Video Transcript
A metallic right circular cone expands as a result of heating such that it remains a right circular cone. The height of the cone decreases at the rate of two centimeters per second, while the radius of the cone increases at the rate of 0.25 centimeters per second. Find the rate of change in the volume of the cone when the radius is two centimeters and the height is nine centimeters. Note that the volume of a cone with radius 𝑟 and height ℎ is 𝑉 equals one over three 𝜋𝑟 squared ℎ.
Let’s begin this question by drawing a little sketch of our right circular cone. And we know that, for this question, this remains a right circular cone. The questions says that the height of the cone decreases at the rate of two centimeters per second. But is there a way to express this mathematically? Well, two centimeters per second is the rate of change of the height with respect to time. Well, we use derivatives to express rates of change. So we express the rate of change of the height with respect to time as dℎ by d𝑡. Now, you might think that this is equal to two. But we need to be really careful here because we’re told that the height of the cone decreases at a rate of two centimeters per second. So in fact, dℎ by d𝑡 is equal to negative two.
We’re also told that the radius of the cone increases at a rate of 0.25 centimeters per second. So let’s also express this mathematically. We can write this as the rate of change of the radius with respect to time. d𝑟 by d𝑡 equals 0.25. And this is going to be positive because this is an increasing rate of change. And we’re going to find the rate of change in the volume when the radius is two centimeters and the height is nine centimeters. We’ve been given a really useful starting point, which is the volume of the cone. 𝑉 equals one over three 𝜋𝑟 squared ℎ. And we want to find the rate of change in the volume. In other words, we want to find d𝑉 by d𝑡, the change in volume with respect to time. So we’re differentiating our function 𝑉 with respect to time.
Well, we can take the constant one over three 𝜋 out of the derivative. But we still need to differentiate 𝑟 squared ℎ with respect to 𝑡. So we’re going to need to do this implicitly. And as well as differentiating implicitly, we also need to notice that this is two functions being multiplied together. So we’re going to have to apply the product rule. Remember that the product rule says that the derivative of the product of the function 𝑓 and the function 𝑔 is 𝑓 multiplied by 𝑔 prime add 𝑓 prime multiplied by 𝑔. So the derivative with respect to 𝑡 of 𝑟 squared ℎ is 𝑟 squared multiplied by the derivative with respect to 𝑡 of ℎ add the derivative with respect to 𝑡 of 𝑟 squared multiplied by ℎ. Well, this is just the product rule applied to the product 𝑟 squared ℎ. So we’re going to need to find the derivative with respect to 𝑡 of ℎ and the derivative with respect to 𝑡 of 𝑟 squared.
Well, differentiating implicitly, the derivative of ℎ with respect to 𝑡 is going to be the derivative of ℎ with respect to ℎ, which is one, multiplied by dℎ by d𝑡. So this just differentiates to dℎ by d𝑡. Similarly, differentiating implicitly, 𝑟 squared with respect to 𝑡 is going to be the derivative of 𝑟 squared with respect to 𝑟, which is two 𝑟 multiplied by d𝑟 by d𝑡. So putting this together, we have that the derivative of the product 𝑟 squared ℎ with respect to 𝑡 is 𝑟 squared dℎ by d𝑡 add two 𝑟 d𝑟 by d𝑡 ℎ. And we can substitute this back into our working so that we now have d𝑉 by d𝑡.
From here, let’s substitute in the values that we know. We have one-third 𝜋 multiplied by 𝑟 squared, which is the radius squared. And we know the radius is two centimeters. And that’s multiplied by dℎ by d𝑡, which we know is negative two. And then, we add two multiplied by 𝑟, the radius, which we know is two. And then, that’s multiplied by ℎ, which is the height. And we know that that is nine. And then, that’s multiplied by d𝑟 by d𝑡, which we know is 0.25. We can work out two squared multiplied by negative two. That gives us negative eight. And we can work out two multiplied by two multiplied by nine multiplied by 0.25, to give us nine.
So our answer is just one over three 𝜋 multiplied by one, which is just 𝜋 over three. But how do we interpret this answer? Well, first of all, note that our answer is positive. So this indicates an increasing rate of change. Also notice that as this is a rate of change in volume with respect to time, our units are centimeters cubed per second.
