 Welcome back to our lecture series Math 12-10, Calculus I for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misildine. In lecture 28, we're going to talk about the idea of related rates. As we talked about in great detail in lecture 27, derivatives are rates of change. The derivative by definition is the instantaneous rate of change of a function. That is you have a quantity that changes relative to some other quantity, and the derivative measures that instantaneous rate of change. Now is one quantity changing at an instant respect to another one. In lecture 28, which we'll also continue talking about this in lecture 29 as well, we talk about this idea of related rates. That is rates, in this case, it's just a synonym. This is just the derivative with respect to time. It's mostly what we're going to see in the situation. Rates here just means derivative and related rates, as this alliterative title says, is that we'll have related derivatives. Let me provide you an example to illustrate the idea behind what related rates are all about. Imagine we are pumping air into a balloon, and just for simplicity's sake, let's assume the balloon has a spherical shape. That's not the usual shape of a party balloon or whatever, or if you want to those balloon benders and make the dogs and such, that's not a sphere. But again, just for simplicity's sake, let's suppose our balloon is that of a sphere, like so. As we're pumping air into the balloon here, so we'll use blue to represent air, that's typically what the sky looks like on planet Earth right here, we're pumping air into the balloon. Just some type of hand pump or maybe we're blowing it with our breath, it doesn't matter. But as air enters the balloon, the volume of the balloon, the volume of the sphere is going to get bigger because the air is going to start causing the balloon to expand outward. But in addition to this expansion in volume, so the volume of the balloon is changing, but likewise the radius of the balloon is also increasing. So after we've pumped air in this for a while, we get this bigger balloon right here. Now our radius has gotten bigger than it was before clearly the volume is bigger as well. So as air is being pumped into the balloon, the volume is changing and the radius is changing. So as these things are changing volume and radius, it's natural to ask about derivatives. So how is volume changing? How is radius changing? Now these two quantities are related to each other clearly as the volume of the balloon gets bigger, so does the radius of the balloon. But the volume is not changing necessarily with respect to radius, that's not the rate of change we'd be probably asking about. Neither is the radius changing with respect to the volume. Instead, both of these quantities, they're changing with respect to a different variable. And in this context, we could think of it as the variable of time. So as time elapses, as we're pumping more and more air into the sphere, as time elapses, the volume gets bigger. So we're asking questions like, well, how does the volume change with respect to time? Okay, but as we're pumping air into the balloon, the radius of the ball also is getting bigger, bigger, bigger. So the radius, DR, I realized my V is poorly written and kind of looks like an R itself. Whoops a daisy. So we have DV over DT and then DR over DT. So the radius of the derivative of the radius, the rate of which the radius is changing with respect to time is also a quantity. But it turns out that these things are derivatives that are related to each other, that knowing something about the change of radius with respect to time also affects the derivative of volume with respect to time. If we know one rate, then we kind of know a lot about the other because they are related to each other and we'll be more explicit what we mean about this in just a second. So in a related rates problem, the idea is to compute the rate of change, aka the derivative of one quantity in terms of the rate of change, that is its derivative of some other quantity, which may be more easily measured. For example, as air is being pumped into the system, we could say something like, oh, we're pumping in three cubic inches of air every second or every minute or whatever. Keeping track of the change of volume given our air pump might be an easy thing to do, but keeping track of the change of radius might be a hard thing. But as the two are related, we can use the change of volume with respect to time to compute the change of radius with respect to time. And so the procedure is to find an equation that relates together the two quantities. In this case, we'd relate together volume and radius. And then we take the derivative of that equation implicitly using the chain rule to make an equation that relates the derivatives of these quantities with respect to time. So that's the general idea. Now let's look at this specifically with some specific numbers here. So let's suppose that air is being pumped into our spherical balloon so that the volume increases at a rate of 100 cubic centimeters per second. And so then we ask how fast is the radius of the balloon increasing when the diameter is at 50 centimeters? So related rates, these type of questions are in fact going to be story problems, but like any other story problem, we can unravel it to find out what the answer is. So let's kind of focus on what are we given and what are we trying to figure out? So the information that was given to us, we know the rate of increase of volume. It's gonna be 100 cubic centimeters per second. So that was given to us. We know how rapidly the volume is increasing probably cause we know how our air pumps working. What we don't know and what we need to find out so the unknown information is gonna be the rate in which the radius is increasing. So we're given information about the volume change. We want to find information about the radius change. So how do we put this together? So the first thing to do with these related rates problems is to recognize what do we know and what do we want to know in terms of derivatives? So if we take, for example, V to represent the volume, the volume of say the balloon, we'll just call it a ball, at some specific time T, all right? So the volume is changing with respect to time. So let V be the volume of the ball at some specific time here. And given that the rate in which it's increasing is centimeters cubed per second, we need to be measuring volume in cubic centimeters and we need to be measuring time in seconds, like so. But we're not given the volume of the balloon. We're given the rate in which the volume is increasing. So notice when you see things like rates, I need you to think about derivatives. So what are words in this story problem that cue us towards, we're talking about derivatives as opposed to other story problems we've seen before. So one big word I want to point out here is the word rate. When you talk about a rate, rate means a ratio and we can see that with the units as well. Derivatives are inherently ratios. They are difference quotients, quotient, ratio, fraction, rate. These are all synonyms in this context. Since we're measuring 100 cubic centimeters per second, we have the units forming a fraction. That tells us we're a ratio. That is a key indicator to us that we're talking about a derivative. So I want you to really nail it into your brain that the word rate in the context of calculus is a synonym for the word derivative. And so derivatives are rates. That's why we call this related rates. Related rates are related derivatives. It's just related rates as a literative. Another key word in this description that indicates that we're talking about derivatives is the word increase. When you talk about words like increasing, decreasing, this will be especially true as we talk about some topics in chapter four. But increasing, decreasing means derivative. If we're talking about a quantity increasing or decreasing, you're talking about its rate of change, which means you're talking about the derivative. So what this tells us is that we know the derivative of volume with respect to time. We see that the derivative of volume with respect to time is equal to 100 centimeters cube per second. And you should indicate it as a derivative, dv over dt. Now, for simplicity and notation, we will probably denote this as just a v prime. The prime here represent the usual derivative, but we're taking the derivative with respect to time. That's important here. Well, what about the thing we need to figure out? We don't need to figure out the radius. We need to figure out how fast the radius is increasing. These words like fast, fast makes you think of velocity. Velocity is the derivative of position. Increasing, like I said before, that is a rate of change type term. So we're asking about the derivative. We wanna know what is the derivative of the radius with respect to time? That's what we need to figure out, but we need to find out a specific moment in time, which they didn't tell us the moment of time. They didn't say like after three seconds, after five minutes. They told us when the diameter was 50 centimeters. Let's unravel that for a second. If the diameter is 50 centimeters, since diameter is twice the length of the radius of a sphere, this tells us that the radius is equal to 25 centimeters. So we need to figure out the radius of the, excuse me, we need to find the derivative of radius with respect to time when the radius is equal to 25. That is the task we're trying to find here. So we're able to unravel this problem with respect to related derivatives here. So we need to find R prime, given V prime, but how are V and R related to each other? The next step, once we've identified the derivatives we know and the derivatives we want, the next step of a related rates problem is to discover what equation relates these quantities together. Not the derivatives, but what relates the quantities together. Well, since we know something about the derivative of volume, and since we need to know something about the derivative of radius, so the equation that we wanna use here is actually the equation of the volume of a sphere. V equals four thirds pi R cubed. Notice how the radius and the volume both come into play in this equation right here. So now we have an equation that relates the quantities. So we have here a related quantity, particularly volume and radius are related to each other. The next step is then to take the derivative of this equation to find the related rates. That is how are the derivatives related to each other. We're gonna take the derivative implicitly. So we're gonna take the derivative of the left-hand side. So we're gonna take the derivative of volume with respect to T. And then we need to do this to the right-hand side as well. What's good for the goose is good for the gander. Take the derivative of both sides with respect to time. So we get four thirds pi R cubed. And then you're gonna continue to compute the derivatives here implicitly. So when you take the derivative of volume with respect to time, that's just gonna become a V prime. There's really no more calculation to do there because we don't have at the moment a specific function of volume with respect to time. We just know volume is volume. So the derivative will just be V prime. We abbreviate that. On the right-hand side, we have this four thirds pi. That's just a constant multiple. That can come out of the derivative process. So the right-hand side is gonna become a four pi over three times R cubed prime. For which we need to take the derivative of R cubed, which is gonna give us a four pi thirds times, the derivative here is gonna be three R squared times R prime. This is a critical part for students as you're working on these related rates problems. Remember, we're not taking the derivative with respect to R. We're not taking the derivative with respect to V. We're taking the derivative with respect to time. We're taking the derivative with respect to this variable T, which the variable T we don't see in this equation, V, the volume, and R, the radius, are functions of time. The radius is increasing with time. The volume is increasing with time, but we don't have those explicit functions in front of us. That's okay. That's what implicit differentiation's all about. We know there's a function relationship with R and time, and V and time, even if we don't have it in hand. We can still figure out the derivative of the radius with respect to time. So don't forget your inner derivative. You need to have an R prime. The most common mistake on these type of problems is you miss the R prime, which after all, that's what we're looking for. So if you don't have the R prime there, then you're never gonna find it because it's not there, you lost it. So simplifying what we have here, notice you do have this multiple of three, this divisor of three, they're gonna cancel out. So writing the equation again, we have that volume, the derivative of volume is equal to four pi R squared times R prime, which we wanna solve for R prime, right? We could divide both sides of the equation by this four pi R squared. What's good for the goose is good for the gander, do it to both sides of the equation. And so we end up with R prime is equal to V prime over four pi R squared. But we don't, so this gives us the general formula for R prime, but we don't want the general formula, there's a specific R prime we're after. We're looking for the derivative of R with respect to time at the moment where the radius is equal to 25. So we can actually plug in for the radius, we're gonna have a 25. So there's a 25 squared in the denominator. What about the V prime on the top? Well, the V prime, remember the change of volume with respect to time, that was a constant 100 cubic centimeters per second. So we actually can plug in volume prime here is just 100 because although the radius of the ball is growing rapidly, the volume is a constant 100 cubic centimeters per second. So this then gives us the quantity we're after. We can simplify this of course. Notice that four times 25 is equal to 100. So this would be, this would simplify as a fraction be one over 25 pi or a probably an approximation would be very appropriate in the situation. So at the moment, the radius is 25 centimeters. We see that the radius will be increasing at a rate of 0.0127 centimeters per second. How do I get the units there? We'll come back to the original derivative. We're measuring radius and centimeters. We're measuring time and seconds. So the derivative will have the unit centimeters per second, which we get as our answer right here. And I wanna indicate to you before we end this video here that the radius is not, even though volume is increasing at a constant rate, the radius is not increasing at a constant rate. How rapidly the radius increases at the, when the circle is small, is very different than how big or how rapidly it increases when it's big. In fact, the bigger and bigger the sphere gets, the harder it is for the radius to expand because we're pumping in the same amount of air. When the balloon is really small, if you pump in 100 cubic centimeters, that's gonna be a huge change of volume there. But when the balloon is really big, pumping in that 100 cubic centimeters per second, that's not gonna make a big difference on the total grand scale of things. That the scale matters a lot. And so we're gonna see that when the, if you were to investigate this formula more and more, is that when the radius is small, then this will actually be a big number, right? Because when you have a fraction, when the denominator gets bigger, that actually makes the whole fraction get smaller. So as r gets bigger, bigger, bigger, bigger, bigger, this derivative will get smaller, smaller, smaller. So as the bigger the balloon gets, the slower the radius is spreading, even though we're increasing at a constant rate. This is actually why babies, how grow their baby clothes so quickly, right? You know, if you've ever seen a newborn, what are those sizes available to them? You have like newborn clothes, zero to three month clothes, three to six months clothes, six to nine months clothes, nine to 12 months clothes, one T is like for the first years of toddler, two T, three T, four T, right? They grow other clothes way faster than grownups, right? You know, some of us have clothes that are years old, right? How, what's the difference? Why do babies have to change their clothes so much? Well, when your little baby is so small, Goo Goo Gaga right here, right? It's gonna, it's growing, right? It's surface area, which is really what the clothes is all about, the fabric you need to clothe that child is growing, let's say it's growing at a constant rate when you're a small baby, well, that change of surface area is gonna be huge. But when you are a much larger adult, let's say you haven't hit full grown yet, so you're still growing, right? Even if you're growing at the same constant rate, which I don't claim that biology whatsoever, the thing is because you're a much larger person, your change of surface area, the amount of skin you have that has to be covered up by the clothes, that's not increasing. It's the same phenomenon we see right here. And so this gives us a very interesting introduction to this idea of related grades that we'll talk about some more in subsequent videos in both this lecture and lecture 29 that we'll see shortly.