posted
Earlier this year I was discussing cooking with my mother and it came to my attention that she doesn't really understand the concepts of mass and volume (certainly not well enough to understand density). It rocks my world every time I think about the number of people who don't know what a derivative is, and discovering that my own mother lacks even the most basic understanding of concepts that are as intrinsic to me as basic arithmetic was pretty earth-shattering.
The derivative is a significant part of how I look at the world. I'm aware that for every person there are core concepts underlying the way they view the world, and that I will never understand or even know about most of those concepts. I'm just trying alleviate the slightest little bit of that by sharing one of my own fundamental concepts, and a very simple one at that. Contrary to what most people think about calculus in general, derivatives are very easy to understand and grasp intuitively.
Math nerds, stay out -- I'm going to use layman's terms and am going to gloss over a number of things that are more or less useless to normal people.
Hobbes, this means you.
So what exactly is a derivative? Well, if you have some quantity, V (say, the amount of water in a pot; V is for 'volume') and that quantity is changing (you are pouring water into the pot), then the derivative of V with respect to time, dV/dt, is the rate of change of V. In other words, it's how fast you're adding water to the pot.
Easy, neh? Let's go over it again, this time with some numbers.
Your pot contains one litre (L) of water initially. So at t = 0, V = 1 L.
You start pouring water into the pot at a rate of 60 millilitres (mL) per minute. That means dV/dt = 60 mL/min.
When you know both the initial quantity and the rate of change, you can work out how much water you have in the pot at any time. Things start to get trickier when the rate of change is not constant. Then you need to start thinking about the second derivative, or the rate of change of the rate of change. To illustrate, let's take an example we're all familiar with: driving.
Position, velocity, acceleration, and the lesser known quantity 'jerk' are all related by derivatives:
Velocity is the rate of change of position. (How fast you're moving.) Acceleration is the rate of change of velocity. (How fast you're speeding up or slowing down.) Jerk is the rate of change of acceleration. (If you stop very quickly, this is what gives you whiplash.)
Let's assign some variables.
Position will be 'x.' Velocity will be 'v.' Acceleration will be 'a.' Jerk will be 'j.'
Therefore:
v = dx/dt
"Velocity is equal to the derivative of position with respect to time," or "Velocity is equal to the rate of change of position." Right?
a = dv/dt
If your car goes from 0-60 mph in 6 seconds, its average acceleration is 10 mph/s. Your velocity changes by ten miles per hour every second. (I say average because your car, as any driver reading this knows, does not speed up at a constant rate. It starts off slowly because it has to overcome inertia.)
j = da/dt
When acceleration isn't constant, you feel a "jerk." Like right when your car comes to a halt, because acceleration in that small time frame is very much not constant.
So if you're cruising along at a constant 60 mph:
v = dx/dt = 60 miles/hour a = dv/dt = 0 j = da/dt = 0
And you can tell this by how you feel in the car. You're just cruising along, you aren't being pressed back into your chair or forward into your seatbelt. When velocity is constant you don't feel any acceleration or jerk.
Let's say you see a red light up ahead, so you start to slow down. Your velocity immediately starts to decrease rapidly, but as you get closer and closer to zero the rate of change decreases. Over the course of your stop, the negative acceleration (deceleration) is not constant: you slow down faster early on and you slow down more slowly later. So jerk comes into play.
Am I making sense? I hope so, because unless you want to start doing math, that's basically it. Posts: 10886 | Registered: Feb 2000
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posted
Although I took two years of physics in high school, we never got to derivatives, although we did a lot of very similar things. I get what they are, though.
I'm interested. How are derivatives always on the top of your mind? Your example is with cooking, but to me as long as the water gets in the pot its transitory state when it's half not in the pot is irrelevent. I assuming this is a bad example of when it would be used, but when does it change your perception?
Posts: 8473 | Registered: Apr 2003
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Teshi, all of those physics formulas you had to learn in high school can be derived in about two seconds with basic calculus. That's why it seems familiar to you.
Derivatives are so intrinsic to me that I don't always think about them consciously. Everything is changing. All the time. There are derivatives (rates of change) everywhere. Most of the time it's velocity (walking, running, driving), but even mundane things like reading (the rate at which I read books fluctuates quite dramatically depending on what else is going on in my life) are things that I think of in terms of derivatives.
Posts: 10886 | Registered: Feb 2000
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posted
See, the problem is that in high school, they only ever taught is what we were doing, not why. We were given rules and numbers and letters and no context as to why this was useful or what it had to do with "real life". I have a feeling more people would be less intimidated if they took the time to explain it to us.
Posts: 2849 | Registered: Feb 2002
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posted
Now explain what happens if the amount of water in the vessel you are pouring from affects how fast water pours out of it, affecting the rate of change for V.
posted
I thought you were going to talk about financial derivatives and I got all excited (being the huge nerd accounting major that I am). Posts: 194 | Registered: Feb 2005
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posted
Actually, I think I could derive this in an idealized case (regularly shaped vessel, water flowing out of a hole in the bottom, not the side). But not with the actual case of pouring from a pitcher or bucket.
quote:Now explain what happens if the amount of water in the vessel you are pouring from affects how fast water pours out of it, affecting the rate of change for V.
That happens with storage tanks in my industry, since they are filled from the top and drained from the bottom. When you're draining a tank, the flowrate from the tank is [edit: related, not proportional] to the amount of fluid remaining in the tank. That's because the fluid has mass, so it is affected by gravity. More fluid in the tank = more weight pressing down = faster flow out the bottom.
Engineers will talk in terms of head, which just means fluid height, really.
[Edit: it's the pressure that's directly proportional to fluid height. I didn't mean to imply that flowrate was directly proportional. It's actually proportional to the square root of the fluid height.]
posted
Still, that was an insightful and well written explanation. Despite the fact that I took calculus and can still calculate the derivative of something, I'd forgotten (if I has ever indeed been taught it) the practical application of such.
Posts: 194 | Registered: Feb 2005
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I understand derivatives, but do not think in terms of them in my daily life. Way too engineery for me, I'm a free spirit. Posts: 7954 | Registered: Mar 2004
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posted
This topic made me remember something I read in one of my e-mails: "Alcohol & calculus don't mix. Never drink & derive." Posts: 18 | Registered: Mar 2005
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posted
That is one of the clearest non-mathematical explanations of derivatives I recall seeing.
The only question is whether it will become an extra credit handout for my chemistry class (we're in the middle of the gas laws) or my physics class (we haven't done anything more than tangentially related recently -- we're talking about light) . . . *ponders*
Don't worry, my understanding of Taylor series is not as intrinsic as my understanding of derivatives. I wouldn't really do that to you.
Posts: 10886 | Registered: Feb 2000
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posted
I gather BtL doesn't much care for Taylor series? *googles*
>_< Oh, those things! <vomit smilie>
quote: You're going to make your kids write essays explaining their understanding of derivatives?
Nope. Extra credit reading assignments in my classes get points by answering fill-in-the-blank type questions for extra credit on the next test. I try to make the reading interesting, and related (at least somewhat) to stuff we are doing (or have done). This will be my second such nabbed-from-Hatrack (not counting articles that people have linked to).
posted
When I was in university we used Taylor series approximations all the time. And of course they're how computers "calculate" stuff like trig functions.
But, being engineers, we almost always truncate after a couple of terms. It's close enough, and makes the calculations nice and easy.
BtL, your post there f*#^ing cracked me up.
quote:Nope. Extra credit reading assignments in my classes get points by answering fill-in-the-blank type questions for extra credit on the next test. I try to make the reading interesting, and related (at least somewhat) to stuff we are doing (or have done). This will be my second such nabbed-from-Hatrack (not counting articles that people have linked to).
At least it's the second one...
Posts: 10886 | Registered: Feb 2000
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posted
Maybe I'm just tired, but I had to think a lot more than I would have expected to sort of understand what you were trying to say. And I did actually get as far as calculus, back in my college days.
Posts: 5771 | Registered: Nov 2000
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Rats. I tried to make it as accessible as possible. Maybe with some tweaking I could make it more readable...
Posts: 10886 | Registered: Feb 2000
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quote:Rats. I tried to make it as accessible as possible. Maybe with some tweaking I could make it more readable...
I am sure you could.
You started losing me when you got into variable rates of change. But, like I said, it might just be cause I'm tired. Posts: 5771 | Registered: Nov 2000
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posted
Twinky, are derivatives also integral to your existence? Do they provide a sense of continuity in your life? Are they truly the summation of all that is important?
Okay, I'd better stop with this series of calculus puns before I reach my limit. Sorry for taking this thread off on a tangent.
That was too much for me at this ungodly hour of the morning.
I should have a fair bit of spare time at work today, so I'll see if I can come up with a better way to explain variable rates of change. Posts: 10886 | Registered: Feb 2000
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Let's say you've got a regularly-shaped tank of some sort that drains from the bottom. Like, for instance, a bathtub. You fill the tub and take a bath, then get out of the tub. It's full, so you start to drain it. Initially the tub drains quickly, but when the tub is nearly empty the last of the water drains more slowly. This happens because of the mass of the water in the tub. At first, the water at the bottom of the tub has the mass of all of the water above it pressing down, but as the tub drains, the water at the bottom has less water above it pressing down, so it flows into the drainpipe more slowly.
The water might drain at 20 L/min initially, but by the end it might only be draining at a rate of 5 L/min. In this case the rate of change of volume with respect to time, dV/dt, is affected by volume itself. For example:
dV/dt = 0.1 * V
In this completely arbitrary case, dV/dt changes as the volume changes. dV/dt is directly proportional to the volume in the tub, meaning that as the volume in the tub decreases the rate of change in volume will also decrease (i.e. the drainage from the tub slows down). This case is a linear relationship, which is not how it would be in the real world, but it'll do for our purposes.
Say that before you start draining, the full tub contains 100 L of water. So right before you pull the plug, at t = 0, the tub contains 100 L of water. It will therefore start draining at a rate of:
dV/dt = 0.1 * (100) = 10 L/min
(For Hobbes' sake, I'll note that the constant, 0.1, has units of inverse minutes.)
But the instant the volume changes, dV/dt will also change. Let's say that shortly after you pull the plug, 82 L of water are left in the tub. Then the drainage rate is:
dV/dt = 0.1 * (82) = 8.2 L/min
So the drainage rate is lower now than it was when you first started draining. By the time you get down to 10 L left in the bottom of the tub, you'll only be draining at a rate of 1 L/min!
I hope that makes a bit more sense.
(Hobbes, if you're feeling adventurous, you're welcome to calculate how long it takes to drain the tub. )
Posts: 10886 | Registered: Feb 2000
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posted
S'ok, no worries. dV/dt = 0.1 * V translates to:
"The rate of change of volume with respect to time is equal to a constant (0.1) times the current volume."
Volume is a function of time, so at t = 1 minute you have a certain volume of water in the tub; the drainage rate at t = 1 minute is equal to a tenth of that volume. The drainage rate at t = 2 minutes or t = 5 minutes is different, because the volume of water in the tub is different at those times.
posted
So it's just telling me that the drainage rate is changing with time, is that right?
Which is what you said in words in your earlier post.
If the constant is 0.1, that means the volume is changing by a 10th at one minute (what you said). So at 2 minutes, it would be a 10th of whatever it was at one minute, or something totally different, because the volume is changing.
Side comment: I hate showing off my math ignorance in public. Posts: 5771 | Registered: Nov 2000
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posted
The thing that may be tripping you up in my bathtub example is that I don't offer a formula for calculating the volume in the tub at any given time. Unless you want to do some calculus (and I don't think you do ) there is no way for you to determine what the volume of water in the tub is at an arbitrary time.
Edit:
quote: I think it would be more fun to calculate the diameter of the pipe.
That'd be up my alley (not to mention choosing the type and gauge of piping most suited for the water circulation system associated with the tub), but it doesn't really work for a differential equation as arbitrary as the one I chose to use in my example.
posted
To put in some specifics, by that formula if the tub has 10 gallons of water in it, it is draining by 1 gallon per unit of time.
And if it has 330 galloons of water in it, it is draining by 33 gallons per unit of time.
Now, what'll really start to get you is, when its at 9 gallons, its only draining by .9 gallons per unit of time. And at 9.9 it was draining at .99 per unit of time. Et cetera.
Posts: 15770 | Registered: Dec 2001
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And besides, my gallons are a completely arbitrary unit used to demonstrate an abstract concept. They're surreal units Posts: 15770 | Registered: Dec 2001
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Let's go over it again, this time with some numbers.
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