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Direct, Joint, and Inverse Variation

  • Use values given in a problem to find the constant of variation. The use this value and other values to find the missing value.

Direct, Joint, and Inverse Variation

y varies directly as x, and when x = 2, y = 8. Find y when x = − 6
  • Recall that k = [y/x] so you have the following
  • k = [8/2] k = [y/( − 6)]
  • You can set - up a proportion, and cross multiply
  • [8/2] = [y/( − 6)]
  • 2y = − 48
y = − 24
y varies directly as x, and when x = 6, y = − 16. Find y when x = − 1.5
  • Recall that k = [y/x] so you have the following
  • k = [( − 16)/6] k = [y/( − 1.5)]
  • You can set - up a proportion, and cross multiply
  • [( − 16)/6] = [y/( − 1.5)]
  • 6y = 24
y = 4
y varies jointly as x and z. When x = 5 and z = 3, y is equal to 45
Find z when x = 2 and y = 18
  • Recall that the joint variation formula is y = kxz so the constant is k = [y/xz]
  • You know the following
  • k = [45/5*3] k = [18/2z]
  • Set up a proportion and solve for z
  • [45/15] = [18/2z]
  • 90z = 270
z = 3
y varies jointly as x and z. When x = 3 and z = 6, y is equal to 36
Find z when x = 4 and y = 40
  • Recall that the joint variation formula is y = kxz so the constant is k = [y/xz]
  • You know the following
  • k = [36/3*6] k = [18/2z]
  • Set up a proportion and solve for z
  • [36/18] = [40/4z]
  • 144z = 720
z = 5
y varies jointly as x and z. When x = 6 and z = 9, y is equal to 42
Find z when x = 5 and y = 40
  • Recall that the joint variation formula is y = kxz so the constant is k = [y/xz]
  • You know the following
  • k = [42/6*9] k = [18/2z]
  • Set up a proportion and solve for z
  • [42/54] = [10/5z]
  • 210z = 540
z = 2.57
y varies directly as x, and when x = 9, y = − 3. Find y when x = − 4.5
  • Recall that k = [y/x] so you have the following
  • k = [( − 3)/9] k = [y/( − 1.5)]
  • You can set - up a proportion, and cross multiply
  • [( − 3)/9] = [y/( − 4.5)]
  • 9y = 13.5
y = 1.5
y varies directly as x, and when x = 27, y = − 1.5. Find y when x = 10
  • Recall that k = [y/x] so you have the following
  • k = [( − 1.5)/27] k = [y/( − 1.5)]
  • You can set - up a proportion, and cross multiply
  • [( − 1.5)/27] = [y/10]
  • 27y = − 15
y = − [5/9]
y varies inversely as x. When x = 32 , y = − 6. Find x if y = − [1/5]
  • Recall that k = xy
  • You have x and y, find k
  • k = (32)( − 6)
  • k = − 192
  • Now that you know k, find x
  • x = [k/y]
  • x = [( − 192)/( − [1/5])] = − 192*( − 5) = 960
960
y varies inversely as x. When x = 10 , y = 12. Find x if y = [1/9]
  • Recall that k = xy
  • You have x and y, find k
  • k = (10)(12)
  • k = 120
  • Now that you know k, find x
  • x = [k/y]
x = [120/([1/9])] = 120*(9) = 1080
y varies inversely as x. When x = 2 , y = 12. Find x if y = 5
  • Recall that k = xy
  • You have x and y, find k
  • k = (2)(12)
  • k = 24
  • Now that you know k, find x
  • x = [k/y]
x = [24/5] = 4.8

*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

Direct, Joint, and Inverse Variation

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
  • Direct Variation 0:07
    • Constant of Variation
  • Graph of Constant Variation 1:26
    • Slope is Constant k
    • Example: Straight Lines
  • Joint Variation 2:48
    • Three Variables
  • Inverse Variation 3:38
    • Rewritten Form
    • Examples in Biology
  • Graph of Inverse Variation 4:51
    • Asymptotes are Axes
    • Example: Inverse Variation
  • Proportions 10:11
    • Direct Variation
    • Inverse Variation
  • Example 1: Type of Variation 12:42
  • Example 2: Direct Variation 14:13
  • Example 3: Joint Variation 16:24
  • Example 4: Graph Rational Faction 18:50

Transcription: Direct, Joint, and Inverse Variation

Welcome to Educator.com.0000

Today we are going to be talking about direct, joint, and inverse variation.0001

Starting out with direct variation: this is a review of a concept that you likely learned in Algebra I.0008

The direct variation equation is y = kx, and this is sometimes expressed as k = y/x.0015

And what this equation shows us is that y varies directly as x.0025

So, there is a relationship between x and y, and it is explained with this constant, which is called the constant of variation.0030

And what it says is that, as y increases, x is also going to increase; or as y decreases, x will also decrease--two things are moving in the same direction.0041

And there are many real-world examples of this.0053

For example, I could say that the amount of time it takes me to cut a lawn varies directly with the size of the lawn.0056

So, the larger the lawn is, the longer it will take me to cut it.0064

And there is some constant, such as the rate: if I can cut two square feet per minute, then that would be0068

the constant of variation--the rate at which I am cutting the lawn.0077

One variable would be the time it takes me, and the other is the size of the lawn.0080

Recall the graphs of direct variation: the graph of a direct variation is a straight line that passes through the origin.0087

And its slope is the constant, k (the constant of variation).0095

Now, you can have a straight line graph, passing through the origin, that moves up to the right.0101

You can also have a straight line graph where y is increasing towards the left.0107

And recall that, when we talked (back in Algebra I) about graphs of linear equations, if the slope is positive,0117

the graph is going to look like this--it is going to go up to the right.0125

If the slope is negative, it is going to go up to the left.0127

And in fact, k is the slope of this line; therefore, if k is positive (it is greater than 0), the graph is going to look like this--0131

whereas, if k is a negative number (the slope is less than 0), the graph will look like this.0144

So, the equation y = kx is actually in slope-intercept form, y = mx + b, where the y-intercept is 0, and the slope is given by k.0151

Joint variation is very similar to direct variation, except now we are working with three variables instead of 2.0169

And we say that y varies jointly as x and z if there is some non-zero constant, k, such that y = kxz (and k is the constant of variation).0175

So again, going back to the lawn analogy: I could say the amount of time it takes me to mow the lawn0188

varies jointly with the size of the lawn and the steepness of the hill that it is on.0194

So, as the size of the lawn increases and the steepness of the hill that it is on increases,0201

the time that it takes me to mow the lawn will increase.0208

So now, I have three things that are maintaining a constant relationship with each other.0210

Inverse variation is different: recall that we say that y varies inversely as x if there is a non-zero constant, k, such that y = k/x.0218

This can also be rewritten as k = xy; and both x and y are not equal to 0, because otherwise this would become 0, and there wouldn't be an equation.0233

An inverse relationship: what we can also look at this as is that y is proportional to the reciprocal of x.0243

Instead of varying directly with it, it is proportional to the reciprocal of it.0252

And again, there are a lot of real-world examples of this--examples from science and math.0257

You could think, in biology, of prey and a predator: for example, the number of rabbits in a forest will vary inversely with the number of coyotes.0263

So, less coyotes means a greater rabbit population, and vice versa.0273

And then, knowing the constant and maybe the population of rabbits would allow you to figure out the population of coyotes.0280

The graph of inverse variation looks very different than the graph of direct variation.0292

Recall the discussion we had in a recent lecture on asymptotes.0298

Asymptotes are lines that the graph will approach, but it will never actually reach or cross.0303

So, the graph of an inverse variation is similar to the graph of a rational function: it has asymptotes.0312

In this case, the asymptotes are actually at x = 0 and y = 0; so, we have, here at y = 0, a horizontal asymptote, and at x = 0, a vertical asymptote.0318

Let's take an example to see what one of these graphs would look like.0335

So, let's let k equal 10; recall that, for inverse variation, k = xy; so 10 = xy.0341

And then, find some values for x and values for y to graph out.0357

So, when x is 1/10, let's rewrite this as y = k/x, so y = 10/...in this case, it would be 1/10.0364

If x equals 1/10, y equals 10/(1/10), so y = 100.0380

If x is 1/5, that is going to be 10 divided by 1/5; y will equal 50.0390

If x is 1, y is 10; if x is 2, 10 divided by 2...y is 5; if x is 10, 10 divided by 10 is 1; this gives me enough points to work with.0398

2, 4, 6, 8, 10...OK, when x is 1/10, y is very large--way, way up here somewhere.0414

When x is 1/5, again, y is still large, but not quite as large.0429

When x is 1, y is 10; when x is 2, y is 5; and when x is 10, y is 1.0437

We can get the idea of this graph, that as x approaches the asymptote, y is going to become very large on this branch of the graph.0448

So, that is what we saw earlier on with asymptotes: that the graph will approach it, but it will not cross it.0466

And we know that the vertical asymptote goes here at x = 0.0473

I also see that this graph is approaching the horizontal asymptote, but it is not going to cross it.0477

This is the expected shape.0485

OK, now, to figure out what is happening when I have negative values for x: again, let's pick some values for x.0488

Let's let x equal -1/10; so again, you are going to get 100, but it is going to be -100 this time.0497

Or if x is -1/5, I am going to get the same absolute value, but it is going to be the negative of this.0504

When x is -1, y is -10; when x is -2, y is going to be -5; when x is -10, in here we are going to get -1.0515

Looking out here at -10, this is going to give me -1 right there; -2--I am going to get -2, -4, -6...right about here.0534

-1...this is going to be way down here at -10.0551

And as x gets closer and closer to 0, y is going to get very large in the negative direction.0560

The shape of this branch of the graph is like this--that the graph is going to approach0568

both the vertical and horizontal asymptotes, but it is not going to reach it.0580

There are two branches to this; and since k is positive, the branches are in these two quadrants.0585

If k is negative, I would actually find the branches in the other two quadrants.0593

If k is less than 0, then I would have a graph with branches in the other two quadrants.0599

This is the graph of inverse variations.0608

Proportions can be used to solve problems involving direct, joint, or inverse variation.0611

So, these are problems where some values are given, and a missing value needs to be found.0618

Recall that, for direct variation, k = y/x.0624

Now, let's assume that we are given two sets of values, (x1,y1)0630

and (x2,y2), that both satisfy this direct variation.0636

This means that, if they satisfy this, y divided by x equals the constant of variation.0641

So, x1/y1 equals the constant of variation, and x2y2 equals the constant of variation.0649

Since k equals x1/y1, and k also equals x2/y2, then these two must equal each other.0658

I can form a proportion, which means that, if I am given a set of coordinates here and then one of these coordinates, I can find the missing value.0665

So, if I am given three of the values, I can find the fourth.0679

And you can use cross-multiplication, as we usually do with proportions.0682

This can also be used to solve inverse variation problems.0692

Recall that k = xy in inverse variations; this is direct, and this is inverse variation.0695

And joint variation is very similar to direct variation, but you would have more unknowns.0704

Here we have inverse variation, k = xy; again, if I am given two sets of values,0712

I can say, "OK, k equals x1/y1; k also equals x2/y2."0719

Therefore, these two must equal each other.0727

Or I could cross-multiply and say, "OK, that is x1/x2 = y2/y1."0734

I could handle this either way; but if I had some coordinates (some values, some ordered pairs),0744

and I was missing one of these, but I had the other three, I could use this proportion to solve for the missing value.0751

And we will look at these types of problems right now in the examples.0759

Example 1: What kind of variation is represented by the equation s = 2πrh?0763

This is actually the equation for the surface are of the side of a cylinder.0769

And they are asking me what kind of variation this is, and what the constant of variation is.0775

Well, I recognize that what I have is: 2 is a constant; π is a constant (because it equals a number0781

that has a certain value that doesn't change); and I have 3 variables, s, r, and h.0787

Well, the only equation that we have gone over so far that has three variables is joint variation.0794

Joint variation has the equation y = kxz; and this fits into this form, so let's look at this: y = 2πrh.0802

So, this is joint variation; and then I am asked, "What is the constant?"0814

Well, both of these are the constant: the constant is the product of these: 2π is the constant.0817

And then, I have three variables: y is s; x is r; and let z equal h.0827

This is a joint variation, where the surface area varies jointly with the radius and height.0845

In this example, we are told that y varies directly as x, and when x = 8, y = 56.0854

And then, we are asked to find y when x equals 9.0860

So, recall that the equation for direct variation is k = y/x.0864

So, given that k = y/x, we can solve for the unknown value, because I know that these values satisfy this equation.0870

So, 56 divided by 8 is one ratio that I have; and I am looking for y when x is 9.0881

Since both of these equal k, and k is a constant, I can form a proportion, 56/8 = y/9, and then solve for y.0900

So, I can cross-multiply, and this gives me 56(9) = 8y.0912

56(9) is 504, equals 8y; 504/8 = y; so this is going to give me y = 63.0918

I solved this using proportions; you could have gone about it a different way.0935

I could have said, just using this first set of values, that I have figured out what k is, because, since k is 56/8, k equals 7.0938

Now, they asked me to find y when x equals 9; I know that k = y/x.0953

That means that y = kx, and I know k, and I have been given a certain value for x, which is 9, so I can solve for y.0960

Either way that you find easier works: using proportions or solving for k, and then going back to the equation for the direct variation.0976

In Example 3, y varies jointly as x and z: when x equals 4 and z equals 3, y is equal to 60.0987

Find z when x equals 7 and y equals 105.0996

So, recall the equation for joint variation: y = kxz.1002

Now, let's figure out the equation for the constant--let's just solve for the constant.1011

This would be...I would divide both sides by xz to get y/xz.1017

Now, I can set up a proportion: I know that k equals y/xz, and I am given this first set of values: x = 4; z = 3; and y is 60.1024

OK, now I am also asked to find z, given these other two values.1044

So, I know that k equals, in the second scenario...y is 105; x is 7; and the unknown is z.1049

So now, I can form a proportion, because since this equals k and that equals k, these two equal each other.1062

So, 60 divided by (4 times 3) equals 105 times 7z.1070

Well, this is 60/12 = 105/7z; 60/12 is just 5, so 5 = 105/7z.1077

I am going to multiply both sides by 7z to get 7z(5) = 105; 7 times 5 is 35, times z equals 105.1091

Dividing both sides by 35, you will find that z equals 3.1103

Again, another way to solve this would have been to just say, "OK, the constant of variation is 60, divided by 12, or 5."1108

Knowing what the constant of variation is, I could have just substituted that here and then solved for z.1116

You can either use the proportion or find the constant of variation and then use the second set of values and find the missing variable.1122

y varies inversely as x: Example 4 is an example of inverse variation.1132

And in inverse variation, we have xy = k.1137

When x is 12 and y is 4, this holds true; so these values satisfy this equation.1143

And we are asked to find x if y equals -1/2.1151

So, let's go about this by finding k: I know that these values satisfy this equation, so 12(4) = k; therefore, k = 48.1154

Then, I am asked to find x if y = -1/2: xy = k.1170

I want to find x; I know that y = -1/2, and I know that k = 48.1178

I am going to multiply both sides by -2 to get that x = -96.1186

So again, this is using the equation for inverse variation, solving for the constant of variation,1197

and then using that constant in the equation with the value of x to solve for the value to solve for the unknown variable, which is x.1202

That concludes this lecture on direct, inverse, and joint variation.1214

Thanks for visiting Educator.com!1219