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Direct Variation

Let's wrap up this unit on rational expressions by looking at variation. Variation is a concept about relating two or more amounts. For example, if you have an hourly wage at a job we can relate the number of hours you work to the amount of money you earn from that job. Let's say your current wage is $10 an hour and if you work 40 hours in a week, you can make $400. We get this simply by multiplying the number of hours we worked. By our current wage. And it turns out there's a unique relationship between the wage and our hours. So what do you think would happen if you worked more hours? Would you make more money, less money, or the same amount?

Direct Variation

You would make more money. If you worked 50 hours, you would end up making $500. This is an example of direct variation, since as one quantity increases, the number of hours, the amount of money we get also increases. In simple terms, the more you work, the more you get paid.

Inverse Variation

Direct variation means that two amounts are related such that if one increases, the other increases. Or, if one decreases, the other decreases as well. This makes sense in terms of working for an hourly wage and earning money. If you work less hours, you earn less money. The second type of variation is inverse variation. If you work more hours, the amount of free time that you would have would decrease. Notice here, one quantity is increasing but the other one would be Decreasing. We could also have the relationship reversed, we could work less hours, which means we would have more free time. In the rest of this lesson we're going to use our knowledge of direct variation and inverse variation to solve some problems. But before we get there, let's see if you understand the difference between the two.

Direct or Inverse Variation

For each pair of these items, I want you to determine if the relationship is direct,or if it's inverse. Really take some time to think about each one of these, and when you're ready put your choices in the appropriate spot. Good luck.

Direct or Inverse Variation

For the first example, if we drive more miles in a rental car the cost will usually go up. Now I know most rental car companies have a flat fee but some also charge per mile and they might charge if you go over a certain number of miles. So in general, if the miles go up the cost also goes up. So that must be direct. The second example would be inverse let's say 2 people live together and rent is $400 this would mean each person pays $200 but if we increase the number of roommates let's say from 2 to 4 then people wouldn't pay $200 each they'd pay less. Now instead of paying $200 each person would just pay $100. So as one quantity increases, the other decreases, inverse. The second case is another example of inverse. As the speed of a vehicle increases, the time that we have to drive that vehicle decreases. And maybe you've never driven a car before. But you could still think about this if you were walking. If we walk slower or drive slower then that means our time would increase to get to our final destination. One quantity is increasing while the other is decreasing. So that's definitely an inverse relationship. This fourth example is an example of direct. If we don't drive for a while, then we can't cover as much distance. So as one decreases, the other would decrease. Likewise, if we drive for a long time then we can cover more ground so this would be direct. For this last example if we make more money then we can spend more money. So that will be a example of the direct. If our arrows point in the same direction then we know it is a direct relationship whereas if the arrows point in opposite directions we know it's an inverse variation.

Direct Variation Practice 1

In our first example, you could work 40 hours and earn $400. Let's say you start a new job that pays a different hourly wage. Now you're going to work 35 hours and earn $630. At this new job, let's say you worked 35 hours and you earned $630. How much could you earn working 50 hours? Whenever we saw problems involving variation we want to set up a proportion. So let's make this really simple. If I worked one hour, I would earn $10 based on my original example. So if I work double that amount of time, I would get double the amount of money. I would earn $20. Notice that if my hours worked increased then my amount of money also increased. This is so that way these two fractions will be equal. And one of the biggest things we want to make sure of when solving variation problems is that we have one fraction involving hours another involving income. We always want to keep our units the same. Whenever we have direct variation, the first amount of one unit will be directly across from the first amount of the second unit. So when I'm filling in this proportion, 35 hours should go here for the amount of hours worked for one And $630 should go here for the income worked for much money could we earn if we worked 50 hours? Be sure you don't include the dollar sign in your answer. I already have that written for you.

Direct Variation Practice 1

The answer is $900. Great work if you figured that one out. We know in our first case we work 35 hours and we earn $630. For the second case we're going to work that income two. Next we want to multiply both sides of our equation by the lowest common denominator or 50 times x. So we'll be left with 35 times x on the left and 630 times 50 on the right. Essentially when we do this we cross multiply. We'll have x time 35 here, and 50 times 630 here. Finally, we divide both sides by 35, and we get x is equal to 900. When you're problem solving, you want to make sense of this number in the end. What were we solving for? Well we were wondering how much we could make, so that means we made $900. I know I included the unit here for you, but make sure you do that in your own problem solving and throughout the rest of your work.

Direct Variation Practice 2

It turns out we can always set up a proportion to solve direct and inverse variation problems. Let's try this one out. For this problem the area of a circle varies directly with the square of it's radius. A circle with a side length of 3 inches has an area of 28 and 278 thousandths inches squared. What is the area of circle with a radius of 4 and 1 tenth inches? Now this problem is very similar to the last one we just did. Except you need to include on more thing. We're not just varying directly, we're varying directly with the square of something. What do you think you're going to do? Try playing around with this one and I hope you'll get it. Good luck. Also you don't want to round while you solve the problem. When you find the final answer, round your answer to the nearest thousandth. And again, I've already included the unit here of inches squared, so just write in your area here.

Direct Variation Practice 2

We would have 52 and 817 1000ths inches squared. This is our correct answer. Fantastic thinking if you got there. Now, I know we haven't seen a problem like this one yet, or that involves the square of a quantity, so don't worry if you struggled. Let's see how we can do it. Because the area varies directly, we want to make sure that we list the first quantities that cross from one another. And notice too that the area doesn't just vary directly with the radius, it varies directly with the square of the radius. So we have to square the first radius. And then, for our second area, it will also vary directly with the square of its radius. Now that we have our proportion set up, let's fill in each of the numbers in the appropriate place. This would be our first area and this would be the radius that goes with it. So we want to put 28 and 278 1000ths here, and 3 here. Next we have the second radius which we know is 4 and 1 10th. So that will go here, and then we'll use x for the area of 2. That's the area we're looking for in our question. First I'm going to square each of these numbers to get this fraction. Next we cross multiply to get this equation. And finally we divide both sides by 9 to get our final answer.

Inverse Variation Practice 1

We've seen a couple of cases of direct variation, so let's look at inverse variation. Earlier we determined that the speed of a vehicle and the driving time were inversely related. If we drive really fast, then we don't need as much time to get to our final destination. If a car takes 20 minutes to travel some distance at 45 miles per hour, how long would it take that same car to travel the same exact distance driving at 60 miles per hour? We want to be very careful though. We don't want to use this proportion. This is for direct variation. This proportion would say that if we drive faster, then it would take longer to get somewhere. But we know this is not true. We know if we drive faster, then our time spent driving decreases. So instead of doing this, we can flip this second fraction over. If we flip the second fraction over, we can see that our speed now would be decreasing. If we drive slower, then it should take longer amount of time to get there. This also fits with our notion of speed and time. If speed increases, time decreases. Likewise if speed decreases, as in this case, then we know that we should expect to drive longer. Time increases. For inverse variation, we always want to put the other quantity in the opposite position. We'll need to flip the fraction that varies inversely. So what do you think, how long would it take us to drive?

Inverse Variation Practice 1

It turns out that it would only take 15 minutes. Nice algebra skills if you got that one correct. We know in the first case, we're driving for 20 minutes at 45 miles per hour. So 20 minutes would be time 1, and 45 miles per hour would be rate 1. We're looking for how long it takes for the car to travel the same distance. So this must be x, our time 2. And for the second rate, we're traveling at 60 miles per hour, so we fill in this with 60. Then, we cross multiply to get 900 equals 60x. We divide both sides by 60 to get x is equal to quantity will be in the numerator while the other will be in the denominator. We usually flip the second fraction to make that happen.

Inverse Variation Practice 2

Let's try another problem on inverse variation using light. The intensity of light varies inveresely as a square of the distance from the source. So think about if we shine the light unto a screen. This screen would have a light spot on it and this would be the intnsity of light on the screen. We measure this intensity on whats called a foot candle, so if a light source is 5 feet from a screen There will be an intensity of 2 foot candles knowing this let's find the intensity on the screen if this light source was moved to 20 feet away. Now you might never have heard of a foot candle before but just know that foot candle is a way that we can measure intensity. Look this up if you want to learn more about light and foot candles. Distance here is just measured in feet, so 5 feet would be a distance, as would 20 feet. So try to figure this problem out and remember that the intensity of light is varying inversely with the square of the distance. Think carefully about the proportion you need to setup and when you're ready put your answer here.

Inverse Variation Practice 2

You should have gotten one eighth of a foot candle. Fantastic work for figuring that one out. We want to start by setting up our proportion using inverse variation. We know the intensity of light varies inversely as the square of the distance. This means that for intensity one, the distance one needs to be squared in the denominator. We flip the second fraction. Now I think it's really easy to make a mistake when plugging in these amounts. So I would make a list of the amounts and then write in the numbers or variables that go with them. We know intensity one is two-foot candles, and that's at a distance of five feet away. So intensity one is two-foot candles, and the distance one is five feet. We're looking for the second intensity. We don't know it so we use x. And this intensity occurs at 20 feet from the light source. So we have 20 feet for our second distance. We can use this list to help us plug in the values to get this proportion. We'll leave two divided by x here, and then we'll have 400 divided by 25 here. We cross-multiply to get 50 equals 400 x, and then we divide both sides by 400. To get x is equal to 1 8th, 1 8th is this equivalent fraction. We simply divide each number by 50. We saw for an intensity of light so we want to make sure we put the correct unit. It was 1 8th foot candles.

Joint Variation

We've seen direct variation and we've seen inverse variation, but these two don't have to be separate. There are times when the quantities can vary jointly. In other words, we'll have more than just two amounts related to each other. An example of this would be simple interest that accrues over time. Simple interest is equal to the principal amount we start with times the interest rate. The amount of interest that accumulates in one year is dependent, or varies directly with these other two values. If the principal amount were to go up, our interest would also go up. Likewise, if we increase the interest rate, our interest would go up as well. So this is a case of joint variation. Because simple interest varies directly with both the principal amount and the interest rate, we could set up a proportion like this. The principal amount and the interest rate associated with our first interest would appear directly across from it. So if we had any five of these quantities, we could solve for the remaining one, or the one that's left out. Let's see if you can use this idea of joint variation to solve a different problem.

Joint Variation Practice 1

This problem deals with mailing a package. The price for mailing a package varies directly with its length and its width. This should make sense, since the larger something is, the more money and work that it would take to ship it across the country or even the world. So if it costs $20 to ship a package that measures 4 feet and weighs 10 pounds. How much would it cost to mail a package that measures 7 and a half feet, and weighs 15 pounds? Write that price here and don't worry about putting your dollar sign. I have it written for you.

Joint Variation Practice 1

The package would cost $56.25. Great work if you found that price. This price also makes sense, since our package weighs more and it's longer in length. So it should definitely cost more than $20 to ship. We want to start by setting up a proportion that has direct variation. The cost varies directly with the length. And the cost varies directly with the weight. The next thing that we want to do is make a list of the things we know. We know the first cost was $20.00, the length was four feet, and it measured 10 pounds. Next we fill in the proportion with the remaining pieces. The second cost, the second length, and the second weight of the package. We make a list of those amounts as well. We're looking for the cost of our package, so that's going to be x. The length of that package is seven and a half feet. And the weight of that package is 15 pounds. Now we're ready to fill in our proportion. Plugging in these amounts into our proportion, we get this equation. I'm going to clean up the right hand side of my equation first. Four times ten is 40. And 7.5 times 15 is 112.5. So I've written this cleaned up part over here. Next we cross multiply to get 2,250 equals 40x and finally we divide both sides by 40 to get our final answer.

Joint Variation Practice 2

With joint variation, though, the variations do not both need to be direct. One quantity might have direct variation, while the other might have inverse variation. For example, if x and y had direct variation, they would both increase, or they would both decrease. And if z varied inversely with x, then if x increased, z would have to decrease. When setting up the proportion, this means that since x varies directly with y, they will be directly across from one another. But since x varies inversely with z, we'll see this z quantity in the denominator. We flip this fraction. So let's assume this is still true. X is varying directly with y, and inversely with z. If the value of x is 10 when y equals 5 and z equals 6. I want you to find the value of x when y is 7 and z is mind in the setup, we had direct variation with y, so this is directly across from each other Whereas we had inverse variation with c. This c is a denominator. Write your answer for x here. You can also think about we're really finding x too, this value. Good luck.

Joint Variation Practice 2

it turns out the value of X would be 84 when Y is & and Z is 1. Nice work if you figured that one out. We know the values for X1, Y1, and Z1 so we plug those in here. We are looking for the X value when Y is 7 and Z is 1, so we plug in 7 and We're looking for x 2, which is x, and we know the second y value is 7, and the second z value is 1. We use these lists to make our substitution. We simplify this right hand side to get 5 42nds. Next we multiply both sides of the equation by the LCD, 42x. Essentially, this is the same as cross-multiplying. So we get to 84.

Variation Practice 1

Try solving this practice problem on variation. You'll want to decide which type of variation it is first, and then set up the correct proportion. Write your final answer here, and don't worry about the unit; I've already included it for you.

Variation Practice 1

A man that weighs 175 pounds should have 35 milligrams of the drug. Great solving if you figured that out. If the proportion gave you a little bit of trouble, let's see how we can set it up together. Because the amount of medicine varies directly with the person's weight, we want to set up this proportion, the amount of medicine or the dosage for the first case Would be directly across from the weight from the first case. 30 milligrams is prescribed for someone who weighs 150 pounds. So this would be the first numbers for our first case. We don't know how much of the drug should be prescribed for a man that weigh 175 pounds. So the amount of the drug, or the amount of medicine 2 is unknown, it's x. But the corresponding weight for that drug dosage is 175 pounds. So we put that here. Finally we cross multiply and then divide both sides by 150. This leaves us with x is equal to 35. Now I really think it's important to make sense of our reason through this answer. Should this be more or less than the 30 milligrams originally? Well we said the amount of medicine varied directly. So that means if the drug dosage increases, the weight increases as well. We know the weight increased from 150 pounds to 175 pounds. So it'd make sense that our drug increased from 30 milligrams to 35 milligrams. If anything, we can just be sure that the second amount is higher, or greater than, the first amount.

Variation Practice 2

Try this second practice problem out. What do you think would be the solution for a? Also be very careful when setting up your proportion. Think about which values are varying inversely and which values are varying directly. Good luck.

Variation Practice 2

And in this problem, a would equal 3. Way to go if you got that one correct. Since a varies inversely with b, we want to make sure that we put the b quantity in the denominator. But for c, a varies directly with it. So we want to put that straight across in the numerator. This is perhaps the trickiest part about setting up variation problems. And you want to make sure you get it right each time. Remember, with inverse relationships. They'll appear in opposite positions, whereas with direct relationships, the variables will appear directly across from one another. So in my first case, a is 12, b is three, and c is eight. So I plug in those values. In the second case, we don't know the value of a, so I leave that written as a in the denominator. But I do know that b2 is equal to six. And c2 is equal to 4. These are the values of b and c in our second case. Now that we have this equation, we multiply these two fractions together to get 48 divided by 12. We cross multiply to get 144 equals 48a, and finally we divide both sides by 48 to get a is equal to 3.

Variation Practice 3

Try this problem out. Here, x is going to vary directly with y and inversely with z. Set up your proportion, and then solve for the value of y when x is 6 and z is 4 based on this other information.

Variation Practice 3

For our third problem, y would be equal to 8. Great work for getting that one correct. You might have had trouble with the set up, so let's see how we can do that. We know x varies directly with y, so they should be directly across from each other in our proportion. But for z, x varies inversely, so z should appear in the denominator. I want to start by plugging in the values of my first case first, x1 is 5, y1 is 5, and z1 is 3. In my second case, I don't know this value of y. It's what we're looking for. So I'd be sure to list y2 as my variable. Or just y. The second value for x is 6, and the second value for z is positive 4. Now we have an equation we can solve with a rational expression. We multiply these two fractions together, to get 20 divided by 3y. Then we cross-multiply to get 15 y times 120. And finally we divide by 15 to get y is equal to 8. Again, I think the setup to this problem is probably the hardest. You want to make sure that when you're varying directly, things are across from one another. Whereas if things vary inversely, they're diagonally across from one another.

Variation Practice 4

Try this fourth problem out. Now, I know this looks really complicated. But it's just about a cylindrical tank. This cylindrical tank is going to hold gasoline in it. The number of gallons in this tank varies directly with the height and the square of this radius. See if you can set up the correct proportion to relate the number of gallons, the height, and the square of the radius. Use all this information in the problem to figure the new height of the gasoline level if these other conditions are true. Good luck.

Variation Practice 4

Since we rounded to the nearest 10th of a foot, our gasoline height, our level, would be 2 and 7 10ths of a foot. Amazing work if you found that one. We start setting up out proportion by looking at the number of gallons, the height, and the square of the radius. We know the number of gallons varies directly with the height of the gasoline level and the square of the radius. Since it varies directly with these two amounts, we want to make sure we list those both in the numerator. And the square part is really tricky. You want to make sure that you square this radius. Once we have this proportion we can plug in the values for each case and solve for the unknown. In one case the tank has a radius of five feet, so we plug in five here. It holds 4,700 gallons, here and has a height of eight feet. So the height one is eight. In the second case, we're looking for the height of the gasoline level so that will be H. At some unknown height, H, the radius will be four feet, so the second radius is four. And the number of gallons that the tank will hold will be 1000. So gallons 2 is 1000. Now, we want to solve this equation for h. We square the 5 and the 4 to get 25 divided by 16. Next, we multiply these 2 fractions together to get 200 in the numerator, and finally, we divide by the coefficient of x to get our final answer. We get a decimal for our answer and we want to round to the nearest tenth. Because the value next to the 6/10 is a 5 we round up. If this value was less then a 5 then we round down and keep our answer as 2 and 6/10th. But for 5 or greater we round up so we get 2 and 7/10. This is the level for our gasoline take that holds a thousand gallons and has a base rate of 4 feet.