Question Video: Finding an Expression That Represents the Rate of Change of a Cube’s Volume Using Related Rates | Nagwa Question Video: Finding an Expression That Represents the Rate of Change of a Cube’s Volume Using Related Rates | Nagwa

Question Video: Finding an Expression That Represents the Rate of Change of a Cube’s Volume Using Related Rates Mathematics

If 𝑉 is the volume of a cube with edge length π‘₯ and the cube expands as time passes, give an expression for 𝑑𝑉/𝑑𝑑.

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Video Transcript

If 𝑉 is the volume of a cube with edge length π‘₯ and the cube expands as time passes, give an expression for 𝑑𝑉 𝑑𝑑.

So first of all, I’m gonna take a look at what we’ve actually given in the question. We know that 𝑉 is the volume and π‘₯ is the edge length. So therefore, we can say that 𝑉 is equal to π‘₯ cubed. So the volume is equal to π‘₯ cubed. And that’s because if we tried to work out the volume of a cube, all we do is we actually cube one of the lengths of the sides. Okay, great, so we’ve got our first little expression there.

So now, the next step would actually to be work out 𝑑𝑉 𝑑π‘₯. So we actually gonna differentiate the value that we had earlier which was that 𝑉 is equal to π‘₯ cubed. So 𝑑𝑉 𝑑π‘₯ can be equal to three π‘₯ squared. And we’ll reach that because if we’re multiplying the exponent by the coefficient, that’s three multiplied by one which gives us three. And then, all we’ve done is we’ve actually reduced the exponent by one because the three minus one which gives us two. So we get three π‘₯ squared.

So now as we’re actually looking to find 𝑑𝑉 𝑑𝑑, we’re now gonna use the chain rule, which states that 𝑑𝑦 𝑑π‘₯ is equal to 𝑑𝑦 𝑑𝑑 multiplied by 𝑑𝑑 𝑑π‘₯. Well, in our case, we’re gonna get 𝑑𝑉 𝑑𝑑 because this is like 𝑑𝑦 𝑑π‘₯, which is going to be equal to 𝑑𝑉 𝑑π‘₯ multiplied by 𝑑π‘₯ 𝑑𝑑. And we get that because if we take a look, our 𝑑π‘₯ terms will actually cancel out because we will have a 𝑑π‘₯ on the numerator, 𝑑π‘₯ on the denominator. So then, we’ll gonna get 𝑑𝑉 𝑑𝑑.

Okay, we already know 𝑑𝑉 𝑑π‘₯ from our previous step. However, we don’t know 𝑑π‘₯ 𝑑𝑑. So therefore, 𝑑𝑉 𝑑𝑑 what it actually means in practice, which is the change in the volume of the cube over time, is equal to three π‘₯ squared 𝑑π‘₯ 𝑑𝑑.

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