11 – The Calculus and Related Topics The student will learn about mathematics leading up to the calculus from ancient times to seventeenth century Europe.

§11-1 Introduction Student Discussion.

§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.

§11-2 Zeno’s Paradoxes Student Discussion.

§11-3 Eudoxus Student Discussion.

§11-3 Method of Exhaustion 370 BC Breaking 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)

§11- 4 Archimedes Student Discussion.

§11-4 Equilibrium Method Knowing the volume of a cone ( 8 r3 /3 ) and of a cylinder (2 r3); to find the volume of a sphere. o r 2r T y = x y = r y 2 = 2xr – x 2 Slice of sphere = (2r x -x2) Δx Slice of cone = x2 Δx Slice of cylinder = r2 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. Hence the volume of the sphere plus the volume of the cone is equal to two time the volume of the cylinder.

§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 2r T 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. V sphere + 8r 3 / 3 = 4r 3 V sphere = 4r 3 /3

§11-5 Beginnings of Integration in Western Europe Student Discussion.

§11-5 Johann Kepler Area of a circle by polygons cut into segment triangles. Area = Σ ½ bh = ½ Cr = Area = Σ ½ bh = ½ Cr = ½ 2π r r = Area = Σ ½ bh = Area = Σ ½ bh = ½ Cr = ½ 2π r r = π r 2. The volume of a sphere was done in a similar manner.

§11 - 6 Cavalieri’s Method of Indivisibles Student Comment S

§11 - 6 Cavalieri’s Method Hemisphere only! VS = VP = r 2 2π r = π r 3 The idea of cutting a volume into slices and adding the areas of the slices. First attributed to Tzu Geng, 5th century China. Hemisphere only! r h 2π r 2π(r – h) P S r h VS = VP = r 2 2π r = π r 3

§11-7 The Beginnings of Differentiation Student Discussion.

§11-7 Fermat’s Sub tangents P: f(x,y) = 0 Q: f(x+e,y(1+e/a)) x e a z y B A R C Let 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. 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.

§11-7 Fermat’s Sub tangents Specific Example: y = x 2 P: f(x,y) = 0 Q: f(x+e,y(1+e/a)) x e a z y B A R C 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.

§11-7 Fermat’s Sub tangents Specific Example: y = x 2 P: f(x,y) = 0 Q: f(x+e,y(1+e/a)) x e a z y B A R C 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.

§11 – 8 Wallis and Barrow Student Discussion.

§11 – 8 John Wallis Etc.

§11 – 8 John Wallis 2 Created a problem due to the fractional exponent. 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.

§11 – 8 Wallis and Newton Let’s consider: Notice Pascal’s triangle and the final results.

§11 – 8 Wallis and Newton 2 Let’s generalize for n = 2m, i.e. n is even: Wallis never saw the patterns, perhaps because he was too interested in his series results - 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.

§11 – 9 Newton Student Discussion.

§11 – 9 Newton From the previous patterns of Pascal’s triangle Newton knew: . . . n=-2 n=0 n=2 n=4 n=6 n=8 times 1 x 2 3 4 - x 3/ 3 6 x 5 / 5 - x 7/ 7 x 9 / 9

§11 – 9 Newton 2 Newton first expanded the table backwards . . . n=-2 times 1 x - 1 2 3 4 - x 3/ 3 6 x 5 / 5 - x 7/ 7 x 9 / 9

§11 – 9 Newton 3 But 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 And thus Newton obtained a new table.

§11 – 9 Newton 4 Newton’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 x - 1 2 3 - x 3/ 3 x 5 / 5 - x 7/ 7 x 9 / 9 Newton was particularly interested in the n = 1 column.

§11 – 9 Newton 5 Newton then concluded: 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!

§11 – 10 Leibniz Student Discussion. By 1700 most of undergraduate calculus had been formulated.

Assignment Papers presented from chapters 7 and 8.