How Many Of You Had To Teach Yourself The Math?

I was being flip. Probably shouldn't have been. But perhaps it gives an excellent foil upon which to write some great thoughts about this.

Some of the best brains, in fact. In case you weren't aware, I've had my high level clearances and my neighbors interviewed in depth about me. So I have at least worked with some of the folks you refer to.

Bingo. One of my very first exposures to electronics came from thick, military manuals on electronics that had been given to me when I was a child -- ones circa 1944 or so. Still have them in a box somewhere.

Not that I was entirely ready to understand them at the time, but I worked at it.

Jon

Thanks for clarifying that! Having read and enjoyed many of your posts in the past, I thought I might suddenly have to start wondering... . I'm relieved to know it's not so.

No, I wasn't aware of that. Very good. And ditto.

Coincidentally, when I was a child, I was given the 1944 ARRL Handbook, by a favorite engineer uncle of mine, who worked on aerospace projects for his whole career. I still have it. Interestingly, it has a slip of paper in it that is his receipt, from his employer, for a patent for the first afterburner, for jet engines. I remember him explaining to me how they were able to keep a ring of individual flames lit, inside a chamber with supersonic airflow.

Thanks again, Jon.

- Tom Gootee

"He who lives in a glass house should not invite he who is without sin."

A bumper sticker I saw: "YOUR BODY IS A TEMPLE. (Mine's an amusement park.)"

Jon,

Thanks for the explanation of sizing a can for maximum volume and minimum area. It was very helpful. I follow most of it, but forget the rules for derivatives. It was a good example of using derivatives to find some optimum value.

I checked a tuna can on my shelf that measures 3.25 diameter by 1.75 inch high, which is not optimum and apparently wastes about 1.6 sq. inch of area. But tuna cans are probably designed for convenience of use and not optimum volume.

Thanks again for the math lesson.

-Bill

The rules take a little getting used to. Most important is that you understand WHY. Then memorize some of the basics. There are two "spaces" to consider -- one is your mental space which allows you to think and understand the where's and whyfor's but where it also isn't strictly necessary for you to know some specific detail that you can go look up somewhere; and the other is your result space where you must be quantitative. A little bit of both is needed -- a good feel for the meanings; and a few memorized details to get you by, as well.

I find that derivatives are rather easy to deal with, with few exceptions that I need to look up these days. On the other hand, the integrals of some things are hard enough to solve by hand and varied enough in their approaches that you often accept a facility with some and just keep a book nearby tabulating many other solutions for you. You then learn to find the ones you need for some problem. Not that you might not be able, on your own and with enough time and creativity, to solve them. But some of these really take a LOT of creativity and facility to figure out and it's handy to tap into someone else's skills, there.

It will take practice and work to gain an intuition. I don't know of any way to short-cut that, though some folks I suppose just tumble naturally to all this like they do to breathing. Not me, anyway. Took years of banging my head against the wall.

By the way, one helpful way to look at these "dx" and "dy" things is to see them as a special kind of variable added to standard algebra. In algebra, your variables can hold any finite value at all. In calculus, these new variables can't hold finite values but only infinitesimal values. Note that there is an infinity of these infinitesimal values, too. Some bigger than others, just like the case with finite values. So you have two kinds of variables in calculus -- finite-holding ones and infinitesimal-holding ones. They can be mixed and multiplied and cancelled and so on.

So when you see dy/dx, just think that this is two variables that only hold really tiny values -- smaller in magnitude than any possible finite value, yet not zero. And that this is a ratio of the two. It may be the case (often is) that the ratio of two infinitesimal values has a finite ratio, too. For example, that dy/dx = 2. That just means that dy is twice the size of dx. That's all. You could also look at it as dy = 2*dx.

Similarly, when you see the equation:

D = S * t

to describe the case showing the relationship of speed, time, and distance, you need to realize that in algebra these variables are finite in size. You normally don't even think about it and may not even have this fact staring you in the face, but you use specific values and plug them in and calculate results. However, these are average values you use. Your car was going an _average_ of 40 MPH, right? The time over which this average was true was, say, 1/2 hour. So the total distance traveled was 20 miles.

But in calculus, you might want exact and precise, moment by moment, knowledge. One more precise way of stating the able is to write:

dD = S(t) * dt

Here, S(t) may not be constant and an average may only approximate, but not be exact. For example, a falling rock (without an atmosphere to interfere) doesn't have any one speed. The speed is constantly changing as it falls. If you pick any particular speed, how long is that truely the exact speed? For an infinitely small time! So you represent this tiny time with dt. How far does the object fall in this infinitely small time? Well, an infinitely small distance, that's how far. So you represent that with dD. So it is the same equation except you don't use averages, you use exact time-dependent values. But it is the same equation as before, except that you have replaced two finite variables with two infinitesimal values, instead.

Of course, it's not practical. You can't really build anything infinitely small and you can't measure infinitely small distances or time bits. So to be practical you need to add a bunch of them together so that they make a finite value, again. How many infinitely small values does that take? An infinite number of them. That's how many. So that is why that integral symbol -- it just means to add up an infinite number of things so that you can get back your finite values.

Taking this falling rock example, let's say it starts at S=0 at time t=0. [S(t=0) = 0, in other words.] What is its speed at some future time? Well, according to the rule for acceleration of gravity, it is S(t)=g*t. Just acceleration times time. So the equation becomes:

dD = g * t * dt

To make anything of this you need to do those infinite sums, and what you do to one side you need to do to the other, so:

Integral( dD ) = Integral( g*t*dt )

On the left side, what happens when you add up an infinite number of infinitely small pieces of D? Well, you get D back, of course. If you divide up D into an infinite number of tiny pieces and then add them all back up again, you just get what you started with. So the equation trivially becomes:

D = Integral( g*t*dt )

Now, as you already know, if you look at this:

Y = ( 5*2 + 5*4 + 5*6 + 5*8 )

You can easily see that we can extract out the 5 and make:

Y = 5 * ( 2 + 4 + 6 + 8 )

Similarly, the same happens with infinite sums, too. If there is a constant multiplier, you can extract it out. Since 'g' is a constant (near Earth's surface, anyway) we can rewrite:

D = g * Integral( t*dt )

Better, eh? Okay, now we have a problem. What is this weird product in there and how do we handle it?

Well, one way to "see" this is that it is a product of two measures, which make up an area. An area? Yes, an area. The rectangle is 't' high, let's say, and 'dt' wide. Okay, so it is pretty narrow indeed. I mean, dt can only hold infinitely small values, right? So it must be pretty narrow. About as narrow as a line, yes?? But an area, just the same.

So what does the first rectangle look like? Well, at t=0, it is 0 high and dt wide. Just next to that one, we place the next rectangle we are adding up. It starts at t=dt and is dt high and dt wide. The next one starts at t=2*dt and is 2*dt high and also dt wide. And so on. If you think about this closely, you will see that as we progress from t=0 towards t = some t we want to get to, that the height continues to climb up until at the end of this we have a height of 't' for the last narrow area. This shape is a right triangle that is 't' wide at the base and 't' high at the end of it. The area of this triangle is obviously (1/2)*t*t or t^2/2. That's the sum. Done.

So the answer is:

D = 1/2 * g * t^2

I'm sure you've seen that somewhere.

But keep in mind you can cancel these new variables like before, move them around, play with them, etc.

In electronics, there is this:

Q = C * V

Q being Coulombs, C in Farads, and V in volts. The charge on the capacitor is simply its capacitance times the voltage. However, a more precise expression is:

dQ = C * dV

Note that all I did was to replace V with dV and was then forced to swap dQ for Q. But let's not stop there. We can divide both sides by a new variable, dt. So that,

dQ/dt = C * dV/dt

But Coulombs divided by time (infinitesimal or finite) is current. So this is just:

I = C * dV / dt

Found in most any electronics book.

You can substitute finite differences for the infinitesimals, so that if you know that there will be a change of 2 volts over 1 millisecond, you can replace dV/dt with the average value of 2000 volts per second and use that to compute the current.

But it's about as facile as that. Just playing with infinitesimal variables, I mean. The choice of dividing by 'dt' above was a practical one. I suppose I could have divided both sides by some other infinitesimal, for example using dX to mean some displacement. But in most common cases, that would not have been very useful. Still, it's up to your imagination and there may be cases where that is important.

Something else before leaving that capacitor equation:

dQ = C * dV

Capacitors usually don't have a fixed capacitance -- as they often change their capacitance against the voltage applied on them. So the function might really look like:

dQ = C(V) * dV

In other words, with a C that varies with voltage. While this equation:

Q = C * V

would be very often inaccurate in such a case, this equation:

dQ = C(V) * dV

is perfectly accurate, throughout.

Calculus lets you express more precision over varying situations.

And being of a recognizable style, which helps people find them quickly.

No problem.

Jon

Bill,

I was given a similar problem in calculus class way back when. IIRC, the sides and ends each had a different cost (per square inch of material), which changed the dimensions from what they would be if you assumed the cost were the same for both.

Mark

Ah. Now there's the next step. Assume that there is a cost function that yields the actual ratio found in a tuna can as optimal. Solve for the currently unknown cost relationship that could have been specified at the outset to get that particular ratio out as optimal.

Only a slight modification required to get there. Anyone want to solve it here?

Jon

That's more like Operational Research or Management Sconce.

This sounds like one of my 400 level economic classes.

Guess not. So I'll start:

As before:

A = top lid area + bottom lid area + side cylinder area = PI*(D/2)^2 + PI*(D/2)^2 + PI*D*H

However, we modify this, so that:

C = cost of top lid area + cost of bottom lid area + cost of side cylinder area

We can assume that this 'cost' is proportional to the area. And that the cost of the top lid and the cost of the bottom lid (per unit area) is the same; but different from the cost per unit area of the cylinder area. So we get something like this:

C = k1*PI*(D/2)^2 + k1*PI*(D/2)^2 + k2*PI*D*H = 2*k1*PI*(D/2)^2 + k2*PI*D*H = k1*PI*D^2/2 + k2*PI*D*H = PI * D * [ k2*H + k1*D/2 ]

However, we can't really wind up figuring out specific values for k1 and k2, but only some ratio of them because that's all that is required by the problem. So let's set up one other variable, m, to be that ratio:

m = k1/k2, or k1 = k2 * m

Then we have:

C = k2 * PI * D * [ H + m*D/2 ]

The volume stays the same:

V = PI*(D/2)^2*H

Thus,

H = (4*V) / (PI*D^2)

Substituting into the cost equation:

C = k2 * PI * D * [ (4*V) / (PI*D^2) + m*D/2 ] C = k2 * [ 4*V/D + m*PI*D^2/2 ]

And,

dC/dD = k2 * [ -4*V/D^2 + m*PI*D ]

Which must be zero, to find the minimum/maximum:

dC/dD = k2 * [ -4*V/D^2 + m*PI*D ] = 0

So,

m*PI*D = 4*V/D^2 m*PI*D^3 = 4*V D^3 = 4*V / (m*PI)

D = CUBEROOT( 4*V / (m*PI) )

For the height, we get:

V = PI*(CUBEROOT(4*V/(m*PI))/2)^2*H = (PI/4) * CUBEROOT(4*V/(m*PI))^2 * H = (PI/4) * CUBEROOT(16*V^2/(m^2*PI^2)) * H = CUBEROOT(PI^3/4^3 * 16*V^2/(m^2*PI^2)) * H = CUBEROOT(PI/4 * V^2/m^2) * H = CUBEROOT((PI/4)*V^2/m^2) * H

Solving this for H gives us:

H = V / CUBEROOT((PI/4)*V^2/m^2) = CUBEROOT( V^3 / [(PI/4)*V^2/m^2] ) = CUBEROOT( V / [(PI/4)/m^2] ) = CUBEROOT( 4*m^2*V/PI )

We also know from the measured can that it has 3.25" diameter and a

1.75" height (given by Bill Bowden.) So this means: 1.75 = CUBEROOT( 4*m^2*V / PI ) 3.25 = CUBEROOT( 4*V / (m*PI) )

Two simultaneous equations with two unknowns. We know that V isn't all that important here, though. Let's solve the last one for V.

V = 3.25^3 * m * PI / 4

Substituting into the other one:

m = 1.75/3.25

Or, in other words, m = .53846 or 7/13ths. Roughly speaking, the lid material must be about half the cost per unit area as the cylindrical side material if this is optimized and ignoring the cost of the can seals. Not unexpected.

Incidentally, the volume is 14.5176 in^3. Plugging these back into the H and D equations, along with this m, will get you back the right values.

Jon

Phaeton,

Below is the URL for a website, "free-ed.net", which I just now happened to accidentally run across, that has free on-line Mathematics courses and tutorials, etc, covering everything from Arithmetic & Pre-Algebra up through Calculus, Differential Equations, and Linear Algebra. I haven't looked at it, closely, yet. But it might be useful to you.

They also have course-material for some areas of basic electronics (and lots of other subjects, too).

Their Mathematics online-course-index is at:

.

Their "main" online-course-index page, for all subject areas, is at:

.

Good luck! I hope that you will persevere.

Mathematics, especially "higher" Mathematics, is one of THE MAIN "basic necessities" for being able to learn how things actually work (or, at least [and usually more-correctly], for *modelling* how things work).

But mathematics is also "its own version of reality", and can actually be used to "explore" OUR reality. It can be used, for example, "to get from a to b", where "b" is a valid result in the physical world, but only "a" was previously known or understood. And ONLY the sequence of equations that was somehow found/developed, which started with "a" and was able to make its way to "b", which then *pointed-out* that "b" should also be true in reality, was used to discover that "b" really IS also the case, "in the real world". Sometimes "b" might be mundane. And sometimes it might be earth-shaking.

Enjoy!

- Tom Gootee

P.S. I forgot to mention that I also found the Navy's "Electrical Engineering Training Series" (NEETS) books on line, in PDF format, downloadable for free, at:

They are supposed to be very good.

- Tom Gootee

Well, at least I USED to be able to download free, from somewhere on that site, or some site that's linked to it, IIRC. But part of the site is down, at the moment. And I can't tell if the free downloads are still available. Sorry. You could also try a web search for NEETS, since they're probably abailable for download from other sites.

- Tom Gootee

I think they even fixed the link to module 15, which was broken on Sept. 4th.

Jon

Yes. The public paid for all this, but now they have to pay again to read it. That's what happens when the RINOs are in charge.

Mathematicians not infrequently will argue that it has it's own reality -- not that it is a tool, but that it is a real, bonfide place all of its own much like our universe. But not the same.

Regardless, it's most important characteristic to me is that it is highly unified and interwoven. It's concepts aren't just random bits of knowledge tossed together because they are "mathy." But all deriving very strongly from just a few basic assumptions, out of which flow inexorably and inescapably the rest. It's all of one piece and there is no way to accept some parts and exclude others. Each part reinforces and strengthens the whole and each other and the whole artifice is very strong as a result of that.

Science mirrors this in its own approach, too. Interlocking fields of study, reinforcing each other and the whole.

This is entirely different and apart from other human systems of "knowledge," such as astrology, tarot cards, tea leaves, pyromancy, and the like, where if any one of them is shown clearly to be false, it affects none of the others at all.

I liken this to the difference between farm walls made out of loose rock (non-science styles of knowledge building) verses reinforcing curves in archways where every shaped stone shares a part of the load and contributes importantly to the overall strength (science knowledge.) You can drive a tractor through a low farm wall without really impacting the strength of any other part of the farm wall, because none of the parts destroyed really contributed any strength to any other part, anyway, or borrowed strength from other parts. They are loose and largely independent of each other. And as a result, the overall farm wall is generally weak, though perhaps still useful. The archway develops an overall strength upon which an entire civilization can be built. And this strength comes from its unity of structure.

Jon

And then there are soda pop cans, which have a different ratio for H/D. Makes me wonder what is so different about making a soda can that the optimum shape would be proportionately so much taller than a tuna can.

Mark

That's calculus for you. It's so useful, that it gets used in areas outside of pure mathematics.

Mark

Soda cans are highly optimized for drawing. They are thinner than they once were, and ISTM they rely on the gas pressure to hold their shape.

Tuna cans are cut and rolled from pieces. I do note that some steel cans now are also drawn - you can only open the top.

Thank you. I was curious.

Mark