WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about variation--both direct variation and inverse variation.
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Variation is a way of talking about how different things relate to each other.
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Depending on the type of the variation, we will know what form the relationship takes.
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However, we should point out that variation isn't connected very closely to functions.
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While this sits in the section on functions, it is only sort of connected to functions.
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We can describe it in the language of functions; but honestly, it is much easier to talk about it with equations.
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It is much easier to talk about variation using equations; so that is what we are going to end up using, that y =...stuff involving x,
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what we are used to from doing lots of algebra for it.
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That is what we are going to see for our variations.
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In any case, variation comes up a lot in a wide variety of real-world situations; so it is important to understand.
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If you are interested in physics, chemistry, economics, astronomy, scientific fields in general...you are going to probably need to be familiar with variation.
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It gets tossed around in those fields; also, it tends to show up on standardized tests a lot--things like the SAT, the ACT, the GRE...
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All of these have a tendency to have one or two questions about variation, either direct or inverse variation.
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So, it is important to know this thing a little if you are going to be taking one of these standardized tests at some point.
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All right, let's take a look at it: **direct variation**: this is the simplest form of variation, but it is also the most common.
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Direct variation says that two things are directly related to each other: if one goes up, the other goes up.
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If the red one goes up, we know that the blue one will also go up.
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Similarly, if the other goes down, if the red one goes down, then we know that the blue one also has to go down.
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They will go at different rates; for example, in this picture we have here, the blue one always goes at a smaller rate,
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and the red one goes at a bigger rate--it goes faster than the blue one.
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But they always go in the same direction; so direct variation is the same direction--one goes up; the other goes up; one goes down; the other goes down.
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They are linked; they move in lockstep.
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We see this kind of relationship in a lot of situations; here is a common, everyday example.
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Say it costs you two dollars to buy a loaf of bread; if you buy 10, 10 would cost $20; and if you bought only 2 loaves, it costs $4.
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The more bread you buy, the more cost; the less bread, the less cost.
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The total cost of the bread and the number of loaves you buy are in direct variation.
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It seems kind of obvious; but direct variation is actually a pretty simple concept to get around.
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It is just two things that are linked: one goes up; the other one goes up; one goes down; the other goes down--direct variation.
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There are lots of different ways to say that two things (let's call them x and y) are in direct variation.
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We could say "x and y are in direct variation"; we could say "x and y vary directly,"
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"y varies directly as x," which is to say "as x would do something," but we don't know what x is doing yet until later on;
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so it is "y varies directly as x"; "x and y are directly proportional," "y is directly proportional to x."
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There are lots of different ways to say it; but they all mean the exact same thing mathematically.
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In mathematics, it means y = k times x, where k is a constant.
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k is just a constant; it is called the proportionality constant or the constant of variation.
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It is the rate at which the two things are connected.
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Remember, with the arrows, the red arrow is bigger than the blue arrow;
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there is some rate that says the red arrow grows faster than the blue arrow, whether it is going up or going down.
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In the previous example about loaves of bread, k would be representing the price of one loaf.
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That is the rate of connection between loaves of bread and the cost of the bread.
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It is the price; price is what is connecting them--that would be our rate k for that loaves of bread example--it would be based on the price.
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All right, **inverse variation**: the idea of inverse variation is the opposite of direct variation--not a big surprise there.
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It says that two things are inversely related to each other; this means that if one goes up, the other one goes down.
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But if one goes down, the other one has to go up.
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They are usually going to go at different rates; once again, they have different growths in the red and blue arrow.
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But they are going in opposite directions; that is the really key idea to get across from inverse variation.
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It is that in inverse variation, they go in opposite directions: if one grows, the other one shrinks; if one shrinks, the other one grows.
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And also, I want to warn you here: inverse variation is not related to inverse functions--it is not connected to inverse functions.
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It is based on the idea of multiplicative inverses, those reciprocals, like 3 and 1/3.
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3 and 1/3 are multiplicative inverses, because they cancel each other out; so that is what we are talking about.
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We are not talking about functions canceling each other out--that kind of inverse function;
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we are talking about the multiplicative inverse; that is where inverse variation is getting its term from.
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All right, this idea comes up a lot less in everyday situations; we aren't going to see it with bread.
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But it is pretty common in the sciences--for example, consider gravity.
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Gravity is an inverse variation effect with distance.
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The force of gravity that Earth exerts is inversely related to the distance from Earth.
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The more distance you have from the earth, the less gravity you experience.
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The closer you are to Earth--the less distance you have--the more gravity you experience.
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Gravity and distance are in inverse variation; one of them goes up; the other one goes down; the first one goes down--the other one must go up.
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And technically, just so we have this on the books officially, it is not actually distance that is the inverse; it is the distance squared; but it is the same idea.
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One is going up; the other one is going down; the first one goes down; the other one goes up.
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Technically, it is d², but you can learn about that in a physics class; don't worry about that right now.
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Inverse variation--there are lots of different ways to call it out, just like direct variation has many names.
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We can say "x and y vary inversely," "y varies inversely as x," "x and y are inversely proportional,"
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"y is inversely proportional to x," or "x and y are in inverse variation."
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And we also will sometimes call the relationship reciprocal proportion (remember, because we are talking about flips, the 3 to 1/3 kind of reciprocal).
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Or confusingly, sometimes, once in a while, you might hear somebody call it indirect variation.
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This is kind of weird, because direct/indirect...I think it is kind of weird; I think "inverse" and "reciprocal" really get the idea across much better than "indirect."
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But once in a while, we will hear people use that; so it is important that we are aware of it.
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In any case, even if it has all of these different names, they all mean the exact same thing mathematically.
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y = k/x; and once again, k is a constant--it serves the exact same purpose as it did for direct variation.
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It gives the rate at which the two things are connected.
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How much is that blue arrow going to grow or go down, depending on how much the red arrow is changing?
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The red arrow and blue arrow, on the last one, where we talked about inverse variation--they didn't go at the same speed; they went at different speeds.
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The red one grew a lot, and the blue one went down a little bit.
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The red one went down a lot, and the blue one went up a little bit; so we have that same rate thing going on with this k.
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And if we wanted to, we could also see this as y = k times 1/x.
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That would be perfectly reasonable, as well; but I prefer k/x, just because it seems a little more compact.
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But they are really just equivalent statements.
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OK, **joint variation**: if we want, we could have multiple direct variations going on at the same time.
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If we want to do this, we use the term joint variation; so we could say "z varies jointly as x and y,"
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which says that z has a direct variation with x and a direct variation with y.
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Or, we could also say that z is jointly proportional to x and y; these are our two ways of saying it.
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But they are both going to mean the same thing: z = k times x times y.
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And once again, k is a proportionality constant; it is the thing that is linking the rates of change between various things.
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Now, really quickly: you might wonder, "Wait a second; if z is connected to x by direct variation, then that would be z = kx."
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But then, we also have direct variation with y, so that is z = ky; so let's use a different letter; let's call them k₁ and k₂.
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So, k₁ and k₂ are both constants; we are going to need different constants, because x and y are different things.
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If we put these together, we would really get z = k₁ times k₂ times xy.
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We can put those constants together, k₁ and k₂; so why is it that is shows up as just k, as opposed to 2 constants?
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Well, remember: if k₁ and k₂ are both constants, then that means that,
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when we multiply them together, k₁ times k₂ is constant, as well.
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So, if k₁ times k₂ is just a constant, then that means...
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let's just give it a new name and say k₁ times k₂ equals just k.
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And that is why we only have to see k there--because it is two constants combined into one constant.
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So, we aren't going to have to worry about keeping track of two separate constants.
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We can just merge it into a single proportionality constant; and that is why we only have the one of them.
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So, we have direct variation; we have joint variation (multiple direct variations going on at once)--
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it is all of the things we are varying with one proportionality constant, k; great.
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We can also combine multiple types of variation; we can stack direct and inverse variation if we have these sorts of relationships going on simultaneously.
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For example, here is a gas law from chemistry and from physics; you can probably learn it in both courses, depending on how the course is taught.
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Given a fixed amount of gas (which is to say, just air, like the kind of air in a room, not "gas" like gasoline/petroleum)--
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given a fixed amount of air or gas in a container, the pressure, P, of the gas
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varies directly as the temperature, T, and inversely as the volume, V.
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So, the inverse thing would show up down here by the "divide by V"; and our direct would be just T.
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So, it is k times T for the direct, and then k/V for the indirect; and once again, we put them together;
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and they just combine, and they become a single constant of variation.
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So, we just get a single constant, k; so we just put the various kinds of variation that we have...we just stack them all together.
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We do them all at once, and we put only one single constant, k; and that will be enough, for the exact same reasons we talked about on the previous slide.
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All right, that is everything; we have everything we need to get to the examples.
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a and b are in direct variation; if a and b are in direct variation, we know that a = kb.
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Or, we could also write it as b = ka; but notice that they will end up having the same effect.
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When a equals 13, b equals 5; so let's plug this in, and let's find out what k is.
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If we want to know what b is when a equals 52, we are going to need to know what k is, to be able to figure that out.
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We start by figuring out what k is: 13 = k(5); so that means that 13/5 = k.
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Great; if that is the case, we can go back, and we have the a = 13/5k; now we plug in a = 52, and we figure it out for a different b.
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If a is 52, then we have 13/5 times k; we multiply by 5 divided by 13 to cancel out that fraction on the right side; we get (5/13)52 =...
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oh, oops, sorry; not k; I made a mistake; a = 13/5 times b; my apologies.
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So, we plug in...now we are trying to figure out what b is; 5/13 times 52...13 actually goes directly into 52;
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52 can be broken up into 13 times 4, so we can come along and cancel this 13, and we just 5 times 4, which is 20.
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So, 20 is what b is when a is 52; great.
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In the next one, x and y are in inverse variation; if they are in inverse variation, we have y = k/x.
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We could also do it as x = k/y; it will have the same effect--we end up having different k's,
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but the important part is that we have this setup of the inverse like this.
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When x = 72, y = 3; once again, we want to figure out what the k is, because the next step is to figure out what x is when y is 24.
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So, we will need that information about what k is.
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So, x we put in as 72; y we put in as 3; so 3 = k/72.
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Multiply both sides by 72; it cancels out the denominator on the right side; 72 times 3 equals k.
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We use a calculator, and we get 216 = k; you probably don't even need a calculator for that one--you might be able to do that one in your head.
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y =...replace the k now; y = 216/x, so now we put in y = 24; 24 = 216/x.
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So, we want to figure out what x is; we multiply the x over on the left side, and we divide by the 24; we get 216/24.
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You probably want to plug that one into a calculator; and we will get x = 9.
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And there we are--our answers; great.
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The next example: p varies jointly as m and n; what does that mean?
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"Jointly as m and n" means that these two are going to be on the right; so p = k (we have to have that proportionality constant), times m, times n.
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When m equals 4 and n equals 8, p equals 2; then we want to find out what n is when p equals 60 and m equals 24.
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So, once again, we need to figure out what that k is, what that proportionality constant is,
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if we are going to be able to figure out what n is in the second half of the question.
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So, we plug in all the values that we know from the beginning; we have p = 2; 2 = k times m at 4, and n at 8; 2 equals k times 32.
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Divide by 32; 2/32 = k; 1/16 = k; great.
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So, with that information, we can now take this; we will create a new thing that tells us the relationship in general.
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p = 1/16 times m times n--great; we know that p is 60; we know that m is 24 for the one where we want to figure out what the n is.
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So, we plug in p = 60; that equals 1/16; times...we don't know what n is, but we do know what m is: m is 24 times n.
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Now notice: there are some common factors between 16 and 24; 24 can be broken down into 8 times 3; 16 can be broken into 8 times 2.
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So, the 8's cancel out, and we are left with 60 = 3/2 times n.
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We can now multiply both sides by 2/3; 2/3 times 60 = n.
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So now, that cancels out the fraction on the right side; we can break down 60 into 3 times 20.
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The 3's cancel out, and we have 2 times 20, which is 40, equals n.
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There is our answer; when p is 60 and m is 24, n has to be 40.
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The next example: all right, this one is a little more complex, because it is a word problem; but we will be able to work through it, actually.
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Hooke's Law connects the force of a spring to its compression; it says that the distance, x, that a spring is compressed or stretched
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from its equilibrium natural length is directly proportional to the force, F, of the spring.
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First, let's understand what that means: the very first thing you should do, if you read a word problem,
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and you are not sure what a word means--go look it up.
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Either (if it is one of the words that you learned because it was in that lesson) go check what that definition meant in your math book,
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or from this lesson; or if it is just a word that you don't know, that didn't show up in the math before, just look it up in a dictionary.
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Equilibrium: say this is a spring--equilibrium is just what it is when it is at rest, when it is not being compressed or stretched.
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It says that the distance, x, a spring is compressed--if we push it in by some amount x, it compresses it.
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And then, that causes the spring to push back with some force, F.
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And this law, Hooke's Law, says that the force, F, is equal to k times that distance.
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It is directly proportional; so it has that proportionality constant, times the thing that it is directly proportional to (direct variation).
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F = kx; now, once again, once we have that part figured out, it is just a normal problem from there on.
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Say a spring is stretched by .1 meters and has a force of 95 Newtons when stretched that far.
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And Newtons is just the unit of force in the metric system--just force.
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And then, what force would it have if stretched by .25 meters?
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Well, .1 meters is just x = 0.1, and force equals 95 for the first part of the problem.
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For the second part of the problem, we have x = 0.25; and we want to find out what the force is: F = ?
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So, we see this diagram; we push the spring in by some amount; and that will cause a force to appear,
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in reaction to that, where the spring is pushing back; and it will depend on the amount we have compressed it.
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If you take a spring in your hands, and you push a little bit on the spring, it pushes a little bit back.
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If you push the spring really, really heavily, it pushes really, really hard back; and that is why we have F = kx.
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How hard it pushes back is directly connected to how much it has been compressed (or stretched).
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So, what force would it have if it was stretched by .25 meters?
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To figure out that, we are going to first have to figure out what k is.
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So, we do that by plugging in our numbers; our force is 95 when it has been stretched by 0.1.
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We divide both sides by 0.1: 90/0.1 = k; so that just moves the decimal place over 1, and we get 950 = k.
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So, that tells us, for this specific spring, the spring that this problem happens to be about--it has a constant of 950.
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It has a coefficient, a proportionality coefficient/constant, of 950 (technically, 950 Newton-meters--it is just that it has units).
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But we don't have to worry about that; 950 = k is perfectly enough.
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So, 950 = k is what it is for this spring; however, this is true, in general, for any spring: F = kx is a very good rule
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for describing any spring that can be compressed or stretched.
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But this specific spring that we are working with has 950.
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A smaller, easier-to-push-around spring would have a smaller k, and a harder, thicker, heavier,
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harder-to-push-around spring would have a larger k.
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So, 950 = k for this one; so now we have F for this specific spring: F = 950x.
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So, what is the x that we have for the second half of the question? x = 0.25, so F = 950(0.25).
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We plug that into a calculator; and 1/4 of 950 is 237.5; what are the units that we are using?
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We were told that the unit of force in the metric system is the Newton; so it is 237.5 Newtons.
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And there is our answer--great!
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All right, the last example: The maximum load a horizontal beam can support, if held up at both ends,
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is jointly proportional to the width of the beam and the square of its depth, while inversely proportional to its length.
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Wow, let's see if we can figure out what that means before we try to do this.
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First of all, we have some horizontal beam; we have some beam, and it is being supported at both ends.
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It says "the maximum load a horizontal beam can support..." so there is some big, heavy weight in the middle of it.
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And it is being supported on both ends; it is being held up just at the ends, and it is able to support an amount.
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The maximum load is jointly proportional to the width of the beam and the square of its depth, while inversely proportional to its length.
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What is length, width, and depth? Well, let's take another look at the beam.
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We could have a beam like this; its depth is how far down it goes.
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Here is d, the depth; we could also talk about what its width is; its width is going to be
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(sorry, that was just me trying to make an arrow pointing at the w, but it ended up looking like a w, as well)
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width...it is jointly proportional to the width; and then finally, the length of it is there.
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Now, this makes sense; it would make sense that it is going to be inversely proportional.
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The farther out the beam gets...we have a beam like this that is the exact same thickness and depth and everything.
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And we support it at the two ends; it makes more sense that it is going to be easier to snap it in the middle.
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The farther and farther we stretch it out on the sides, the easier it is going to be to snap it in the middle.
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On the other hand, if it is all the same thing, and it is a very short thing, and it is very stout,
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it is going to be able to support a lot more load, if it is supported very, very closely, like this.
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So, that makes sense; it also makes sense that, if it is wider (there is more stuff there), then it is going to be able to support more.
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And its depth...it is going to be able to support more.
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Now, it turns out that it is not just the depth, but the square of the depth.
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So, we have to integrate that into our formula; so let's figure out how we can turn this into just a formula in math.
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The maximum load is equal to...the first thing we have to do is put in that proportionality constant, k.
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k times...what is it? It is jointly proportional to the width of the beam and the square of its depth,
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so times the depth squared, because it is the square of the depth; divided by (because it is inversely proportional to) the length.
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This gives us a formula for figuring out maximum load.
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We would have to know what k is to be able to use this formula; but we have a formula for doing that.
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All right, now let's continue with the problem and see if we can use that to figure out the rest of these questions.
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OK, so now that we have our maximum load formula here, if a given beam can support a maximum load of 750 kilograms,
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how much could it support if its length is tripled, or its width is doubled, or its length is doubled, its width is halved, and its depth is tripled?
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Wait a second; we don't have any information!
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The way we did all of these previous problems was that they gave us enough to figure out k,
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and then we used k to figure out the rest of these problems.
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We don't have any specific numbers to work with; so what are we going to do?
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Well, the first thing we need to do is not get scared; we have this problem, so there is a good chance that we can solve it.
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So, let's just try the things that we normally try with word problems.
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Let's try to just name things that we aren't sure of, at least.
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We don't have a specific number for the length that it is; a given beam must have a length; it must have a width; and it must have a depth, at first.
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We don't know what they are, but we can still give them names.
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This is a great thing to do--to just give names to the things that you don't know.
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So, let's say, right from the beginning, that we will name its initial width w<font size="-6">i</font>, its initial depth d<font size="-6">i</font>, and its initial length l<font size="-6">i</font>.
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These all end up being the initial width, depth, and length, respectively.
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Now, what do we know when we use the initial width, depth, and length, respectively?
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We know that 750, the maximum load, is equal to k (we don't know what k is), times w<font size="-6">i</font>, times d<font size="-6">i</font>², over l<font size="-6">i</font>.
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Well, wait; I still can't figure out what k is.
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So, what are we going to do? Once again, don't get scared yet.
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Let's actually try some of these out: the first one is if its length is tripled.
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So, if its length has tripled, we are going to have a different thing than l<font size="-6">i</font>; but it is going to be connected to l<font size="-6">i</font>.
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If we triple the length of l<font size="-6">i</font>, it is going to be 3l<font size="-6">i</font>, so it is going to be 3 times l<font size="-6">i</font>.
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What is it going to be if we have k times d<font size="-6">i</font> times w<font size="-6">i</font>, over l<font size="-6">i</font>(3l<font size="-6">i</font>)?
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Well, we don't know what that is; but oh, that looks a lot like this; that is just 1/3--we can pull that 1/3 out.
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We have 1/3 times (k times w<font size="-6">i</font> times d<font size="-6">i</font>)², over l<font size="-6">i</font>.
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We already know what this is; that is just 750, so it is 1/3 times 750; 1/3 times 750 is 250.
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What is the unit we are working with? We are working with kilograms as our unit.
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So, the maximum load, if we were to triple the length of this beam, would be 250 kilograms.
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Great; so now we have an understanding of how this is working out.
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We just plug it in, and we can use what we have here; we can use the information that we have.
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We don't have to know all of the numbers; knowing just one of them is enough, because we have a general form that it is working in.
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When the width is doubled, it would be 2 times the initial width, w<font size="-6">i</font>; so that would be k times 2w<font size="-6">i</font> times d<font size="-6">i</font> squared, over l<font size="-6">i</font>.
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So, we pull that 2 out; we have 2 times (k times w<font size="-6">i</font> times d<font size="-6">i</font>) squared, over l<font size="-6">i</font>.
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We plug in what we know that all is; that is 750, so it is 2(750), or 1500 kilograms; that would be the maximum.
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Great; on the last one, this is a lot of things: length is doubled (that would mean 2l<font size="-6">i</font>);
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width is halved (which would mean 1/2w<font size="-6">i</font>); and depth tripled would mean 3d<font size="-6">i</font>.
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OK, so how does this come out? We have k times w; what w are we using? We are using 1/2w<font size="-6">i</font>.
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Times d...what is our new d? It is 3d<font size="-6">i</font>; and this is important.
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Remember, we are not plugging in for just 3d<font size="-6">i</font> squared; we are plugging in all of this--it is what all of the new depth is.
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And the new depth is 3d<font size="-6">i</font>, so it goes in in parentheses: (3d<font size="-6">i</font>)², so that 3 is going to get squared, as well.
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Divided by 2l<font size="-6">i</font>: so k times 1/2...let's pull the 1/2 down; so we will get w<font size="-6">i</font> up here,
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but it will be divided by 4l<font size="-6">i</font> times 9d<font size="-6">i</font>².
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So, we can pull out the coefficients, and we will get 9/4k times w<font size="-6">i</font> times d<font size="-6">i</font>², over l<font size="-6">i</font>.
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We know what that one is; that one is 750; so 9/4 times 750...
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plug that into a calculator, and we get 1,687.5 kilograms; and there we are.
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One thing I would like to point out: notice that depth, by far, matters the most, because it is depth squared.
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So, you can get a lot more load by just having a larger depth.
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You have to increase width a lot more than increasing depth, because depth gets squared.
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This is why, if you have ever worked in roofing, or anything where you see a beam supporting a long horizontal length,
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if you get the chance to look up inside of an attic or inside of a roof,
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at what is actually holding up the structures, you will notice that the beams aren't flat like this.
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They are supported like this, so that they can have the most depth, because it is the depth squared that makes the strength.
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So, they are always supported on a long, deep kind of axis, because that gives them the most strength.
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So, if you have done construction, you have actually seen this before.
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You have seen something where you realize that what you are seeing there is mathematics in effect in the real world--pretty cool.
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All right, I hope variation makes sense; and we will get started on the next section in the next lesson.
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All right, see you at Educator.com later--goodbye!