Video: Further Direct and Inverse Variation

Learn to tackle direct and inverse variation problems in which one variable is directly or inversely proportional to the square, cube, or some other power or root of the other variable (for example, if 𝑦 varies inversely as βˆ›π‘₯).

17:58

Video Transcript

Before watching this video, you should already be familiar with simple direct variation. We are gonna be looking at relationships where one variable is directly proportional to a square or cube or root or some other power of another. And we’ll also be looking at inversely proportional relationships.

First though, a quick summary of the basics of direct proportionality. This means 𝑦 is directly proportional to π‘₯. But you might also encounter phrases like 𝑦 varies directly with π‘₯ or 𝑦 varies directly as π‘₯ or even 𝑦 is a simple multiple of π‘₯. And we can present this in an equation like this: 𝑦 is equal to π‘˜ times π‘₯, where π‘˜ is just a constant, a number. And we call π‘˜ the constant of proportionality or the constant of variation. And if we were to plot 𝑦 versus π‘₯, it would always look like a straight line. And it would always pass through the origin. So if the graph doesn’t look like that, then it’s not an example of direct proportionality. Let’s just quickly have a look at a question that involves direct proportionality.

For example, 𝑦 is directly proportional to π‘₯. When π‘₯ is equal to five, then 𝑦 is equal to 12. Find the value of π‘₯ when 𝑦 is equal to seven.

So we’d always start off a question like this by saying if 𝑦 is directly proportional to π‘₯, then 𝑦 is some number times π‘₯. That’s an equation which will encapsulate that relationship between 𝑦 and π‘₯. And the question tells us specifically that when π‘₯ is equal to five, then 𝑦 is equal to 12. So we can substitute those values in for 𝑦, π‘₯ in order to work out the value of π‘˜. So 12 is equal to π‘˜ times five. Now if I divide both sides of my equation by five, I can get π‘˜ on its own on the right-hand side. So dividing by five and then cancelling five on the numerator and the denominator on the right-hand side leaves me with just π‘˜. And π‘˜ in this case is equal to 12 over five.

So now I know what the equation is that encapsulates the relationship between 𝑦 and π‘₯: 𝑦 is equal to 12 over five times π‘₯. And I can use that equation to find out the value of π‘₯ when 𝑦 is equal to seven. So seven is equal to 12 over five times π‘₯. Now if I multiply both sides by five, that means I can cancel five from the right-hand side. And then if I divide both sides by 12, I can cancel the 12 off the right-hand side. And that means that π‘₯ would be 35 over 12. So my answer would be when 𝑦 equals seven, then π‘₯ is equal to 35 over 12.

So that sums up simple direct proportionality. But sometimes one variable is directly proportional to the square or the cube or some other power or root of another. For example, if I have a cube which is π‘₯ by π‘₯ by π‘₯ and I double the length of each side, then the surface area won’t be doubled. It will be multiplied by two squared. So it will be multiplied by four. And the volume wouldn’t be doubled. It would be multiplied by a factor of two cubed, so that’s times eight.

So if we called the surface area 𝐴 and the volume 𝑉, the surface area is directly proportional to the square of the side length, and the volume would be directly proportional to the cube of the side length. In fact, the constant of proportionality for the area would be six and for the volume would be one. So there is direct proportionality, but it’s not quite as simple as we saw before. With the area, the area is directly proportional to the square of the side length. And with volume, the volume is directly proportional to the cube of the side length.

Now if I plotted the area against the side length of the cube, we will get a graph that looks something like this. And clearly, that’s not a straight line relationship although it does go through the origin. So we can safely say that the area is not directly proportional to the side length. But if I plot the area against the square of the side length, that is a straight line relationship and it also goes through the origin. So the area is directly proportional to the square of the side length. So let’s have a look at some questions then.

We’ve got three situations here. And we want to express each of these relationships in an equation involving π‘˜, the constant of proportionality. So in the first one, 𝑦 varies directly as the square of π‘₯. Now we can write that as 𝑦 is directly proportional to π‘₯ squared. And as an equation, that means 𝑦 is equal to some constant times π‘₯ squared. Now the second one is 𝑦 is directly proportional to the cube root of π‘₯. So we can write the symbol like this: 𝑦 is directly proportional to. Now the cube root of π‘₯ is basically written like this. And as an equation, 𝑦 is equal to some constant times the cube root of π‘₯. And the last one, 𝑦 varies directly with the cube of π‘₯. Well, 𝑦 is directly proportional to π‘₯ cubed. And in its equation format, we write that as 𝑦 is equal to some constant times π‘₯ cubed.

Now, 𝑦 varies directly as the fifth power of π‘₯. Using π‘˜ to represent the constant of variation, write an equation for 𝑦 in terms of π‘₯.

Well, if 𝑦 varies directly as the fifth power of π‘₯, we can write 𝑦 is directly proportional to π‘₯ to the power of five. And in equation format, that means that 𝑦 is equal to the constant of variation, π‘˜ in our case, times π‘₯ to the power of five. So this question is as simple as writing out that equation, in that format.

And now we’ve got a question that tells us that 𝑦 is directly proportional to the square of π‘₯. And when 𝑦 is 32, then π‘₯ is equal to four. And we’ve gotta find the value of 𝑦 when π‘₯ equals 10.

So as always, start off by writing the nature of the proportional relationship: 𝑦 is directly proportional to the square of π‘₯. And that means that 𝑦 is equal to some constant of proportionality times the square of π‘₯. And the question told us that when π‘₯ is equal to four, then 𝑦 is equal to 32. So if we put those values into that equation, we’ll be able to work out the value of π‘˜. So 𝑦 is 32 and π‘₯ is four. So 32 is equal to π‘˜ times four squared, which means that 32 is equal to π‘˜ times 16. So if I divide both sides of my equation by 16, I can isolate π‘˜ on the right-hand side, which means that π‘˜ is equal to 32 divided by 16, which is two.

So now I can write up my equation: 𝑦 is equal to two times π‘₯ squared. And now, we’ve gotta use that equation to find the value of 𝑦 when π‘₯ is equal to 10. So that means that 𝑦 would be equal to two times 10 squared, which is 𝑦 is equal to two times 100. So, our answer is when π‘₯ is equal to 10, then 𝑦 is equal to 200.

Now in this question, we’re told that 𝑦 varies directly as the cube root of π‘₯. And we’re also told that when the value of π‘₯ is eight, then the value of 𝑦 is 10. So we’ve gotta find the value of π‘₯ when 𝑦 is equal to eight and one-third.

So again, let’s write down that statement of proportionality: 𝑦 is directly proportional to the cube root of π‘₯. And in its equation format, that means that 𝑦 is equal to the constant of proportionality times the cube root of π‘₯. Now the question told us that when π‘₯ is eight, then 𝑦 is 10. So let’s substitute those values into that equation to work out the value of π‘˜. So that means that 10 is equal to π‘˜ times the cube root of eight. Well, the cube root of eight is two. So 10 is equal to π‘˜ times two. Now if I divide both sides of my equation by two, I can cancel off the twos on the right-hand side just to leave myself with π‘˜. And 10 divided by two is five. So π‘˜ is five. So I can put that back into my equation.

So the relationship between π‘₯ and 𝑦 is summed up by the equation 𝑦 is equal to five times the cube root of π‘₯. Now I can use that equation to find out the value of π‘₯ when 𝑦 is equal to eight and a third. Well, actually, I’m gonna express eight and a third as a top-heavy fraction, 25 over three, because that’ll probably help me with my calculations. So replacing 𝑦 with 25 over three in that equation, I’ve got 25 over three is equal to five times the cube root of π‘₯. Well, if I divide both sides of my equation by five, then I can cancel the fives on the right-hand side, just leaving me with the cube root of π‘₯.

And on the left-hand side, I’ve got 25 over three divided by five. Well, I’m gonna express that five in its fraction version, five over one, because I’m doing a fraction division. And when I divide fractions, I flip the second and-and change it into a multiply. So that’s gonna be the same as 25 over three times one over five. So dividing by five is the same as multiplying by one over five. Now I can cancel the five on the bottom and the top. So I’ve got the cube root of π‘₯ is equal to five over three. So now I’m gonna cube both sides of that equation to find out what π‘₯ is equal to. Now the cube of the cube root of π‘₯ is just π‘₯, the cube of five is 125, and the cube of three is 27.

But in the question, they told us that 𝑦 was equal to eight and a third. They gave us a mixed number. So, really, we should give our answer as a mixed number. So our answer is when 𝑦 is eight and a third, then π‘₯ is equal to four and 17 27ths.

So we’re seeing that 𝑦 is somewhere- sometimes directly proportional to π‘₯, and sometimes it’s directly proportional to some power of π‘₯. But sometimes, it’s inversely proportional to one of those things. Take gravity for example, as you move further away from an object, the magnitude of the force that its gravity exerts on you decreases. In fact, it varies inversely proportionally to the square of the distance from the object. So the magnitude of the gravitational force is directly proportional to one over the distance squared. Now we can deal with inverse proportion questions in much the same way that we dealt with ordinary direct proportion questions, except the equation of proportionality looks a little bit different.

For example, if 𝑦 is inversely proportional to π‘₯, we write this as 𝑦 is directly proportional to one over π‘₯. And the equation is 𝑦 is equal to the constant of proportionality times one over π‘₯ or simply as 𝑦 is equal to π‘˜ over π‘₯. And if 𝑦 is inversely proportional to the square of π‘₯, that means that 𝑦 is directly proportional to one over π‘₯ squared. And in its equation format, that’s π‘˜ times one over π‘₯ squared or just π‘˜ over π‘₯ squared is equal to 𝑦. And if 𝑦 was inversely proportional to the cube root of π‘₯, we can write that like this, which would give us the equations either 𝑦 is equal to π‘˜ times one over the cube root of π‘₯ or 𝑦 is equal to π‘˜ over the cube root of π‘₯.

Now again, there are different ways of expressing inverse relationships. For example, 𝑦 varies inversely as the cube of π‘₯. And that means that 𝑦 is directly proportional to one over the cube of π‘₯ β€” π‘₯ cubed. And 𝑦 is inversely proportional to the square root of π‘₯ means 𝑦 is directly proportional to one over the square root of π‘₯. Or if you want to get really wordy, the pressure, in atmospheres, in a glider varies inversely as the square root of its height above sea level, in yards. If we define a couple of variables, 𝑝 for the pressure in atmospheres and β„Ž for the height above sea level in yards, then we can express that relationship as 𝑝 is directly proportional to one over the square root of β„Ž. And as an equation, 𝑝 is equal to the constant of variation divided by the square root of β„Ž.

Now it’s also worth noting that the graphs for inversely proportional relationships aren’t the same as graphs for directly proportional ones. So, for example, if 𝑦 is directly proportional to one over π‘₯, the graph would be 𝑦 equals the constant of proportionality times one over π‘₯. And you can see that as π‘₯ gets larger and larger, then some constant number divided by a bigger and bigger number is just gonna get closer and closer to zero. And as π‘₯ gets closer and closer to zero, we’ve got a number divided by smaller and smaller and smaller number that’s gonna get bigger and bigger and bigger. If π‘₯ was actually zero, we’d have a number divided by zero which would technically be infinity. So that explains the shape of this graph. Let’s have a look at a few questions then.

𝑦 is inversely proportional to π‘₯. When π‘₯ is equal to three, then 𝑦 equals six. Find the value of 𝑦 when π‘₯ equals eight.

So we can express that proportionality by 𝑦 is directly proportional to one over π‘₯. Or as an equation then, 𝑦 is equal to the constant of proportionality times one over π‘₯ or 𝑦 is equal to π‘˜ over π‘₯. And we’re also told that when π‘₯ is three, 𝑦 is six. So again, just like before, we can substitute those values into our equation to work out what π‘˜ is. So π‘₯ is three and 𝑦 is six gives us six is equal to π‘˜ over three. Multiplying both sides by three, we’ve got 18 is equal to π‘˜.

So let’s put that back into our original equation, which means that 𝑦 is equal to 18 over π‘₯. So now we gotta use that to find the value of 𝑦 when π‘₯ is equal to eight. Well, that means that 𝑦 is equal to 18 over, in this case, eight. And 18 over eight simplifies down to two and a quarter. So when π‘₯ is equal to eight, then 𝑦 is equal to two and a quarter.

Now in this question, 𝑦 varies inversely with the square root of π‘₯. When π‘₯ is equal to 25, then 𝑦 equals four. Find the value of π‘₯ when 𝑦 is equal to two.

So our proportionality can be expressed this way: 𝑦 is directly proportional to one over the square root of π‘₯. 𝑦 varies inversely with the square root of π‘₯. And as an equation, that means that 𝑦 is equal to π‘˜ times one over the square root of π‘₯ or 𝑦 is equal to π‘˜ over the square root of π‘₯, where π‘˜ is our constant of proportionality. So let’s use the information we were given in the question to find the value of π‘˜.

Putting π‘₯ equals 25 and 𝑦 equals four gives us four is equal to π‘˜ over the square root of 25. So that’s four is equal to π‘˜ over five. Now if we multiplied both sides by five, we’ve got π‘˜ is equal to 20, which means the equation describing the relationship between π‘₯ and 𝑦 is 𝑦 is equal to 20 over the square root of π‘₯. And plugging 𝑦 equals two into the equation, we can see that when 𝑦 equals two, two is equal to 20 over the square root of π‘₯. So if I multiplied both sides of that equation by the square root of π‘₯, I’ve got two root π‘₯ is equal to 20. Now if I divide through by two, I got the square root of π‘₯ is equal to 10. So squaring both sides gives me my answer that when 𝑦 is equal to two, then π‘₯ is equal to 100.

Now in this question, well instead of using π‘₯ and 𝑦, we’ve used 𝐴 and 𝐡. 𝐴 varies inversely with 𝐡 plus five. If 𝐴 is equal to one when 𝐡 is equal to five, then find the value of 𝐡 when 𝐴 equals 10.

So the first thing that we have to do is write down an expression of proportionality. 𝐴 varies inversely with, that means that 𝐴 is directly proportional to one over something, and it varies inversely with 𝐡 plus five. So what it’s inversely proportional to is 𝐡 plus five. And we can write that in its equation format: 𝐴 is equal to π‘˜ times one over 𝐡 plus five, which can be written as 𝐴 equals π‘˜ over 𝐡 plus five. Now let’s substitute in the values that we know.

𝐴 is one when 𝐡 is five. This means that one is equal to π‘˜ over five plus five. And five plus five is obviously 10. So if I multiply both sides by 10, I can see that π‘˜ is equal to 10. And the relationship between 𝐴 and 𝐡 is summed up in this equation: 𝐴 is equal to 10 over 𝐡 plus five. Now we’ve gotta find the value of 𝐡 when 𝐴 is 10. So let’s put 𝐴 is equal to 10 in that equation.

So 10 is equal to 10 over 𝐡 plus five. Then multiplying both sides by 𝐡 plus five, I’ve got 10 lots of 𝐡 plus five or 10 times 𝐡 plus five is equal to 10 because the 𝐡 plus fives were cancelled on the right-hand side. Now dividing both sides by ten, I can cancel the tens on each side to give me 𝐡 plus five is equal to one. Now subtracting five from both sides gives me my answer: 𝐡 is equal to negative four.

So we’ve seen that directly proportional relationships can be with different powers of one of the variables or they can even be inverse relationships.

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