WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to take a look at a very interesting application of our rational.
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We will look at variation and proportion.
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We will have to do a little bit of work just to explain what I mean by the word variation.
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We will break this down into few other things.
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We will look at direct variation, inverse variation, and combined variation.
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This particular section is filled with lots of different word problems all of them involving variation.
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Let us see what we can do.
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When you hear that word variation we are talking about a connection between two variables,
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such that one is a constant multiple of the other.
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You will see some nice handy formulas that will highlight how one is a constant multiple of the other.
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If you are looking for little bit more intuition on the situation then you can grasp on to that.
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Intuitively a variation can be thought of as a special connection and the two basic types are direct and inverse.
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In this connection when one quantity goes up then the other one would go down.
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In that one I'm talking about an inverse variation.
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If I’m thinking about those two quantities when one goes up and the other one goes up as well, that will be an example of a direct variation.
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It is how one affects the other one and they could be moving in the same direction our going in the opposite directions.
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To model this type of situation that has variation in it, you can end up setting up a proportion.
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Be very careful on where you put each of the individual components and make sure you line them up correctly.
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To model a direct variation, you could use the following proportion x1 / y1 = x2 / y2.
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The way want to interpret those subscripts that you are seeing on everything is that all of the values with a subscript of 1 involved one situation.
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Everything with the 2 involves a second situation.
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Since the x are both on the top that will be from one particular type of thing and y will be from another type of things.
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We will line those up to make sure that they at least agree.
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It seems a little vague but let us get into an example
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and you will see how I do set this up using a proportion and how everything does line up.
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This one says that an objects weight on the moon varies directly as its weight on earth.
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Neil Armstrong, the first person to step on the mood weighed 360 pounds on Earth, when he is on the moon he was only 60 pounds.
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The question is if we take just an average person who weighs 186 pounds, what will their weight be if they go to the moon?
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We have that this is a direct variation because it pointed out and says it is a direct variation.
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Let us just think a little bit and see if that makes sense.
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Intuitively in a direct variation when one quantity goes up then should be the other one.
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When one quantity goes down, then so should the other one.
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That is what is going on with our moon weight.
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If I decide to go on and get heavier and heavier on Earth, that same thing is going to happen on the moon.
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I will get heavier and heavier on the moon.
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This is a type of direct proportion.
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Let us see what we can set up.
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x1 / y1 = x2 / y2
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To help out I’m going to highlight two things, we will go ahead and keep our Earth weights on top
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and we will go ahead and put our moon weights on the bottom.
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Let us look at our first situation with our Neil Armstrong.
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He weighed 360 pounds on Earth, and only 60 pounds on the moon.
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360 on Earth and 60 on the moon.
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I will compare that with our average person there.
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They weigh 186 pounds on the Earth, but we have no idea how much they weigh on the moon.
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I'm going to put that as my unknown.
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Everything on the tops of these fractions represents an earth weight and everything on the bottom represents a moon weight.
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You can see that we have developed one of these rational equations and now we simply have to solve it.
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This one is not too bad if you go ahead and reduce that fraction on the left, it goes in there 6 times.
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We can multiply both sides by x.
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I have 6x = 186.
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Finally dividing both sides by 6 I will get that x =31.
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This reveals how much that average person, the 186 pounds would weigh when they are on the moon.
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There are lots of other types of situations that could involve a direct proportion and they might not come out and say its direct proportion.
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Look for clues along the way to help out.
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This one says that a maintenance bill for a shopping center containing 270,000 ft² is $45,000.
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What is the bill of the first store in the center that is only 4800 ft²?
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Let us think of what type of a variation this should be.
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If I have a large store as a part of the shopping mall and I will have to end up paying a much larger maintenance or bill because of my larger store.
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If I have a small store then my bill should be very small.
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Those quantities are moving in the same direction.
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I’m going to say this is an example of direct variation.
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We will set it up using our proportion.
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We want to keep things straight.
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What do each of these quantities represent?
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Let us say the top will be our square footage and we will make the bottom the bill.
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How much it cost?
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Let us go ahead and substitute some of this information we have.
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If I’m looking at the entire shopping center I know how big it is and I know the bill for the entire place.
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270,000, 45,000 and now I can look at my much smaller store, it only has 4800 ft²
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and this one we have no idea what the bill is so I will leave that one as my unknown.
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I think we can do a little bit of simplifying with this one as well.
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I will divide these two and we get 6 and I need to just solve this rational equation which is not too bad as long as we multiply both sides by x.
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I have 6x = 4800.
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Divide both sides by 6 and then I think we will have our solution x =800.
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Even if they do not come out and tell you what type of variation is going on here,
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look at the values present to see if things are moving in the same direction, or in opposite directions.
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For inverse variation we want to make sure that if one quantity goes up, the other one goes down.
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One way that we can model it now with our proportions is to set up like this.
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We will go ahead and use us a subscript for one situation and we will put them on opposite sides of our equal sign.
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We will get that inverse relationship.
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When one goes up, the other one will go down.
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For our second situation we will put these on two opposite sides of the equal sign.
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One thing that will not change is that both of these x will represent the same type of thing
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and both of these y will still represent the same type of thing.
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We can definitely note that the things are on opposite sides of the equal sign for these given situations.
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For my inverse variation problem I will look at the current in a circuit and we are told that it varies inversely with the resistance.
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If I have a current that is 30 amps when the resistance is 5 ohms, find the current for resistance of 25 ohms.
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I have a lot of things in here and we just want to keep track of each of them.
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Let us see if we can highlight each of these situations.
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The current is 30 amps when the resistance is 5.
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We will be looking for the current when the resistance is 25.
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Let us get our proportion out here.
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I’m using these little subscripts and put them on opposite sides.
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Onto the first situation, we will keep everything on the top.
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Let us say that is our resistance, we will put everything on the bottom as the current.
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Our first situation, we know that the current is going to be 30 when the resistance is 5.
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In that situation I will put my 5 and 30.
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For the other situation I want to figure out what the current is.
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I have no idea what that is but I know that the resistance is 25 ohms.
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This is my initial proportion that I have set up.
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All we have to do is go through the process of solving it.
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This one is not too bad.
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Let us multiply both sides by 30x and see what is left over.
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30 × 5, since the x will cancel out equals 25 × x since the 30 cancel out.
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Okay looking pretty good.
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Also 5 × 30 =150 equals 25x and we can simply divide both sides by 25 to get our answer.
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The current is 6 amps.
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When dealing with these variation problems always look at your solution and see if it makes sense in the context of the problem.
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One thing that I'm noticing in here is that I start off with a resistance of 5 and a current of 30.
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One thing I'm doing with my current is that I'm lowering it, and I'm taking it from 30 down to 25.
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Since I have an inverse relationship that is going to have an effect on our amps is the current.
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That should bring it up.
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Originally I started with 5 and I can see my answer is 6.
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It did go up and things are working in the right direction.
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Since earlier, we said that it variation is relationship when one is a constant multiple of the other.
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We can model this using some nice formulas.
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For direct variation you can use the formula y = k × x.
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For your inverse variation, you could use y = k ÷ x.
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We have two variables in each of these equations that we are looking at connecting are the x and y.
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The k in each of these is known as our constant of proportionality.
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It connects how they are related.
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A lot of problems that you will end up doing with these variations, if you want to use these formulas
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then you will end up going through three steps to figure out what is going on.
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You might first go ahead and identify what type of variation you are using so that you can take one of these formulas.
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Then you will end up using a little bit of known information to go ahead and solve for that constant of proportionality, k.
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Once you know what k is, then you can usually figure out some new information by substituting it into the formula.
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I got a couple of examples where we are going through these 3 steps.
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Let us first just see an example of using an inverse of variation using one of these formulas.
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This one says that the speed of water through a hose is inversely proportional to a cross section of the area of the hose.
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If a person places their thumb on the end of the hose and decreases the area by 75%, what does this do the speed of the water?
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Let us think about what we got here.
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We are looking for a connection among the cross area section of the hose and the speed of the water.
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We are told that they are set up inversely so when one goes down, the other one should go up.
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For decreasing the area then we should expect it to increase the speed of the water.
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In fact we can be a little bit more precise, but despite playing around with those formulas for a little bit.
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Our relationship is that the speed of the water is inversely proportional to a cross section of the area.
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Maybe x = k/a.
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I'm going to decrease the area by 75%, one way that I could model that is by simply multiplying our area by what is left over, 25%.
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Let us relate this to the original before I manipulated it.
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25 is the same as ¼ I'm dividing by ¼.
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When you divide by a fraction that is the same as multiplying.
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This is multiply by 4.
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Now we can see the difference between the original situation.
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The area decreased by 75%.
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They are almost exactly the same but in the second situation, it is 4 times as much.
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It will actually increase the speed by 4 times.
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There are lots of other types of variations that you can go ahead and work into a problem.
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Some of these will look similar to previous types that we covered.
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Let us go through how each of these are connected.
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If I say that y varies directly to some power of x then there are some k that connects those two things.
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You will notice that it looks like our direct proportion only since we are dealing with an nth power,
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like a 3rd power 4th power that we put that exponent right on the other variable.
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Our y varies inversely as the nth power looks much like our inverse relationship as well.
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But since we are doing it to the nth power, notice how we put that exponent right on the other variable.
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In this one is a new one that involves a few more variables, this is our joint variation.
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y varies jointly as x and z so more than one variable now.
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If there some sort of positive constant such that y = k × x × z.
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In this one the k is our constant and the variables are being connected, the y, x, and z.
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You can start mixing and matching these various different types of variation.
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The reason why that we do not want to start mixing and matching them is we can get some more complicated models.
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With these more complicated models we are hoping to be able to more accurately model what is happening in a real life situation.
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Usually it does a better job than any one variation could do by itself.
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The key when putting all of those different types of variations together is to look for clues in the actual problem itself.
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Here is a formula that I have it is T = k × x × z³ /√y .
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Here is how I could describe that problem using the language of these variations.
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The k right here is my constant of proportionality and we are looking to get it connect together the t, x, z, and y.
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Here is what I can say, the t varies jointly with x and z³.
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It is telling me that I have my constant of proportionality but the x and z should end up in the top and the z should be cubed.
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t also varies inversely with the √y.
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Since it is an inverse relationship I will put that one in the bottom and make sure to put in that square root.
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Watch for those keywords so we can figure out where we should put things.
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Let us try some of these variation problems and watch as I walk through the three steps of
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starting with some template solving for our constant of proportionality, k and the last part figure out some new information.
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What do I know so far?
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I know that y varies jointly as x and z.
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Just on that little part, let us go ahead and make a nice formula.
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y varies jointly as x and z.
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y = 12 when x = 9 and z = 3.
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I can use that information right there to go ahead and solve for my constant of proportionality.
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y =12, x = 9, z= 3
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The only unknown I have in there is that k value.
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We will combine things I will get 27 divide both sides by 27 and then maybe reduce that by dividing the top and bottom by 3.
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Now that we know much more I can end up revising our formula.
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y = xz and now I can say that the k value is this 4/9.
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Now that we know about our constant of proportionality, let us find z when y = 6 and x = 15.
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We will substitute those directly in there and you will see the only unknown I have left will be that z.
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Y= 6, 4/9 × 15 and we are not sure what z is now we have to solve for it.
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A little bit of work.
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I think I can cancel out an extra 3 here.
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I have 6 = 20/3 × z we could multiply both sides by 3.
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I will get rid of those 3, 18 = 20z.
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Lastly let us go and divide both sides by 20.
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18 ÷ 20 =z
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We will just go ahead and put that into lowest terms.
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Divide the top and bottom by 2 and I know that 9/10 is equal to z.
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If you follow along we start with some sort of template for our variation, we solved for our unknown constant of proportionality k.
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We have used it to figure out some new information about the other variables.
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For this last type of example let us look at one that is a little bit more of a word problem.
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This one says that we have the maximum load of the beam and how much you can support varies jointly
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with the width of the beam and the square of its depth, but it varies inversely as the length of the beam.
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Some of that initial information there is just giving us a formula for how everything is connected.
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We will definitely set up the equation that will model that situation.
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Let us continue on.
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Assume that a beam that is 3 feet wide, 2 feet deep, 30 feet long and it can support 84 pounds.
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What is the maximum load a similar beam can support if it is 2 feet wide, 5 feet deep and 100 feet long?
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A lot of information is floating around there.
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Let us start off by seeing if we can set up this formula.
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The maximum load that a beam can support varies jointly as the width of the beam and the square of its depth.
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Since this is a jointly, we are talking about k × width × the square of its depth or d².
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It varies inversely with the length, I will put the length on the bottom.
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This part here is our formula that we will end up using.
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Assuming that a beam is 3 feet wide, 2 feet deep and 30 feet long, and it can support 84 pounds,
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let us use that to figure out our unknown constant sitting right here that k.
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I know it can support 84 pounds, we do not know k.
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3 feet wide, I put that in for width, 2 feet deep and it is 30 long.
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The only unknown is that k.
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Let us go ahead and simplify it just little bit in here.
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I have 3 × 2² and 2² is 4 = 12
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I can get that k a little bit more by itself if I multiply both sides by 30, 84 × 30.
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Let us go ahead and put those together and see that would be 2520.
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Divide both sides by 12.
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That is where we have been able to figure out what that unknown constant is.
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Now that I have that I can go back to the original formula.
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M = 210w × d² / l
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We will use this new formula to figure out a little bit of known information.
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We will look at it for the other beam where we are looking for its maximum load if we know its width, depth, and its length.
00:25:19.800 --> 00:25:24.900
Let us give it a try.
00:25:24.900 --> 00:25:35.800
That formula we developed was its maximum load = 210 × width × d² / length.
00:25:35.800 --> 00:25:40.900
Let us put in our information.
00:25:40.900 --> 00:25:48.800
For this new beam, its width is 2 feet and its depth is 5.
00:25:48.800 --> 00:25:51.400
It also has a length of 100.
00:25:51.400 --> 00:25:58.200
As soon as we simplify all that we will be able to figure out what its maximum load should be.
00:25:58.200 --> 00:26:13.400
210 × 2 = 420 × 25 ÷ 100
00:26:13.400 --> 00:26:33.400
Let us do some simplifying here 420 × 25 = 10500 ÷ 100 and looks like this reduces to 105.
00:26:33.400 --> 00:26:38.800
This new beam can support a maximum of 105 pounds.
00:26:38.800 --> 00:26:42.400
Being able to recognize various different types of variation
00:26:42.400 --> 00:26:47.500
and combine them all together is essential for some more complicated problems.
00:26:47.500 --> 00:26:51.600
Remember that if you have those nice formulas then you can go through 3 steps.
00:26:51.600 --> 00:26:56.800
One, build a formula or equation that models the situation, solve for your unknown k
00:26:56.800 --> 00:27:02.400
and finally use some new information to figure out some more information about the problem.
00:27:02.400 --> 00:27:04.000
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