User:Addemf/sandbox/Who Invented Calculus?/Net Change Is Area

Recall that our goal is to understand free fall motion. We have rejected the constant velocity hypothesis, and now are proposing the constant acceleration hypothesis.

However, testing the constant acceleration hypothesis is hard. We know that constant acceleration implies linear velocity. But the challenge now is to try to determine the position function, if the velocity is linear.

We will consider this problem generally, for arbitrary acceleration functions, and how they determine a position function. In several lessons later, we will return to the question of free fall motion.

Approximations of Change
Consider, for example, an object which has acceleration function


 * $$ a(t) = 2 $$

By the same sort of analysis from the previous lesson, it is clear that the corresponding velocity function is


 * $$ v(t) = 2t$$

Our goal is to find the position function. Although it is hard to do so directly, let's consider "chopping this up" and getting an approximation on each piece.

For example, let us imagine that for the first second of motion, the acceleration is approximately zero. Throughout the first second, we will pretend as though the velocity is constantly zero, with the understanding that we will be getting an under-approximation for the motion of the object.

We know that when $$t=1$$ the velocity is $$v(1) = 2$$. Therefore throughout the next second, we assume that the velocity is constantly 2.

And we proceed so-on. On each one-second interval, we assume that the velocity is a constant value. On each interval, we can find the change in position, and get an approximation of the total position.

For instance, by proceeding like this, let us find the approximation of $$x(3)$$. Throughout the first one-second interval the velocity is approximated as zero, and so we estimate that there is no change in position. Throughout the second interval, the velocity is approximately 2, so the net change is 2. Throughout the third interval, the velocity is approximately 4, so the net change is 4.

In total, we estimate that the net change is therefore
 * $$0+2+4 = 6$$

Of course this is a rough estimate, and guaranteed to be too small. But we can get a better estimate by making the time-intervals smaller.

You may already be worried that we will only ever approximate the answer and never get the true answer. That would feel rather disappointing.

But it turns out that what we are soon to see, is a first example of how you can actually use approximations to arrive that precise answers! This is, in fact, the fundamental concept of all of calculus.

A Better Approximation
Let's walk through one more process like the one above, but getting a better approximation.

This time, rather than approximating the velocity for an entire second, let's approximate it for half of a second.

So for the first half-second, the velocity is approximately $$v(0)=0$$. For the second, it's $$v(1/2) = 1$$. And so on. Let's summarize all of this in a table.

Because the first interval lasts only half a second, then the net change in position is $$0\cdot 0.5= 0$$. For the second interval, it's $$1\cdot 0.5$$.

And so on. In general, since each interval lasts a half-second, then one finds the net change in position by multiplying velocity by 0.5. Therefore we could augment the table further and write

Finally, to get the approximate position, we sum together all of the net changes in position.


 * $$0+0.5+1+1.5+2+2.5 = 7.5$$

Repeat the above exercise, this time using time intervals that last only a quarter of a second.

Nothing about the above idea requires that velocity be linear.

Consider a velocity function, $$v(t) = t^2$$. Approximate the net change over the time interval from $$t=0$$ to $$t=2$$ using intervals of length a half a second.

You may be wondering why we kept choosing under-approximations in the previous exercises. There is no particular reason, and we just as well could make over-approximations.

For example, when the velocity function is $$v(t) = 2t$$, then to approximate $$x(3)$$ using one-second intervals, we could have used the velocity when $$t=1$$ to approximate the velocity over this interval. That would mean that, on the first interval, we approximate the velocity as 2.

On the second interval we would approximate the velocity as $$v(2)=4$$ and on the third, as $$v(3)=6$$.

Use these numbers to find the approximate net change for this example.

Approximations in a Graph
Let us make a graphical representation of just what we're doing when we make these approximations.



To the right is a graph of $$v(t)=2t$$. It shows a yellow region along the t axis from 0 to 1, representing the idea that we approximate the velocity as zero on this interval.

It also shows a yellow rectangle running along 1 to 2 on the t axis, and then up to a height of 2. This represents the idea that we approximate the velocity as 2 on this interval.

And it shows a rectangle running along 2 to 3 on the t axis.

Notice that the areas of the rectangles correspond to the calculations that we made previously, for the net change on each interval!

For example, on the first interval the area is 0. And when we computed the net change here, we computed $$0\cdot 1 = 0$$.

On the next interval the area is $$2\cdot 1=2$$ and that is the same as the net change calculation on this interval.

And on the next interval the area is $$4\cdot 1=4$$, again the same as the net change calculated earlier.

Therefore the total area of all the rectangles is 0+2+4 = 6, which again is the same as what we calculated for the net change.

Show the graph of $$v(t)=2t$$ and show the rectangles representing approximations, where the intervals all last for half a second.

What you should notice in the exercise above is that, yet again, the net change in position is the same as the combined area of the rectangles that you draw underneath the graph!

And as you draw these intervals smaller and smaller, two things happen: The result for the position becomes more accurate, and the area that you compute becomes closer to the exact area underneath the velocity function.

Net Change Is Area
The above observation is in fact a generally correct result: For any velocity function, the area under the curve is the same as the net change in position!

Take for example, a more exotic velocity function like the on you see to the right.



This velocity function is visibly more complicated, and we couldn't really even guess at its formula. Still, we may imagine that it is the velocity of some object, and it seems that the velocity is initially a little more than 2. Let's say $$v(0)\approx 2.1$$. This is about the height of the first rectangle.

After one second passes, it seems like maybe $$v(1)\approx 4.6$$, which is about the height of the second rectangle. These are, of course, rough guesses based on the graph, but it's good enough to illustrate the current point.

After two seconds pass, maybe $$v(2)\approx 5.3$$, and after three seconds pass, $$v(3)\approx 5$$.

We can use these approximations to obtain an approximate displacement. The displacement over the first second is velocity times time, $$2.1\cdot 1$$. Likewise for the remaining intervals. Combining all of these into a net approximate displacement,


 * $$ \Delta_{t=0}^{t=4} x \approx 2.1\cdot 1 + 4.6\cdot 1 + 5.3\cdot 1+5\cdot 1 = 17$$

As before, this is exactly the same calculation one would make to find the combined area of the rectangles in the graph. So yet again, the approximate net displacement is the approximate area under the velocity curve!

Moreover, as one makes the duration of the time intervals smaller, one gets a better approximation of the displacement. And simultaneously, this same thing makes the rectangles a better approximation to the area under the velocity curve.

For the velocity function
 * $$v(t)=t^2$$

approximate the displacement $$\Delta_{t=0}^{t=2}x$$, first using two time intervals, and then using four time intervals. In each case, approximate the constant velocity on each interval, using the time at the start of the interval.

Similarly, approximate the area under the graph of $$v(t)$$ from $$t=0$$ to $$t=2$$, using two rectangles and then four rectangles. In each case, determine the height of the rectangle by using the least value of t in each interval.

Observe at the end that these two approximations are computed in exactly the same way.

The Area Under a Curve
Because it seems as though the area bounded by a curve will be important for our investigations, let us introduce some notation and vocabulary for it.



You can see a diagram of the area under an arbitrary curve to the right. We use the notation


 * $$S = \int_a^b f(x)\ dx$$

to denote the area labeled S in the diagram. We refer to $$\int_a^b f(x)\ dx$$ as the "definite integral of f from a to b."

We are not yet in a position to formally define just what this means and how it's calculated. But the intuition should be clear, in the same way that we intuitively understand both the area of a rectangle and the area of a circle.

The definite integral $$\int_a^b f(x)\ dx$$ denotes the area between the curve of f, the x (or horizontal) axis, and the lines $$x=a$$ and $$x=b$$.

Consider the graph to the right, of the function $$f(x)$$. This function is
 * $$f(x) = \sqrt{1-x^2}$$

(1) Show that the graph of this function is the top half of a circle of radius 1, centered at the origin.

(2) Graph this function and indicate in the diagram the area bounded by the curve.

(3) Find $$\int_{-1}^1 f(x)\ dx$$.

(1) Find $$\int_{-2}^2 \sqrt{2-x^2} \ dx$$.

(2) Find $$\int_0^1 x\ dx$$.

(3) Find $$\int_1^2 x \ dx$$.

(4) Find $$\int_0^2 x \ dx$$.

Consider the function to the right, which is composed of joined lines and portions of circles.