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Home , Chord (geometry), Method of exhaustion, Parabola, Quadrature (mathematics)

JOHN ABBOTT ' of the II COLLEGE and the (SCIENCE)

Carefully study the text below and attempt the exercises at the end. You will be evaluated on this material by writing a 30 to 45 minute test (which may be part of a larger class test). This test will be worth 10% of your class mark and may include questions drawn from the exercises at the end. This activity will contribute to your attainment of the Science Program competency: To put in context the emergence and development of scientific concepts. B 1. CONIC SECTIONS The Greeks originally viewed (and also , el- lipses and ) as conic sections. Imagine rotating a straight about a vertical line that intersects it, thus obtaining a A S circular (double) . A horizontal intersects the cone in a height . When the plane is slightly inclined, the section becomes an . As the intersecting plane is inclined more towards the ver- P tical, the ellipse becomes more elongated until, finally, the plane becomes to a generating line of the cone, at which Given a parabolic segment S with base AB (see the above fig- the section becomes a parabola. If the intersecting plane is in- ure), the point P of the segment that is farthest from the base is clined still nearer to the vertical, it meets both branches of the called the of the segment, and the () distance cone (which it did not do in the previous cases); now the of from P to AB is its height. (The vertex of a segment is not to be intersection is a . confused with the vertex of the parabola which, as you will recall, is the intersection point of the parabolawith its axis of symmetry.) Archimedes shows that the of the segment is four-thirds that hyperbola of the inscribed triangle AP B. That is, ellipse the area of a segment of a parabola is 4/3 times the area of the triangle with the same base and height. (Exercise 1 asks you to check Archimedes’ result in a very simple case.)

circle 3. PRELIMINARIES ON PARABOLIC SEGMENTS parabola By the time of Archimedes, the following facts were known concerning an arbitrary parabolic segment S with base AB and hyperbola vertex P (see the figure below). P1. The line at P is parallel to AB. P2. The straight line through P parallel to the axis of the parabola 2. ARCHIMEDES’THEOREM intersects AB at its midpoint M. P3. Every QQ′ parallel to AB is bisected by PM. A segment of a convex curve (such as a parabola, ellipse or hy- P4. With the notation in the figure below, perbola) is a region boundedby a straight line and a portion of the PN NQ 2 curve. = PM MB 2 (Equivalently, PN = (PM/MB 2) · NQ 2. In modern terms, this says that, in the pictured oblique xy-coordinate system, the of the parabola is y = λx2, where λ = PM/MB 2.) y B

M Q In his book Quadrature of the Parabola, Archimedes gives two methods for finding the area of a segment of a parabola (previous A N x had unsuccessfully attempted to find the areaofa Q′ segmentofa circle and ofa hyperbola). Thesecond of thesemeth- ods, which we discuss below, is based on the so-called “method P of exhaustion.” axis 2 B Archimedes quotes these facts without proof, referring to ear- lier treatises on the conics by and Aristaeus. (You are asked to prove the first three properties, mostly by modern methods, in exercises2, 3 and 4. Using propertyP2 to find the vertex, you will A then, in exercise 5, verify Archimedes’ theorem in another special B′ case. Exercise 6 will then guide you through a modern proof of the general case.) P 4. PART 1: THE METHOD OF EXHAUSTION A′ axis To find the area of a given parabolic segment S, Archimedes constructs a sequence of inscribed P0, P1, P2, . . . , that The area of this is of course greater than the area fill up or “exhaust” S. The first P0 is the inscribed tri- of the parabolic segment S. Since it also equals twice the area of AP B with AB the base of segment S and P its vertex. To triangle AP B (why?), it follows that the area of this triangle is construct the next polygon P1, consider the two smaller parabolic more than half the area of the parabolic segment. The remaining segments with bases PB and AP ; let their vertices be P1 and P2, area, which equals M0 because P0 = △AP B, must therefore be respectively, and let P1 be the polygon AP2P P1B. less than half the area of S: 1 B M0 < 2 area(S) Now consider the two triangles (△AP2P and △P P1B) that are added in the next step to form polygon P1. The above argu- ment, applied to the two smaller parabolic segments with bases A P0 AP and PB, shows that the of these triangles are more than P1 half the areas of the two segments. It therefore follows that 1 M1 < 2 M0 P2 P B Continuing in this way, we see that 1 1 M2 < 2 M1,M3 < 2 M2,..., 1 and in general Mn < 2 Mn−1. It is now easy to prove that A P1 limn→∞ Mn =0 (see exercise 7), and therefore (1) follows. P1 5. PART 2: FINDING THE AREA OF Pn

P2 P At each step in the construction of the polygons P0, P1, P2, . . . , we add triangles to the previous polygon: a single tri- We continue in this way, adding at each step the triangles in- angle (△AP B) begins the process, then two triangles (△AP2P scribed in the parabolic segments remaining from the previous and △P P1B) are added in the next step, then four triangles are step. As seems clear from the above figures, the resulting poly- added, etc. Let a0, a1, a2, . . . , be the total areas of the triangles gons P0, P1, P2, . . . , exhaust the area of the original parabolic added at each step. Thus segment S. In fact, Archimedes carefully proves this by showing a0 = area(△AP B), that the difference between the area of S and the area of Pn can a1 = area(△AP2P ) + area(△P P1B), be made as small as one pleases by choosing n sufficiently large. In modern terms, this simply means that and so on. In the second part of his proof, Archimedes finds the area of the polygons P0, P1, P2, . . . , by evaluating the sum (1) lim area(Pn) = area(S) n→∞ (2) a0 + a1 + a2 + · · · + an = area(Pn) (but it is important to realize that Archimedes, like all the ancient Greek mathematicians, had no limit concept). The key step is to show that the total area of the triangles added To prove (1), we let at each step is equal to 1/4 the total area of the triangles added at the previous step. In other words, Mn = area(S) − area(Pn) for n =0,1,2, . . . 1 1 (3) a1 = 4 a0, a2 = 4 a1,..., and show that limn→∞ Mn = 0. Consider the parallelogram ′ ′ ′ 1 ABB A circumscribed about the segment S, whose sides AA and in general an = 4 an−1. We will describe Archimedes’ proof ′ 1 and BB are parallel to the axis of the parabola, and whose base that a1 = 4 a0, leaving the general case to exercise 8. ′ ′ A B is tangent to the parabolaat P (and therefore parallel to AB, We want to show that the sum a1 of the areas of triangles by property P1). AP2P and P P1B is 1/4 that of △AP B. Apply property P2 to both the original parabolic segment S and the smaller segment with base PB: we obtain two lines parallel to the axis of the parabola, one going through P and intersecting AB at its mid- point M, and one going through P1 and intersecting PB at its 3 midpoint Y . Let M1 be the intersection point of this second par- Returning now to the area of polygon Pn (see (2) above) and allel line with AB. Then M1 is the midpoint of MB because using (3), we have finally the triangles YM1B and PMB are similar. Finally, let V be the 1 1 1 area(P )= a0 + a0 + a0 + · · · + a0 intersection with PM of the line through P1 parallel to AB (so n 4 42 4n is a parallelogram). VMM1P1 In other words, the areas a0, a1, a2, . . . , added at each step in B the construction of Pn form a geometric sequence with common axis M1 ratio 1/4, and area(Pn) is the sum of the first n +1 terms of this M sequence.

6. PART 3: CONCLUSION OF ARCHIMEDES’ PROOF Y Now that the area of P has been determined, it follows from A n the conclusion of Part 1 that the difference between the area of the P1 parabolic segment S and the sum V 1 1 1 a0 + a0 + a0 + · · · + a0 4 42 4n P2 P can be made as small as one pleases by choosing n sufficiently large. In modern terms, 1 1 1 Applying property P4 (with N = V and Q = P1) and noting (9) area(S) = lim a0 + a0 + a0 + · · · + a0 1 n→∞ 4 42 4n that V P1 = MM1 = MB, we have   2 and Archimedes now seeks to determine this limit. 2 P V V P1 1 He begins by deriving the identity = 2 = PM MB 4 1 1 1 1 1 4 (10) 1+ + + · · · + + · = so PM = 4P V (and, therefore, VM = 3P V ). Two conse- 4 42 4n 3 4n 3 quences follow from this. First, since P1M1 = VM, we have which is a restatement of the formula you learned for the par- (4) P1M1 =3P V tial sums of a geometric (see exercise 10). As Archimedes 1 1 shows, (10) follows from the observation that Second, because M1B = 2 MB, YM1 = 2 PM (by similar tri- 1 1 1 4 1 1 again), so that + · = = · 4k 3 4k 3 · 4k 3 4k−1 (5) YM1 =2P V for we can then sum the terms on the left side of (10) by repeat- Now (4) and (5) imply that P1Y = P V , so in fact edly adding the last two terms: YM1 =2P1Y 1 1 1 1 1 1+ + 2 + · · · + n + · n Now consider the two triangles PM1B and P P1B. They have 4 4 4 3 4  1 1 1 1 1 the same base PB and, because YM1 = 2P1Y , a simple argu- =1+ + 2 + · · · + n−1 + · n−1 ment with similar triangles shows that the height of △PM1B is 4 4 4 3 4  = . . . twice the height of △P P1B (both heights being relative to the common base PB). Therefore 1 1 1 1 =1+ + + · 4 42 3 42 (6) area(△PM1B) = 2area(△P P1B)   1 1 1 Similarly, triangles PMB and PM1B have the same base PB =1+ + · 4 3 4 and, since MB = 2M1B, the height of △PMB is twice that of 1 4 =1+ = △PM1B, so 3 3 (7) area(△PMB) = 2area(△PM1B) From a modern perspective, Archimedes’ theorem is now a It follows from (6) and (7) that simple consequence of (9) and (10): 1 1 1 1 (8) area(△P P1B)= 4 area(△PMB) area(S)= a0 · lim 1+ + 2 + · · · + n n→∞ 4 4 4  An argument similar to the one in the last two paragraphs 4 1 1 · − · shows that = a0 lim n n→∞3 3 4  1 4 area(△AP2P )= 4 area(△AP M) = a0 · ( 3 − 0) 4 Combining this with (8) then gives = 3 area(△AP B) area(△AP2P ) + area(△P P1B) No doubt Archimedes intuitively obtained the answer 4/3 in a = 1 area(△AP M)+ 1 area(△PMB) similar way but, rather than taking limits explicitly, he completed 4 4 the proof by showing that the two alternative conclusions = 1 area(△AP B) 4 area(S) < 4 area(△AP B) 1 3 so a1 = 4 a0, as desired (see exercise 9 for a shorter proof). 4 and 6. Prove Archimedes’ theorem by modern methods as follows. 4 area(S) > 3 area(△AP B) Using property P2, we can label the x-coordinates of A, B and P both lead to a contradiction, and so must be false. This approach as in the figure below.

(whose details we will omit) was in fact typical of Greek proofs y by the method of exhaustion. B

7. EXERCISES

1. Use integration to verify Archimedes’ theorem for the segment 2 bounded by y = x2 and the line y = 1. (Determine the vertex of y = kx the segment and show that the inscribed triangle has area 1. Then A integrate to verify that the segment has area 4/3.) 2. Property P1 follows easily from a theorem you learned in Cal- culus I. Which one? (Rotate the parabolic segment until its base P AB is horizontal. What can you then say about the vertex P ?) p − r p p + r x 3. Prove property P2 by modern methods as follows. Introduce r r a rectangular xy-coordinate system centred at the vertex of the parabola, as in the figure below. Assume first that k =1.

y (a) Use the method of exercise 5(b) to show that the area of tri- B angle AP B is r3. (b) Show that the equation of line AB is y =2px − p2 + r2. M (c) Find the area of the segment by integrating. (You should get a 2 4 3 y = kx value of 3 r , thus proving Archimedes’ theorem for k =1.) A (d) If k =6 1, what changes do you need to make to the calcula- tions in parts (a), (b) and (c)?

P 7. (a) Let M0, M1, M2, . . . , be any sequence of positive num- 1 x bers such that Mn < 2 Mn−1 for n = 1, 2, . . . . Show that axis limn→∞ Mn = 0. This result, which is really the crux of the method of exhaustion, is originally due to Eudoxus, a pre- In these coordinates, the equation of the parabola has the form decessor of Euclid’s. (For the proof, start by showing that n y = kx2 and its axis of symmetry is the y-axis. Assume first that 0

8. SOLUTION OUTLINES (c) The area of the parabolic segment AP B is 3 1. The inscribed triangle has vertices at (±1, 1) and at the origin 2 32 (2x +3 − x ) dx = 3 (the vertex of the segment). Since it has width 2 and height 1, its Z−1 area is 1. The segment has area which is indeed 4/3 times the area of triangle AP B. 1 2 2 2 4 6. (a) A = (p − r, (p − r) ) and B = (p + r, (p + r) ), so the (1 − x ) dx = 3 2 2 Z−1 midpoint of AB is M = (p,p + r ) because 1 2 2 2 2 as required. 2 (p − r) + (p + r) = p + r 2. Rotate the parabolic segment until its base AB is horizontal. Since P = (p,p 2), it follows that the length of PM is r2. As Because the vertex P is the point of the segment that is farthest in the previous problem, the area of triangle AP B is half the from AB, it corresponds to an extreme value (a maximum or a area of a parallelogram with ‘base’ equal to r2 and ‘height’ minimum, depending on whether the segment is above or below equal to the horizontal distance 2r between A and B. In other AB). By Fermat’s Theorem (section 4.1 of Stewart), the tangent words, 1 2 3 line at P must be horizontal, in other words, parallel to AB. area(△AP B)= 2 r · 2r = r ′ 3. (a) Since the derivative y = 2x, the slope of the tangent line (b) Since the derivative y′ = 2x and P = (p,p2), line AB has at P is 2p. On the other hand, the slope of AB is slope 2p by property P1. (Alternatively, compute the slope b2 − a2 directly from the coordinates of A and B given in part (a).) = b + a b − a Using the point-slope form By Property P1, these two slopes are equal, so y − y1 = m(x − x1) 1 2 2 2p = a + b =⇒ p = 2 (a + b) with (x1,y1)= M = (p,p + r ) and m =2p then gives 2 2 2 2 (b) The straight line through P parallel to the axis of the parabola y = p + r +2p(x − p)=2px − p + r is vertical, so it intersects AB at a point with x-coordinate (c) The required 1 2 (a + b), in other words, at its midpoint. p+r (c) If k =6 1, then the two expressions for the slope in part (a) (2px − p2 + r2 − x2) dx are scaled by a factor of k and one finds, as before, that Zp−r 1 p = 2 (a + b). can be computed in two ways. The first is to integrate directly: ′ p+r 4. Any segment of the given parabola whose base Q Q is parallel 2 2 2 to will clearly also have vertex . Applying Property P2 to 2px dx = p (p + r) − (p − r) =4p r AB P Z − ′ ′ p r  the segment with base Q Q, we conclude that PM intersects Q Q p+r at its midpoint, as required. (r2 − p2) dx =2r(r2 − p2) Z − 5. (a) A = (−1, 1), B = (3, 9), P = (1, 1), M = (1, 5). p r p+r y 2 1 3 3 2 3 2 x dx = 3 (p + r) − (p − r) = 3 r +2p r Z − B p r  The area of the segment is therefore 2 3 2 2 3 2 4 3 4p r +2r − 2p r − 3 r − 2p r = 3 r The second way is to note that the integrandcontains a perfect M ′ B , 2px − p2 + r2 − x2 = r2 − (x − p)2, and to introduce the substitution t = x − p. The area of the segment is then A P p+r r 2 2 2 2 x r − (x − p) dx = (r − t ) dt Z − Z− p r  r t=r 3 1 3 ′ =2r − 3 t A  t=−r 3 2 3 4 3 (b) The length of PM is 4. Referring to the above figure, the =2r − 3 r = 3 r ′ ′ circumscribed parallelogram ABB A has ‘base’ equal to 4 (d) If k =6 1, then the y-coordinates of P and M, the length of (the length of PM) and ‘height’ equal to 4 (the horizontal PM, and the area of triangle AP B are each scaled by a fac- distance between A and B). Its area is therefore 16, which tor of k. The same is true of the y-coordinates of all points on is twice the area of △AP B. Triangle AP B therefore has the line AB, and of the integrand in part (c). area 8. (Alternatively, add the areas of triangles AP M and PBM, and note that these triangles have the same base PM and equal heights.) 6

7. (a) We have 9. NOTE ON PROPERTY P4 1 M1 < 2 M0, For completeness, we outline a modern proof of property P4. 1 1 This property is equivalent to the assertion that M2 < 2 M1 < 4 M0, 1 1 PN PM M3 < 2 M2 < 8 M0, = NQ 2 MB 2 n and in general M < M0/2 . The result then follows from n for every chord QQ′ parallel to AB, so it suffices to show that the Squeeze Theorem (section 8.1 of Stewart). 2 ′ n the ratio PN/NQ remains constant as QQ varies. Introduce (b) Show that 0 < Mn < r M0 and use the Squeeze Theorem, 2 n xy-coordinates as in exercise 3, assume k = 1 in y = kx , and noting that limn→∞ r =0 since 0