## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Variation & Proportion

- Variation is a connection between two variables such that one is a constant multiple of the other.
- Some of the major types of variation are
- Direct variation y = kx
- Inverse variation y=k/x
- Joint variation y=kxz
- Variation can be modeled using proportions, or simple formulas. Make sure you label what the variables represent, and to line them up carefully.
- Variation can also be used to connect variables when they are squared, or taken to a root. Watch for clues in the wording of the problem to help identify these.
- Multiple types of variation can be combined into a single problem. For these make sure variables connected directly, or jointly end up in the numerator. Variables connected inversely should be put in the denominator.

### Variation & Proportion

_{1}= 15 when y

_{1}= 10 find x

_{2}when y

_{2}= âˆ’ 6

- x
_{1}y_{1}= x_{2}y_{2} - ( 15 )( 10 ) = x
_{2}( âˆ’ 6 ) - [150/( âˆ’ 6)] = x
_{2}

_{2}= âˆ’ 25

_{1}= 7 when y

_{1}= 13, find y

_{2}when x

_{2}= 11

- x
_{1}y_{1}= x_{2}y_{2} - ( 7 )( 13 ) = 11y
_{2}

_{2}

_{1}= 14 when y

_{1}= 7 find x

_{2}when y

_{2}= âˆ’ [1/4]

- ( 14 )( 7 ) = x
_{2}( âˆ’ [1/4] )

_{2}= âˆ’ 392

_{1}= 8 when y

_{1}= 4 find y

_{2}when x

_{2}= âˆ’ [4/6]

- ( 8 )( 4 ) = x
_{2}( âˆ’ [4/6] ) - ( âˆ’ [6/4] )( 8 )( 4 ) = x
_{2}

_{2}= âˆ’ 48

- Identify equivalent form to xy = ky = [12/x]
- Make a table of points
x line 1 2 3 4 5 y line 12 6 4 3 [12/5]

- Identify equivalent form to xy = ky = [20/x]
- Make a table of points
x line 1 2 3 4 5 y line 20 10 [20/3] 5 4

- Identify equivalent form to xy = ky = [39/x]
- Make a table of points
x line 1 4 7 10 13 y line 39 [39/4] [39/7] [39/10] 3

- Identify equivalent form to xy = ky = [( âˆ’ 24)/x]
- Make a table of points
x line âˆ’ 12 âˆ’ 8 âˆ’ 4 4 8 12 y line 2 3 6 âˆ’ 6 âˆ’ 3 âˆ’ 2 - Graph utilizing points

- Identify equivalent form to xy = ky = [1/3x]
- Make a table of points
x line âˆ’ 3 âˆ’ 2 âˆ’ 1 1 2 3 y line âˆ’ [1/9] âˆ’ [1/6] âˆ’ [1/3] [1/3] [1/6] [1/9] - Graph utilizing points

- Identify equivalent form to xy = ky = [28/x]
- Make a table of points
x line âˆ’ 24 âˆ’ 16 âˆ’ 8 8 16 24 y line âˆ’ [7/6] âˆ’ [7/4] âˆ’ [7/2] [7/2] [7/4] [7/6]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Variation & Proportion

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Objectives 0:10
- Variation and Proportion 0:34
- Variation
- Inverse Variation
- Direct Variation
- Setting Up Proportions
- Example 1 2:27
- Example 2 5:36
- Variation and Proportion Cont. 8:29
- Inverse Variation
- Example 3 9:20
- Variation and Proportion Cont. 12:41
- Constant of Proportionality
- Example 4 13:59
- Variation and Proportion Cont. 16:17
- Varies Directly as the nth Power
- Varies Inversely as the nth Power
- Varies Jointly
- Combining Variation Models
- Example 5 19:09
- Example 6 22:10

### Algebra 1 Online Course

### Transcription: Variation & Proportion

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at a very interesting application of our rational.*0002

*We will look at variation and proportion.*0006

*We will have to do a little bit of work just to explain what I mean by the word variation.*0012

*We will break this down into few other things.*0017

*We will look at direct variation, inverse variation, and combined variation.*0019

*This particular section is filled with lots of different word problems all of them involving variation.*0025

*Let us see what we can do.*0032

*When you hear that word variation we are talking about a connection between two variables,*0036

*such that one is a constant multiple of the other.*0041

*You will see some nice handy formulas that will highlight how one is a constant multiple of the other.*0044

*If you are looking for little bit more intuition on the situation then you can grasp on to that.*0050

*Intuitively a variation can be thought of as a special connection and the two basic types are direct and inverse.*0055

*In this connection when one quantity goes up then the other one would go down.*0063

*In that one I'm talking about an inverse variation.*0069

*If Iâ€™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.*0075

*It is how one affects the other one and they could be moving in the same direction our going in the opposite directions.*0084

*To model this type of situation that has variation in it, you can end up setting up a proportion.*0094

*Be very careful on where you put each of the individual components and make sure you line them up correctly.*0100

*To model a direct variation, you could use the following proportion x1 / y1 = x2 / y2.*0107

*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.*0116

*Everything with the 2 involves a second situation.*0126

*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.*0133

*We will line those up to make sure that they at least agree.*0142

*It seems a little vague but let us get into an example*0148

*and you will see how I do set this up using a proportion and how everything does line up.*0151

*This one says that an objects weight on the moon varies directly as its weight on earth.*0156

*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.*0163

*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?*0174

*We have that this is a direct variation because it pointed out and says it is a direct variation.*0186

*Let us just think a little bit and see if that makes sense.*0192

*Intuitively in a direct variation when one quantity goes up then should be the other one.*0196

*When one quantity goes down, then so should the other one.*0201

*That is what is going on with our moon weight.*0204

*If I decide to go on and get heavier and heavier on Earth, that same thing is going to happen on the moon.*0207

*I will get heavier and heavier on the moon.*0214

*This is a type of direct proportion.*0217

*Let us see what we can set up.*0223

*x1 / y1 = x2 / y2*0224

*To help out Iâ€™m going to highlight two things, we will go ahead and keep our Earth weights on top*0231

*and we will go ahead and put our moon weights on the bottom.*0238

*Let us look at our first situation with our Neil Armstrong.*0245

*He weighed 360 pounds on Earth, and only 60 pounds on the moon.*0250

*360 on Earth and 60 on the moon.*0255

*I will compare that with our average person there.*0267

*They weigh 186 pounds on the Earth, but we have no idea how much they weigh on the moon.*0270

*I'm going to put that as my unknown.*0277

*Everything on the tops of these fractions represents an earth weight and everything on the bottom represents a moon weight.*0280

*You can see that we have developed one of these rational equations and now we simply have to solve it.*0287

*This one is not too bad if you go ahead and reduce that fraction on the left, it goes in there 6 times.*0292

*We can multiply both sides by x.*0301

*I have 6x = 186.*0308

*Finally dividing both sides by 6 I will get that x =31.*0316

*This reveals how much that average person, the 186 pounds would weigh when they are on the moon.*0327

*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.*0338

*Look for clues along the way to help out.*0344

*This one says that a maintenance bill for a shopping center containing 270,000 ft ^{2} is $45,000.*0348

*What is the bill of the first store in the center that is only 4800 ft ^{2}?*0356

*Let us think of what type of a variation this should be.*0361

*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.*0366

*If I have a small store then my bill should be very small.*0378

*Those quantities are moving in the same direction.*0382

*Iâ€™m going to say this is an example of direct variation.*0385

*We will set it up using our proportion.*0392

*We want to keep things straight.*0403

*What do each of these quantities represent?*0405

*Let us say the top will be our square footage and we will make the bottom the bill.*0409

*How much it cost?*0421

*Let us go ahead and substitute some of this information we have.*0424

*If Iâ€™m looking at the entire shopping center I know how big it is and I know the bill for the entire place.*0427

*270,000, 45,000 and now I can look at my much smaller store, it only has 4800 ft ^{2}*0435

*and this one we have no idea what the bill is so I will leave that one as my unknown.*0453

*I think we can do a little bit of simplifying with this one as well.*0459

*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.*0464

*I have 6x = 4800.*0481

*Divide both sides by 6 and then I think we will have our solution x =800.*0488

*Even if they do not come out and tell you what type of variation is going on here,*0497

*look at the values present to see if things are moving in the same direction, or in opposite directions.*0502

*For inverse variation we want to make sure that if one quantity goes up, the other one goes down.*0510

*One way that we can model it now with our proportions is to set up like this.*0516

*We will go ahead and use us a subscript for one situation and we will put them on opposite sides of our equal sign.*0521

*We will get that inverse relationship.*0528

*When one goes up, the other one will go down.*0531

*For our second situation we will put these on two opposite sides of the equal sign.*0535

*One thing that will not change is that both of these x will represent the same type of thing*0541

*and both of these y will still represent the same type of thing.*0546

*We can definitely note that the things are on opposite sides of the equal sign for these given situations.*0550

*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.*0562

*If I have a current that is 30 amps when the resistance is 5 ohms, find the current for resistance of 25 ohms.*0572

*I have a lot of things in here and we just want to keep track of each of them.*0581

*Let us see if we can highlight each of these situations.*0586

*The current is 30 amps when the resistance is 5.*0589

*We will be looking for the current when the resistance is 25.*0596

*Let us get our proportion out here.*0605

*Iâ€™m using these little subscripts and put them on opposite sides.*0611

*Onto the first situation, we will keep everything on the top.*0615

*Let us say that is our resistance, we will put everything on the bottom as the current.*0624

*Our first situation, we know that the current is going to be 30 when the resistance is 5.*0635

*In that situation I will put my 5 and 30.*0641

*For the other situation I want to figure out what the current is.*0647

*I have no idea what that is but I know that the resistance is 25 ohms.*0651

*This is my initial proportion that I have set up.*0657

*All we have to do is go through the process of solving it.*0661

*This one is not too bad.*0664

*Let us multiply both sides by 30x and see what is left over.*0666

*30 Ã— 5, since the x will cancel out equals 25 Ã— x since the 30 cancel out.*0673

*Okay looking pretty good.*0688

*Also 5 Ã— 30 =150 equals 25x and we can simply divide both sides by 25 to get our answer.*0690

*The current is 6 amps.*0708

*When dealing with these variation problems always look at your solution and see if it makes sense in the context of the problem.*0722

*One thing that I'm noticing in here is that I start off with a resistance of 5 and a current of 30.*0729

*One thing I'm doing with my current is that I'm lowering it, and I'm taking it from 30 down to 25.*0736

*Since I have an inverse relationship that is going to have an effect on our amps is the current.*0744

*That should bring it up.*0751

*Originally I started with 5 and I can see my answer is 6.*0753

*It did go up and things are working in the right direction.*0757

*Since earlier, we said that it variation is relationship when one is a constant multiple of the other.*0764

*We can model this using some nice formulas.*0769

*For direct variation you can use the formula y = k Ã— x.*0773

*For your inverse variation, you could use y = k Ã· x.*0779

*We have two variables in each of these equations that we are looking at connecting are the x and y.*0785

*The k in each of these is known as our constant of proportionality.*0791

*It connects how they are related.*0795

*A lot of problems that you will end up doing with these variations, if you want to use these formulas*0799

*then you will end up going through three steps to figure out what is going on.*0806

*You might first go ahead and identify what type of variation you are using so that you can take one of these formulas.*0811

*Then you will end up using a little bit of known information to go ahead and solve for that constant of proportionality, k.*0818

*Once you know what k is, then you can usually figure out some new information by substituting it into the formula.*0825

*I got a couple of examples where we are going through these 3 steps.*0832

*Let us first just see an example of using an inverse of variation using one of these formulas.*0840

*This one says that the speed of water through a hose is inversely proportional to a cross section of the area of the hose.*0846

*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?*0855

*Let us think about what we got here.*0867

*We are looking for a connection among the cross area section of the hose and the speed of the water.*0868

*We are told that they are set up inversely so when one goes down, the other one should go up.*0876

*For decreasing the area then we should expect it to increase the speed of the water.*0879

*In fact we can be a little bit more precise, but despite playing around with those formulas for a little bit.*0887

*Our relationship is that the speed of the water is inversely proportional to a cross section of the area.*0893

*Maybe x = k/a.*0900

*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%.*0905

*Let us relate this to the original before I manipulated it.*0923

*25 is the same as Â¼ I'm dividing by Â¼.*0928

*When you divide by a fraction that is the same as multiplying.*0936

*This is multiply by 4.*0943

*Now we can see the difference between the original situation.*0947

*The area decreased by 75%.*0956

*They are almost exactly the same but in the second situation, it is 4 times as much.*0966

*It will actually increase the speed by 4 times.*0971

*There are lots of other types of variations that you can go ahead and work into a problem.*0978

*Some of these will look similar to previous types that we covered.*0982

*Let us go through how each of these are connected.*0984

*If I say that y varies directly to some power of x then there are some k that connects those two things.*0990

*You will notice that it looks like our direct proportion only since we are dealing with an nth power,*0997

*like a 3rd power 4th power that we put that exponent right on the other variable.*1004

*Our y varies inversely as the nth power looks much like our inverse relationship as well.*1013

*But since we are doing it to the nth power, notice how we put that exponent right on the other variable.*1020

*In this one is a new one that involves a few more variables, this is our joint variation.*1029

*y varies jointly as x and z so more than one variable now.*1034

*If there some sort of positive constant such that y = k Ã— x Ã— z.*1039

*In this one the k is our constant and the variables are being connected, the y, x, and z.*1046

*You can start mixing and matching these various different types of variation.*1057

*The reason why that we do not want to start mixing and matching them is we can get some more complicated models.*1064

*With these more complicated models we are hoping to be able to more accurately model what is happening in a real life situation.*1071

*Usually it does a better job than any one variation could do by itself.*1079

*The key when putting all of those different types of variations together is to look for clues in the actual problem itself.*1085

*Here is a formula that I have it is T = k Ã— x Ã— z ^{3} /âˆšy .*1092

*Here is how I could describe that problem using the language of these variations.*1100

*The k right here is my constant of proportionality and we are looking to get it connect together the t, x, z, and y.*1107

*Here is what I can say, the t varies jointly with x and z ^{3}.*1116

*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.*1123

*t also varies inversely with the âˆšy.*1133

*Since it is an inverse relationship I will put that one in the bottom and make sure to put in that square root.*1138

*Watch for those keywords so we can figure out where we should put things.*1145

*Let us try some of these variation problems and watch as I walk through the three steps of*1151

*starting with some template solving for our constant of proportionality, k and the last part figure out some new information.*1155

*What do I know so far?*1167

*I know that y varies jointly as x and z.*1168

*Just on that little part, let us go ahead and make a nice formula.*1173

*y varies jointly as x and z.*1177

*y = 12 when x = 9 and z = 3.*1184

*I can use that information right there to go ahead and solve for my constant of proportionality.*1191

*y =12, x = 9, z= 3*1198

*The only unknown I have in there is that k value.*1206

*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.*1210

*Now that we know much more I can end up revising our formula.*1228

*y = xz and now I can say that the k value is this 4/9.*1236

*Now that we know about our constant of proportionality, let us find z when y = 6 and x = 15.*1244

*We will substitute those directly in there and you will see the only unknown I have left will be that z.*1253

*Y= 6, 4/9 Ã— 15 and we are not sure what z is now we have to solve for it.*1259

*A little bit of work.*1270

*I think I can cancel out an extra 3 here.*1272

*I have 6 = 20/3 Ã— z we could multiply both sides by 3.*1276

*I will get rid of those 3, 18 = 20z.*1292

*Lastly let us go and divide both sides by 20.*1298

*18 Ã· 20 =z*1301

*We will just go ahead and put that into lowest terms.*1306

*Divide the top and bottom by 2 and I know that 9/10 is equal to z.*1310

*If you follow along we start with some sort of template for our variation, we solved for our unknown constant of proportionality k.*1316

*We have used it to figure out some new information about the other variables.*1325

*For this last type of example let us look at one that is a little bit more of a word problem.*1331

*This one says that we have the maximum load of the beam and how much you can support varies jointly*1336

*with the width of the beam and the square of its depth, but it varies inversely as the length of the beam.*1343

*Some of that initial information there is just giving us a formula for how everything is connected.*1352

*We will definitely set up the equation that will model that situation.*1358

*Let us continue on.*1364

*Assume that a beam that is 3 feet wide, 2 feet deep, 30 feet long and it can support 84 pounds.*1365

*What is the maximum load a similar beam can support if it is 2 feet wide, 5 feet deep and 100 feet long?*1371

*A lot of information is floating around there.*1378

*Let us start off by seeing if we can set up this formula.*1381

*The maximum load that a beam can support varies jointly as the width of the beam and the square of its depth.*1385

*Since this is a jointly, we are talking about k Ã— width Ã— the square of its depth or d ^{2}.*1395

*It varies inversely with the length, I will put the length on the bottom.*1408

*This part here is our formula that we will end up using.*1413

*Assuming that a beam is 3 feet wide, 2 feet deep and 30 feet long, and it can support 84 pounds,*1421

*let us use that to figure out our unknown constant sitting right here that k.*1426

*I know it can support 84 pounds, we do not know k.*1432

*3 feet wide, I put that in for width, 2 feet deep and it is 30 long.*1440

*The only unknown is that k.*1448

*Let us go ahead and simplify it just little bit in here.*1451

*I have 3 Ã— 2 ^{2} and 2^{2} is 4 = 12*1455

*I can get that k a little bit more by itself if I multiply both sides by 30, 84 Ã— 30.*1463

*Let us go ahead and put those together and see that would be 2520.*1473

*Divide both sides by 12.*1483

*That is where we have been able to figure out what that unknown constant is.*1488

*Now that I have that I can go back to the original formula.*1494

*M = 210w Ã— d ^{2} / l*1498

*We will use this new formula to figure out a little bit of known information.*1506

*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.*1512

*Let us give it a try.*1520

*That formula we developed was its maximum load = 210 Ã— width Ã— d ^{2} / length.*1525

*Let us put in our information.*1536

*For this new beam, its width is 2 feet and its depth is 5.*1541

*It also has a length of 100.*1549

*As soon as we simplify all that we will be able to figure out what its maximum load should be.*1551

*210 Ã— 2 = 420 Ã— 25 Ã· 100*1558

*Let us do some simplifying here 420 Ã— 25 = 10500 Ã· 100 and looks like this reduces to 105.*1573

*This new beam can support a maximum of 105 pounds.*1593

*Being able to recognize various different types of variation*1599

*and combine them all together is essential for some more complicated problems.*1602

*Remember that if you have those nice formulas then you can go through 3 steps.*1607

*One, build a formula or equation that models the situation, solve for your unknown k*1611

*and finally use some new information to figure out some more information about the problem.*1617

*Thank you for watching www.educator.com.*1622

1 answer

Last reply by: Professor Eric Smith

Mon Oct 19, 2015 11:45 AM

Post by Khanh Nguyen on October 17, 2015

In practice question #7, in the answer, it says quardrants.

In practice question #8, in the answer, it says quadrant instead of quadrants.

Could you please fix it?

:^D