Archimedes quadrature of the parabola and the method of exhaustion

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1 JOHN OTT COLLEGE rchimedes quadrature of the parabola and the method of ehaustion CLCULUS II SCIENCE) Carefull stud the tet below and attempt the eercises at the end. You will be evaluated on this material b writing a 30 to 45 minute test which ma be part of a larger class test). This test will be worth 0% of our class mark and ma include questions drawn from the eercises at the end. This activit will contribute to our attainment of the Science rogram competenc: To put in contet the emergence and development of scientific concepts.. CONIC SECTIONS The Greeks originall viewed parabolas and also circles, ellipses and hperbolas) as conic sections. Imagine rotating a straight line about a vertical line that intersects it, thus obtaining a circular double) cone. horizontal plane intersects the cone in a circle. When the plane is slightl inclined, the section becomes an ellipse. s the intersecting plane is inclined more towards the vertical, the ellipse becomes more elongated until, finall, the plane becomes parallel to a generating line of the cone, at which point the section becomes a parabola. If the intersecting plane is inclined still nearer to the vertical, it meets both branches of the cone which it did not do in the previous cases); now the curve of intersection is a hperbola. hperbola ellipse height Given a parabolic segment S with base see the above figure), the point of the segment that is farthest from the base is called the verte of the segment, and the perpendicular) distance from to is its height. The verte of a segment is not to be confused with the verte of the parabola which, as ou will recall, is the intersection point of the parabola with its ais of smmetr.) rchimedes shows that the area of the segment is four-thirds that of the inscribed triangle. That is, the area of a segment of a parabola is 4/3 times the area of the triangle with the same base and height. Eercise asks ou to check rchimedes result in a ver simple case.) S hperbola circle parabola 2. RCHIEDES THEORE segment of a conve curve such as a parabola, ellipse or hperbola) is a region bounded b a straight line and a portion of the curve. 3. RELIINRIES ON ROLIC SEGENTS the time of rchimedes, the following facts were known concerning an arbitrar parabolic segment S with base and verte see the figure below).. The tangent line at is parallel to. 2. The straight line through parallel to the ais of the parabola intersects at its midpoint. 3. Ever chord QQ parallel to is bisected b. 4. With the notation in the figure below, N = NQ 2 2 Equivalentl, N = / 2 ) NQ 2. In modern terms, this sas that, in the pictured oblique -coordinate sstem, the equation of the parabola is = λ 2, where λ = / 2.) In his book Quadrature of the arabola, rchimedes gives two methods for finding the area of a segment of a parabola previous mathematicians had unsuccessfull attempted to find the area of a segment of a circle and of a hperbola). The second of these methods, which we discuss below, is based on the so-called method of ehaustion. Q ais N Q

2 2 rchimedes quotes these facts without proof, referring to earlier treatises on the conics b Euclid and ristaeus. You are asked to prove the first three properties, mostl b modern methods, in eercises 2, 3 and 4. Using propert 2 to find the verte, ou will then, in eercise 5, verif rchimedes theorem in another special case. Eercise 6 will then guide ou through a modern proof of the general case.) 4. RT : THE ETHOD OF EXHUSTION To find the area of a given parabolic segment S, rchimedes constructs a sequence of inscribed polgons 0,, 2,..., that fill up or ehaust S. The first polgon 0 is the inscribed triangle with the base of segment S and its verte. To construct the net polgon, consider the two smaller parabolic segments with bases and ; let their vertices be and 2, respectivel, and let be the polgon We continue in this wa, adding at each step the triangles inscribed in the parabolic segments remaining from the previous step. s seems clear from the above figures, the resulting polgons 0,, 2,..., ehaust the area of the original parabolic segment S. In fact, rchimedes carefull proves this b showing that the difference between the area of S and the area of n can be made as small as one pleases b choosing n sufficientl large. In modern terms, this simpl means that ) lim n area n) = areas) but it is important to realize that rchimedes, like all the ancient Greek mathematicians, had no limit concept). To prove ), we let n = areas) area n ) for n = 0,, 2,... and show that lim n n = 0. Consider the parallelogram circumscribed about the segment S, whose sides and are parallel to the ais of the parabola, and whose base is tangent to the parabola at and therefore parallel to, b propert ). ais The area of this parallelogram is of course greater than the area of the parabolic segment S. Since it also equals twice the area of triangle wh?), it follows that the area of this triangle is more than half the area of the parabolic segment. The remaining area, which equals 0 because 0 =, must therefore be less than half the area of S: 0 < 2 areas) Now consider the two triangles 2 and ) that are added in the net step to form polgon. The above argument, applied to the two smaller parabolic segments with bases and, shows that the areas of these triangles are more than half the areas of the two segments. It therefore follows that < 2 0 Continuing in this wa, we see that 2 < 2, 3 < 2 2,..., and in general n < 2 n. It is now eas to prove that lim n n = 0 see eercise 7), and therefore ) follows. 5. RT 2: FINDING THE RE OF n t each step in the construction of the polgons 0,, 2,..., we add triangles to the previous polgon: a single triangle ) begins the process, then two triangles 2 and ) are added in the net step, then four triangles are added, etc. Let a 0, a, a 2,..., be the total areas of the triangles added at each step. Thus a 0 = area ), a = area 2 ) + area ), and so on. In the second part of his proof, rchimedes finds the area of the polgons 0,, 2,..., b evaluating the sum 2) a 0 + a + a a n = area n ) The ke step is to show that the total area of the triangles added at each step is equal to /4 the total area of the triangles added at the previous step. In other words, 3) a = 4 a 0, a 2 = 4 a,..., and in general a n = 4 a n. We will describe rchimedes proof that a = 4 a 0, leaving the general case to eercise 8. We want to show that the sum a of the areas of triangles 2 and is /4 that of. ppl propert 2 to both the original parabolic segment S and the smaller segment with base : we obtain two lines parallel to the ais of the parabola, one going through and intersecting at its midpoint, and one going through and intersecting at its

3 3 midpoint Y. Let be the intersection point of this second parallel line with. Then is the midpoint of because the triangles Y and are similar. Finall, let V be the intersection with of the line through parallel to so V is a parallelogram). 2 ais V ppling propert 4 with N = V and Q = ) and noting that V = = 2, we have V = V 2 = 4 so = 4V and, therefore, V = 3V ). Two consequences follow from this. First, since = V, we have 4) = 3V Second, because = 2, Y = 2 b similar triangles again), so that 5) Y = 2V 2 Y Now 4) and 5) impl that Y = V, so in fact Y = 2 Y Now consider the two triangles and. The have the same base and, because Y = 2 Y, a simple argument with similar triangles shows that the height of is twice the height of both heights being relative to the common base ). Therefore 6) area ) = 2 area ) Similarl, triangles and have the same base and, since = 2, the height of is twice that of, so 7) area ) = 2 area ) It follows from 6) and 7) that 8) area ) = 4 area ) n argument similar to the one in the last two paragraphs shows that area 2 ) = 4 area ) Combining this with 8) then gives area 2 ) + area ) = 4 area ) + 4 area ) = 4 area ) so a = 4 a 0, as desired see eercise 9 for a shorter proof). Returning now to the area of polgon n see 2) above) and using 3), we have finall area n ) = a a a a 0 In other words, the areas a 0, a, a 2,..., added at each step in the construction of n form a geometric sequence with common ratio /4, and area n ) is the sum of the first n + terms of this sequence. 6. RT 3: CONCLUSION OF RCHIEDES ROOF Now that the area of n has been determined, it follows from the conclusion of art that the difference between the area of the parabolic segment S and the sum a a a a 0 can be made as small as one pleases b choosing n sufficientl large. In modern terms, 9) areas) = lim a 0 + n 4 a a ) a 0 and rchimedes now seeks to determine this limit. He begins b deriving the identit 0) = 4 3 which is a restatement of the formula ou learned for the partial sums of a geometric series see eercise 0). s rchimedes shows, 0) follows from the observation that 4 k k = k = 3 4 k for we can then sum the terms on the left side of 0) b repeatedl adding the last two terms: ) = ) =... = ) 4 2 = ) 3 4 = + 3 = 4 3 From a modern perspective, rchimedes theorem is now a simple consequence of 9) and 0): areas) = a 0 lim + n ) = a 0 lim n = a ) ) = 4 3 area ) No doubt rchimedes intuitivel obtained the answer 4/3 in a similar wa but, rather than taking limits eplicitl, he completed the proof b showing that the two alternative conclusions areas) < 4 3 area )

4 4 and areas) > 4 3 area ) both lead to a contradiction, and so must be false. This approach whose details we will omit) was in fact tpical of Greek proofs b the method of ehaustion. 6. rove rchimedes theorem b modern methods as follows. Using propert 2, we can label the -coordinates of, and as in the figure below. 7. EXERCISES. Use integration to verif rchimedes theorem for the segment bounded b = 2 and the line =. Determine the verte of the segment and show that the inscribed triangle has area. Then integrate to verif that the segment has area 4/3.) 2. ropert follows easil from a theorem ou learned in Calculus I. Which one? Rotate the parabolic segment until its base is horizontal. What can ou then sa about the verte?) 3. rove propert 2 b modern methods as follows. Introduce a rectangular -coordinate sstem centred at the verte of the parabola, as in the figure below. ais = k 2 In these coordinates, the equation of the parabola has the form = k 2 and its ais of smmetr is the -ais. ssume first that k = and write = a, a 2 ), = b, b 2 ) and = p, p 2 ). a) Show that p = 2 a + b). Compute the slope of the tangent line at using i) calculus and ii) propert.) b) Eplain wh the result of part a) proves propert 2. c) If k, what changes do ou need to make to the calculation in part a)? 4. Eplain wh propert 3 follows from 2. Hint: what is the verte of the parabolic segment Q Q?) 5. Consider the parabolic segment bounded b = 2 and the line = a) Sketch the segment and find the points, and. Use propert 2 to find the verte of the segment; see also eercise 3a).) b) Let be the midpoint of. Find the length of and use it to compute the area of triangle. c) Find the area of the segment b integrating and check rchimedes theorem. p r p p + r r r ssume first that k =. = k 2 a) Use the method of eercise 5b) to show that the area of triangle is r 3. b) Show that the equation of line is = 2p p 2 + r 2. c) Find the area of the segment b integrating. You should get a value of 4 3 r3, thus proving rchimedes theorem for k =.) d) If k, what changes do ou need to make to the calculations in parts a), b) and c)? 7. a) Let 0,, 2,..., be an sequence of positive numbers such that n < 2 n for n =, 2,.... Show that lim n n = 0. This result, which is reall the cru of the method of ehaustion, is originall due to Eudous, a predecessor of Euclid s. For the proof, start b showing that 0 < n < 0 /2 n.) b) Given an positive number r <, show that the conclusion of part a) still holds if the positive numbers 0,, 2,..., satisf n < r n for n =, 2, Show that a = 4 a 0 implies a 2 = 4 a, and convince ourself of the general case a n = 4 a n. 9. In eercise 6, ou showed that for the parabola = k 2, the area of triangle is kr 3. Use this to give a second proof that a = 4 a rove 0) using the formula for the partial sums of a geometric series: + r + r r n = rn+ r. Write a short summar of the three parts of rchimedes proof.

5 5 8. SOLUTION OUTLINES. The inscribed triangle has vertices at ±, ) and at the origin the verte of the segment). Since it has width 2 and height, its area is. The segment has area as required. 2 )d = Rotate the parabolic segment until its base is horizontal. ecause the verte is the point of the segment that is farthest from, it corresponds to an etreme value a maimum or a minimum, depending on whether the segment is above or below ). Fermat s Theorem section 4. of Stewart), the tangent line at must be horizontal, in other words, parallel to. 3. a) Since the derivative = 2, the slope of the tangent line at is 2p. On the other hand, the slope of is b 2 a 2 b a = b + a ropert, these two slopes are equal, so 2p = a + b = p = 2 a + b) b) The straight line through parallel to the ais of the parabola is vertical, so it intersects at a point with -coordinate 2 a + b), in other words, at its midpoint. c) If k, then the two epressions for the slope in part a) are scaled b a factor of k and one finds, as before, that p = 2 a + b). 4. n segment of the given parabola whose base Q Q is parallel to will clearl also have verte. ppling ropert 2 to the segment with base Q Q, we conclude that intersects Q Q at its midpoint, as required. 5. a) =, ), = 3, 9), =, ), =, 5). b) The length of is 4. Referring to the above figure, the circumscribed parallelogram has base equal to 4 the length of ) and height equal to 4 the horizontal distance between and ). Its area is therefore 6, which is twice the area of. Triangle therefore has area 8. lternativel, add the areas of triangles and, and note that these triangles have the same base and equal heights.) c) The area of the parabolic segment is )d = 32 3 which is indeed 4/3 times the area of triangle. 6. a) = p r, p r) 2 ) and = p + r, p + r) 2 ), so the midpoint of is = p, p 2 + r 2 ) because p r) 2 + p + r) 2) = p 2 + r 2 2 Since = p, p 2 ), it follows that the length of is r 2. s in the previous problem, the area of triangle is half the area of a parallelogram with base equal to r 2 and height equal to the horizontal distance 2r between and. In other words, area ) = 2 r2 2r = r 3 b) Since the derivative = 2 and = p, p 2 ), line has slope 2p b propert. lternativel, compute the slope directl from the coordinates of and given in part a).) Using the point-slope form = m ) with, ) = = p, p 2 + r 2 ) and m = 2p then gives = p 2 + r 2 + 2p p) = 2p p 2 + r 2 c) The required integral 2p p 2 + r 2 2 )d can be computed in two was. The first is to integrate directl: 2pd = p p + r) 2 p r) 2) = 4p 2 r r 2 p 2 )d = 2rr 2 p 2 ) 2 d = 3 p + r) 3 p r) 3) = 2 3 r3 + 2p 2 r The area of the segment is therefore 4p 2 r + 2r 3 2p 2 r 2 3 r3 2p 2 r = 4 3 r3 The second wa is to note that the integrand contains a perfect square, 2p p 2 + r 2 2 = r 2 p) 2, and to introduce the substitution t = p. The area of the segment is then r 2 p) 2) d = r r = 2r 3 r 2 t 2 )dt [ 3 t3 ] t=r t= r = 2r r3 = 4 3 r3 d) If k, then the -coordinates of and, the length of, and the area of triangle are each scaled b a factor of k. The same is true of the -coordinates of all points on the line, and of the integrand in part c).

6 6 7. a) We have < 2 0, 2 < 2 < 4 0, 3 < 2 2 < 8 0, and in general n < 0 /2 n. The result then follows from the Squeeze Theorem section 8. of Stewart). b) Show that 0 < n < r n 0 and use the Squeeze Theorem, noting that lim n r n = 0 since 0 < r <. 8. The conclusion a = 4 a 0 is valid not onl for the original parabolic segment with base ) but also for all the smaller segments that arise in the construction of the polgons 0,, 2,.... Thus if we appl this result to the two parabolic segments with bases and, we find that, of the four triangles added to to construct 2, two have combined area equal to 4 area 2), while the other two have combined area equal to 4 area ). It follows that a 2 = 4 a. Similarl, if we appl a = 4 a 0 to the four parabolic segments with bases 2, 2, and, we obtain a 3 = 4 a 2, and so on. 9. ppl the stated result from eercise 6 and propert 2 to the two parabolic segments with bases and : we conclude that triangles 2 and each have area equal to Therefore, k 2 r)3 = 8 kr3 a = area 2 ) + area ) = 4 kr3 = 4 area ) = 4 a 0 9. NOTE ON ROERTY 4 For completeness, we outline a modern proof of propert 4. This propert is equivalent to the assertion that N NQ 2 = 2 for ever chord QQ parallel to, so it suffices to show that the ratio N/NQ 2 remains constant as QQ varies. Introduce -coordinates as in eercise 3, assume k = in = k 2, and write = p, p 2 ), Q = q, q 2 ). The equation of QQ is then and so Therefore, q 2 = 2p q) N = p, 2pp q) + q 2 ) = p, p 2 + p q) 2 ) N = p q) 2 NQ 2 = p q) 2 + 4p 2 p q) 2 so N NQ 2 = + 4p 2 does not depend on the choice of QQ, as desired. If k, we have = p, kp 2 ), Q = q, kq 2 ) and N NQ 2 = k + 4k 2 p 2 which again is independent of the choice of QQ.) 0. SUGGESTIONS FOR FURTHER REDING 0. If r = 4, r n+ r and 0) follows. = r rn+ r = Edwards, C. H., Jr. The Historical Development of the Calculus. New York: Springer-Verlag, 979. The principal source for preparing this assignment. Contains much else about rchimedes and the development of calculus. Heath, T. L. The Works of rchimedes. Cambridge: Cambridge Universit ress, 897. vailable as a reprint from Dover ublications.) The complete tet of Quadrature of the arabola, as translated b Heath, is also available on the web: cass/ archimedes/parabola.html. Netz, Reviel and William Noel. The rchimedes Code. London: Weidenfeld & Nicolson, recent popular work on rchimedes. See also