AMS :: Feature Column from the AMS
Reading, calculating and plotting data from tests in Langley's wind tunnels To meet the increasing demand of the war effort and offset the loss of . on Monroe Methods for Algebra, a guide to using calculating machines in. The volume of a typical human brain is cm3 and its weight is about times Readers who know high school algebra should be able to determine that the Artists met this challenge with a theory of perspective and mathematicians the differences in how humans might do calculations, visualize a polyhderon. There are a number of reasons why algebra holds an important place in the curriculum. Second it shows that a calculation that can be done for particular numbers can .. Since this foundation took the human race many centuries to construct it is When they initially meet equations they are learning arithmetic and come.
Now, if you know-- they're giving us in this problem, they're giving us the rate. If we think about the scalar part of it, they're telling us that that is 3 meters per second. And they're also telling us the time. Or, sorry, they're not telling us the time. They are telling us the distance, and they want us to figure out the time. So they tell us the distance is meters.
And so we just have to figure out the time. We're dealing with the rate or speed and distance. So we have 3 meters per second is equal to meters over some change in time.
And so we can algebraically manipulate this. We can multiply both sides times time. Multiply time right over there. And then we could, if we all-- well, let's just take it one step at a time. So 3 meters per second times time is equal to meters because the times on the right will cancel out right over there.
And that makes sense, at least units-wise, because time is going to be in seconds, seconds cancel out the seconds in the denominator, so you'll just get meters. So that just makes sense there.
So if you want to solve for time, you can divide both sides by 3 meters per second. And then the left side, they cancel out. On the right hand side, this is going to be equal to divided by 3 times meters. That's meters in the numerator. And you had meters per second in the denominator. If you bring it out to the numerator, you take the inverse of this.
So that's meters-- let me do the meters that was on top, let me do that in green. Let me color code it. And now you're dividing by meters per second. That's the same thing as multiplying by the inverse, times seconds per meters. And so what you're going to get here, the meters are going to cancel out, and you'll get divided by 3 seconds. So what is that? So this is going to be This part right over here is going to be And it's going to be seconds.
That's the only unit we're left with, and on the left hand side, we just had the time. So the time is seconds. Sometimes you'll see it. And just to show you, in some physics classes, they'll show you all these formulas.
But one thing I really want you to understand as we go through this journey together, is that all of those formulas are really just algebraic manipulations of each other.
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So you really shouldn't memorize any of them. You should always say, hey, that's just manipulating one of those other formulas that I got before. And even these formulas are, hopefully, reasonably common sense. And so you can start from very common sense things-- rate is distance divided by time-- and then just manipulate it to get other hopefully common sense things. So we could have done it here.
So we could have multiplied both sides by time before we even put in the variables, and you would have gotten-- So if you multiplied both sides by time here, you would have got, on the right hand side, distance is equal to time times rate, or rate times time.
And this is one of-- you'll often see this as kind of the formula for rate, or the formula for motion.
So if we flip it around, you get distance is equal to rate times time. So these are all saying the same things. And then if you wanted to solve for time, you could divide both sides by rate, and you get distance divided by rate is equal to time. And that's exactly what we got. Distance divided by rate was equal to time.
So if your distance is meters, your rate is 3 meters per second, meters divided by 3 meters per second will also give you a time of seconds. If we wanted to do the exact same thing, but the vector version of it, just the notation will look a little bit different.
And we want to keep track of the actual direction. So we could say we know that velocity-- and it is a vector quantity, so I put a little arrow on top.
Velocity is the same thing as displacement. Let me pick a nice color for displacement-- blue. As displacement-- Now, remember, we use s for displacement.
We don't want to use d because when you start doing calculus, especially vector calculus-- well, any type of calculus-- you use d for the derivative operator. If you don't know what that is, don't worry about it right now.
That looks as if it is going to be a pain to calculate. Maybe your initial response is to get out a calculator and plug in all the numbers from 1 to to get the answer.
Solving for time
The famous German mathematician Carl Friedrich Gauss — was faced with this problem in school. The teacher knew that it would keep them quiet for quite a while. This is what Gauss did. At this point, Gauss then added the two sets of numbers. At first sight this would appear to make things harder but wait, you need to group the numbers in the right way.
All the numbers in brackets on the right hand side of the equation add to And there are brackets, one for each of the numbers from one to Well the left hand side is two T s. Clearly this is one up to young Carl Friedrich. But Gauss may have seen this geometrically. Geometrically we have T 5 represented in two ways. First as columns of blocks starting from one and going up to 5. And then as these blocks inverted.
Adding these two lots of blocks together gives five columns of 6 blocks. In this final picture there are 30 blocks. But this is twice as many blocks as we started with. So there were 15 blocks at the start.
But things gets better. The same two methods will work for any triangular number. Try it for yourself. Work out T 10 or T or T It certainly beats adding all of the numbers from 1 to 10 or from 1 to or from 1 to The same method we have just used will work for any triangular number — no matter how big or how small.
And now this is where the power of Algebra starts to kick in. This method will give us the value of any triangular number! To see this, we first replace the in T by n. Look at T n. What does it mean? That means that we can get a formula for T n. Add the two sets of numbers and group them in the right way. And there are n brackets, one for each of the numbers from 1 to n. Well the left hand side is two T n s.
The importance of this is that it gives us a formula for T n. We can find the n-th triangular number for any value of n we like. So you want T 10? How about T ? And what about T ? Can you see the power of Algebra here? It would have worked. But the same method works for every triangular number. And Algebra allows us to encapsulate this method in the most general case. Oh but what if we wanted to know what triangular number had the value ?
Now there are many ways to solve this equation. For instance, you could use trial and error. If you drop a few values in you should eventually be able to find a number that would satisfy this equation. Perhaps it might be better to use a table and approach the problem more systematically. We can even look at the factors of and see if any of them fit the pattern of the left hand side of the equation.
However, Algebra has made a living out of solving equations. First, we have used letters to make communication more efficient and less prone to error. By using T nespecially, we have saved ourselves a lot of words. These words may well have got in the way of any calculations that we might have been trying to use. While this may not appear to be much of a saving, try working the above example through without using any symbols at all.
Not only will the lack of symbols make it hard to follow but it will also make it tedious too. Second, we were able to take a calculation that we could do for and show that it could be done for any whole number. The process we had carried out withwe were able to perform with the variable n. Third, we were able to solve an equation for an unknown, when we found which triangular number was equal to Now being able to solve equations is extremely important. Historically, solving equations has led to new types of numbers.
But history pushed further in this direction by inventing abstract number-like systems. These were able to operate on objects that are not numbers.
Things like rotations and symmetries can now be manipulated in ways that are similar to the way that we manipulate numbers. Even more surprisingly, these abstract systems have useful applications too.
How did Algebra develop historically? Well initially, problems that we might think of being arithmetical or even algebraic, were solved without the use of any special signs for unknowns or for arithmetic operations.
As we have suggested above, this must have been fairly tedious. Even as late as Sir Isaac Newton 17th Centurymuch more language was used than is used today. It is impossible for a cube to be written as the sum of two cubes or a fourth power to be written as the sum of two fourth powers or, in general, for any number which is a power greater than the second to be written as the sum of two like powers. This makes it very difficult to understand. There was a change for the better in about AD, when Diophantus came on the scene.
He used letters to stand for unknowns. But although this was progress there was still a long way to go. The concern of algebraists from Diophantus, right through to the 16th Century, was to find unknowns — to solve equations. So although they developed many useful techniques these were not encapsulated using unknowns as we did in the previous section when we added the first n numbers.
However, mathematicians did work hard throughout this period to develop ways to solve equations. He was responsible for us being able to add the first n integers in a neat way. So now we have letters being used for two types of variable.