148 



KNOWLEDGE 



[Deo. 16, 1881. 



I) as cciitri', cuttinff OL, 01/ in i-, i' rcBpoclively, and join Jr, Kr' 

 which ore obviounly perp. to OA (trianjflc AqO PC|unl in all rcBpocts 

 to trianKlp rJV. Ac.) Thi-n trinnplc c'f/'i/ is in the limit obviously 

 isOBCclcH, q'o pcrp. to qc, iind e'q ^2f"i='2KJ. 



OA 

 Now area PM^PN. ^11 = 3^. jrrz eq. 00.=;. nqe' 



= qN. 



OA 



ON 

 ON 



OA 



KJ 



ON Oq Oq 



= 2qA. KJ=2rJ. AV=2 root. rK 

 This being true of all such areas, as OL pa.sses from OA to OL 

 and onwards, it follows (proceeding to the limit) that 



area PNA = 2 area rJA 

 and entire space between OA, Oa produced indefinitely, and the 

 curre APb carried on iudeliuitely 



= 2 quadrant AraO 

 — ■!• semicircle OP A 

 Therefore, entire area between the " Witch of Agnesi " and its 

 asymptole is eqHal to foui- times the area of the circle from which 

 the curve is formed in accordance with its definition. 



It may be noticed that wo can thus (after a fashion) square the 

 circle, withont>arrying on tlio curve to infinity. For, let CP perp. 

 to OA meet the witch in fi, and lot OP produced meet the quadi-ant 

 in c. Then we know that 



Area APhG = 2 area AcN. 



= 2 (sector JOc -triangle cO.V). 

 = quadrant ^Ou — triangle 0.4 T. 

 = circle .4iJ0— its inscribed square. 

 = va' - a'. 

 So that if APbC be regarded as detei-mined by this mechanically 

 constructed curve, na' — a' is determined, too ; or since a? is known 

 wn' the area of the circle OPA is determined. 



In passing, it may bo noted that from the above construction 

 follows a ready demonstration of tho property of the tangent, esta- 

 blished analytically above. For the limit PP' is a straight lino, and 

 the tangent of the angle which PP" makes with OA is ultimately 



tho — :- of our analysis. 



Now tan. P'PL = 



Pi 

 Pi 

 q'e' 



LL' 



nq 



PN 

 qN 



P'Pl = - 



0L_ 



Oq 



OL 



04 ^ 



ON^ Oq 

 9% ^'^ , 



717 qN ' 



Cq. P£ 



qN qN 



,.pir 



- £2- 



~(qN)'-' 



This in the same relaliim as wc have already obtained ; for if a 

 tangent at /' meets AO prwhiccd, in a point «•, we have 



py DC „,, . 



(qSl 



OC Bd.J 



tan. r'PI. = - rr- = 



•• G. P." considers that he has solved tho very difficult siimd*- 

 tancons equations '" 



r^ + y.^a' (\) v' +:« = !(' (ii) s« + ry » r' (iii) 

 by showing, what is in truth obvious, that ' 



x'- a'j = y'— 6'!/ = 2'- c':. 

 " It is now reduced to a cubic " (wo fail to see this) '' which might!' 

 bo solved in a variety of ways. For instance, if wc suppose x=tiy,' 

 we get : — 



(Where we have written h^, "G. P." writes 1, through some error.) 

 From which by substituting in (i) (ii) and (iii), wc find n, and' 

 hence .t, y, and t. It is evident G. P. has not tried substituting in 

 (i) (ii) and (iii). If ho had, he would have found, first, that sub- 

 stitution in (i) or (ii) would suffice [(iii) giving only an identity], 

 and that the resulting equation in n would be of the Cth degree ! 



[5] — DlKFERENILlL CaLCULCS. 

 tho relation 



MATHEMATICAL QUERIES. 



" G. P." next asks why, if we take 



2 + J 



and differentiate with respect to J-, we get' 



COS. ir= — . 



instead of cos. r- 



(2 + .r)> 



as by trigonometry. "Whence the contradiction?" Simply be- 

 cause " G. P." has applied the differential calculus the wrong way. 

 His relation assigns a definite value to x and to sin .r, and that 

 being so, we cannot differentiate. Take a simpler relation, and we 

 may find the differential calculus, applied in this way, contradicting 

 common sense. Thus take .1=1, and differentiate both sides with 

 respect to j, and we get 1 = 0. 



[G] — In'Dktermi.v.vte Eqc.wiox. — Is it possible to solre 

 1 = 12 + 2;/ ? 

 to find the value of x and y, without assumiri'j any numerical value. 

 — J. A. L. R. [There must always be some assumptions in the 

 solution of iudetcrmiuato equations. — En.] 



[7] — Could yon kindly favour me with a solution to the follow- 

 ing rider on II. Book of "Euclid": — "The angle A of a triangle 

 ABC is a right angle, and D is the foot of the perpendicular from 

 A on BC ; DM, DN, are perpendicular on AB, AC, respectively ; 

 show that the angles BMC, BXC are equal."— Nemo Impixe. 



[As a filler on Book A*I., thus : — 



Triangle BAG similar to triangle AND, 



.-. BA: AC :: AN:ND ( = AM). 

 Hence BA.AM = AC.AN, wherefore points B, M, N, C, lie on 

 circumference of a circle, 



.-. angle BMC=anglc BNC. 

 But the problem can be as readily solved without the aid of any 

 properties beyond those in Book III., thus : — Join MN. Then, 

 triangle .\MN being in all respects equal to the triangle NDA, the 

 angle .\MN is equal to the angle ADN, and then fore to the angle 

 ACD (.\CD and ADN being obviously equal, because each with the 

 angle DAN makes up a right angle). Hence the angles BMN and 

 BCN together are equal to tho angles BMN and AMN together, 

 or to two right angles. Therefore B, M, N, C are points on tho 

 circnmferonco ef a circle, and therefore as above 



Angle BMC = angle BNC. —En.] 



