 Alright, let's focus a little bit more on the rate of change. So again, the derivative has two main interpretations. One of them, the primary one, is that it corresponds to the rate of change of a function at a particular place. The other one is that the derivative is the slope of the line tangent to the graph of the function at a particular place. And here we'll focus on the first interpretation, the derivative as a rate of change. So, for example, the position of a particle at time t is given by, as some function, t cubed minus 20t squared plus 12t, meters from a fixed position. And we want to find out where the particle is and how rapidly it's moving. Now let's read our function very carefully. The position of the particle is given by this expression, and that's going to be measured in meters from a fixed position. Now my first question, where is the particle? So the question is, where is the particle? Well, that's the type of answer that can be answered by a question like five meters away. And again, if we look at the dimensions, we want a value whose units are units of distance. So we want a value that is given in meters, which is going to be an s of t value. So I'll just evaluate s of t. I know what it is. I want to find out s of t when t is equal to three seconds. I'll substitute that in. I'll do the arithmetic and find my answer minus 127, and this is a value that's measured in meters. So we do want to include our units there, and that tells us this answer here is a distance and is a reasonable answer to the question, where? On the other hand, our second question, how rapidly is it moving? How rapidly? Well, that suggests we're looking at a rate of change, and more importantly, when you ask the question how rapidly, that's a velocity of some sort. So we'd expect an answer to be in, say, meters per second or something like that. And it suggests we're looking for the derivative, but we will check what the units are. So let's find our derivative first. That's going to be 3t squared minus 40t plus 12, and I want to find out what the value of the derivative is at t equal to three. I'll substitute in. After all the dust clears, I get value minus 81, and importantly, the derivative, s prime, is the same as the differential notation ds over dt, and since s of t is measured in meters, then s prime of t is going to be measured in meters over seconds, meters per second, and those will be our units. Now the important check here is whether or not the values that we find make sense as answers to the question. So here we might want to look, where is the particle? Well, minus 127 meters. Again, that's meters from a fixed position. 127 meters away from our fixed position, that's a pretty good answer to the question where. Might not be correct, maybe we miscalculated back here, but it is at least a reasonable answer to the question, where do you find something? 127 meters backwards is how we might interpret that negative. On the other hand, this question here, how rapidly is it moving? Minus 81 meters per second. That seems to be a reasonable answer to the question, how rapidly something is moving, because that indicates some sort of velocity. Backwards in the case of the minus, meters per second tells you how quickly that is. Before you throw the box around the answer and say, here is my answer to this question, before you throw that box around these answers, it is always worth verifying that the units of the answer correspond to the expected units that you would get for the question. Let's take a look at another example. The population of a country in year t is given by 315 times 1.03 t minus 2010 million persons. This actually is a rough model for the population of the United States. How rapidly will the population be growing in the year 2015? So again, the suggestion is that this is a rate of change. We want to find the rate of change when t equals 2015. And since we have a function of t, we want to find the derivative with respect to t. That's a derivative with respect to t. And so I can differentiate that. Again, a little bit of analysis goes a long way. The expression here, if I substitute in the t value, minus 1.03 raised to that constant multiple, this is the derivative of a constant multiplied by something. Well, I can handle the constant no problem. So the something I do have to worry about here, this is something raised to a power. So I'm differentiating a function that is something raised to a power. Now the problem is the only thing I know how to differentiate in that respect is e to the. So I'll apply a little bit of algebra and rewrite this as e to the something. So this is, again, still haven't differentiated anything. I've just rewritten our function as e to the something. And again, this is an e to the something. So I'll factor out that constant multiple. Don't have to worry about that. And then e to the, well, drop everything except for the last thing that I do. And now I have derivative of e to the world's easiest derivative is the same thing times, don't forget the chain rule, derivative of whatever I dropped out. We apply the kindergarten rule. If you take something out, make sure you put it back in. All of that stays the same. I don't change anything there, but I do need to find the derivative of this thing here. Again, that's a constant times something. Log of 1.03 is a constant. So constant times function. The derivative is constant times the derivative of the function, which is just going to be 1. And a little cleanup will be in order. This first expression here. Remember that just came from rewriting our function. So this first part of our derivative is just the same as our original function. So I'll substitute that in. And again, the check to make sure that you're on the right direction as far as the derivative is concerned is that you should see an echo of the original function in what we claim to be the derivative. And here we do see that echo. So we're probably heading in the right direction. We're looking for the units. So our population is measured in millions of persons and T is measured in years. So the derivative, the rate of change of the population with respect to time, that must have units of millions of persons per year. And so we can fill in that unit. And our T value, 2015, is the year that we're looking at. So T is 2015. So we can compute, substitute those in, and after all the dust settles, 10.79 million persons per year. And so we ask the important question. Is that your final answer? 10.79 million persons per year. And before we say yes and check the box, hit the buzzer, whatever, we do want to go back to the original question. We're looking at how rapidly is the population growing. Now, we should always verify that our units will at least be relevant to the question. And there's two possible units here. Millions of persons, because that's the units of our functions. And that doesn't really make sense as a how rapidly question. That makes sense as an answer to a question like what is the population, so and so many millions of persons. But as an answer to a how rapidly question, million persons doesn't make sense as an answer. On the other hand, million persons per year does make sense as an answer to a question, how rapidly is the population growing? So I may have the number wrong. 10.799 might not be the actual value because I've made a mistake in my arithmetic. But my units tell me that I should have been looking at the derivative and assuming that I got the derivative correct, then the derivative is going to answer the question how rapidly is the population growing? So 10.79 million persons per year. Yes, yes, that is my final answer.