Basic Physics: Part 0, Section 4: Derivatives

Previously in this series, we covered algebra, trigonometry, vectors and vector multiplication. Now (after more delay than I would have liked) it's time to tackle the elephant in the room--calculus.

No, please don't close this tab! I swear, it's not as scary as you've been told. If you made it through trig and vectors (which, if you are reading this I assume you have) you've really made it through more mind bending stuff than we will need to cover here.

Why cover calculus at all? Aren't there algebra-based physics courses at every university? Yes, yes there are. And anyone who has taught physics with only algebra, trig and vectors will tell you it's actually harder to teach physics without reference to derivatives and integrals. Newton invented/discovered calculus so he could describe his theory of gravity and motion (his notation was abysmal, though). Calculus is the mathematics of change. Algebra is the mathematics of stability. And physics is really boring if nothing ever moves.

Now, depending on when you went to school, learning calculus may have been reserved for the students who were good at math, or who hadn't been told that "math wasn't for them". I am here to tell you this is like telling students who are going to live in another country that they don't need to learn the past or future tense, they can get along just fine with the present tense. Technically, this is true in a lot of cases, but it limits their ability to get everything out of their trip. Try to think of calculus in this way--not as some strange new kind of math, but just a different tense in this language.

We'll begin where most calculus textbooks begin with derivatives. Calculus has a very intuitive explanation of derivatives: they are the slopes of lines. That's it. What makes derivatives interesting is that they give you the slope at any point along a line*. You will generally hear the included caveat that the line must be smooth and continuous, but this isn't a calc class and I'm not going to show you any equations which are not differentiable (capable of having their derivative taken), so we aren't going to worry about that here.

Let's start with the simplest case, a straight line going through the origin of our coordinate system:

In this case, the slope of the line is going to be the same everywhere, and we can find the slope using the tried and true "rise over run" method. In moving 4 units to the right, the line has moved 3 units up, so our slope \(a\) is $$ a = 3/4 = .75 $$ So far so good. Nothing scary or uncomfortable  to date. A little algebra, that's all, and a little reading off a plot. Now, what if we were just given the equation for this line, in the slope-intercept form encountered in algebra class: $$ y = ax$$ $$y = .75 x$$
Still not too bad. And if I had presented this to your first, you probably could have told me the slope of this line just from this--the coefficient of \(x\) gives the slope, so \(.75\). Congratulations, you just took your first derivative without knowing it!

So, if derivatives are that easy, you ask your computer suspiciously, why is there an entire semester of calculus dedicated to it, hmm? Well, two reasons. First of all, because there are way more complicated kinds of lines than straight lines, and second of all, no one dedicates an entire semester to derivatives. They usually also teach limits (proto-derivatives) and numerical integration (proto-integrals) in the same semester. Derivatives are usually 4-5 weeks of the semester, a lot of that learning special cases.

What if I gave you the line with the equation $$ y = x^2 + 3, $$ would you know what it's slope is? It looks similar to the linear equation in slope-intercept form, but you probably have a feeling that the \(x\) being squared complicates things. And it does, since \(x^2\) is a parabola.

Parabola!

Now, you could draw tangent lines at a sampling of points along the parabola, and find the slope of those tangent lines, plot those slope values and approximate the slope of \(y = x^2 + 3\) and you would find that it came close to \(2x\). I can't speak for everyone here, but I find doing that unbelievably boring. Some algebra teacher once made me do that once and  it was tiresome to say the least.

But calculus and the tool of differentiation gives us a much better way.  Remember, mathematicians do not "invent" new kinds of math to torture students and non-mathematicians. They develop new techniques because the old way was inefficient or tedious or just didn't work all that well. Calculus is  a great example of this. Rather than calculating a bunch of individual slope points and extrapolating what we think the slope is, we can find the exact slope with a few simple rules, and a little new notation.

Let's look at our parabola again. So we have the equation $$y = x^2 +3$$ which describes the line itself. If we want to say that we are looking at the equation for the slope of that line we can write it in Leibniz notation as $$ \frac{dy}{dx}$$ which is nice and concise (there is also Lagrange notation and Newton notation). But Leibniz is nice for beginning with because it has a nice math to english translation: the change in \(y\) over the change in \(x\). This is the more formal way to say "rise over run" and is more generally applicable. Also, now that we are finding the slope of the parabola everywhere we call it a "derivative", and we find it by the process of "differentiation".

To find this, we need two rules. The first rule is formally known as the "elementary power rule" but I just learned it as "this is how you do it". For a function \(f(x)\) that has the form (i.e., it looks like or follows the pattern of) $$ f(x) = c x^n $$ where \(c\) is a constant, \(x\) is the variable and \(n\) is a real number (usually integer, but not necessarily) the derivative can always be found in the following manner: $$\frac{df}{dx} = c*n*x^{n-1} $$ If you are wondering what that \(f(x)\) is doing here, since I kinda just started using it, think of it as a way to label a generic equation. You could keep saying \(y=\) such and such, but then it's not always clear which \(y\) you're talking about. If you instead use the notation of Letter(variable) it lets you label both the equation uniquely (function f, function g, function h) and specify which letter is acting as your variable (x, y, z). Neat, huh?

That's it. That is the most basic rule and definition of the derivative. For the special case where there is no variable, just a constant, the derivative of a constant is \(0\). So, to summarize,

1. Given a function \( f(x) = c x^n\), the derivative is \(\frac{df}{dx} = c*n*x^{n-1}\).
2. Given a constant function \(f(x) = c\), the derivative is \( \frac{df}{dx} = 0 \)

So, let's apply these rules to the equation for our parabola.
$$y = x^2 + 3$$
$$\frac{dy}{dx} = (2) x^{(2-1)} + 0$$
$$\frac{dy}{dx} = 2x^1 = 2x$$

And so we find in three lines of calculus the  same answer that a bunch of line drawing and measuring and plotting got you. Let's try another one, that's a little longer.

And really funky looking on a graph.

$$f(x) = 3 x^{5} - 2 x^{2} + x^{-3} $$
$$\frac{df}{dx} = 3*5 x^{(5-1)} - 2*2 x^{(2-1)} + -3 x^{(-3-1)} $$
$$\frac{df}{dx} = 15 x^{4} -4 x - 3 x^{-4} $$

Longer, but still not too bad, right? See, I told you calculus wasn't the terror it was made out to be. One more rule and we've knocked out all the differential calculus we'll need for both physics 1 and physics 2. This rule is called the "chain rule" and it covers almost every other situation we could face outside of a calculus book or more advanced physics. What it is, really, is a short cut when your variable of interest is buried inside a parenthetical expression, instead of having to bother to separate it out by algebra (if it can be separated by algebra at all).

Let's start with something that we could mess around with algebra and get it into a form that our first two rules apply. Let's begin with the equation $$g(x) = (x+2)^2$$ By using the FOIL method, we  find that this could also be stated as $$g(x) = x^2 + 4 x + 4$$ Using the two rules laid out above, we find that it's derivative is $$\frac{dg}{dx}= 2x + 4$$ Now we have something to check the chain rule against.

Displaced parabola!

The chain rule is a way to approach these things methodically, working from the outside in. You start by treating everything inside the parentheses as a block. It does not matter how complicated it is inside the parentheses, or how simple. Treat it all as though it were just the variable. So step one of the chain rule gives us $$\text{ Step 1: } \frac{dg}{dx} = 2 (x+2)^{2-1}$$
Now you take the derivative of what's inside the parentheses, and multiply that result by the result of Step 1. $$\text{Step 2: } \frac{dg}{dx} = 2(x+2)^{1} (1+0) = 2x+4$$
Lo and behold, it's the same result. Now for something this simple is using the chain rule worth it? Maybe, maybe not. But what about something that I don't know how to FOIL, like $$h(x)= (x+2)^{-1/2} =\frac{1}{\sqrt{x+2}} $$

How do you FOIL a square root?! Tell me!

Let's try the chain rule on this and see if it doesn't save us having to look that one up in an obscure algebra text.
Step 1: Ignore what's inside parentheses, take the derivative as if (blah blah) = variable.
$$\frac{dh}{dx} = (-.5)(x+2)^{(-.5 - 1)} $$
Step 2: Take the derivative of what's inside the parentheses, multiply it by Step 1.
$$ \frac{dh}{dx} = -.5(x+2)^{-1.5} (1)$$
Step 3: Simplify if necessary
$$\frac{dh}{dx} = -.5 (x+2) ^(\frac{-3}{2}) = \frac{-1}{2 (x+2)^{\frac{3}{2}}}$$

I can guarantee that that was easier than trying to FOIL a square root. But what about something really nasty, like THIS

Honestly had no idea what this would look like before I graphed it

Behold, the rollercoaster that is $$ k(x) = (x^3 + 2)^{-.5}$$ Surely my nasty, terrifying calculuses gets horrifying and complicated now, heh? Stupid physicistses.

Um, nope. Not really. Let's take a look, shall we?
Step 1: Ignore what's inside parentheses, take the derivative as if (blah blah) = variable.
$$\frac{dk}{dx} = (-\frac{1}{2})(x^3+2)^{(-.5 - 1)} $$
Step 2: Take the derivative of what's inside the parentheses, multiply it by Step 1.
$$ \frac{dk}{dx} = (-\frac{1}{2})(x^3+2)^{-1.5} (3x^2)$$
Step 3: Simplify if necessary
$$\frac{dh}{dx} = (\frac{-3x^2}{2})(x^3+2)^{-1.5}  = \frac{-3x^2}{2 (x^3+2)^{\frac{3}{2}}}$$

Still just 3 bite sized steps.

Ah ha, you say, but what if there are parentheses inside the parentheses? What if I have a russian nesting doll of a problem?

You just repeat step 2 until you run out of parentheses inside parentheses. But I honestly can't say that I've ever seen that happen.

And that's nearly all you really need to know about differential calculus to conquer introductory physics! Hopefully you can see, at least a little bit, why physicists and mathematicians love it. It's like upgrading from a hand drill to a power drill. Or a sheet of sandpaper to a power sander. It might take a little getting used to, but it is a very powerful tool in our toolbox and one that will open up the rules of the physical universe to us in a way that algebra just can't. Because as I said in the beginning, the physical world is dynamic and changing, and algebra is the math of the static and stable.

There are two "special cases" that aren't really special cases that we will need, and they are very easy to use, but a bit lengthy to explain, so I'll cover them in a separate section, partly because they are both really cool, and partly because this post is already pretty long.

If anything is still unclear, or even a little foggy, let me know in the comments and I'll do my best to explain! And I hope to see you next time for integration!

* there are a few significant exceptions to this, which we don't have to be concerned with here. If you are interested in knowing more about these exceptions, brownian motion is a particularly interesting case being continuous everywhere and differentiable nowhere.

Basic Physics: Part 0, Section 2: Vectors and Coordinate systems

In the previous sections, I cover some basic algebra topics and necessary trig functions.

In this section, I'm going to lay out some basics of coordinate systems and vectors. I'm pairing these concepts because vectors without coordinate systems are a little esoteric for this series, and coordinate systems are necessary, but easily dealt with, at least compared to things like trig functions.

These two concepts are needed because when we talk about a problem and set out to solve it, we need a way to describe where things are, where they are going and how they are getting there. The combination of vectors and coordinate systems allows us to know exactly what we are referring to  and what it's relation is to anything else that might be relevant to the problem. Without this tool, problems in more than one dimension can easily become a hopeless jumble.

Let's start with coordinate systems. There are many possible coordinate systems of which  11 of which are commonly used and  only 1 of which we need for right now. That coordinate system would be the Cartesian coordinate system, apocryphally realized by M. Rene Descartes  (he of "I think therefore I am" fame) as he lay in bed watching a fly buzzing above him. It was also realized by M. Fermat, though he failed to publish it. If you ever had to plot things by hand in a math class, you have used the two-dimensional cartesian coordinate system! The cartesian coordinate system can be thought of as a grid system in 3 dimensions, that lets you specify a  location based 3 numbers, one for each dimension. You can think of it as giving someone a latitude, longitude and altitude. You have given them all the information they need to locate a particular spot on planet earth. (I will officially note that the earth is NOT a cartesian system, since the lines of longitude are not parallel but intersect. But for small distances, say NYC, it is a decent approximation).

Formally, a location in any coordinate system  is the intersection of three orthogonal planes. Orthogonal, for our purposes, means that the lines/planes intersect at \( 90^{\circ}\) to each other. Think of the walls in your house. Your walls (hopefully!) intersect your floors at right angles most of your walls will intersect each other at right angles, unless you have a very interesting house. So your walls are orthogonal to each other and they all orthogonal to the floor.

An example is probably the easiest way to show this. Let's start with a basic three-dimensional (3D) cartesian coordinate system:

In the picture above, I have drawn the same coordinate system from two slightly different perspectives. The top one you are staring down the barrel, so to speak, of the z-axis, looking at the x-y plane straight on. In the bottom one the picture has been rotated \(45^{\circ}\) about the y-axis so you can see along the z-axis as well. This becomes very useful if you are talking about things in 3D, while the top one is fine if you are only worried about two dimensions.

Another word about terminology and notation. What is an axis, and why are those letters wearing hats? As with a lot of math-stuff, it comes down to the dual needs for conciseness and precision. Let's start with the hats. When you draw a coordinate system for a problem and you label the axis, you are defining your directions. It's as if you are creating a mini universe and saying "this is East/West, this is North/South and this is Up/Down". But rather than label things in poetic victorian manner as "easterly direction" mathematicians and their ilk like to label things with letters. So "easterly direction" becomes "x-direction", but that's still too wordy. So 'direction' becomes 'axis', and that can get further shortened with vector notation as \( \hat{x} \) said "x-hat".  So an axis defines the direction of your coordinate system, but it also serves as a point of reference, much like the equator, the Greenwich meridian, and sea level  serve as reference points for finding places on the earth. So if you are on the x-axis, you are not moving in a y-direction or a z-direction. If you want to give a location in the coordinate system, you can notate it either as \( \left\langle a, b, c \right\rangle \) or you can use vector notation, to get a little ahead of ourselves: \( a \hat{x} + b \hat{y} + c \hat{z} \). The latter notation is preferably simply because it is more flexible, as we shall see.

Getting back to our example. Let's say we want to find a point \( \left\langle 3, 2, 2 \right\rangle\) (\( 3 \hat{x} + 2 \hat{y} + 2 \hat{z} \) ). For the moment we don't care what the units are. We start by locating the \( x = 3\) plane, that is, the plane that contains every point of the form \( \left\langle 3, b, c \right\rangle \) where \( b \), \(c\) are every real number. Then our diagram looks like this

with the red dot noting the point where the plane intersect the x-axis  in the bottom view, since it's a little hard to see. Next, we locate the \( y = 2\) plane.

The blue dot notes it's intersection in the y-axis in the bottom diagram because again, it's hard to tell. It's much easier to see in the top image, but there' a reason why the bottom diagram is actual preferable in some ways. This can be most easily seen when we try to add in the last point, the \( z = 2\) plane to give us our three-plane intersection.

Amazing what you can do with a basic drawing program and a little insanity.

The top image doesn't really allow us to visualize that last necessary dimension. You can mentally add it, but you can't draw it into the top one. The bottom one you can see the last plane and pinpoint their intersection (marked with a black dot here). 

That's pretty much all there is to coordinate systems. They let you pick a frame of reference so you can locate things in a mini-universe for the purpose of problem solving. What I find particularly neat is that you can place your coordinate system anywhere you like and the problem will still be solvable. It may be easier to solve from a computational standpoint if you center it nicely, but you don't have to. Why this is the case is something that I'll get into more when we start doing physics properly. 

Now, on to vectors. If a coordinate system gives you a frame of reference, vectors are what let you move around in that frame of reference and deal with more than just static problems. Now, what they are precisely requires linear algebra and is way outside the scope of this series, so we are going to stick to just definitions and not get into the nitty gritty. So, here goes.
 

A vector pairs a quantity with the direction that quantity is in, going or pointing to. A vector has both "magnitude" (quantity) and "direction". So long as you can describe a quantity as having these two qualities, you can express it as a vector*. We've already shown how we can describe position as a vector. You can also describe velocity as as a vector. "He's going 80 miles per hour" gives you a speed (a magnitude). "He's going 80 miles per hour due north" gives you a magnitude and a direction. We'll get more deeply into what physical quantities can be described using vectors in the first section of Part 1. 

For now, let's just work with two arbitrary vectors and see what we can do with them. As discussed in the algebra post, we'll use letters to stand in for numbers that we can plug in later. 

$$\vec{v} = a \hat{x} + b \hat{y} + c \hat{z}$$

$$\vec{w} = d \hat{x} + e \hat{y} + f \hat{z}$$

The little arrow above \(v\) and \(w\) indicates that they are vectors. In math everything has its own shorthand because you never know when you will want to deal with something in its entirety, or just don't want to write out the whole thing for the umpteenth time. 

So, what can we do to these things? Well, we can add them. The trick is that you can only combine things attached to like 'hats'. So you combine all the x-hat components, all the y-hat components and all the z-hat components, but you can't combine x-hat components with non-x-hat components. So 

$$\vec{v} + \vec{w} = a \hat{x} + b \hat{y} + c \hat{z}+d \hat{x} + e \hat{y} + f \hat{z}$$

$$\hspace{10 pt} = (a+d) \hat{x} + (b+e) \hat{y} + (c+f) \hat{z}$$

Subtraction works the same way:

$$\vec{v} - \vec{w} = a \hat{x} + b \hat{y} + c \hat{z}-d \hat{x} - e \hat{y} - f \hat{z}$$

$$\hspace{10 pt} = (a-d) \hat{x} + (b-e) \hat{y} + (c-f) \hat{z}$$

At this point you are probably wondering about multiplication and division, since addition and subtraction have been relatively straight forward. The answer is that there are two types of multiplication for vectors, and no types of valid division. Why this is starts getting into linear algebra and "outside the scope of this course". So I'm going to ask you to trust me on this one, because it's absolutely true even if I can't show you right now why it's true. They are also rather more involved than vector addition/subtraction, so I am going to move them to their own post so we can really take our time with them. 

I hope that this all was clear. If it wasn't, please let me know in the comments!

*There are also a few things that we'll get to over the course of this series that you wouldn't think you could describe as vectors, but they behave identically to the ones we deal with here. 

Basic Physics: Editorial Consortium

The next promised post on trigonometry is in the final polishing stages, but in the meantime I would like a post to mention several people who have graciously agreed to help me in this endeavor to bring the first year of a physics majors schooling in physics to a non-math, non-science types audience.

 I know full well that as a grad student in physics I am a very bad judge of what is and isn't understood or common knowledge. Teaching has helped rein me in enormously, but my students are assumed to have at least basic calculus knowledge. So I anticipated myself having a problem recognizing what needed more explanation, what was over-explained or even patronizing. I don't want to be the detective novel criminal who spells all the easy words wrong and all the hard words right (in reverse, kinda). Dear Husband, my usual editor, is too well versed in math to much use in this particular arena, so I reached out to some other family members, specifically my mother, my sister, and my brother, to help make sure I do this right. I asked them to do this because each of them brings something that I felt I really needed on what I am dubbing my Editorial Consortium.

My mother is in the demographic group, you might say, that always gives me deer-in-the-headlights or horrified looks when I say I do physics and protest it was too hard for them. Though very talented, she has not directed her talents in a STEM field direction. She is, however, the only reason that I can do long multiplication or division and light years ahead of me in mental arithmetic (also cooking, social skills, language, and checkbook balancing). She is also a natural copyeditor of high standards who is not shy of letting me know when I have fallen short of the mark.

My sister, hereafter to be referred to as Sylvia, Historian Extraordinaire, just graduated college with an absurd amount of honors with a major in History and a minor in French, her thesis work (yes, thesis for undergrad) being on Dorothy L. Sayers. She has a good math background, but hasn't used it much, having no call to do calculus as a literary historian. Her one and only basic physics class was the same one in high school that inspired me to physics. She is also representing a group that I want to reach--younger adults--and she would know if a reference is too obscure. She is also incredible at calling me out for being obtuse and/or patronising.

Last, but not least, is my brother, who will start high school in the fall. I included him for three reasons. First of all, he has had all of the math that I claim is required to understand the blog, but has never taken a physics class in his life. He's interested in the sciences, but he is yet untainted by misconception and bad teaching (other than my own). Second of all, it turns out he inherited Mother's copy editing skills and is very good at noting my inconsistent use of single and double quotation marks. Thirdly, I'm curious if the explanations are clear enough for younger persons who might be interested, but don't have much of a background. The flip side of my mother, so to speak.

They have all agreed to read, edit and comment every post that I write in this series. Between them all I think there is a fair shot that I will do what I set out to do. But I won't know if I am actually succeeding unless you, the reader, let's me know. You are the other part of this Editorial Consortium. If something is not clear, if I mess something up or forget something or just plain gloss over with the hated "the reader can obviously see", let me know! There is a comments link below each post. I'd love your feedback.

Basic Physics Part 0, Section 0: Algebra

[This post is the first in a series intending to teach basic physics concepts in a blog format.]

As I mentioned in my introductory post, math is the language of physics. Physics cannot realistically be understood or done without math. While advanced physics requires some advanced math, basic first-year type physics requires some relatively basic math and math concepts. The first math topic that I want to cover is one that everyone who graduated high school should have covered at some point: algebra. 

Algebra has a kind of strange reputation. Among STEM people, it's the boring math that you needed to do to do the REAL math, or at least the non-boring stuff. It carries the same emotional connotations as diagramming sentences. Among non-STEM people, it's the boring math that they forced you to do and you never ever used again.

Until I really got into teaching and my research, I was mostly of the opinion that algebra was best left to machines. It was tedious and beneath my dignity to spend hours and pages rearranging symbols. When I started teaching, I began to understand the subtle power of algebra to make or break a solution. When I finally started to understand my research, I saw not only its power, but its beauty. Algebra is a tool that allows order to arise out of chaos.

To do the kind of physics this series is going to look at, you really only need 2 major algebra skills: the FOIL method, and some equation manipulation skills. The quadratic equation can come in handy, but that is one time that I am ok using a math program for because it doesn't pop up as frequently.

But before we get to that, I think some terminology definition is in order. When I speak of a "variable" I am referring to a symbol that can take on any value on the real number line (i.e., any where between negative infinity and infinity) within the confines of the equation and/or is the quantity we are solving for. A coefficient is a symbol that has a fixed value for that particular problem. Most physics texts I've seen and used have the convention that any letter from p-z can be used as a variable, while letters a-m are used as coefficients. The letter 'n' is a special case because it is typically used for integer numbers only. The letter 'd' is sometimes used as a variable because it's just so convenient to use it to stand for 'distance'. The letter 'o' is never used, because in handwritten notes it can all too easily look like a zero. A constant, for our purposes, is a symbol that has a fixed value that does not change from problem to problem. For example, \( \pi = 3.14159...\) no matter what problem we are doing. A 'term' is a catchall, just denoting that a symbol stands for something, without specifying type.

Now, on to algebra!

FOIL Method

The FOIL method (First Outside Inside Last) is one of the first things I was taught in algebra class, way back in 7th grade. It's basically a method for multiplying mathematical expressions together in a way that doesn't let you double multiply or leave something out.  If you are multiplying just two terms together, say \(a\) and \(b\), its easy to know when you got it all.
$$ (a)(b)  = ab$$
But what if you don't have just two items, but two expressions, \( (a+b) \) and \( (c+d) \) ? FOILing the two expressions makes sure you do all available multiplications without double counting.  You multiply the first terms from each expression, here \( a \) and \( c \), then the outside ones, here \( a\) and \( d\). Then you do the inner ones, \( b\) and \( c\), and the last ones from each expression, \( b\) and \( d\). Thus
$$ (a+b)(c+d) = ac + ad + bc + bd $$
This method can be logically extended to cover expressions with more than two terms, with the corresponding result being proportionately longer.

When I first learned this, it seemed incredibly useless. Why on earth would I need such a simple method? The answer is 'everywhere in physics'. From the simplest two-body problems to the most complex problems I've worked on for research, FOILing turns up again and again and again. Becoming not just  proficient, but a master at this technique has been crucial to my work. It is something that my students consistently underestimate, to their detriment, every semester.

Manipulating Equations

This isn't so much a single method as the Rules of Engagement for math. Equations are pretty flexible, but there are some rules. The underlying principle to these rules is that you have to do the same thing to each side of the equation, and you have to do it to everything on each side. For example, lets say we have this equation $$ 5 x + 2 y = 6, $$ and we want to solve for \( y\). We can start by subtracting \( 5x\) from each side like this $$ 5x + 2y - 5x = 6 - 5x$$ where you can see we have explicitly taken \( 5x\) from each side and thus have not changed the equation. By adding the same thing to both sides, we have effectively added zero, just like if you add a one pound weight to either side of a balance scale, it won't change position.  So now we have the equation $$ 2y = 6 - 5x,$$ but we still have not completely isolated \( y \). So now we have to divide both sides by 2, which is the coefficient of the variable \( y\). $$ \frac{2y}{2} = \frac{6 - 5x}{2}$$ Again, it is important to note that we have done exactly the same thing to both sides of the equation and in the case of division or multiplication we have applied that change to every term. $$y = \frac{6}{2} - \frac{5 x}{2}$$ $$ y = 3- \frac{5}{2}x$$ is the correct solution in this case. Do not, I repeat, DO NOT make the mistake I see so often, which is to only apply the division to one (usually convenient) term. The following 'solution' is wrong for this problem: \( y = 3- 5x\)

In certain cases, this also involves remembering the Order of Operations: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. The Multiplication/Division and Addition/Subtraction orders are less critical, since they are just two sides of the same coin. Division is the same as multiplying by a fraction, subtraction is the same as adding a negative number. But the parentheses-> exponents->multiplication/division->addition/subtraction ordering is inviolate. It is impossible, outside of sheer fluke, to get a correct answer if you do not abide by this rule.

And that's the basics of algebra that you may have forgotten  (accidentally or on purpose) that you need for physics, other than the kind that you, honestly, do intuitively. Next week, we'll cover some basic trig[onometry] that everyone should know.