1

11 – The Calculus and Related Topics

The student will learn about

mathematics leading up to the calculus from ancient times to seventeenth century Europe.

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§11-1 Introduction

Student Discussion.

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§11-1 Introduction This chapter is devoted to a brief account of the origins and development of the important concepts of the calculus, concepts that are so far reaching and that have exercised such an impact on the modern world that it is perhaps correct to say that without knowledge of them a person today can scarcely claim to be well educated.

Integration developed, followed by differentiation followed by the relationship between the two as inverse operations.

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§11-2 Zeno’s Paradoxes

Student Discussion.

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§11-3 Eudoxus

Student Discussion.

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§11-3 Method of Exhaustion 370 BCBreaking up areas/volumes into polygons/polyhedron.

Archimedes’ area of a quadrature of a parabola.

K = ΔABC + ΔABC/4 + ΔABC/42 + …

= ΔABC ( 1 + 1/4 + 1/42 + … )

= 4 ΔABC / 3

Note : S = a1 / (1 – r)

4

3

2

1

-2 -1 1 2

C

E

A

B

D

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§11- 4 Archimedes

Student Discussion.

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§11-4 Equilibrium MethodKnowing the volume of a cone ( 8 r3 /3 ) and of a cylinder (2 r3); to find the volume of a sphere.

Slice of sphere = (2r x -x2) Δx

Notice that if x = r, as it is in the cylinder then twice the moment of the cylinder is equal to the moment of the sphere plus the cone.

Slice of cone = x2 ΔxSlice of cylinder = r2 Δx

o r 2rT

y = x

y = r

y 2 = 2xr – x 2

Hence the volume of the sphere plus the volume of the cone is equal to two time the volume of the cylinder.

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§11-4 Equilibrium Method - continued

Knowing the volume of a cone ( 8 r3 /3 ) and of a cylinder (2 r3) to find the volume of a sphere.

o r 2rT

y = x

y = r

y 2 = 2xr – x 2Hence the volume of the sphere plus the volume of the cone is equal to two time the volume of the cylinder.

V sphere + 8r 3 / 3 = 4r 3

V sphere = 4r 3 /3

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§11-5 Beginnings of Integrationin Western Europe

Student Discussion.

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§11-5 Johann KeplerArea of a circle by polygons cut into segment triangles.

Area = Σ ½ bh =

The volume of a sphere was done in a similar manner.

Area = Σ ½ bh = ½ Cr =Area = Σ ½ bh = ½ Cr = ½ 2π r r =Area = Σ ½ bh = ½ Cr = ½ 2π r r = π r 2.

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§11 - 6 Cavalieri’s Method of Indivisibles

Student Comment

S

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§11 - 6 Cavalieri’s MethodThe idea of cutting a volume into slices and adding the areas of the slices. First attributed to Tzu Geng, 5th century China.

r

r h

r h

h

2π r

2π(r – h)

P

Hemisphere only!

VS = VP = r 2 2π r = π r 3 3

1

3

2

S

rh

22 hr

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§11-7 The Beginnings ofDifferentiation

Student Discussion.

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§11-7 Fermat’s Sub tangentsLet P be a point on a curve. Consider a second point Q on the curve. As Q approaches P it can be treated as a point on the tangent to the curve at P.

P: f(x,y) = 0

Q: f(x+e,y(1+e/a))

x

ea

z

y

BA

R

C

Our goal is to find the distance a which will describe the tangent at P.

ΔABP ΔPRQ so a/y = e/z or z = ey/a giving a y value at Q of y(1+e/a)

Substituting the coordinates of point Q into the original equation and allowing e to assume the value 0, allows us to solve the resulting equation for a in terms of x and y of the point P.

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§11-7 Fermat’s Sub tangentsSpecific Example: y = x 2

Our goal is to find the distance a which will describe the tangent at P.

f (x,y) = x 2 – y = 0 and

f [(x+e, y(1+e/a)] = (x+e) 2 – y(1+e/a) = 0

x 2 + 2ex +e 2 – y - ye/a = 0

But x 2 – y = 0 and e goes to 0 so

2x - y/a = 0 and a = y/2x = x/2

Then at P(3,9), a = 3/2

Note that this means that the x-intercept or the tangent is 3 – 3/2 = 3/2.

P: f(x,y) = 0

Q: f(x+e,y(1+e/a))

x

ea

z

y

BA

R

C

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§11-7 Fermat’s Sub tangentsSpecific Example: y = x 2

Confirming our answer (x-intercept) of the tangent at 3/2 by calculus.

y’ = 2x, so at P(3,9) m = 6

The line through (3,9) with m = 6 is

y = 6x – 9.

This line has an x-intercept at x = 9/6 = 3/2.

P: f(x,y) = 0

Q: f(x+e,y(1+e/a))

x

ea

z

y

BA

R

C

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§11 – 8 Wallis and Barrow

Student Discussion.

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§11 – 8 John Wallis

Etc.4

1

4

xdxx

1

0

41

0

2

6

1n

1

1n

xdxx

1

0

1n1

0

2

n2

1xdxx1

0

1

0

2

0

2

1

2

xdxx

1

0

21

0

2

2

3

1

3

xdxx

1

0

31

0

2

4

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§11 – 8 John Wallis 2

Created a problem due to the fractional exponent.

?dxx1

0

2

1

1n

1

1n

xdxx

1

0

1n1

0

2

n2

However, there were square root algorithms to check ones work and calculations.

It ultimately led to what Wallis really wanted to do in considering f (x) = 1 – x 2.

4dx)x1(

1

0

2

12

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§11 – 8 Wallis and NewtonLet’s consider:

1x1dxx11

0

1

02

02

3

2

3

11

3

x1x1dxx1

1

0

31

02

22

15

8

5

x1

3

x2)x(1dxx1

1

0

531

02

42

35

16

7

x1

5

x3

3

x3)x(1dxx1

1

0

7531

02

62

Notice Pascal’s triangle and the final results.

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§11 – 8 Wallis and Newton 2Let’s generalize for n = 2m, i.e. n is even:

m

0i

i1

02

n2

1i2

1

i

mdxx1

Wallis never saw the patterns, perhaps because he was too interested in his series results -

9

8

7

8

7

6

5

6

5

4

3

4

3

2

1

2

2

Newton did the same thing but instead of 1 as the upper limit of the integral he used x getting the same results but seeing the pattern.

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§11 – 9 Newton

Student Discussion.

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§11 – 9 NewtonFrom the previous patterns of Pascal’s triangle Newton knew:

. . . n=-2 n=0 n=2 n=4 n=6 n=8 . . . times

. . . 1 1 1 1 1 . . . x

. . . 1 2 3 4 . . . - x 3/ 3

. . . 1 3 6 . . . x 5 / 5

. . . 1 4 . . . - x 7/ 7

. . . 1 . . . x 9 / 9

. . . . . . . . . . . . . . . . . . . . . . . . . . .

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§11 – 9 Newton 2Newton first expanded the table backwards

. . . n=-2 n=0 n=2 n=4 n=6 n=8 . . . times

. . . 1 1 1 1 1 1 . . . x

. . . - 1 0 1 2 3 4 . . . - x 3/ 3

. . . 1 0 0 1 3 6 . . . x 5 / 5

. . . - 1 0 0 0 1 4 . . . - x 7/ 7

. . . 1 0 0 0 0 1 . . . x 9 / 9

. . . . . . . . . . . . . . . . . . . . . . . . . . .

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§11 – 9 Newton 3But we want (1 – x 2) 1/2 . Newton ignored the restriction that n must be even and used the formula for binomial coefficients when n was odd. I.e. the fourth entry in the n = 1 column is

16

1

321

221

121

2/1

142

1

And thus Newton obtained a new table.

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Newton was particularly interested in the n = 1 column.

§11 – 9 Newton 4Newton’s new table.

. . . n=-3 n=-2 n=-1 n=0 n=1 n=2 n=3 n=4 n=5 n=6 . . . times

. . . 1 1 1 1 1 1 1 1 1 1 . . . x

. . . - 1 0 1 2 3 . . . - x 3/ 3

. . . 1 0 0 1 3 . . . x 5 / 5

. . . - 1 0 0 0 1 . . . - x 7/ 7

. . . 1 0 0 0 0 . . . x 9 / 9

. . . . . . . . . . . . . . . . . . . . . . . . . . .

16

358

152

3

128

315

128

3516

58

32

12

1

8

1

16

1

128

5

2

3

8

3

16

1

128

3

2

5

8

15

16

5

128

5

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2

1

8

1

16

1

128

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§11 – 9 Newton 5Newton then concluded:N=1

1 ...

9

x

128

5

7

x

16

1

5

x

8

1

3

x

2

1xdxx1

97532

1x

0

2

Which for x = 1 gives and infinite series for the area of ¼ of the circle or ¼ .

He checked his results with a square root algorithm, the binomial theorem, squaring both sides, and by using formal long division to obtain an infinite series for (1 + x) – 1.

Newton was tremendously excited with this new tool. The Binomial Theorem became a mainstay of his newly developing calculus.

He also calculated the log 1.2 to 57 decimal places!

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§11 – 10 Leibniz

Student Discussion.

By 1700 most of undergraduate calculus had been formulated.

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Assignment

Papers presented from chapters 7 and 8.