Video: Finding the Rate of Change of the Radius of an Expanding Circle given the Rate of Change of Its Area Using Related Rates

The area of a circular disc is increasing at 1/5 cmΒ²/s. What is the rate of increase of its radius when the radius is 6 cm? Use πœ‹ = 22/7 to simplify your answer.

08:25

Video Transcript

The area of a circular disc is increasing at one over five centimeters squared per second. What is the rate of increase of its radius when the radius is six centimeters? Use πœ‹ equals 22 over seven to simplify your answer.

Let’s focus on the first sentence for the moment. The area of a circular disc is increasing at a fifth of a centimeter squared per second. This is the given in the question. It sets up the scenario that the question is based on. See if you can imagine what this scenario would look like. Personally, I imagine a black dot slowly getting bigger. Okay, so how do we move from this sentence, or from the picture in our head, to something more mathematical.

Well, let’s define some variables starting with the area of the disc, which we’ll call capital 𝐴. And what we’re told about 𝐴? We’re told that it is increasing at a fifth of a centimeter squared per second. This is a rate of increase. The unit centimeter squared per second give that away. And we know that the right way to think about the rates of increase, or rates of change, mathematically is using derivatives. The rate of increase of the area is then 𝑑𝐴 by 𝑑𝑑, where, of course, 𝑑 represents time. And we’re told that this is a fifth. We don’t need to add the units centimeters squared per second here. But you should convince yourself that makes sense, that the quantity 𝑑𝐴 by 𝑑𝑑 has these units.

Now, let’s move on to the second sentence. What is the rate of increase of its radius when the radius is six centimeters? This sentence tells us what we are required to find. And again, we’re talking about the rate of increase, or rate of change, of something, this time, the radius of the disc. So if we call π‘Ÿ the radius of the disc, then the rate of increase of the radius, which we are required to find, is π‘‘π‘Ÿ by 𝑑𝑑. Well, actually, π‘‘π‘Ÿ by 𝑑𝑑 is a function that will give us the rate of the increase of the radius at any time 𝑑 or for any radius π‘Ÿ. We only want to know its value when the radius π‘Ÿ is six centimeters which corresponds to evaluating this derivative at π‘Ÿ equals six.

The last sentence is just about simplifying our answer. So we’ll ignore it for now. We’ve been given one rate, the rate at which the area increases. And we’re asked to find the value of another rate, the rate at which the radius increases. And these aren’t two randomly-selected rates, like the rate at which the continents are drifting apart and the rate of increase of money in your bank account. They are related rates. Related how? Well, we know that the area of a circular disc 𝐴 is related to the radius of that disc π‘Ÿ by the formula 𝐴 equals πœ‹π‘Ÿ squared. This is a missing ingredient that we need to solve this question.

Okay, so now that we have the ingredients, let’s see if we can find π‘‘π‘Ÿ by 𝑑𝑑. We do this by using the chain rule. We’ve been given the value of 𝑑𝐴 by 𝑑𝑑 in the question. And so, it’s a really good idea to use this in our chain rule. Now the question is, what is the other derivative? Well, it’s going to be 𝑑 something by 𝑑 something else. And how do we work out what those somethings are? Well, on the left-hand side, we’ve got a π‘‘π‘Ÿ at the top and we don’t have one on the right. So pretending these are fractions for the moment, the numerator of the second derivative should be π‘‘π‘Ÿ. On the denominator, we already have a 𝑑𝑑. But we need to cancel the 𝑑𝐴 in the numerator of the first derivative on the right-hand side. So the denominator is 𝑑𝐴.

Take a second now to check that this really is a chain rule. And now we’re ready to substitute. We substitute a fifth for 𝑑𝐴 by 𝑑𝑑, to find that π‘‘π‘Ÿ by 𝑑𝑑 is a fifth times π‘‘π‘Ÿ by 𝑑𝐴. Now, how do we find π‘‘π‘Ÿ by 𝑑𝐴, the rate of change of the radius of the disc with respects to the area of the disc? Well, we use the relation we have between the area 𝐴 and the radius π‘Ÿ. 𝐴 equals πœ‹π‘Ÿ squared. And there are various methods we could use to find π‘‘π‘Ÿ by 𝑑𝐴 from this relation. We could differentiate with respects to π‘Ÿ, finding that 𝑑𝐴 by π‘‘π‘Ÿ is two πœ‹π‘Ÿ. And then, we could use the fact that π‘‘π‘Ÿ by 𝑑𝐴, which is what we’re looking for, is one over 𝑑𝐴 by π‘‘π‘Ÿ, to show that π‘‘π‘Ÿ by 𝑑𝐴 is one over two πœ‹π‘Ÿ.

If you didn’t know that π‘‘π‘Ÿ by 𝑑𝐴 is just the reciprocal of 𝑑𝐴 by π‘‘π‘Ÿ or, more generally, that 𝑑π‘₯ by 𝑑𝑦 is the reciprocal of 𝑑𝑦 by 𝑑π‘₯, then you could find the same result by differentiating 𝐴 equals πœ‹π‘Ÿ squared implicitly with respects to 𝐴. 𝑑𝐴 by 𝑑𝐴 is one. And we can apply the chain rule on the right-hand side. And we can differentiate πœ‹π‘Ÿ squared with respects to π‘Ÿ, as before. And now, it’s a simple matter of rearranging to find π‘‘π‘Ÿ by 𝑑𝐴. Again, it’s one over two πœ‹π‘Ÿ. So let’s substitute this for π‘‘π‘Ÿ by 𝑑𝐴. Simplifying, we get one over 10πœ‹π‘Ÿ. Before we evaluate this expression for π‘‘π‘Ÿ by 𝑑𝑑 at π‘Ÿ equals six to find our final answer, let’s first think if this expression is reasonable.

We’ve shown that the rate of change of the radius of the circular disc π‘Ÿ with respects to time, as the area of the disc increases at a constant rate of a fifth of a centimeter squared per second, is one over 10πœ‹π‘Ÿ. As the radius π‘Ÿ of the disc must be positive, this one over 10πœ‹π‘Ÿ must also be positive. The rate of change of the radius with respect to time is positive. And so the radius is increasing. Hopefully, this fits with the picture we have in our head. As the area of the disc is increasing, its radius must be increasing too.

But, notice that although the area is increasing at a constant rate of a fifth of a centimeter squared per second, the radius is not increasing at a constant rate. The rate at which the radius is increasing depends on the radius π‘Ÿ. π‘‘π‘Ÿ by 𝑑𝑑 gets smaller as π‘Ÿ increases. And so although the radius is increasing and so the circular boundary of the disc is moving further and further away from the centre, the speed at which it does so is decreasing. Strictly speaking, it isn’t necessary for you to understand all that to answer the question. But it’s always a good idea to think.

Back to the problem at hand, we found π‘‘π‘Ÿ by 𝑑𝑑. But what is required is π‘‘π‘Ÿ by 𝑑𝑑 at π‘Ÿ equals six, which if you recall is the rate of increase of the radius of the disc when the radius is six centimeters. So let’s clear some room to substitute. Substituting six for π‘Ÿ, we get one over 10πœ‹ times six, which is one over 60πœ‹. And we’re told in the question to use the approximation that πœ‹ equals 22 over seven. So we make this substitution. Simplify by multiplying the numerator and the denominator by seven. And performing the multiplication in the denominator, we get seven over 1320.

But, seven over 1320 what? What are the units? Well, this is the rate of change of the radius with respects to time. And the time is given in seconds. And as the area was given in centimeters squared, the radius is given in centimeters, so we have units of centimeters per second. Interpreting this result in context then, when the area of a circular disc is increasing at a constant rate of a fifth of a centimeter squared per second, the rate of increase of the radius of the disc when the radius is six centimeters is seven over 1320 centimeters per second. Well, at least, using the approximation that πœ‹ equals 22 over seven.

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