our original agreement still stands: CRT is not exactly old-school Marxist, but it is certainly Neo-Marxist. The entire point of this post scriptum is to tell you that you were onto something when you first proposed that CRT is more Hegelian than Marxist, more than either of us at the time realized.

[end transmission]

On the first night, she surveyed the smokey battlefield.
Sitting on the horizon she could make out the faint outline of the enemy stronghold.
The objective from Command was clear. Liberate the fortress within four days.
However, tomorrow was as uncertain as the ominous form in the distance, nevermind day four.
With a determined huff, she dons her helmet and takes up arms.
As she takes her shield, the visage of her fellow cadets reflects off of its brilliant, mirrored surface and catches her eye.
Men she had studied with. Men she had trained with. Men she will shed blood with.
The weight of worry for their creature safety tore at her heart. Be well, my friends.
In lockstep they descend into the chaos awaiting them below, for which they knew survival was not guaranteed.
Steel liberating blood from flesh, the battlefield quickly devolves into an abattoir. The girl sustains a hard blow to the chest.
Gritting her teeth through the haze of pain, she charges forward under heavy shield, clearing a path to a garrison.
The rendezvous point offering respite in a reality overwhelm with obscenity and disorder.


On the second night, she scanned the bloodied battlefield.
Well within reach, the sought after structure simultaneously inspired both dread and hope.
The objective from Command was clear. Liberate the fortress within four days.
Yesterday now seeming as unreal of a prospect as tomorrow, the present became unbearably immediate.
With a hardened countenance, she affixes her pauldrons and takes up arms.
As she takes her shield, the visage of her troops reflects off of its brilliant, mirrored surface and catches her eye.
Savagely they had fought. Savagely they had killed. Savagely they had died.
The weight of duty towards her comrades bore down on her conscience. I will protect you, my brothers.
In lockstep they traversed the fog awaiting them, for which they accepted survival was not guaranteed.
Fire liberating courage from spirit, the scene descends into Abaddon. The girl catches a blade across her eye.
Blind with blood and fury, she charges forward, leading her men into a trench short of the fortress gate.
The blackened Earth offering tenuous shelter from the murderous calculations of the enemy.


On the third night, she studied the unassailable architecture.
A stone’s throw away, the brutish, foreboding citadel threatened to rob her of all morale.
The objective from Command was clear. Liberate the fortress within four days.
Brooding clouds gave backdrop to a lustrous moonrise not meant for her.
With a weary sigh, she dons her tattered cape and takes up arms.
As she takes her shield, the visage of her own blood reflects off of its brilliant, mirrored surface and catches her eye.
Sacrificed was her body. Sacrificed was her will. Sacrificed will be herself.
The weight of excellence demanded from them steeled resolve. Per angusta ad augusta.
In lockstep they ascend from the muddy embankment and towards a grand, fatalistic end.
Accession liberating pneuma from weathered husk, one by one they ascend to Maon. The enemy gates begin to rise.
Senses sharpened with action and purpose, she takes notice and bounds towards the objective.
With one last great expenditure of force, she dives shield-first into the fortress. The entrance clangs shut with resounding finality.


On the fourth morning, she gasps as her consciousness finds her anew; light reclaiming what the dark had stolen.
The objective from Command was clear. Liberate the fortress within four days.
As she takes her shield, her own wretched visage reflects off of its brilliant, mirrored surface and catches her eye.
Surrounding her in every direction was the infinite vastness of nothingness. She was alone.

 

Si enum comprehendis, non est Deus.

Saint Augustine. Patrologia Latina, Vol. 38. 1841. 

[begin transmission 3/?]

This post will be a slight deviation from the main subject I was addressing in this series of posts.
It was my intent to briefly explain systems theory and dynamics before getting into optimal control.
However, in my elaboration re: dynamics, the question of how to solve differential equations came up.
Possessing a deep appreciation for history of maths and engineering, I can't help but gush about the Laplace transform.

History Lesson

Let us begin w/ a brief discussion of the background of the Laplace transform. If it wasn't already apparent to you, the technique is named after none other than Pierre-Simon Laplace, the French mathematician and scientist. The history of this technique is somewhat mysterious, as the entire Enlightenment era had tons of great thinkers (and their even greater egos) rivaling each other, often times converging onto the same kinds of ideas. As a result giving credit where credit is due can be a contentious affair. But for the most part, yes, Pierre-Simon Laplace is widely credited w/ being the primary developer of the technique, using it mostly in his work on probability.

What I absolutely love about the history of the Laplace transform is that it was developed in the 18th century--remained in relative obscurity, being confined to the dusty libraries of mathematicians' minds--only to make a strong resurgence in the 20th century when the demands of warfare forced it into the mainstream. This is why, outside of teasing of course, I never give my mathematically or theory-inclined contemporaries too much of a hard time. As outlandish or worthless some abstraction may seem at the time, one that makes you think "When the hell am I ever going to use that?", one can never be too certain when a need may arise that grants a once useless concept newfound utility. Certainly Laplace couldn't have envisioned his mathematics being used to operate the fire-control systems directing the behemoth canons aboard the naval destroyers of WWII. So that ostensibly stupid idea your math friend dreamt up after his fourth pint of beer? It may just save the day in a century or two's time.

Okay, okay. That's enough history and fangirling. Let's get back to technical matters.

What Is It Though?

Technically-speaking, the Laplace transform is an integral operation that takes a time-domain function and transforms it into the complex s-domain. These words mean absolutely nothing to you, so let us unpack these concepts. An integral operation is like any other mathematical operation--just like addition, subtraction, or taking a derivative, only it involves integrals. The Laplace transform involves integration, as demonstrated by its formal definition:

\begin{equation} \mathscr{L}(f(t)) = F(s) = \int_{0}^{\infty}f(t)e^{-st}dt \tag{1}\end{equation}

When you take your time-domain function f(t) (that is, a function that has its independent variable t for time; virtually all real-world signals) and plop it into the Laplace transform above, you get its s-domain representation. What is the s-domain? It is the complex frequency-domain. What is the complex frequency-domain? Well, first understand the frequency-domain. To do this, take Fourier's theorem to heart: every periodic time-domain signal can be decomposed into sinusoidal components. These sinusoids have a frequency (how many times per time unit they complete a cycle, often expressed in Hertz) associated with them. What the frequency-domain representation of a signal reveals is 'how much' of the signal is found across the frequency spectrum. Here's a figure of a simple sinusoid that illustrates this:

Figure 1. A simple sine wave

The complex frequency-domain, the s-domain, is complex-valued , meaning it has two components: a real part and an imaginary part. It assumes the form:

\begin{equation} s = \sigma \pm j\omega \tag{2}\end{equation}

Remember when you learned about imaginary numbers and thought to yourself "When will I ever use this?". Well...here it is, coming back to haunt you. By the way, want to know another way to determine whether someone is a mathematician or an engineer? Ask them to write down a complex number. Mathematicians favor using "i" to denote imaginary numbers; engineers use "j". The imaginary part, the part to the right of the plus minus, refers to the frequency portion of the signal. The sigma part refers to the decay/growth of the signal. You might've noticed in Equation 1 that there is an exponential; this is exactly what the sigma part represents. Intuitively, this makes sense: real-world signals can and do indeed grow or decay over time. If sigma is positive, the signal grows; if sigma is negative, it decays. Those that are paying close attention, or have a background in signal processing will be able to appreciate right now that an ever-growing signal is something that is not...not very desirable.

And that's a very brief introduction into Fourier and complex analysis. I didn't expect to get too deeply into it, yet here we are. In truth you really don't have to understand this stuff to be able to solve differential equations using the Laplace transform, but a thorough understanding helps you gain a reliable intuition as an engineer. For example, had you not known that the s-domain conveys information regarding the growth/decay of a signal, you would be able to read a root-locus plot (a tool often used in classical control theory analysis) and determine that your system is underdamped, resulting in a large overshoot or unstable response.

Application of the Laplace Transform

It's about time we solve our mass-spring-damper mathematical model. Here it is, in case you had forgotten:

\begin{equation} M\ddot y + b \dot y + ky = r(t) \tag{3}\end{equation}

It is our wish to obtain y--the response of the system in the form of a vertical displacement--subject to an applied force r(t). Let's say that r(t) is a Heaviside function, also known as a step function u(t); this is reasonable, as it is often times the case in the real world that a constant force is applied to a system at any given moment in time then it is unapplied. Mathematically:

\begin{equation} u(t) = \begin{cases} 0 \quad if \quad t < 0 \\ 1\quad if \quad t > 0 \end{cases}\tag{4}\end{equation}

Graphically:

Figure 2. Heaviside function

So, how do we apply the Laplace transform? Well, you perform the operation term-by-term to Equation 3 and solve the integral. Honestly, solving integrals is such a pain in the ass, and we leave that kind of busy work to mathematicians. Being the jocks of academia engineers, instead we memorize the Laplace transform of several functions. They are as follows:

\begin{equation}f(t) = u(t), \quad \mathscr{L}(f(t)) = \frac{1}{s} \tag{5}\end{equation}
\begin{equation}f(t) = e^{at}, \quad \mathscr{L}(f(t)) = \frac{1}{s-a} \tag{6}\end{equation}
\begin{equation}f(t) = t^{n}, n=1, 2, 3..., \quad \mathscr{L}(f(t)) = \frac{n!}{s^{n+1}} \tag{7}\end{equation}
\begin{equation}\mathscr{L}(f(t)) = F(s) \tag{8} \end{equation}
\begin{equation}\mathscr{L}(\dot f(t)) = sF(s) - f(0) \tag{9}\end{equation}
\begin{equation}\mathscr{L}(\ddot f(t)) = s^{2}F(s) - sf(0) -\dot f(0) \tag{10}\end{equation}

Now that we have our handy table to refer to, and keeping in mind that M, b, and k are merely constants, applying the Laplace transform to Equation 3 yields:

\begin{equation}M\mathscr{L}(\ddot y) + b\mathscr{L}(\dot y) + k\mathscr{L}(y) = \mathscr{L}(u(t))\tag{11}\end{equation}

Using the table above:

\begin{equation}M(s^{2}Y(s)-sy(0)-\dot y(0)) + b(sY(s) - y(0)) + kY(s) = \frac{1}{s}\tag{12}\end{equation}

What the hell. What are these y(0) terms? These are initial conditions. When solving differential equations, it is necessary that these are given or assumed. Because we're utilizing a frequency-domain approach, our system is assumed to be using zero'd out initial conditions. Applying our initial conditions yields:

\begin{equation}Ms^{2}Y(s) + bsY(s) + kY(s) = \frac{1}{s}\tag{13}\end{equation}

A little bit of algebraic manipulation:

\begin{equation}Y(s) = \frac{1}{s(Ms^{2}+bs+k)}\tag{14}\end{equation}

We're almost there! Problem is, we didn't declare the parameters when we formulated the problem in the previous post. We'll take care of that right now; let's suppose mass M is equal to 1, coefficient of friction b is equal to 8, and spring constant k equal to 15. Of course these are easy-peasy numbers that work out nicely but are highly unrealistic. Equation 14 becomes:

\begin{equation}Y(s) = \frac{1}{s(s^{2}+8s+15)}\tag{15}\end{equation}

Notice anything different about our dynamic equation? Noti--well, yes, very astute of you. All of the t's have been replaced by s's. But beyond that, and this is fairly key, notice that our differential equation has now been transformed into a simple algebraic equation. This is what makes the Laplace transform fairly attractive. There are no derivatives or integrals to be found here, so it's now much easier to solve. So let's solve it, utilizing the method of partial fraction decomposition. All that this method does is take a rational expression and decomposes it into constituent fractions. The set up looks something like this:

\begin{equation}\frac{1}{s(s+3)(s+5)}=\frac{A}{s}+\frac{B}{(s+3)}+\frac{C}{(s+5)}\tag{16}\end{equation}

All we did here was take the polynomial in the denominator and factored it out. Then we equated it to a sum of fractions and declared new constants A, B, and C; one for each factor. The idea here is to now solve for these constants. Next, we multiply both sides by the denominator:

\begin{equation}s(s+3)(s+5) \frac{1}{s(s+3)(s+5)}=(\frac{A}{s}+\frac{B}{(s+3)}+\frac{C}{(s+5)})s(s+3)(s+5)\tag{17}\end{equation}

Cancelling out some terms and simplifying the expression yields:

\begin{equation}1 = A(s+3)(s+5) + B(s)(s+5) + C(s)(s+3)\tag{18}\end{equation}

The next step in our approach is to set s equal to something to reduce this expression. If we set s = -3, so that the terms grouped w/ A and C go to zero, we get:

\begin{equation} 1 = -6B\Rightarrow\ B = \frac{-1}{6}=-0.166\tag{19}\end{equation}

Similarly, if we set s = -5 so that the terms grouped w/ A and B go to zero:

\begin{equation} 1 = 10C\Rightarrow\ C = \frac{1}{10}=0.1\tag{20}\end{equation}

Finally, setting s = 0, so that the terms grouped w/ B and C go to zero:

\begin{equation} 1 = 15A\Rightarrow\ A = \frac{1}{15}=0.066\tag{21}\end{equation}

Repopulating Equation 16 w/ the constants we solved for:

\begin{equation}Y(s)=\frac{0.066}{s}-\frac{0.166}{(s+3)}+\frac{0.1}{(s+5)}\tag{22}\end{equation}

Finally, at this stage we apply the inverse Laplace transform (simply going the other way in our handy little table), we can go from the s-domain back to the time-domain; we transform Y(s) back into y(t):

\begin{equation}\mathscr{L^{-1}}(Y(s))=\mathscr{L^{-1}}(\frac{0.066}{s})-\mathscr{L^{-1}}(\frac{0.166}{(s+3)})+\mathscr{L^{-1}}(\frac{0.1}{(s+5)})\tag{23}\end{equation}

\begin{equation}y(t)=(0.066 - 0.166e^{-3t}+ 0.1e^{-5t})u(t)\tag{24}\end{equation}

As a reminder, b/c I know we sort of went through a lot here that it is all too easy to lose the plot, we solved the differential equation--Equation 3--by finding y(t) that satisfies it. That is the mathematical interpretation. Practically, Equation 3 was the mathematical model for our mass-spring-damper system; y(t) is the response of the system (remember, in the form of a vertical displacement) subject to an applied force modeled by u(t).

If you made it this far, congrats. You've made it one month into an introductory course in classical control theory. Keyword here is classical. These kinds of mathematics, as I mentioned before, were in vogue in the early half of the 20th century; I don't want to say that these methods are completely outdated, as they are still used today, but what you must understand is that they are fairly limited. If there is anything at all that you take away from this post is that the Laplace transform method used here only applies to linear, time-invariant systems. You might think to yourself "Wow! How useless! Most real-world systems and phenomenon are non-linear and time-varying in nature.". That last statement is true, but the s-domain approach still has some utility, as several non-linear systems can be linearized w/o losing too much fidelity, and depending on the time scale under consideration time-variance can be made negligible.

If these notions make you feel uncomfortable, if they make you roll your eyes and scoff, I recommend you revise your attitude or get out of engineering. Although the discipline does demand a fair amount of precision, you must come to terms w/ dropping any notion of perfectionism; approximations, tolerances, and trade-offs rule the day when it comes to dealing w/ real-world problems. The world is a complex, messy place, and neat, tidy little solutions rarely, if ever, occur.

...And I'm catching myself before I get too rant-y. Right. So if the frequency-domain approach is seen as an antiquated method of solving differential equations involved in classical control theory, what is the alternative? Glad you asked. Descriptively enough, it's known as modern control theory, or state-space control. This is what's in vogue these days and w/ good reason, as it enables for the analysis of ALL kinds of systems: linear/non-linear, time-invariant/time-varying, single-input-single-output/multi-input-multi-output, deterministic/stochastic etc. Moreover, it doesn't involve any exotic transformations; you can conduct your analysis and design in the time-domain. I don't think I'll get into it, but I'll cover a few concepts related to the field, as they're prerequisite towards understanding the real good stuff: optimal control.

[end transmission 3/?]

[begin transmission 2/?]

As promised, the day has finally come. We're covering dynamics. God help us all.
I'd like to preface this explanation w/ the qualification that dynamics is a very general term.
Just like systems, dynamics can assume a mechanical, electrical, biological, financial, etc. flavor.
The entire point of dynamics is to describe how these types of systems change over time, using mathematical formalisms.

Sounds crazy, right? You're trying to capture the overwhelming complexity of the world in mere symbols.
With good reason that dynamics is a field of study in itself, w/ people dedicating their lives in the pursuit.
Myself, by no means am I an expert, but in my profession I have to have an acute understanding of them.
It's a piece of the greater picture, and a very important piece at that, but it is not the entire story.

Anyways, let's get to it.

Normal Mode: Dynamics. But First, Babby Mode: Algebra

Okay, so I'm kind of torn as to where to begin. I'm assuming most of my readers are literate in basic algebra, and understand the concept of a function. But for completeness sake, let me remind you that a function is merely an equation that establishes a relationship between an input variable and an output variable (ignore multivariate functions for now). You put a quarter into the machine and receive a gumball. Here's a simple function:

\begin{equation} y(x) = x + 2 \tag{1}\end{equation}

You can take your input variable, x, and determine the value of y, the output variable. Suppose that x = 2, you can conclude that y = 4. If you conclude here that y is anything else other than four--and you're not a pure mathematician armed w/ a formal proof--please take your critical theory idiocy elsewhere. For the rest of you, onwards.

Functions can be expressed graphically. For a univariate function such as Eq. 1, this only takes two axes to describe:

Figure 1. Graph of y(x) = x + 2

I apologize if this is all so painfully rudimentary and unnecessary, but I read something like 40% of Americans can't read basic graphs. So here I hope it makes sense that, if you assume a value of x = 2, you can see that y = 4; if you assume a value of x = 3, y = 5, etc. String a bunch of these x and y value pairs together and you get the graphical representation of our function.

Normal Mode: Dynamics. No, no. Hold on. First Calculus.

But okay, you probably knew all of this. Let's graduate from algebra and tackle calculus. The central concept of calculus is that of the derivative: namely, that for every function (ignore discontinuous functions) there exists another function that describes its sensitivity to change in input variables. The common notation for a derivative is dy/dx. To illustrate the concept of the derivative, consider the gumball example above: suppose you put in two quarters--how would this affect the number of gumballs spat out? What if you insert three quarters? Four? A normal gumball machine will yield one gumball per quarter, so you get two, three, and four, respectively. Mathematically:

\begin{equation} y(x) = x \tag{2}\end{equation}

Graphically:

Figure 2. Graph of y(x) = x

Where y is the number of gumballs spat out, x is the number of quarters inserted. To graphically determine the derivative of this function, take any sample point for x, let's say 3, and increment it by one unit. Then observe how y changes accordingly. If you go from x = 3 to x = 4, then y behaves accordingly, going from y = 3 to y = 4. Expressing this change as a ratio defines the function's derivative:

\begin{equation} \frac{dy}{dx} = \frac{y(4) - y(3)}{4 - 3} = \frac{4 - 3}{4 - 3} = \frac{1}{1} = 1 \tag{3}\end{equation}

The derivative for our function is simply 1. Again, remember, a derivative describes a function's sensitivity to change in input. x, our input, is being changed, and as a result y changes. For every one unit change in x, y changes by one unit as well.

 Now, suppose that the gumball manufacturing supply chain is disrupted by a global pandemic, lowering the supply of available gumballs. Also suppose that God awful Democratic policy has disincentivized gumball factory workers from working--by offering them more in unemployment than they'd otherwise earn by working--resulting in an additional decrease in the supply of available gumballs. Since gumballs are now scarce and demand hasn't proportionally decreased but remained constant or even increased, the price of gumballs has now increased. Now gumballs cost four quarters a piece. Mathematically:

\begin{equation} y(x) = 0.25x \tag{4}\end{equation}

And graphically:

Figure 3. Graph of y(x) = 0.25x

Following the same procedure as before, we take x = 3, increment it by one unit, and observe the change in y. We then express that as a ratio to obtain the derivative:

\begin{equation} \frac{dy}{dx} = \frac{y(4) - y(3)}{4 - 3} = \frac{1 - 0.75}{4 - 3} = \frac{0.25}{1} = 0.25 \tag{5}\end{equation}

And yes, I botched the mathematical modeling here on the first cut, how embarrassing. I caught it on my own.

So far we've covered two basic examples of determining a function's derivative. These functions are linear, so determining them are child's play. We've obtained them w/ the help of graphs; this is cumbersome. A good student that paid attention in calculus class or an engineer worth their salt knows derivatives of several functions by heart; it's akin to memorizing one's times tables. Such functions are as follows:

\begin{equation} y(x) = cos(x), \frac{dy}{dx} = -sin(x)\tag{6}\end{equation}

\begin{equation} y(x) = sin(x), \frac{dy}{dx} = cos(x)\tag{7}\end{equation}

\begin{equation} y(x) = tan(x), \frac{dy}{dx} = sec^2(x)\tag{8}\end{equation}

\begin{equation} y(x) = ln(x), \frac{dy}{dx} = \frac{1}{x}\tag{9}\end{equation}

\begin{equation} y(x) = e^x, \frac{dy}{dx} = e^x\tag{10}\end{equation}

And if they're not memorized, there are several readily implementable heuristics available to obtain them, such as the power rule, chain rule, product rule, quotient rule...etc. We won't cover how these heuristics are derived because I don't remember how they are derived; that's left for mathematicians to fawn over because, much like how one doesn't need to understand the Carnot cycle to drive a car, an engineer doesn't need to understand the derivation of every single one of their mathematical tools to successfully employ them.

...Why on Earth did we use gumballs as an example? Let's illustrate an example that is easily comprehensible yet readily translatable to an engineering context. For your consideration, the scenario of going for a run. Why a run? Because we can describe a run in terms of spatial position, velocity, and acceleration. These three quantities are directly related to each other by, you guessed it, their derivatives. Specifically, by their time derivatives. What is a time derivative? It's the derivative of a function w/ respect to time. Want to know how you can tell an engineer from a mathematician? Ask a mathematician to write down the symbol for a derivative, and they'll ask "with respect to what?". Ask an engineer the very same thing and they'll instinctually write down something like dy/dt. This is because, in most dynamics problems, we're concerned w/ how systems change over time. An engineer is concerned w/ the sensitivity of a particular function as time changes.

Right, so the time derivative of a function that describes position yields velocity; take the time derivative of velocity and you get acceleration. This ought to make intuitive sense: you describe a point in space via length units, such as a meter; I am 5 meters ahead of you. You describe a velocity in terms of length units per time units, such as m/s; I am running at a speed of 4.5 m/s; for every increment of 1 time unit, my position changes by 4.5 length units. Accelerations are reported as...well, a variety of other units, but keeping w/ meters and seconds, as m/s^2; I am accelerating 2 m/s^2; for every increment of 1 time unit, my velocity changes by 2 velocity units.  These three quantities are used frequently in classical mechanics types of problems.

Personally I like to go for a three mile run every other day, keeping up a pace of about six minutes per mile. In engineering units (meters and seconds), this translates to a total of 4827 meters, ran in 1080 seconds. Therefore, my velocity is, on average, 4.47 meters per second. Below are graphical representations of the run, the first describing position:

Figure 4. Position Graph

So, at time 100 seconds (t = 100) I'll be 447 meters along the track, at time 101 seconds (t = 101) I'll be at 451.5 meters along the track, at 102 seconds (t = 102) 455.9 meters along the track, etc. Easy enough, right?

As mentioned before, my velocity, on average is 4.47. This is confirmed by the position graph, as moving 1 time unit increases my position by 4.47 meters. So if we take the derivative of the position at every time point, we obtain the following velocity graph:

Figure 5. Velocity Graph

It's a straight, flat line. What the heck. Why? Well, that's because we're running at a constant velocity of 4.47 m/s. There are no changes in the velocity, hence it is flat. Why did I show you this, this is boring. It is because it helps one grasp what an acceleration, the second time derivative of position, is. Much like how velocity describes change in position per time unit, acceleration describes change in velocity per time unit. Here, because our velocity is constant (it doesn't change at all, but remains 4.47 m/s), acceleration is zero. Hence, our acceleration graph looks like this:

Figure 6. Acceleration Graph

The second, equally important concept of calculus is that of the antiderivative. It is simply the inverse of the derivative, and is mathematically implemented via the integral operation. As such, taking the time integral of acceleration yields velocity; taking the time integral of velocity yields position. Thus, differentiating and integrating is commonplace in dynamics, as it yields several variables of interest. Much like derivatives, the antiderivatives of common functions are memorized like your times tables, and for those that aren't so easy to obtain, there are heuristics to follow to determine them.

Alright, I hope that I haven't lost too many of you. These things are really elementary to understand, but I'm sure my ham-fisted attempts at explaining them are making them seem much more complicated than they need to be. Part of that is because I've been dealing w/ these kinds of things for a thousand years, so tons of information is taken as for granted and lots of small, nuanced details are omitted in explanation. One of such nuances include notation; let's get square that circle here quickly.

Normal Mode: Dyna--No, Dot Notation

Let's take the following function as an example. Suppose it is our equation that describes our position at any given time (hence t):

\begin{equation} y(t) = t^2 + 5t + 11 \tag{11}\end{equation}

Quick note on notation (heh): I like to represent functions w/ their arguments. Here, rather than just leaving it as y = blah blah blah, I like to write y(t) = blah blah blah. This reminds me that y is a function of time. One more note here, but notice that this function is non-linear. All equations that we've been dealing w/ before have been linear, making them trivial to differentiate or integrate. This one is a second degree polynomial (second because the 'highest' power contained by a term is 2). Although this equation is non-linear, it is still trivial to differentiate and obtain our velocity equation. It is obtained using the heuristic known as the power rule. For every t term, you take its power, multiply the term by it, and decrease the power by one. So, term-by-term:

\begin{equation} \frac{d}{dt}(t^2) = 2t \tag{12}\end{equation}

For the next term, t is implicitly raised to the power of 1, so:

\begin{equation} \frac{d}{dt}(5t) = 5 \tag{13}\end{equation}

And constants--terms that have no t--simply evaluate to zero and can be omitted:

\begin{equation} \frac{d}{dt}(11) = 0 \tag{14}\end{equation}

Putting all of our terms together, we obtain our velocity equation:

\begin{equation} \frac{dy}{dt} = 2t + 5 + 0 = 2t + 5\tag{15}\end{equation}

To obtain the second derivative, our acceleration equation (notice the difference in notation), apply power rule once more:

\begin{equation} \frac{d^2y}{dt^2}=\frac{d}{dt}(2t+5) = 2\tag{16}\end{equation}

I don't like this notation, I think it's too busy. So, allow me to introduce you to dot notation. Dot notation is much more compact and cuter. The first derivative:

\begin{equation} \dot y = 2t+5\tag{17}\end{equation}

And the second derivative:

\begin{equation} \ddot y = 2\tag{18}\end{equation}

...Okay. Armed w/ a fundamental understanding of functions, their time derivatives/integrals, and dot notation, we can now tackle dynamics in earnest.

...May I Say It Now? Normal Mode: Dynamics

Right. As stated before, dynamics seeks to capture the behavior of a system of interest and represent that behavior mathematically, to varying degrees of complexity. That representation is known as a model. Typically a model is expressed via a series of differential equations.  What are differential equations? They are simply equations that have derivatives in them. This should make sense, since systems in the natural world evolve over time, so you must be able to express how that system's variables change w/ respect to time (the time derivative).

Let's not get caught up on superfluous examples. Let's go straight to an engineering example. Let's step through the process a mathematical modeler would undertake to obtain dynamic equations. Consider the following spring-mass-damper mechanical system, commonly found in automobile shock absorbers:

Figure 7. (a) Schematic of spring-mass-damper system. (b) Free-body diagram of spring-mass-damper system.

On the left is a simplified schematic representation of the system. Pertinent features are an object of mass M, walls that have a coefficient of friction b, and a spring with a spring constant k. Notice that these terms, M, b, and k are intrinsic properties of their corresponding entities. Thus, they are parameters, and not variables. They do not change with respect to time (though they can, and that would introduce more complexity to our model). On the right is what's called a free-body diagram of the same system; it is meant to demonstrate all of the forces acting on an object. Here we can see that the object of mass M is subject to three forces: a frictional force by, a spring force ky, and an applied force r(t). The fourth vector y (a quantity that specifies a magnitude and a direction, such as forces, typically designated as an arrow) refers to a displacement along the y-axis (the object moves strictly vertically, up and down).

\begin{equation} M\ddot y + b \dot y + ky = r(t) \tag{19}\end{equation}

This is known as a second order differential equation. Second order because, you guessed it, the highest derivative found here is 2. This particular equation--obtained by analyzing the pertinent forces involved in a system and applying the appropriate physical principles--describes the displacement y of the mass M, subject to the three aforementioned forces. And that's the core of what a modeler does, really. They examine the system under question, determine the pertinent forces at play, and use physics to formulate a series of dynamical equations that collectively form a model. This part is incredibly tricky as, once again, you're trying to capture the enormous amount of complexity in the world and represent it mathematically. Fidelity, that is, how accurately your model represents real-world phenomenon, increases as complexity increases.

This particular example is merely a toy; in reality that b parameter would probably be non-uniform--it's value changing depending on the position of mass M. Or worse yet, perhaps it does not vary just according to the position of mass M, but also according to the temperature of the system. Thus that parameter b is no longer a simple constant, but becomes a function. And not only is it a simple function, but a multivariate function, depending on position and temperature. Those two quantities, position and temperature are not simple constants either; they change w/ time. Hence, they need to be modeled w/ differential equations too. If you're following me so far, you can easily see how things can get hairy, as mathematical models of increasing fidelity incorporate more differential equations that are often times coupled. This is both the bane and beauty of dynamical systems: the quantities that differential equations describe are often times interdependent on each other. 

Great, so now we have this mathematical model in the form of a second order differential equation. What now? The answer is, of course, to solve it. What the heck are we solving for? I like to conceptualize this part as analogous to solving for a variable in an algebraic equation. When you're solving an algebraic equation, such as x + 2 = 5, you apply some simple arithmetic to arrive at x = 3. You're solving for a scalar; that is, a simple number. When you're solving differential equations, you're not solving for a scalar, but rather a function; the very same thing that we defined at the start of this post. However, very much unlike algebraic equations, sometimes differential equations can't be solved w/ pen and paper, following a set procedure to solve them. This is known as an analytical method. No, almost all of the mathematical models that describe real-world complex systems must be solved via numerical methods--in short, by computer simulation.

Numerical approaches to solving differential equations is an ENTIRE discipline in it's own right. As far as I'm concerned, this is the fundamental mathematical basis for what will probably be referred to in the future (if not already) as "simulation theory". No, not the hypothesis that we're living in a simulation. I mean the formal discipline on how to construct simulations. But anyway, that's enough of that digression. Back to our example.

Lucky for us, our little toy model of a mass-spring-damper system is amenable to an analytical solution. There is a set, procedural, pen-and-paper way to work it out. We don't have to (but we very well could) simulate it in order to solve it. Though, now I have to wonder whether or not I should get into the nitty-gritty and introduce you all to the Laplace transform. I think I will, but this post has gone on long enough. We'll pick it up in the next one.

[end transmission 2/?]