WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today we are going to be talking about direct, joint, and inverse variation.
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Starting out with **direct variation**: this is a review of a concept that you likely learned in Algebra I.
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The direct variation equation is y = kx, and this is sometimes expressed as k = y/x.
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And what this equation shows us is that y varies directly as x.
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So, there is a relationship between x and y, and it is explained with this constant, which is called the constant of variation.
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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.
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And there are many real-world examples of this.
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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.
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So, the larger the lawn is, the longer it will take me to cut it.
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And there is some constant, such as the rate: if I can cut two square feet per minute, then that would be
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the constant of variation--the rate at which I am cutting the lawn.
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One variable would be the time it takes me, and the other is the size of the lawn.
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Recall the graphs of direct variation: the graph of a direct variation is a straight line that passes through the origin.
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And its slope is the constant, k (the constant of variation).
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Now, you can have a straight line graph, passing through the origin, that moves up to the right.
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You can also have a straight line graph where y is increasing towards the left.
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And recall that, when we talked (back in Algebra I) about graphs of linear equations, if the slope is positive,
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the graph is going to look like this--it is going to go up to the right.
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If the slope is negative, it is going to go up to the left.
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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--
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whereas, if k is a negative number (the slope is less than 0), the graph will look like this.
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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.
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Joint variation is very similar to direct variation, except now we are working with three variables instead of 2.
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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).
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So again, going back to the lawn analogy: I could say the amount of time it takes me to mow the lawn
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varies jointly with the size of the lawn and the steepness of the hill that it is on.
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So, as the size of the lawn increases and the steepness of the hill that it is on increases,
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the time that it takes me to mow the lawn will increase.
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So now, I have three things that are maintaining a constant relationship with each other.
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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.
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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.
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An inverse relationship: what we can also look at this as is that y is proportional to the reciprocal of x.
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Instead of varying directly with it, it is proportional to the reciprocal of it.
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And again, there are a lot of real-world examples of this--examples from science and math.
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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.
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So, less coyotes means a greater rabbit population, and vice versa.
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And then, knowing the constant and maybe the population of rabbits would allow you to figure out the population of coyotes.
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The graph of inverse variation looks very different than the graph of direct variation.
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Recall the discussion we had in a recent lecture on asymptotes.
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Asymptotes are lines that the graph will approach, but it will never actually reach or cross.
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So, the graph of an inverse variation is similar to the graph of a rational function: it has asymptotes.
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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.
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Let's take an example to see what one of these graphs would look like.
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So, let's let k equal 10; recall that, for inverse variation, k = xy; so 10 = xy.
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And then, find some values for x and values for y to graph out.
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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.
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If x equals 1/10, y equals 10/(1/10), so y = 100.
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If x is 1/5, that is going to be 10 divided by 1/5; y will equal 50.
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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.
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2, 4, 6, 8, 10...OK, when x is 1/10, y is very large--way, way up here somewhere.
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When x is 1/5, again, y is still large, but not quite as large.
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When x is 1, y is 10; when x is 2, y is 5; and when x is 10, y is 1.
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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.
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So, that is what we saw earlier on with asymptotes: that the graph will approach it, but it will not cross it.
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And we know that the vertical asymptote goes here at x = 0.
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I also see that this graph is approaching the horizontal asymptote, but it is not going to cross it.
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This is the expected shape.
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OK, now, to figure out what is happening when I have negative values for x: again, let's pick some values for x.
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Let's let x equal -1/10; so again, you are going to get 100, but it is going to be -100 this time.
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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.
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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.
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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.
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-1...this is going to be way down here at -10.
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And as x gets closer and closer to 0, y is going to get very large in the negative direction.
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The shape of this branch of the graph is like this--that the graph is going to approach
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both the vertical and horizontal asymptotes, but it is not going to reach it.
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There are two branches to this; and since k is positive, the branches are in these two quadrants.
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If k is negative, I would actually find the branches in the other two quadrants.
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If k is less than 0, then I would have a graph with branches in the other two quadrants.
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This is the graph of inverse variations.
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Proportions can be used to solve problems involving direct, joint, or inverse variation.
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So, these are problems where some values are given, and a missing value needs to be found.
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Recall that, for direct variation, k = y/x.
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Now, let's assume that we are given two sets of values, (x₁,y₁)
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and (x₂,y₂), that both satisfy this direct variation.
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This means that, if they satisfy this, y divided by x equals the constant of variation.
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So, x₁/y₁ equals the constant of variation, and x₂y₂ equals the constant of variation.
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Since k equals x₁/y₁, and k also equals x₂/y₂, then these two must equal each other.
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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.
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So, if I am given three of the values, I can find the fourth.
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And you can use cross-multiplication, as we usually do with proportions.
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This can also be used to solve inverse variation problems.
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Recall that k = xy in inverse variations; this is direct, and this is inverse variation.
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And joint variation is very similar to direct variation, but you would have more unknowns.
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Here we have inverse variation, k = xy; again, if I am given two sets of values,
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I can say, "OK, k equals x₁/y₁; k also equals x₂/y₂."
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Therefore, these two must equal each other.
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Or I could cross-multiply and say, "OK, that is x₁/x₂ = y₂/y₁."
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I could handle this either way; but if I had some coordinates (some values, some ordered pairs),
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and I was missing one of these, but I had the other three, I could use this proportion to solve for the missing value.
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And we will look at these types of problems right now in the examples.
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Example 1: What kind of variation is represented by the equation s = 2πrh?
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This is actually the equation for the surface are of the side of a cylinder.
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And they are asking me what kind of variation this is, and what the constant of variation is.
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Well, I recognize that what I have is: 2 is a constant; π is a constant (because it equals a number
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that has a certain value that doesn't change); and I have 3 variables, s, r, and h.
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Well, the only equation that we have gone over so far that has three variables is joint variation.
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Joint variation has the equation y = kxz; and this fits into this form, so let's look at this: y = 2πrh.
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So, this is joint variation; and then I am asked, "What is the constant?"
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Well, both of these are the constant: the constant is the product of these: 2π is the constant.
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And then, I have three variables: y is s; x is r; and let z equal h.
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This is a joint variation, where the surface area varies jointly with the radius and height.
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In this example, we are told that y varies directly as x, and when x = 8, y = 56.
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And then, we are asked to find y when x equals 9.
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So, recall that the equation for direct variation is k = y/x.
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So, given that k = y/x, we can solve for the unknown value, because I know that these values satisfy this equation.
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So, 56 divided by 8 is one ratio that I have; and I am looking for y when x is 9.
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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.
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So, I can cross-multiply, and this gives me 56(9) = 8y.
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56(9) is 504, equals 8y; 504/8 = y; so this is going to give me y = 63.
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I solved this using proportions; you could have gone about it a different way.
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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.
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Now, they asked me to find y when x equals 9; I know that k = y/x.
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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.
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Either way that you find easier works: using proportions or solving for k, and then going back to the equation for the direct variation.
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In Example 3, y varies jointly as x and z: when x equals 4 and z equals 3, y is equal to 60.
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Find z when x equals 7 and y equals 105.
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So, recall the equation for joint variation: y = kxz.
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Now, let's figure out the equation for the constant--let's just solve for the constant.
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This would be...I would divide both sides by xz to get y/xz.
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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.
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OK, now I am also asked to find z, given these other two values.
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So, I know that k equals, in the second scenario...y is 105; x is 7; and the unknown is z.
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So now, I can form a proportion, because since this equals k and that equals k, these two equal each other.
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So, 60 divided by (4 times 3) equals 105 times 7z.
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Well, this is 60/12 = 105/7z; 60/12 is just 5, so 5 = 105/7z.
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I am going to multiply both sides by 7z to get 7z(5) = 105; 7 times 5 is 35, times z equals 105.
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Dividing both sides by 35, you will find that z equals 3.
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Again, another way to solve this would have been to just say, "OK, the constant of variation is 60, divided by 12, or 5."
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Knowing what the constant of variation is, I could have just substituted that here and then solved for z.
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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.
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y varies inversely as x: Example 4 is an example of inverse variation.
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And in inverse variation, we have xy = k.
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When x is 12 and y is 4, this holds true; so these values satisfy this equation.
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And we are asked to find x if y equals -1/2.
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So, let's go about this by finding k: I know that these values satisfy this equation, so 12(4) = k; therefore, k = 48.
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Then, I am asked to find x if y = -1/2: xy = k.
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I want to find x; I know that y = -1/2, and I know that k = 48.
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I am going to multiply both sides by -2 to get that x = -96.
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So again, this is using the equation for inverse variation, solving for the constant of variation,
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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.
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That concludes this lecture on direct, inverse, and joint variation.
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