How Many Of You Had To Teach Yourself The Math?

I think the one piece of advice I could give anyone is 'don't be afraid of maths & don't give up on it'. I failed maths at High School miserably 'cos I didn't get on with the lecturers but degrees in electronics, maths & post grad nuclear physics later it still fascinates me. For the electronics & physics maths was a tool that achieved an end but it was only when my curiosity got the better of me and I took a maths degree that I found out the 'why' of it all !! Some of the most interesting maths only needs a pencil, paper & dram of good malt to get the brain cells really shook up ........ well that's what I think, for what it's worth !!!

If you can find books written by and for the armed services they are often the best, because if they can teach math to soldiers they can teach it to anyone. They probably won't go very far, however.

As someone else mentioned, please do feel free to come back here and ask specific questions (or, general questions), whenever you need a boost.

Good luck!

- Tom Gootee

Yes, I can only imagine how the need for solving ordinary differential equations play in the battlefield or relate to commitments for honor and truth. ;)

Jon

Ever tried to do gun firing calculations? That's what got a lot of computing started.

Wasn't Eniac built specifically for that purpose?

You are overlooking the fact that soldiers in tech school have a very strong motivation to learn. If they flunk out, they are transferred to the infantry as foot soldiers, armed only with an M-16 or similar weapon.

I believe so. Also, Zuse's computers were built to solve warplane design problems. Even Babbage's efforts were to produce books for the Admiralty.

There were also a lot of mechanical computers used in bombsights, torpedo firing calculations and the like.

Yes, but it's all very confusing. A real world example would be helpful.

Is there a real world example of using 2X, (the derivative of X^2) to obtain some useful information from the function y=x^2 ?

-Bill

Not exactly, but try to figure out the maximum size of package that can be mailed, considering that the maximum size is 108 inches in combined length and girth (distance around the thickest part). Note that I did not say the shape has to be rectangular or cylindrical.

To expand on the above comment....

When a curve rises and then falls, or falls and then rises, it must pass through a point where the slope is zero. This fact is one of those things taken advantage of, using derivatives. By no means is this the only use. Just one of many.

An example is finding the case where you minimize the amount of sheet metal used to make a can for soup or tuna. If you know the volume you need to enclose, that is.

A can is a right circular cylinder and has two ends (top lid and bottom lid) and body cylinder itself that must be made out of the sheet metal. Let's call the height H, the cylinder's diameter D, the volume V, and the total area of sheet metal required A. For these purposes, lets ignore the bit of metal used to seal the edges or form flaps or lips and just focus on the simplified surface area and make that what we want to minimize (for cost reasons, let's say.)

The area is:

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

and after a little algebraic manipulation,

A = PI * D * [ H + D/2 ]

Hopefully, Bill, you can get this far without too much difficulty. At this point, we have two variables, H and D, and there are an infinite number of possible combinations that could be used to achieve any particular A. Worse than that, we aren't even given the area, but instead given some volume. So let's look at the equation for the volume:

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

Well, that's not too bad. Except it also has those two variables, H and D, and for any given volume we can still have an infinite number of possible H's and D's to achieve any particular V.

In thinking just a moment about this before going further, you should be able to think about and realize that there are two incredible extreme cases. One where the area of the lids is about zero and the height must be very, very high in order to achieve some volume V. The other where the height is about zero and the area of the lids must be very, very large to achieve some volume V. In this two extremes, it should be fairly easy to see that a lot of sheet metal would be needed and that it would be a lot more than the optimum. So you should be able to guess that the amount of metal goes from a very large number when the height is near zero down to some smaller value where the height is reasonable and then back up to a very large amount as the height nears infinity. We know the answer must be in-between.

Imagine plotting such a curve of area versus, for example, just the height of the can. It would go from infinity at H=0, down to some minimum value at some H we don't yet know, then back up to infinity when H=infinity. All this with the volume V remaining constant, of course. Remember that when a curve rises and then falls, or falls and then rises, it must pass through a point where the slope is zero. This must apply here, just by thinking about it.

So what we want to do is find the point where the area is a minimum. That says that the A= equation mentioned above needs to be minimized and suggests that to do so we just need to find where the slope is zero.

But we are still stuck with those two pesky variables, H and D. So let's get rid of one, right now.

To do that, take the volume equation and solve for H, instead (or you could do that for D as that's another path that will get you to the same place.) It is:

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

Now we can substitute that into the A= equation, thus:

A = PI * D * [ (4*V) / (PI*D^2) + D/2 ]

Again, some re-adjusting with typical algebra skills yields:

A = 4*V/D + PI*D^2/2

Now, there is just the one variable, D. At this point we can use calculus rules to produce the derivative:

dA = (-4*V/D^2 + PI*D) dD, or dA/dD = (-4*V/D^2 + PI*D)

This represents the slope. Just set it to zero and solve or inspect it.

There it is. For any volume, this will tell you what the diameter of the lids should be.

So let's look at the height. If D is the value above, what would the height then be. Well, we can plug in the D= equation we just generated back into the volume equation:

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

Solving this for H gives us:

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

Note that this H value is the same as the D value we also worked out. In other words, the optimum is found when both D and H are the same. A kind of "square can" so to speak. At least in profile.

This could have been solved differently, as well. We could have set up a ratio, H/D, as a new variable and then worked out the equations in terms of that H/D variable. In the end, we would have also set that derivative equation to zero and solved and, I predict, would have found a value of 1 as the answer.

In fact, this is the differential equation in that case (I'm skipping all the drudgery here.) I'll use X = H/D:

dA = PI*CUBEROOT([4*V/(PI*X)]^2)*(X-1)/(3*X) dX

Setting this to zero, is:

Upon inspection, you can see that this is zero if X = 1 simply because the term (X-1) there will be zero forcing the rest to go to a zero value. Also, if V=0, then the CUBEROOT will go to zero and the equation will be zero then, too. But I suppose that is kind of an obvious possibility. But the main point is that you can see that (X-1) term in there and that tells us that when X=1 the area equation will be at a minimum. Confirmation of what we discovered otherwise.

Similar reasoning using partial differentials and a basic description of an error term allows one to develop the standard "least squares fit" method used to fit a line to a set of data points taken in the face of measurement error. There are many such applications using this idea. And this is only one of many ideas using the derivative, each of them with many interesting applications.

So it is important.

Jon

Well, as others have noted, a lot of the money behind early computing was for guns :)

A shell, when fired, describes a parabolic trajectory (ignoring drag and other side effects for now), so what angle must the gun barrel be at to fire a shell a particular distance?

The starting point is to set dy/dx = 0 at x = distance / 2. This will yield Y (the maximum altitude of the shell) and may now be used to solve for dy/dx where x = 0. I remember a question like that from my GCE O level mathematics paper (not GCSE :)

The actual calculation is complicated by having to calculate the velocity of the shell of course, which is also a derivative (it will undergo negative acceleration).

A more common one for engineering is designing a simple (very simple) directional antenna, such as found in simple radar.

A parabolic antenna has a focal point at Y= 0.5 (normalised) and has a shape described by y=x^2. The antenna feed should be placed at the focal point. So if you're trying to figure out the dimensions of a parabolic antenna, the function d(x^2)/dx comes in very handy :)

Cheers

PeteS

By the time you get it to the battlefield, it's been reduced to slide rules with odd geometries. I guess I'm not sure if they use electronic computers now, but they still train the slide rules.

What do you want to know about y=x^2? There's nothing real-world useful about mathematics until it's applied to something-- until it's used to model some system that you care about. Maybe you have a contour map and want to know the slope of a hill, and therefore design parameters of a machine that brings stuff up and down it. The contour might not follow y=x^2, but if you know that m=dy/dx you can work it out for whatever the contour might be.

A spring has a force of F=-kx, where x is the displacement from equilibrium, k is the spring constant, and the minus sign means the force is in the opposite direction of the displacement. That's Hooke's law. When you stretch it, the spring stores an energy of E=(1/2)kx^2. That may or may not be useful, depending on your interest in the force and energy of a stretched spring. But because the energy stored is linear in spring constant but quadratic in displacement, you can show that stretchy bungees can absorb more energy, e.g. from a falling rock climber, than less stretchy ropes that have a higher tensile strength.

In general, force and potential energy are related by F=-dE/dx. Or F=-grad E in three dimensions.

And for position x, velocity v, and acceleration a, we have v=dx/dt and a=dv/dt. For a falling rock climber, if acceleration is constant, a=-g, then velocity is v=-gt and position is x=-1/2 gt^2 (assuming all initial values are zero). Then you can ask, given that the rock climber fell a distance h, how much energy must the bungee absorb to hold him?

The batteries don't fall out of the slide rules!

Slide rules are resistant to electronic warfare and nuclear EMP, too. That might be less of a problem now that the paradigm has shifted from World War III to brushfires and terrorists, but it never hurts to know the basics.

Jon,

As others have said, here: The development of weapon systems has been a very large factor in motivating many important advances in technology and mathematics, throughout history.

And, actually, Jon, the U.S.'s armed forces are so vast that there are many among them who are trained far beyond differential equations. At the least, they do have people who must try to keep up with the defense contractors' engineers, after all. And some of them are quite good. (And they apparently also have LOTS of people just to write technical manuals.) The military's actual combatants probably only constitute a one- or two-digit percentage of all military personnel.

There are also quite a few good U.S. Military manuals about electronics, by the way.

Just FYI.

- Tom Gootee

The last figures were that one out of seven soldiers ever see combat. The rest are in all other aspects of supplying their needs and doing their best to keep them alive when they do have to go into battle.

That reminds me of my high school Physics teacher, Victor Nixon (now deceased). He was one of only two graduates in Physics, at Indiana University, the year he got his degree (i.e. it was a very LONG time ago). He once told us of his time in the U.S. Navy. He was a human computer for the gun-firing calculations. As he told it, he would sit, down inside a ship, somewhere, waiting for someone to yell target measurements down to him, through a metal tube. His job was to mentally do the calculations and yell the gun-aiming data back up through the tube, "immediately". Among other things, he mentioned that he had been taught, by the military, a method for doing long division of large numbers, very quickly, in his head. At the time, I was too uninterested to listen to his explanation of the method, unfortunately.

I, myself, had my very first exposure to Calculus when I was attempting to come up with a method for being able to lob projectiles (i.e. rocks, apples, et al) at certain neighborhood kids, very accurately, with a homemade catapult. The librarian was quite surprised when I, age 8, told her that I needed to find out how to calculate parabolic trajectories. But she dutifully found a textbook that covered the subject, and presented it to me. Unfortunately, it wasn't until I was a freshman electrical engineering student, about ten years later, that I finally learned exactly what I had wanted to know.

Actually, ignoring friction and the effects of the earth's curvature and rotation, projectile kinematics and dynamics are almost trivial, once you realize that the vertical and horizontal components are independent of each other. i.e. SEPARATE the vertical (to gravity) and horizontal components of the muzzle-velocity vector (by mutiplying speed by sine and cosine) and write two separate (one-dimensional) equations. Then solve them simultaneously. One possibly-clarifying example (still assuming the simplifications noted above): If you fire a rifle horizontally, its bullet will strike the ground at exactly the same time as a bullet that you simply drop from the same height. But, how far away will the rifle's bullet be, when it strikes the ground? For any firing angle, you can express the time-of-flight using the vertical components of the initial velocity and position, with the familiar second-order equation for a mass in a single gravitational field. And the horizontal component of the initial velocity goes into the basic first-order distance=rate x time equation. And Voila.

Now try to do it while your gun and your (intelligent) target are both moving and accelerating in three dimensions. Your gun is aboard an aircraft whose sub-structures are also bending and twisting and heaving, and which might be vibrating so much that it's screaming like a banshee, and you have almost-supersonic airflow effects to worry about, and the gun is a 6000-round-per-minute cannon that, when fired, might cause the whole airframe to pitch by tens of degrees (for all of which, you do have sensor data being fairly-rapidly sampled). The system you design will fly the aircraft automatically, and fire the gun.

(That reminded me of a tidbit that surprised me, when I first heard it: Once in a while, in the old days, often many minutes later, they'd be shot down when they flew into their own bullets.)

Of course, if the other guy is in a position to shoot at YOU, it might help if you've already had a peek at one of their fire-control computers and found out that it can't solve a certain type of equation, and have programmed your flight-control computers to fly your aircraft in a path that's decribed by that particular type of equation, hehehehehe....

See? Mathematics is fun!

- Tom Gootee