Science

The Physics Philes, lesson 2: Math for Miles!

In which math is relearned, more books are purchased, and heads are banged.

Last week was my first full week of lessons. It nearly killed me. OK. Not really. But I have been sleeping really well.

As I wrote last week, I’m really not a math person. Which makes my choice of physics as a topic a bit curious. But I was convinced that, provided with an explanation. I was humming along fine until I came across this phrase:

“…on a graph of position as a function of time…”

Whut?

I also ran across some problems that required me to know and execute derivative equation rules. I had no idea what to do. I thought I could make it work. But I can’t. Not on my own. So I did what any self-respecting student would do. I bought a study book.

It turns out that the warnings issued by many a math teacher are true: Math is, indeed, cumulative knowledge. If you don’t grasp a basic concept, it’s hard to move on to more advanced topics. Since it’s been about 10 years since I’ve had a math class (and I’ve never had calculus) I’ve forgotten most of what I used to know. So I have to start from the beginning. And by “the beginning,” I mean algebra. This project has morphed into my own part-time class schedule: physics and calculus. So in addition to discussing a physics concept every week I’ll discuss some math that I am either relearning or learning for the first time.

Let me apologize if this is incredibly boring for you. I didn’t fully realize when I started how much algebra, geometry, and trigonometry I had forgotten. The examples I use are taken from my study book, because I want to make sure: a. I got the answer write and b. the problem is set up correctly. I’ll get to the physics in a bit.

Graphs seem to be an important part of physics so far, likewise with calculus. So I needed to brush up on my linear equations.

Linear equations always have three components: two variable terms and a constant term. Standard form of a linear equation will look like this:

Ax + By = C

OK. That looks familiar. I seem to remember solving equations like that years and years ago. I wonder if I can still hack it. I’ll put this equation in standard form:

3x – 4y – 1 = 9x + 5y – 12

I’ll first need to subtract 9x from both sides:

(3x – 9x) – 4y – 1 = (9x – 9x) + 5y – 12

-6x – 4y – 1 = 5y – 12

Next, I’ll need to subtract 5y from both sides:

-6x (-4y – 5y) – 1 = (5y – 5y) – 12

-6x – 9y – 1 = -12

-6x – 9y – 1 = -12

Hey Mr. 1! What are you doing on that side of the equation? Get over there with Ms. 12:

-6x – 9y – (1 + 1) = (-12 + 1)

-6x – 9y = -11

Whoops! See that -6x? That’s supposed to be a positive number in standard form. I better multiply each side by -1:

-1 (-6x – 9y) = -1 ( -11)

6x + 9y = 11

Ah yes. It’s coming back to me. Slowly, and I need a lot more practice, but it’s there. A rich vein of algebra-ore waiting to be mined. But there is oh-so-much more. Next I need to relearn the two major ways to create the equation of a line.

The first is the slope-intercept form: y = mx + b, where m represents the slope of the line and b represents the y-intercept of the line. Solve for y if the slope of the line is -3 and the y-intercept is 5:

y = -3x + 5

See what I did there? I just plugged in the values. Easy peasy lemon squeezy. The second major way to create the equation of a line is point-slope form. You can use this form when you only have a point and a slope, but no y-intercept. If we have point (x1, y1) and slopem, the equation will be:

y – y1 = m(x – x1)

The x and y with subscripts represent the point coordinates. Let’s try one. Find the equation of the line through point (0, -2) with slope 2/3. First, let’s plug in the values we have into the above equation:

y – (-2) = 2/3(x – 0)

y + 2 = 2/3(x – 0)

Oh noes! Fractions. It’s possible that I got the mechanics of this next part wrong. If I did, please tell me in the comments. Now, we multiply each side by 3 to get rid of the fraction:

3(y + 2 = 2/3(x – 0)

3y + 6 = 2(x – 0)

Let’s get that right side simplified.

3xy + 6 = 2x

Subtract 3x from both sides:

3y + 6 -3y = 2x – 3y

6 = 2x – 3y

Ta-da! I did it! I think. Right? Someone please check my work.

That’s not quite all the math I retaught myself this week, but you came here for some physics. So that’s what you’re going to get.

This week was dominated by velocity and acceleration. This is how we can talk about the motion of an object. Today I’ll be discussing average velocity and average acceleration (as opposed to instantaneous velocity and instantaneous acceleration. Mostly because the instantaneous stuff completely confounds me.) First, velocity!

Average velocity is the change in position of an object divided by the change in time. To find the average velocity (represented by v), all we have to do is divide the change in position (represented by x) by the change in time (represented by t). Or, in math-speak:

v = Δx/Δt

Whoa there. What’s with those shapes? That’s the Greek letter delta, used in algebra to represent “change in.” Let’s say we have a starting place of x0 and an ending point of xf. To find Δx, we just subtract xf from x0. We do the same with Δt. We subtract the final time (tf) from the initial time (t0). Like this:

v = xf – x0 / tf – t0 = Δx/Δt

Hm. That might not be very clear. Let’s do an example. Let’s say you’re on your bike. You start from rest and ride for 10 s. In that 10 s you go 50 m. What is your velocity? All we have to do is plug the values into equation:

v = 50 m/10 s = 5 m/s

Pretty easy, right? Once you get past all the Greek letters and subscripts, it’s a pretty straight forward concept. We can also use this equation to solve for the position of the object by solving for x.

Now that we have average acceleration down, let’s figure out average acceleration. Acceleration is the change in velocity divided by the time in which that change takes place. We come across our good friend Δ again:

a = vf – v0/tf – t0 = Δv/Δt

OK! Let’s plug in some values! Let’s say that you’re at a stop light. When the light turns green you accelerate. After 10 s the car is moving at 50 m/s. What is the acceleration?

a = (50 m/s – 0 m/s) / (10 s – 0 s) 

a = 50 m/s / 10 s = 5 m/s2

Notice the difference between the answers for velocity and acceleration? For velocity, the unit is m/s. For acceleration, the unit is m/s2.  That’s how you know when we’re talking about velocity and when we’re talking about acceleration.

I mentioned above that I was having a little trouble with instantaneous acceleration and instantaneous velocity. But my existing math experience does not encompass derivative equations, which is evidently necessary to calculate instantaneous acceleration and velocity. This is cause for great concern; I’m afraid I won’t be able to catch up on the math in time to figure out the physics. I know we have a bunch of smart-pants readers. You will get 100 gold stars if you can explain the concept of limits and derivative equation rules to me in a useful way. The challenge is issued. Will you accept? (Please accept!)

This is, of course, only a fraction of what I actually covered this week. Did I make any mistakes? Should I have gone about something in a different manner? Let me know in the comments!

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Mindy

Mindy

Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

8 Comments

  1. June 19, 2012 at 12:00 am —

    If you wanted to use the point-slope form to find the standard form of the equation then 2x-3y=6 is right, also you could have just multiplied 2/3 to x and 0, would have been shorter.

    I never studied physics so I didn’t look at that.

    If you’re looking for some educational resources to help you relearn some math here some AWESOME sites. Everything I list is absolutely FREEEEEEEE, and HIGH QUALITY education!

    http://www.khanacademy.org/ (Over 3200 FREE videos on Math, science, and even Art! Literally goes from 1+1=2 to differential calculus, and everything in between. He teaches the math very informally, focusing alot on intuitively understanding what you’re doing. Very entertaining, Highly Recommend! Plus I suspect Salman Khan is one of us, i.e. an atheist/skeptic.)

    http://www.openculture.com/freeonlinecourses (Over 500 FREE online courses (from Modern Poetry to Astrobiology) in video or audio format, the Math courses are mainly from Harvard, Berkeley, MIT, and Princeton)

    http://www.openculture.com/free_textbooks (150 Free textbooks! Typically people want the most up to date textbooks which I don’t think these are, but it should be okay for Math textbooks, because, you know, Math doesn’t really change.)

    There’s also stuff like “MIT opencourseware” (I never bothered with this because i don’t know how to use it, maybe you’ll have better luck lol.) and http://www.udacity.com/ which has a free beginners physics class beginning June 25th, not sure if you need that though.

    Anyways hope you find something useful in all of that. (Alternatively you could just torrent a thousand textbooks on Math and Physics. Just kidding…. Or am I? Dun dun dun)

  2. June 19, 2012 at 12:52 am —

    Challenge accepted!
    So, limits.
    Limits are a useful way of looking at equations when they get really big or really small. So generally, the way we use limits is by letting x get as close as possible to a value without actually letting it reach that value and seeing what happens. For instance, in the equation y=x, if we let x get really really big, then y gets really really big as well. So we can assume than when x goes to infinity, so does y. (Although assuming things in math does get a bit tricky sometimes.) What happens with the equation y=1/x? Well, in this example, as x gets big, y gets small. So if x goes to infinity, y goes to zero. But what happens if x gets really small? X can never equal zero. The fancy math Powers That Be decided that 1/0 is undefined. That is, it doesn’t exist. But what if x were to simply *approach* zero? There’s nothing in the rules about that. So we let x approach 0, and we notice that y gets really big. So we can conclude that in the *limit* as x approaches zero, y goes to infinity.
    Limits are useful in other ways, too. Take for example the equation y=1/(x-2). X can never equal two, because then the equation would be y=1/0, and that’s not allowed. But suppose we wanted to find out what happens around x=2? Well, we could calculate the value of y in the limit as x approaches 2. What we would find is that the value of y approaches infinity as x approaches 2. In this way, we can use the concept of limits to understand certain things about equations that we wouldn’t be able to understand otherwise.
    And now, on to derivatives. This explanation might seem a bit roundabout, but stick with me.
    Let’s say you have an equation and you need to find the slope. Easy, right? You pick two points (x1,y1) and (x1,y2) and then use the slope equation m=(y2-y1)/(x2-x1). Problem solved! Well, not quite. You see, that equation only works for lines, and I never said that this equation was a line. In fact, it’s a very curvy curve. “But wait!” you say. “If it’s a curve, then the slope changes with the value of x. How am I supposed to find the slope when it’s not constant?” Very good question. The answer is you can, but it’s a little complicated, so we’ll get to it later. For now, let’s consider the slope at a single point. Now you might think that you still can’t find the slope because it’s not a line, but you can at least pretend that it is one. I’m going to let that sink in for a moment. Pretend that a curve is a line. This probably violates a dozen math rules or something, but maybe if we pick two points that are fairly close together, nobody will notice. So we’ll let (x1,y1) be the point that we’re trying to find the slope at, and we’ll let (x2,y2) be a point not too far away. We plug them into the slope equation, and voila! We have a slope that might not be too far away from the actual value!
    But surely we could do better. What if we moved our second point a little closer to the first? That would certainly make our approximation a little more accurate. But why stop there? Why not move the second point as close as possible? Why not, indeed? So how close can we get? The answer is infinitely close. And now (hopefully) you see where I’m going with this. If we use the concept of limits, we can calculate the value of the slope as the limit as (x2,y2) approaches (x1,y1). And what we get is something called the derivative.
    The derivative can be thought of as the slope of an equation at a particular point, but more generally, it’s the rate of change of the y variable at a specific x value. Now granted, both of those definitions mean pretty much the same thing, but the second is more useful when you’re talking about instantaneous velocities and accelerations. If velocity is the change in position over time, then instantaneous velocity is the derivative of the position equation at a specific point. Likewise, acceleration is the change in velocity over time so the instantaneous acceleration is the derivative of the velocity equation.
    Now, it turns out that it’s easier to apply derivatives to general equations than to specific numbers, so that’s what we’ll be dealing with here. First, this next section will involve some actual math but most of it shouldn’t be very complicated. Second is a bit of terminology. We indicate the derivative of a function by putting an apostrophe after it. For example, the derivative of the function y is denoted y’ (pronounced “y prime”). You may also see it written dy/dx. That means a derivative of y with respect to x. There’s also one other way to denote derivatives that involves putting a dot over the y, but I’ll only use the apostrophe method. Now that that’s out of the way, time to actually do some derivatives.
    There are four main rules that you need to learn in order to be able to do derivatives. They’re not really all that hard, but they will require a bit of practice. The first is known as the “Power Rule.” Let’s say you’re trying to take the derivative of the function: x^3+x^2+x. That may look complicated, but notice that each term is of the form x^n, where n is some number. This is where the Power Rule comes into play. The Power Rule says that if you encounter a term of the form x^n, you simply multiply the whole term by n and reduce the power by 1. The derivative of x^n is nx^(n-1). So in the example above, the derivative of x^3+x^2+x^1 is 3x^2+2x+1x^0. Anything raised to the 0th power is 1, so this simplifies to 3x^2+2x+1. Not bad, right?
    This also works for negative powers. Say you have x^-2. Multiply the term by -2, then reduce the power by 1. -2-1 is -3, so the derivative of x^-2 is -2 x^-3. Similarly, this works for non-integer powers as well. What if we had x^(3/2)? Then the derivative would be (3/2)x^(1/2). Simple!
    This next rule is probably the most difficult to understand. It’s called the “Chain Rule.” What if we had the equation (x^3)^2? Notice that it’s a power raised to a power. We might be tempted to simply say that (x^3)^2 equals x^6 (algebra) and use the Power Rule to get 6x^5, but we’re not going to do that. Instead, we’re going to use the Chain Rule. The Chain Rule says that when you have an equation nested inside another equation like those weird Russian dolls, you evaluate them as separate equations and multiply them all together. So if we were to set x^3 equal to some other variable, let’s say y, then the equation would look like y^2. Here you have two equations, the y^2 equation, and the equation the y represents. So first you take the derivative of each equation separately, and then you multiply them all together. The derivative of y^2 is 2y, that’s just the Power Rule. The derivative of y is y’, and that’s equal to 3x^2. So the answer is (2y)(3x^2). But remember that y=x^3, so our final answer is (2x^3)(3x^2), which simplifies to 6x^5. This was the answer we got before, and it shows that math does, indeed, work.
    Let’s look at another example. Try finding the derivative of the equation (x^4+x^2+2)^3. Well, the first thing to do is substitute y for x^4+x^2+2. The derivative of y^3 is 3y^2, and the derivative of y is y’ is 4x^3 +2x. (The derivative of a constant is always zero.) Multiplying those together, we get (3y^2)(4x^3 +2x). Plugging in x^4+x^2+2 for y gives us 3(x^4+x^2+2)^2(4x^3 +2x). It may take some practice, but soon you’ll be doing these almost instinctively.
    The next two rules are rather simple, and deal with what happens if you take the derivative of two functions that are being multiplied or divided together. The first is called the “Product Rule.” Let’s say we have two functions, y and z. Y is equal to 3x^2+2, and z is equal to x^3. What’s the derivative of yz? Well, the Product Rule says that the derivative is equal to y’z+z’y. So, it’s the derivative of the first times the second, plus the derivative of the second times the first. We can calculate y’ as 6x, and z’ as 3x^2. Therefore, the derivative of yz is equal to (6x)(x^3)+(3x^2)(3x^2+2). That’s not too complicated.
    What happens if we want to calculate the derivative of y/z? Then we use the “Quotient Rule.” The Quotient Rule says that the derivative of y/z is equal to (y’z-z’y)/z^2. Because it involves subtraction, the order is really important on this one. Many a math test has been failed because of mixing up the order. Probably. Honestly, I rarely use this one, because it’s often easier to rewrite y/z as y(z^-1) and use the Product Rule. Still, it’s helpful to know. If we use the values from our last example, the derivative of y/z is [(6x)(x^3)-(3x^2)(3x^2+2)]/(x^3)^2. That’s a lot of math. I’ll give you a moment to read through it. Now perhaps you can see why nobody likes the Quotient Rule.
    You’ll notice that all of these equations are in general forms. That is, taking the derivative of a general equation produces another general equation. The equation y’ is the value of the slope, not at one value of x, but for all values of x. Pretty cool, right? If you wanted to find the slope at a specific point, just plug in the value of that point for x.
    But this raises another question. If the derivative of an equation is another equation, is it possible to take the derivative of a derivative? In fact it is. This is called the second derivative, and is denoted by two apostrophes: y”. Or like this: (d^2)y/dx^2. Or with a y with two dots above it. Second derivatives have a physical meaning too, but it’s much more complicated. Although, you may notice that acceleration is the second derivative of position. So there’s that. You can also get third derivatives too, and fourth derivatives, and so on, sometimes to infinity.
    And that’s it! Well, almost. There are other functions besides powers of x. Functions like exponentials, and logarithms, and trig functions, and other wonky stuff, and all of them have their own derivative rules. If you want me to go through them, I can, but you could also just look them up in the back of your calculus textbook.
    Derivatives are complicated little things. The amount of math in this comment will probably be a little overwhelming, but I’m sure you’ll get used to it soon. In addition to limits and derivatives, I’d recommend learning more about higher order polynomials, especially quadratics (if you haven’t already!) which take the form y=ax^2+bx+c. From there, you might try learning a bit about trig. You’re going to need it in a few months.
    BTW, all of your math was right, and your understanding of the concepts of velocity and acceleration was very good. You’ll make a great physics student!
    Lastly, if there’s anything you’d like clarified, if you’d like something covered in more depth, or you want me to go over anything else, don’t be afraid to ask!

    • June 19, 2012 at 2:14 pm —

      You, Sir/Madam, have won the internet for today. I’m (obviously) not in the same position as Mindy, but that was enormously helpful

    • June 20, 2012 at 12:42 am —

      Seconded. That is absolutely amazing.

    • June 24, 2012 at 10:46 am —

      Oh. Em. Gee. GravitysWings, you win the Internet. I will be referencing this comment over and over as I get deeper in to this class. Thank you!

  3. June 20, 2014 at 10:20 am —

    ” I got the answer write and b. ” write = right

  4. June 20, 2014 at 10:47 am —

    “To find Δx, we just subtract xf from x0. We do the same with Δt. We subtract the final time (tf) from the initial time (t0). Like this:v = xf – x0 / tf – t0 = Δx/Δt”
    In your equation you I believe you are subtracting initial from final.

  5. June 20, 2014 at 10:49 am —

    “Now that we have average acceleration down, let’s figure out average acceleration.” Now that we have average velocity down, let’s figure out average acceleration.

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