Calculus: The Pebble Problem

This lesson has not been reviewed.

Please purchase the lesson to review.

Optimization
Related Rates
The Pebble Problem Page [1 of 3]
Well, where are we? We've just taken a big chunk of time working through some serious max and min problems where we're giving some sort of story and we want to optimize something, either make something very, very small, make something very, very large. And we saw that thinking about the derivative as a slope of a tangent line allows us to find those points where, in fact, you may have a low or a high point because that function comes down, levels off and goes back up, or it goes up, levels off and comes down.
I would now like us to return to the notion of a derivative as a rate, as an instantaneous rate of change, in fact, and take a look at what are sometimes referred to as related rate problems. Now these are questions that are interesting. Unlike the other ones where we were given something to optimize and maximize, here what we're trying to do is actually figure out how things are changing and how things are moving, how things are growing, how things are shrinking. And so, the problems, and the reason why they're called related rates usually involve the following kind of issue.
Something is moving and that causes a related thing to change. And the question is if we know how fast the first thing is changing, how can we use that information to figure out how this other related thing is changing. That really is the fundamental notion of related rates. And the key thing to always remember is that if you're doing, or thinking about an issue that involves a rate or movement of any kind, or growth, what is the variable that is mysterious? What is the variable that's independent? It's always time. Time is the variable that is the variable we will differentiate with respect to always. So it's a good thing to remember that we'll always be taking derivatives with respect to time because we're looking for how things are changing with respect to time. That's what it needs to change.
Okay. Let me try to inspire this by looking at our first question. This is a classical question, and it involves, actually, looking at a lake. So, imagine if you will a beautifully still blue lake. Now those of you who, who are from the city may not have these things recently, but you can certainly visualize it in your mind's eye and your imagination is beautiful, and glass, like glass. There's not even a zephyr in it. And then you take a stone, just sort of impish, and what you do is you drop the stone in the lake. And I'll do this for you right now if you pretend that this white thing is a lake. Well, it wouldn't bounce off the lake unless you are from the city, but, but actually what would happen is it would go and it would create this ripple effect. It would create a very beautiful ripple effect. And what it would really do, actually, is create concentric circles that would start to grow and emanate out, and so on. And it would grow out, and grow out. It would be really beautiful.
In fact, I can show you what this looks like. Here's a picture of what happens when you drop something, and you see the drop happened right there. And you get a little back splash and then you can see these beautiful concentric circles that are growing, and growing, and growing.
All right, well let me tell you something about my particular drop. When I dropped the stone in, I happened to have been able to compute the fact that the radius of these circles -- the circles that are growing out like this -- the radius, of course, as we can see, is increasing, because the radius is getting larger and larger. You can see how the radius is growing. Here it's very, very large. Here it's even larger, and larger, and larger. They're growing and moving outward. It turns out that that radius is increasing at a rate of six inches per second. That means when I drop it in, in one second I already have a circle that has a radius six. In two seconds I now have a circle that's radius twelve inches, and so forth.
And now what I would like us to figure out is how fast is the area of the circles changing. You'll notice that as these circles grow, certainly the area is growing as well, and what I would like us to figure out is the following. If I know how fast the radius is growing, how can I figure out how fast the area is growing, let's say, two seconds after I drop the stone? So, I drop the stone, bloop. I count thousand one, thousand two. At that instant, how fast is the area of the circles made by the, by the ripple changing?
So that's the first question I want us to think about, and so let's take a look at how we might actually attack this question. Well, the recipe is always going to be to revert back to our method for solving all of life's problems. So, the first thing we want to do is to understand what is being asked of us. Let's think about the question again and make sure we understand what we're being told. So, what is being asked? What we want to do is we want to find out how the area of the circle is changing. So we want to find out, want to find how the area of the circle is changing. And changing when? Changing when -- let's say time equals, equals two seconds. So, that's what we want to find.
Now, that's all written out in words. What I want us to do is now think about how we'd write that down in symbols. Now this is where, actually, something new is about to happen. We have to think about how to analyze this. So, let me give area of the circle a name. Let me call it A. Let's let a equal the area of the circle. So what then do I want to find out? I want to find out how the area is changing. So that's a rate. So I'm asked to find a rate of change of the area in terms of time. So that actually represents some derivative, and it's the derivative of what with respect to what? Well, if we think about it, I believe it's the derivative of area with respect to time. And so that would actually be written as that we want the derivative of area with respect to time. So that symbol, remember, means I want to find the derivative of area with respect to time, how the area is changing with respect to time. So I want to find that at the moment when t equals two, so when t equals two. That is what we want. So, that's the first thing is to understand what we're being asked. And let me point out, again, this is really something new. This is where we have to think about this notion of a rate as a derivative. Okay. Well, that's what we want.
The next step is to figure out what we know. Well, what do we know? Well, one thing we know is we know how the radius is growing. Remember that we were told that the radius is increasing at a rate of six inches per second. So, how can I write that? So, we know that the radius is increasing at a rate of six inches per second. So how can I write that in terms of Calculus? Let's think what that's saying. It's saying that the rate of change of radius with respect to time is six. Six inches per second. So that would be what? Well, that would be the derivative of radius with respect to time, how radius is changing in terms of time. And so that would be, if we call radius r, that would be the r d t. So, the r d t would be six inches per second. So what we know is that the r d t is six inches per second, and why do we know that? Well, because that's exactly what this says in Calculus terms. It's saying the rate of change of radius with respect to time is six. And what we were told is that the radius is increasing at a rate of six inches per second. That means that it's changing with respect to time at a rate of six.
Okay. Well, that's what we know. This is what we want. And so now, we come to the last, and the most important step and that is to find the connection. You see, I know how one thing is changing and I want to see how another thing, another related thing is changing. That's why we call these related rates. I see the rate of this. I want the rate of this, which is related. Somehow, it's got to be related. If the radius is changing, it makes sense that the area should be changing as well.
So what's the connection between these two things? Well, I know how radius is changing. I want how area is changing. I need a relationship that links area and radius together. Can you think of one? It's the formula for a circle. Area equals pi r squared. There is the relationship between the things I know something about, I know how r is growing, and something I want. I want to know how a is growing. Here's the relationship. So there's the connection between those two things.
Okay. Well, now let's see if we can actually solve this question, answer this question. What would we do? Well, what we need to do here is look at this expression, this is a relationship, and I want to see how this allows me to conclude something about how things are changing. So what I'd like to do now is differentiate this. So I want to take the derivative of this. Now what do I take the derivative of this with respect to? Should I take the derivative with respect to x? Well, no, that doesn't make much sense, there are no x's here. Maybe with respect to r? Maybe with respect to a? Well, what we have to do is see what's the information we seek. We want to see how things are growing with respect to time. So I should differentiate this whole thing with respect to time.
So I want to now differentiate the whole thing with respect to time. And this is going to involve implicit differentiation, which we talked about before - implicit differentiation. So, I'm saying take the derivative with respect to time of the whole thing, equal sign and all. What do we get? Well, I do the left-hand side and then I consider the right-hand side. Now we have to consider these things separately. Each of these is going to require a little teeny implicit differentiation, by the way.
All right. Well, let's take a look at this first piece right here. What's the derivative of area with respect to time? Well, that would be the derivative of area with respect to time. We have a name for that. That is ____ the area, the derivative of area with respect to time. It's the da/dt. So this derivative is, in fact, da/dt. Right? This is saying find the derivative of area with respect to time. Well, we call that da/dt.
Let's take a look now at the right-hand side. Remember pi is just a constant number, so that's just a constant in front of r squared. How do I take the derivative of this with respect to t? I have to use implicit differentiation because there's no t there; it's an r. So I look at this and say, "Well, the derivative will be pi times, and this is blop squared." The derivative of blop squared is two blop multiplied, and I peeled off the two, multiplied by the derivative of r with respect to t. And what is the derivative of r with respect to t? We have a name for that. We call that drdt.
So, notice how we're using implicit differentiation here to actually figure out the derivative of this whole expression with respect to t. All this related rate problem will require implicit differentiation, differentiating with respect to t.
Well, this is great news! Look at this! This is a fantastic fact to discover and I hope that you realize the power of this fact because look, I'm trying to figure out what this equals, that's what I want. I know what this equals. So now, this connection between the two variables, after differentiating, allows me to get a relationship between the rate that I know and the rate that I want. This is great news. In fact, we're almost done, because what do I know? I know that the r d t equals six. So this number is just six. That rate is six.
Now what is r? I don't know what r is though, but I was asked to find out what this rate is when t equals two. What is the radius when t equals two? Well, since the radius is increasing at a rate of six inches per second, after the first inch - after the first second, it's six inches, and then after the second second, it's six more inches away, so it's twelve inches. So, in fact, the radius when t equals two is twelve inches. So, when t equals two, we know that the radius would be twelve inches.
I'm trying to find the adt when t equals two, so this value at that moment will be twelve. Well, I'm done. I put in twelve for r, and six for the r d t, and what do I see? Well, what I see is that the adt equals pi times two times twelve, at that moment, times this rate, which is six. So, what does that equal? Well, that equals - well, twelve times twelve, which is a hundred and forty four pi. And, by the way, what are the units here? Well, it's area, so that's going to be inches squared per second, because it's a rate.
And so, in fact, we just found out that the adt equals a hundred and forty four pi. Pi is around three point one four, so you can actually multiply three point one four by one forty four, and you get a sense that the area is going very, very, very quickly. The radius is growing at six inches.
So, that's a first example of related rates. And the theme here to notice is the following. First of all, you have to very carefully write down what you want. Figure out very carefully what you know and then figure out a connection that links those things together. Differentiate to get a connection between the rate and then plug in all the information to actually get your answer.
By the way, it's sort of interesting that it's growing so fast, but it sort of makes sense when you look back at the original picture. Because, in some sense, the radius is growing only in one dimension; whereas, the area is actually expanding in sort of two directions - two dimensions, and so it should be growing much, much faster. And that's what we're seeing here.
Okay. Well, stay with us and we're going on a little cavalcade ride through all sorts of examples of related rates. Get you having a feel for it and have some fun. See you up next. Bye.