 Imagine we have some type of rod or piece of wire or just just something that you might be a really, really long object like a pipe or a stick or a pole, what have you. We say that a rod, which you know about something like this here on the screen. We'll just call it a rod for the sake of terminology here. A rod is homogeneous if its density is consistent throughout so it has the same density whatsoever. So like imagine you go to like your favorite home supply store or something. You pick up a long copper wire or a long PVC pipe or maybe doing plumbing or whatever. Again, something really long like a long pole. The idea of a homogeneous rod is that it's the same density throughout. It's not thicker at one part versus the other. In contrast, like a non-homogeneous rod might be something like this, right? When you look at it, its density, just it could get thicker in some places, wider at other places, something like this. We have this homogeneous rod right here and then this non-homogeneous rod right here. So the homogeneous rod, we could actually calculate the density of that rod fairly easy. So what we would do is we could first put this rod on a scale of some kind. So we take our bathroom scale right here. We put the rod on and it's like, oh, the rod weighs three pounds, something like that. Great. So we can figure out the mass of the object. We can also pull out our measuring tape, right? And we can measure that, oh, this thing turned out to be 10 feet long. And so then we can figure out the length of this rod right here. And so because the rod is homogeneous, we can then take the ratio between the mass of the object, divided by the length of the object. So if we said like three pounds, what did I say, over 10 feet, right? We would then get this 0.3 pounds per foot as the linear density of this rod. And when it comes to density, one typically denotes that using the Greek letter rho. It kind of looks like a P, but it's more like an R sound rho. And so we can measure this linear density of this uniform rod fairly easy. Now for the sake of scientific measurements, we will actually measure our linear density using kilograms per meter. But you can use any measurement of mass versus any measurement of length and you'd be just fine right here. So measuring density is a very easy thing if you have this homogeneous rod. But what happens when you have this non-homogeneous rod? Measuring density becomes a much more complicated issue, right? You know, if you look at, for example, this segment right here versus this segment right here, measures this segment here, the density could be very, very different at these different sectors of the rod. And as such, computing its instantaneous density at a moment or the so-called linear density at a moment becomes a little bit more complicated and calculus is going to be necessary in that situation. So for this example, consider the function m to be equal f of x where the variables have the following meaning. So we're going to pick some location which we call x equals zero. So some idea of zero length, right? So it's just a marker. We could put x equals zero in the middle of the rod. We could put it on the right-hand side. We could put it on the left-hand side. We could put it anywhere we want. It's just some reference here. So this is like our starting point. Again, just need some type of reference. So for the sake of it, we'll just be like, oh, we'll put x equals zero on the left side of the rod. No big deal. So the measurements x, like x equals one, would represent that we're one meter away from the starter marker. x equals two would mean we're two meters away from the marker. x equals three would mean we're three meters away from the marker. That's what the variable x here is measuring. It's measuring the length in meters from some fixed point associated to this rod. Then the function m, which is a function of x here, this will be the mass function. So if we looked at, for example, x equals one here, f of one, this is going to give us the mass from this sector right here. So it calculates the mass of this portion of the rod. So it's like, oh, with this rod, if you look at one meter, that's going to weigh, say, two kilograms with this rod. So f of one equals two kilograms. It's the mass of that sector. But if we want to measure, like, say, x equals two, then f of two would measure the mass of the rod from x equals zero all the way down to x equals two. So you'd get that bigger chunk right there. Well, what does something like delta m mean for this function right here? If we take f of x two, that would be the mass from x equals zero to x equals two. You see that right there. If we take f of x one, that would be the mass from this point to that point right there, their difference then would then take off the mass of the portion they overlap, which we see the overlap is going to be right here. That's what's taken away. So f of x two minus x one, that would measure the mass of the rod from x one to x two. So delta m in this situation is measuring the mass then of the rod between the two markers we have right here. That's going to give us delta m. Delta x would have the same interpretation that we've seen before. It's just the difference of x one and x two. So if you take the difference there, that would measure the length of this sector. So delta x is measuring the length of this sector of the rod, and then delta m measures the mass of that sector of the rod. So if we take the ratio delta m over delta x, which would explicitly would look like f of x two minus f of x one over x two minus x one. That would be the average rate of change of the mass function with respect to length, or what we call more affectionately, the average density. Because unlike the homogeneous rod, which is completely uniform right, the density is the same no matter where you're at. The density of this non homogeneous rod depends where you are on the rod. And so different sectors, different intervals of the rod will have different average densities. So this average density is computing well, if the rod was homogeneous, what would be the uniform density at that, at that sector, right? Now if we allow the length of the rod to go to zero, that is the just the sector we're looking at. If we allow x two to get closer and closer and closer to x one, you're getting these smaller and smaller and smaller intervals. That is, if we allow delta x to go to zero, the length of the sector gets smaller and smaller smaller. What happens to the average density? Well, the average density, if we take the limit as delta x approaches zero here as x two gets closer and closer and closer to x one. Then this would give us the derivative because we're taking the limit of a difference quotient. This would give us the derivative dm over dx. This is what we call the linear density of the rod or just drop the adjective and call it density. Whenever you're working on these problems about instantaneous rates of change and average rates of change. If I drop the adjective and just say rate of change by default, that means the instantaneous rate of change. So if I ever say density by default, that means the instantaneous density, the linear density of the rod at that moment. We could talk about what's the density at a specific location of the rod here. What's the density at that marker right there? So dm over dx, the change of mass with respect to length would be the instantaneous density of a rod. So imagine we have a rod whose mass function is given by the formula m equals the square root of x. So what I want you to kind of imagine there is that at the very beginning of our picture, at the very beginning of our rod, which like we did before, we're going to use the left hand side. We're going to use the left hand side here to be x equals zero. If you have a rod whose mass function was given as the square root of x, you'd actually get something like this. So at the very beginning, because as you move to the right, the mass is always increasing. The longer the rod gets, the mass increases. But the thing is with the square root function, there's this initial like explosion of mass increase. But as you go farther along, the rate in which the square root function is increasing, it gets slower and slower and slower as you go higher and higher and higher. And that's as that that would suggest that you're offering less and less and less mass, the more length you add here. So it's quite reasonable that a rod, a non-homogeneous rod could have a mass function equal to square root of x. You get something that looks like this. All right. What would be the average density of our rod on the interval 1 to 1.2? Well, that would just mean we could take x to equal 1, we take x to equal 1.2, and we then compute the average rate change for this function on that interval. So delta m over delta x, this would be the square root of 1.2 minus square root of 1 on top. And then on the bottom, you get 1.2 minus 1. The denominator is much easier to do, right? Because 1.2 minus 1 will just be 0.2. That's the length of this portion of the rod. The mass is a little bit more tricky because I take the square root of 1.1, subtract that from 1, square root of 1 of course is 1. In the end, that is going to give you approximately 0.48 kilograms per meter. So this tells us that on our picture, this sector of the rod has an average density of 0.48 kilograms per meter. What's the instantaneous, what's the linear density when x equals 1? Well, that comes from taking the derivative. The derivative here, so the derivative of mass with respect to length would equal the density function. We need the derivative of mass with respect to length and evaluate at x equals 1. Well, given that our function is the square root of x, if we take the derivative with respect to x by the usual power rule, you'd get 1 over 2 times the square root of x. Evaluate that at x equals 1, you end up with 1.5 or 0.5 kilograms per meter. And so the density at the moment x equals 1 is that rho is equal to 0.5. Of course, the average density we got on this sector when you're really close to 1 was 0.48. And we anticipate that the average density should approximate the instantaneous or the linear density of the rod. Now, the point of all of this is this is very similar to what we did at the end of Chapter 2, talking about the velocity problem. We can use average velocity to approximate the instantaneous velocity. The difference now is that the calculation of the derivative, I didn't even show any of the details. I was just like, use the power rule, and that's how we got it. So all of the difficulty of the derivative calculation is now, in some sense, behind us. And we still have to do it. We're still practicing it, but it's not just this crazy hard problem anymore. It's a lot easier because of what we've already done in Chapter 3. So we can do the same type of scientific analysis we did earlier with derivatives, but we're no longer burdened by the calculation of the derivative itself. And so this is the true power that we see from what we've been doing in Chapter 3. We can now compute derivatives effectively and use them to analyze things about, in this case, the density of a rod.