What is Calculus? Calculus is perhaps the most powerful body of analytic tools ever devised. Calculus adds the limit process to elementary algebra and analytic geometry. With this expanded mathematical system, we may compute various new quantities rather easily. Analytic Geometrical Computation Calculus Computation 1 The slope of the tangent line to any given smooth curve C at a specified point, P. P
C
where A . . . . . .
along C . 2 The length of any given smooth path from point A to point B. 2 The length of the polygonal path from point A to point B. Here, B 1 The slope of the tangent line to a circle or another conic section at a specified point, P. P B A | | |
| |
| . . . |
| |
| and the approximation improves continuously as the number of sample points selected along the path increases. In fact, |
|
|
|
|
| . . . |
|
|
|
as n increases and each segment length in the sum 0. Analytic Geometrical Computation 3 The area interior to any simple smooth closed curve S. 3 The area of the polygon P. Area ( P ) = Theorem: Any n‐sided polygon can be partitioned into (n‐2) non‐overlapping triangles using only the vertices of the polygon in the partition. Calculus Computation Area ( S ) In fact, . . . . . .
A S as the length of the base of the i th inscribed rectangle R i 0 for each i 1 , 2, . . . n . We may determine the slope at a specified point along a curve, the length of a spiral curve, and the area bounded by a smooth closed curve by applying the limit processes known in calculus as differentiation and integration to appropriate functions. Although Isaac Newton and Gottfried Leibnitz are recognized as the two primary founders of calculus during the scientific revolution in the early 1700s, the notion of the limit process goes back at least as far as the ancient Greeks. The Method of Exhaustion and the Limit Process
The method of exhaustion is a technique for finding the area of a shape by
inscribing within it a sequence of polygons whose areas converge to the area
of the containing shape. If the sequence is correctly constructed, the
difference in area between the nth inscribed polygon and that of the
containing shape will become arbitrarily small as n becomes large. As the
space between the inscribed polygon and the containing shape becomes
arbitrarily small, then the possible values for the area of the containing shape
are systematically "exhausted" by the lower bound polygonal areas
successively established by the sequence members.
This idea originated with Antiphon of Athens in the 5th century BC. The
method of exhaustion is seen as a precursor to the methods of modern
calculus. The development of analytical geometry and rigorous integral
calculus in the 17th-19th centuries (in particular the development of the limit
definition) subsumed the method of exhaustion so that it is no longer explicitly
used today to solve geometrical problems.
Bryson of Heraclea (ca. 450 BCE - ca. 390 BCE) was an ancient Greek
mathematician and sophist who along with his contemporary Antiphon, was
among the first to inscribe a polygon inside a circle, find the polygon's area,
double the number of sides of the polygon, and repeat the process, resulting
in a lower bound approximation for the area of a circle. "Sooner or later (they
figured), ...[there would be] so many sides that the polygon ...[would] be a
circle". Bryson later followed the same procedure for polygons circumscribing
a circle, resulting in an upper bound approximation for the area of a circle.
With these calculations Bryson was able to approximate π and place lower and
upper bounds on π's true value. But due to the complexity of the method, he
was only able to calculate π to a few digits.
Circle Area
= π x r2
We will probably never know exactly who first discovered
that the ratio between the area of a circle and the area of a
square having side length equal to that of the circle’s
radius is a constant. Nor will we ever know who first tried
to calculate this ratio. However, it is well documented that
the Babylonians used the value (4/3)4 for π nearly 4000
years ago, Archimedes showed that: 223/71 < π < 22/7
more than 2000 years ago, and Zu Chong-Zhi established
355/113 as an extremely accurate estimator of π in the fifth
century A.D.
Note:
4
0 < ⎛⎜ 4 ⎞⎟ – π < 1.9 x 10 ‐2 ⎝3⎠
0 < 22 – π < 1.3 x 10 ‐3 7
355
0 < 113
– π < 2.7 x 10 ‐8 Antiphon’s Method of Exhaustion Applied to
Finding the Area of the Circle of Radius r .
The area of the nth inscribed
polygon, A n , converges to
the area, A, of the bounding
circle of radius r as n
increases without bound.
That is:
An → A
as n → + ∞
First Approximation:
The inscribed square of
area A 1 :
A1
4
=
1 2
r ⇒
2
A 1= 2 • r 2
Second Approximation:
The inscribed octagon of
area A 2 :
A2
8
=
1 2
o
r sin( 45 )
2
A2 = 2 2 • r 2
0
Proof_________________________________________________________
In the figure below, observe that
Area( ΔABC ) =
It is also evident from the green Δ that
sin θ =
1
a•h
2
h
b
0
(1)
θ
h = b sin θ
(2) .
A
Therefore, upon substituting from (2) into (1), we have that
Area( ΔABC ) =
Thus,
A2
8
=
b
1
a • b sin θ . ( * )
2
2
1
1
r • r sin (45o) = r 2 •
2
2
2
h
c
θ
C
a
A2 = 2 2 • r 2 .
________________________________________________________
B
Third Approximation:
The inscribed 16-gon of
area A 3 :
A3
16
0
=
1 2
o
r sin( 22.5 )
2
A 3 = 4 2- 2
• r2
Proof_______________________________________________________
In the half-angle formula:
sin 2 θ =
we set θ = 22.5
Thus,
(3)
θ
to obtain :
sin2 ( 22.50 ) =
1
( 1 – cos 450 )
2
sin2 ( 22.50 ) =
1
1 ⎛ 2 − 2 ⎞⎟
2− 2
2
(1–
) = ⎜⎜
=
2
4
2
2 ⎝ 2 ⎟⎠
0 sin ( 22.50 )
So
0
1
( 1 – cos 2θ )
2
A3
16
=
=
2− 2
2
1
r • r sin (22.5o) by (*) .
2
A 3 = 8 r2 sin (22.5o) = 4 2 - 2
• r2
_______________________________________________________
Fourth Approximation:
The inscribed 32-gon of
area A 4 :
A4
=
32
0
1 2
o
r sin( 11.25 )
2
A 4 = 8 2- 2 +
2
Proof____________________________________________________
Utilizing th e half-angle formula ( 3 ) once more, we find that:
sin 2 ( 11.250 ) =
=
1
(1–
2
sin 2 ( 11.250 ) =
1 ⎛⎜
1−
2⎜
⎝
so
and
Thus,
sin ( 11.250 ) =
A3
32
=
θ
1
( 1 – cos 22.50 )
2
2−
1 − sin 2 22.5 o ) =
2 + 2 ⎞⎟
=
⎟
2
⎠
2−
1
2
⎛
⎜1 −
⎜
⎝
1−
2 − 2 ⎞⎟
⎟
4
⎠
2+ 2
4
2+ 2
2
1
r • r sin (11.25o) by (*) .
2
A 4 = 16 r2 sin (11.25o) = 8
2−
2+ 2
• r2
____________________________________________________
• r2
Continuing in this way, we find that :
A 5 = 16
A 6 = 32
A 7 = 64
2-
2-
2-
2+
2+
2+
2+
2+
• r2
2+ 2
2+
2+
• r2
2
2+
2
• r2
•
•
•
So now it is evident that
⎛
o
⎞
180
A n = 2n • sin ⎜⎜ n ⎟⎟ • r 2
⎝ 2
⎠
→
π
2+
2+
• r2
as
n → +∞
where
⎛
o
⎞
180
2n • sin ⎜⎜ n ⎟⎟ = 2(n - 1) •
⎝ 2
⎠
2-
b
Note: there are ( n – 1 ) nested radicals here
2+
. . .
2
as
→
π
n → +∞