In today's video, we go through an extended example involving data points from the velocity curve of an accelerating vehicle to estimate the displacement or distance traveled. This exemplifies the idea of using the derivative to get information about the original function. As you'll see, the calculation or estimation of areas under curves is the key to moving backwards so to speak from the derivative. The underlying reason is that derivatives are alleged to instantaneous rates, which are limits of fractions. To undo a fraction, you expect multiplication to be involved, and of course, multiplication of real numbers is related to and may be defined in terms of areas of rectangles. This is our first foray into integral calculus. Two key words will become familiar to you are integration and antidifferentiation. In common language, to integrate, means to put things together into a coherent whole. The word anti-differentiate is not part of common language and only used by mathematicians, and it means to undo differentiation in the sense that we can make precise. That is to go backwards from derivatives to the functions from which they came. For us, putting things together will have a technical meaning of finding areas, which involves multiplication and converting rates into quantities, which is some kind of process by which we unpack the derivative. The central theme is to use information about the derivative to gain information or insight concerning the original function. Think of all that hard work in the previous two modules where we created derivatives. Now, we going in the reverse direction, developing techniques for undoing the derivative. This leads naturally into setting up and solving differential equations, which are equations in which at least one derivative appears somewhere. The study of differential equations will be for you a natural continuation following on from this introductory course in calculus. Working backwards from the derivative it includes for example, knowing the velocity, the instantaneous rate of change of displacement. To get back to or find displacement, or knowing acceleration the instantaneous rate of change of velocity to get back to or find velocity, or knowing the rate of absorption of a drug, to determine the amount of the drug that's been absorbed, or knowing the rate of growth of the colony of bacteria to determine or predict size of the population, or knowing the rate of spread of a disease to estimate or predict the number of people that will become infected. Let's focus now on perhaps the most important motivating example. Velocity denoted by v. The derivative of displacement denoted by x, both of which are functions of time t. We can keep in our minds that v is dx/dt. If we have enough information about v and even without a formula for dx/dt, then we can find or estimate displacement x by finding areas under the velocity curve. Here's a typical velocity curve, and we're interested in the displacement over a particular interval of time as t passes say from a to b. We move up the curve from t equals a and t equals b to create a region in the plane colored here in green, bounded by the curve the horizontal axis and the vertical lines through a and b. It turns out that this green area represents the change in position, the displacement x of the object of this particular time interval. To try to understand this phenomenon, we first look at the simplest case when the velocity v is constant, say c. So, that the velocity curve is a horizontal line passing through c on the vertical axis. We move up to the curve from t equals a and t equals b and obtain a rectangular region colored green in this diagram. To make it more concrete, let's measure the velocity in say kilometers per hour and time t in hours, and you could imagine that we're traveling in a car alongside a straight road. Note that in the diagram, the scales on the horizontal and vertical axes are typically different, and can change according to the numbers involved. For example, if c equals 60, a equals five and b equals nine, then we'd be traveling at 60 kilometers per hour for four hours. So, the distance traveled will be four times 60 equal to 240 kilometers, represented exactly by the area of the green rectangle of width four units and height 60 units. In general, the width of the rectangle is b minus a and the height is the constant c. So, the green area is the product c times b minus a. Exactly representing the number of kilometers traveled by multiplying the constant velocity by the number of hours spent traveling. Of course in practice, the velocity fluctuates. Let's go through a typical example. To estimate the distance traveled by a car that is accelerating over an interval of 10 seconds so that its speed keeps increasing. We will assume the car is traveling along a straight road in one direction. So, we can use the words velocity and speed interchangeably. The velocities were measured every two seconds, and recorded here in meters per second to keep the units consistent and to simplify the calculation that follows. Note, to put this in context, in case you're used to measuring speeds in kilometers per hour, that 90 kilometers per hour converts to 25 meters per second. You can see from the table that the car must have been going really fast after about five or six seconds. We can introduce a horizontal axis for time t and the vertical axis for velocity v and plot the points from the table. We can join the points to get what we expect will be a reasonable approximation to the true velocity curve of the car. Now, in the first two seconds, the car accelerates from five to 14.5 meters per second. So, it's always traveling at a speed greater than or equal to five meters per second much of the time considerably faster than that. Any car moving at a constant speed of five meters per second would travel a shorter distance, which is represented by the green rectangular area in the diagram. This green area sitting beneath the curve then becomes a lower bound for the distance traveled by the accelerating car over the first two seconds. Similarly, we can draw a rectangle for the period from two to four seconds. This area represents the distance traveled by a car moving at the constant speed of 14.5 meters per second. This again is a lower bound for the distance traveled by the accelerating car, which is always moving at a speed greater than or equal to 14.5 meters per second over this interval. We can continue drawing in rectangles beneath the curve for time t equals four to six seconds, t equals six to eight seconds and t equals eight to 10 seconds. We get a lower bound on the actual total displacement of the accelerating car by adding up the areas of all these green rectangles or columns sitting beneath the curve. All of the columns have width two units. So to get each successive shaded area, we just multiply the heights by two. As the car is accelerating, its speed keeps increasing throughout. So we just take the velocities in the table as heights of the rectangles in succession from five all the way up to 31 and this sum quickly evaluates to 200. Thus, a lower bound for the distance traveled by the car becomes 200 meters. The previous estimate used rectangles sitting beneath the curve. If we use upper bounds for the velocities on each of the sub intervals, then we build rectangles that sit above the curve. In this diagram, we enlarge the green area in each case to the blue area, so that the combined blue and green areas form columns that encompass all of the area under the curve. The area of each larger rectangle or column now represents the distance traveled by moving at a constant speed greater than or equal to the speed of the accelerating car on any particular interval in two seconds. By adding up all of the areas, we should get an upper bound for the total distance traveled. Again the width of each column is two units and heights now moved from 14.5 and upper bound for the first interval successively through the heights in the table finishing at 32.5 and upper bound for the last sub interval. These areas all add up to give 255. Hence, we estimate an upper bound for the true distance travelled by our car to be 255 meters. Putting this together, we found a lower bound and an upper bound and note that the true distance traveled is somewhere in between. The number halfway between these two bounds seems likely to be a reasonable estimate and that number is just the average, which turns out to be 227.5. Hence finally, we estimate that the car traveled about 227.5 meters over those 10 seconds. In fact, we expect this average to be a slight underestimate. As you can see from the sketch that the curve appears to be concave down. I'll try to explain the significance of the concavity. Consider any rectangle, split it into two pieces and color the low part of the rectangle green and the upper part blue. Now, draw a diagonal across the upper part upwards from left to right and coloring pink everything below this diagonal. This pink area, is the green area plus half of the blue area, which can be expressed as one half of the green area plus the combined green and blue areas. In other words, the pink area is the average of the small and large areas from the previous two diagrams. The shape colored in pink is an example of a trapezium in geometry. In our example with an accelerating car, in a particular interval of two seconds, we had a green rectangle sitting below a fragment to the velocity curve; the combined green and blue rectangles sitting above the curve. And there's this trapezium shape, which we have colored in pink and as we've just explained, the pink area is the average of the areas of the upper and lower rectangles. Let's add to this diagram, a curved fragment joining the points of the diagonal that creates the trapezium and draw it so that it's concave down as in our physical example. Because of the concavity, there's a tiny piece of area not included by the pink shading, which we've colored beige. The pink area, the area with trapezium therefore becomes a slight underestimate to the area under this curve fragment. Nevertheless, the trapezium becomes an excellent approximation for the area if the curved fragment is almost a straight line. This leads to a technique known as the trapezoidal rule for approximating areas under curves using trapeziums, which you can read about if you wish. In the case of our velocity curve, you can see visually that though concave down overall, the individual fragments of the curve are close to being straight lines. So, we expect this average, the upper and lower bounds, which is the aggregate of the averages of the areas of the lower and upper rectangles to be very good if not excellent approximation to the true distance traveled by the car. The previous estimate was based on a table with six data points. Clearly, if we had more data, then we expect to be able to improve our estimate. Suppose that the velocity was measured every second instead of every two seconds. Here's a new table with the original data, but including measurements taken also for each second in-between. As before, we can plot all of the points and join them with a smooth curve. Again, we build rectangles to sit beneath the curve and then shade in the areas in green. We can build rectangles that sit above the curve by extending the lower rectangles and shade in the additional areas in blue. As before, we can calculate lower and upper bounds to the area under the true velocity curve. Note that the width of each rectangle or column is one unit, since the time intervals are now only one second each. Adding up the green areas gives a lower bound of 216. Whilst combining the green and blue areas gives an upper bound of 243.5. We can again take the average to produce a revised estimate of 229.75 meters for the distance traveled by the car. Because the curve appears to be concave down, we still expect this to be a very slight underestimate. The curve fragments appear to be very close to a straight line segments on each of these, now very short time intervals. So we expect to have produced an excellent estimate of the distance traveled. In today's video, we began with some general remarks concerning moving backwards, so to speak, from rates to original quantities or, in mathematical language, using the derivative to gain information about the original function. Moving backwards from velocity to displacement becomes a prototype for this phenomenon. We saw that the connection involves finding areas under curves. We discussed in detail how we could use a table of values for the velocity to estimate the distance traveled by an accelerating car over a given time interval. We use rectangular approximations that sit both below and above the velocity curve to get lower and upper bounds for the distance traveled. By taking the average of these bounds, we appear to get a very good if not excellent approximation to the true displacements. Taking averages is in fact related to using trapezoids to estimate the area under curve. Please read the notes and when you're ready please attempt the exercises. Thank you very much for watching and I look forward to seeing you again soon.