If ancient Greeks were more accepting towards
the idea of infinity, there would not be any
debate between who discovered calculus, as
it would probably be the ancient Greek mathematician
Archimedes, who was born 2 millennia before
both Newton and Leibniz. This is because his
development of the method of exhaustion was
a quintessential precursor to modern-day calculus.
The method of exhaustion is typically used
to calculate the area or the volume of a figure
that is not completely enclosed by straight
lines. There are two variations of this method,
although very similar. We use the example
of calculation of pi to illustrate both variations.
We work on a unit circle, so pi is the area
of the circle. Firstly, we chop off some smaller
regions off the figure,
and calculate the area of the remaining figure.
Then add some of the chopped region back and
calculate the new area. This approximation
should be closer. Then we can repeat this
process as many times as we want to get as
close an approximation as desired. This is
basically identical to the modern definition
of limit of a sequence, just not written explicitly
in modern notations.
The second variation is to add something to
the intended figure and calculate its area.
Then chop some of the added regions and calculate
the area again. So in this variation, we approach
pi from above as opposed to the previous method
which approaches pi from below. Although it
sounds like it acts only as a tool to calculate
the area of a curved figure, more importantly
it is used to define the area of such a figure
in the ancient times.
What Archimedes did was to calculate the area
enclosed by a parabola and a line, which is
called a parabolic segment. In this video,
we are recreating what Archimedes would do
if he were to calculate this integral, which
means the area of the yellow region. It would
be beneficial to reflect this area across
the x-axis due to reasons explained later.
Archimedes work was to calculate the red region,
but we can see that these two areas add up
to a triangle, and so to get the value of
the integral, we just have to subtract the
red region, which is what Archimedes’ work
was all about, from the area of the triangle.
Archimedes method of exhaustion on a parabolic
segment starts off with a novel way to calculate
the area of the triangle. We first cut the
triangle vertically into two different triangles.
Regarding h as the base of the two triangles,
then the corresponding heights would be these
horizontal lengths, so we have this expression
as the area. If we now denote s as the sum
of the horizontal heights, then we have a
simplified formula of the area of the triangle.
Here, h can be thought of as the vertical
height, and s can be thought of as the horizontal
span. Things will be getting more complicated,
so feel free to pause the video at any point
from now on.
Now, the overall idea is to first inscribe
a triangle in a parabolic segment. This triangle
should have the x-coordinates of its vertices
evenly spread out. More precisely, the vertex
in the middle should have x-coordinate the
average of the other two. Then we add the
orange areas. Again, the middle vertex is
not chosen at random, and it should have the
x-coordinate as the average of the other two.
Then we repeat this process indefinitely.
The amazing thing about this method of exhaustion
is that if we denote the area of the largest
triangle as A, then the sum of the orange
areas would be a quarter of A, then the sum
of the yellow areas would be a quarter of
that of the orange ones, which means one-sixteenth
of A, and so on. So if we know the value of
A, we should know the area of the parabolic
segment. But now we want to figure out why
these orange areas sum up to a quarter of
A. To explain that, we use physics and treat
the parabola as the trajectory of an object
flying through the air under gravity. In physics,
we usually analyse the projectile motion by
separating the velocity vector into horizontal
and vertical components. Here, the blue component
does not change, while the green component
changes at a constant rate. Then this yellow
vector represents the average direction the
object travels throughout the journey. Since
the green component changes uniformly with
time, while the horizontal component doesn’t
change, the average throughout the journey
happens midway. This means this yellow average
velocity vector should be the same as the
red velocity vector midway through the journey.
This implies that the red tangent and the
yellow line are parallel. Notice that this
point is the third vertex when we draw the
largest inscribed triangle.
Now consider the orange triangle, and the
lengths of these two yellow lines. Think of
this red line as the trajectory of the object
if gravity suddenly disappears. Then any vertical
difference between the red line and the white
parabola should be caused by gravity only.
And in physics, we have this equation of motion,
which is applicable here because there is
no difference in velocity initially. This
equation means the vertical distance between
the red line and the white parabola is proportional
to the time elapsed squared. In this specific
case, the time elapsed during these two intervals
are the same, so the ratio between the lengths
of the yellow lines would be 1 to 4, because
of this squaring proportionality.
Finally, we just need to realize this pair
of similar triangles. Then this length here
should be half that of 4 times l, and since
we previously discussed that this length is
l, which means the vertical height of the
orange triangle is of length l. The vertical
height of the red triangle is 4 times l, because
the yellow and red lines are parallel. And
by construction, if the horizontal span of
the red triangle is 2 times w, the horizontal
span of the orange triangle would be w. As
a summary, the orange triangle has a quarter
of vertical height, and a half of the horizontal
span compared with the red triangle. Thus
the area of the orange triangle is an eighth
that of the red triangle. By running the same
argument on the other side, the other orange
triangle would also have an eighth of the
area of the red triangle, and so the total
area of the orange triangles would be a quarter
that of the red one.
Running a similar argument, the total of the
yellow areas would be a quarter of the total
of the orange areas, and so on. So the area
of the parabolic segment is A times an infinite
geometric series with 1 as the first term
and a quarter as the common ratio. But for
this geometric series, we have an elegant
method to deduce the sum. In this diagram,
the side length of the largest square is 2.
Then the largest red square should have area
1, then the second largest have area of a
quarter, and so on. Since the two yellow parts
and the red part are identical in area, and
the area of the whole square is 4, the red
part has area a third of the entire figure,
which means the geometric series sums up to
four-thirds. That means the area of the parabolic
segment is four-thirds that of the largest
inscribed triangle.
Recall why we have to do all these. It’s
for the integral of x^2 from 0 to 1. Since
both the horizontal base and the vertical
height are 1, the area of the triangle formed
should be one half. And for the red parabolic
segment, it would be four-thirds that of the
orange area. We know the area of the triangle
can be calculated using this formula as discussed
earlier. The vertical height of this triangle
is a quarter deduced from the difference in
these two coordinates, while the horizontal
span is simply 1. So we can calculate this
expression, and obtain the answer as one third.
The reason why we reflected the parabola so
that it opens downwards is to allow physical
intuition which we have used extensively to
arrive at this result.
The parabolic segment area calculation is
based on a treatise by Archimedes called Quadrature
of the Parabola, and I have drastically simplified
Archimedes’ work because his work did not
rely on physics or vectors, but purely geometric.
If you want to see Archimedes work in its
full glory, I’ll put a link in the description.
If you enjoyed this video, give it a thumbs
up. Don’t forget to leave a comment and
subscribe to Mathemaniac with notifications
on! See you in the next video! Bye!
