Lesson Video: Direct and Inverse Variation Mathematics

In this video, we will learn how to recognise direct and inverse variation involving different powers and roots of π‘₯.

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Video Transcript

In this video, we will learn how to recognize direct and inverse variation, involving different roots and powers of π‘₯. We will begin by defining what we mean by direct and inverse variation or proportion. In direct variation, as one number increases, so does the other. Direct variation is also called direct proportion. If two quantities 𝑦 and π‘₯ vary directly, we can say that 𝑦 is proportional to π‘₯. The 𝛼 symbol shown means is proportional to. The ratio or proportion of the two quantities must be equal to some constant π‘˜. This is more commonly written as 𝑦 is equal to π‘˜ multiplied by π‘₯. We have multiplied both sides of the equation by π‘₯. π‘˜ is known as the constant of proportionality, and its value will change for each question.

In inverse variation, as one number increases, the other decreases. This is also known as inverse or indirect proportion. If we once again consider two variables π‘₯ and 𝑦, if 𝑦 is inversely proportional to π‘₯, we can write that 𝑦 is proportional to one over π‘₯. This time, the product of the two variables is equal to a constant 𝑦 multiplied by π‘₯ is equal to π‘˜. Making 𝑦 the subject, we have 𝑦 is equal to π‘˜ over π‘₯ or π‘˜ divided by π‘₯. As previously mentioned, in this video, we will look at direct and inverse variation with regards different powers and roots of π‘₯. This means that 𝑦 will be proportional to π‘₯ to the 𝑛th power or one over π‘₯ to the 𝑛th power.

This will lead us into solving equations of the type 𝑦 equals π‘˜ multiplied by π‘₯ to the 𝑛th power or 𝑦 equals π‘˜ divided by π‘₯ to the 𝑛th power. Our first two questions will involve direct variation.

Given that 𝑦 varies directly as π‘₯ squared, write an equation for 𝑦 in terms of π‘₯ using π‘˜ as a nonzero constant.

If two variables vary directly, we can say that they are in direct proportion to one another. This means that as one value increases, so does the other. Our two variables are 𝑦 and π‘₯ squared, which means that 𝑦 is directly proportional to π‘₯ squared. This also means that the ratio or quotient of the two variables is equal to some constant π‘˜. 𝑦 divided by π‘₯ squared is equal to π‘˜. We can make 𝑦 the subject of this equation by multiplying both sides by π‘₯ squared. 𝑦 is therefore equal to π‘˜ multiplied by π‘₯ squared.

When dealing with questions of this type, we usually go from the proportion expression to our final answer. We replace the proportion symbol with an equal symbol and a constant π‘˜. If 𝑦 varies directly as π‘₯ squared, then 𝑦 is equal to π‘˜ multiplied by π‘₯ squared.

Given that π‘₯ is proportional to 𝑦 cubed and π‘₯ equals 81 when 𝑦 is equal to three, what is π‘₯ when 𝑦 is equal to four?

In this question, we’re dealing with direct proportion or variation. We are told that π‘₯ varies directly with 𝑦 cubed. This can be rewritten as the equation π‘₯ is equal to the constant π‘˜ multiplied by 𝑦 cubed. Dividing both sides of this equation by 𝑦 cubed gives us the constant π‘˜ is equal to π‘₯ divided by 𝑦 cubed. We are also told that when π‘₯ is equal to 81, 𝑦 is equal to three. This means that we can calculate the value of π‘˜ by dividing 81 by three cubed. Three cubed is equal to 27, as three multiplied by three is nine and multiplying this by three gives us 27. This means that π‘˜ is equal to 81 divided by 27. There are three 27s in 81. Therefore, π‘˜ is equal to three.

If we didn’t spot this, we could have firstly canceled the fraction by dividing the numerator and denominator by nine. This would leave us with nine divided by three, which we know is equal to three. Alternatively, we might have noticed that 81 is equal to three to the fourth power. And dividing this by three to the third power or three cubed would give us three to the power of one, which is three. Substituting this value of π‘˜ back into our equation gives us π‘₯ is equal to three 𝑦 cubed. We now need to calculate the value of π‘₯ when 𝑦 is equal to four. This gives us π‘₯ is equal to three multiplied by four cubed. Four cubed is equal to 64. And multiplying this by three gives us 192. If π‘₯ is proportional to 𝑦 cubed and π‘₯ equals 81 when 𝑦 is equal to three, then π‘₯ is equal to 192 when 𝑦 is equal to four.

We will now look at a couple of questions involving inverse variation.

Given that 𝑦 varies inversely as π‘₯ squared, write an equation for 𝑦 in terms of π‘₯ using π‘˜ as a nonzero constant.

We recall that when two variables vary inversely, we can say that they are inversely proportional. If 𝑦 is inversely proportional to π‘₯ squared, then 𝑦 is proportional to one over π‘₯ squared. We also know that when two variables are inversely proportional to one another, that product is equal to some constant π‘˜. 𝑦 multiplied by π‘₯ squared is equal to π‘˜. Dividing both sides of this equation by π‘₯ squared gives us the equation 𝑦 is equal to π‘˜ divided by π‘₯ squared.

In general, when dealing with questions of this type, we go from our proportional expression straight to the equation in the answer. We replaced the proportional symbol with equals π‘˜. We know that π‘˜ multiplied by one over π‘₯ squared can be written as π‘˜ over π‘₯ squared.

We will now use our knowledge of inverse variation to match the equation with the correct statement.

If 𝑦 is proportional to one over the square root of π‘₯, then which of the following is true? Is it (A) π‘₯ is directly proportional to 𝑦? (B) π‘₯ is directly proportional to 𝑦 squared. (C) π‘₯ is inversely proportional to 𝑦. (D) π‘₯ is inversely proportional to 𝑦 squared. Or (E) π‘₯ is inversely proportional to 𝑦 cubed.

We know that the expression 𝑦 is proportional to one over the square root of π‘₯ is the same as 𝑦 is inversely proportional to the square root of π‘₯. When two variables are inversely proportional to one another, as one increases, the other decreases. This means we can immediately rule out options (A) and (B) as in direct proportion, as one variable increases, the other also increases. In this question, as 𝑦 increases, the square root of π‘₯ decreases and vice versa. If we consider the expression we were given, we can begin by multiplying both sides by the square root of π‘₯. This means that the square root of π‘₯ multiplied by 𝑦 is proportional to one.

Dividing both sides of this by 𝑦, we get that the square root of π‘₯ is proportional to one over 𝑦. We can then square both sides of this such that π‘₯ is proportional to one over 𝑦 squared. When squaring a fraction, we can square the numerator and denominator separately. We can therefore conclude that π‘₯ is inversely proportional to 𝑦 squared. The correct answer is option (D).

In our final question, we will solve a real-life problem involving variation.

The height of a right circular cylinder β„Ž varies inversely with the square of its radius π‘Ÿ. If β„Ž equals 93 centimeters when π‘Ÿ equals 7.5 centimeters, determine β„Ž when π‘Ÿ is equal to 1.5 centimeters.

We know that if two variables vary inversely, as one increases, the other decreases. This means that in this question, β„Ž and π‘Ÿ squared are inversely proportional. This can be written as β„Ž is proportional to one over π‘Ÿ squared. This can be rewritten as an equation using the constant of proportionality π‘˜ such that β„Ž is equal to π‘˜ divided by π‘Ÿ squared. Multiplying both sides of this equation by π‘Ÿ squared gives us β„Ž multiplied by π‘Ÿ squared is equal to π‘˜. When dealing with inverse proportion or variation, our two variables will have a product equal to some constant π‘˜.

We are told that when the height of the cylinder is 93 centimeters, the radius is 7.5 centimeters. This means that we can calculate the value of π‘˜ by multiplying 93 by 7.5 squared. 7.5 squared is equal to 56.25. Multiplying this by 93 gives us a value of π‘˜ equal to 5231.25. We can substitute this constant back into our equation such that β„Ž is equal to 5231.25 divided by π‘Ÿ squared. We want to calculate this value of β„Ž when π‘Ÿ is equal to 1.5. 1.5 squared is equal to 2.25. This means that β„Ž is equal to 5231.25 divided by 2.25. Typing this into the calculator gives us 2325. The height of the cylinder when the radius is 1.5 centimeters is 2325 centimeters.

There is a slightly quicker way of calculating the value of β„Ž without working out the constant π‘˜. We begin by considering the fact that the product of the height and the radius squared must be equal to some constant π‘˜ for any height and radius in this cylinder. This means that in our first scenario, with a height of 93 centimeters and a radius of 7.5 centimeters, we have 93 multiplied by 7.5 squared. In our second scenario, we have β„Ž multiplied by 1.5 squared as the radius is 1.5 centimeters. We can then divide both sides of this equation by 1.5 squared. Once again, typing this into the calculator gives us an answer for β„Ž equal to 2325. This confirms that this is the height of the cylinder when the radius is 1.5 centimeters.

We will now summarize the key points from this video. We can express the direct variation or direct proportion of two variables π‘₯ and 𝑦, as follows. 𝑦 is proportional to π‘₯ or 𝑦 is equal to π‘˜ multiplied by π‘₯, where π‘˜ is a nonzero constant. In the same way, if 𝑦 is inversely proportional to π‘₯, we can write this as 𝑦 is proportional to one over π‘₯ or 𝑦 is equal to π‘˜ divided by π‘₯. We saw in this video that the two equations can be rewritten such that the constant of proportionality π‘˜ is equal to the quotient or product of the two variables.

In this video, we saw direct and inverse variation with different powers and roots of π‘₯. When dealing with problems of this type, the relationship holds. If 𝑦 is proportional to π‘₯ to the 𝑛th power, then 𝑦 is equal to π‘˜ multiplied by π‘₯ to the 𝑛th power. In the same way, if 𝑦 is proportional to one over π‘₯ to the 𝑛th power, then 𝑦 is equal to π‘˜ over π‘₯ to the 𝑛th power. We also saw in this video that by substituting given values of the variables π‘₯ and 𝑦, we can calculate the value of π‘˜ in abstract and real-life problems.

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