May 18, 1883.] 



♦ KNOWLEDGE 



301 



^ur iHatt)tmatical Column* 



COLLECTING the results we have obtained in the last two 

 lessons, so far as they add to the number of our known 

 integrals, wo hare 



xvx'—a' a' x 



ijr = :; — o sin ~' - 



^ Z a 



/-'.TZl 





-,i- dx = 



2 



"2 log^a+v'jr'-u^J 

 + -5 log ( a:+ ^' x--,r ) 



y = log I a; + vx'—o-j 



These results should be committed to memory, because these 

 logarithms are of frequent occurrence. As for the first, it is easily 

 remembered from tho relation which, as already pointed out, it 

 bears to a circular area. The next two are similarly related to tho 

 areas APKN and CA'K'X in Fig. 1. Also, if AK, AK' bo 

 joined, it mil be seen how the first part of the one represents the 

 triangle C K N, while the first part of the other represents tho 

 triangle C K' N — the logarithmic parts represent respectively the 

 areas A P K C and A' P' K' C. " These areas, then, are represented 

 by the fourth and fifth of the above tabulated integrals. 



It remains, then, to be ascertained whether the curved surfaces, 

 ACK, A'C K', can be at once and so readily shown to be repre- 

 sented by the fourth and fifth integrals, that a convenient aid to 

 the memory may be obtained from the relation. Obviously, the 



only way to get these areas is to proceed by polar formute. Thus, 

 let A P Q K be part of a rectangular hyperbola, C the centre, C A 

 the axis. Then we know* that if C P = r, and A C P = 



r»= 



Cos'S-sin'Q Cos29 



Now, if P C Q =rf 6, we have, elementary area P C Q 



r' ,„ aVO 



= -<J9=- -- 



2 2 cos 20 



and if K C A = o, we have 



area APK = 2!/"-^ 

 2y„ cos 29 



Thus to integrate /- and/- is our next care. 



y cos 29 y sin 29 



Take the second first, as more likely to be easy. We have 



/- dO f dd r cos 9 dB 



J^a2Jd^ J Z sin 9 cos d^ J 2 sin 9 cos '9 



y tau( 



Whence wo obtain / very easily. Thus, 



J cos 29 



V dB 



Put = -. -0 ; BO that 7T= — 1 



* We give, in fact, the well-known polar equation of the rect- 

 angular hyperbola ; but it can be deduced at once from the relation 

 already given : i/«=i' — u'; putting !/ = r sin 9, and x = )- cos 9. 



/- <J9 /■ 1 dfl r d» 



1 1 /T \ 



= -^ log tan <t> =—2 '°g **" Vt"" ' 



= 2 log cot (^-0) 

 Ilcnco area A P K C 



= -J[log cot(^-«)] (v logcot J =u ) 



=-jlogcot KCN 

 a' CN 



a» C N' 



' C N . N K 

 (CK )' 

 ■ CG' 



o» CN 

 =¥^°ScG 



a? CN .XK 



'"■=T'''S (NK)-^" 



a- (A G)= 



or =^ log ^^^y 



a' AG 

 or =2- log -^ 



Wo have seen already that AQK C, fig. 3 = G A KX. Hence 

 tho result just obtained shows that we may write — 



Area GAKS = — log^ or =4- log — 

 2 «! 2 1/1 



(putting NK = yi)- This is, of course, obvious, since {niY='X\V\r 

 that is, «i_.^ 



V\ «i 

 But it is convenient to note the two forms. 



Putting a = -, or carrying the radius vector on to coincidenco 

 4 

 with the asymptote C Z, we find that the area between the 

 asymptote, the semi-axis C A, and the curve A Q K to infinity 



= — log (cot 0) = infinity. 



We may here add to our list of known integrals these : — 



/Ji - log tan? and /JL = logcot (^-^,) 

 ysinO 2 ycosW V-i 2/ 



/j!£_ = llogtane; and /_i£_ = l log cot (^ - e) 

 y sin 20 2 yco3 292 V^ 



/■JL.=\ log tan !i^; and/-i!i-=l logcot (--^) 

 yBin«9 n 2 y cos ii0 n \i i/ 



(j^ur CI) ess Column, 



By Mephisto. 



THE INTERNATIOXAL TOURNAMENT. 



Since our last report Steinitz sustained an important defeat 

 by Blackbume on the 11th inst. From tho game which we publish 

 below it will be seen that Steinitz adopted the same defence of 

 3. P to KKt3 as in his gamo with Mackenzie and Zukertort, but 

 this time with fatal result. This, however, did not come as a sur- 

 prise to many, who had no faith in that line of play. Blackbum& 

 has raised his score to six, and he 1ms two more draws to play off. 

 Zukertort does not seem to be much affected by his only defeat of 

 last week by Steinitz. He defeated Sellman, Bird, and Eosenthal 

 in very good style, and heads the list with a score of ten. He has 

 to meet English and Winawer before the conclusion of the first 

 round, and it is possible he may have to yield the palm of victory 

 to one or other of these two strong players. Nevertheless, he 

 onght to come out at the head of the list at tho first round. 

 Tschigorin has done very well indeed, baring lost but three games 

 out of ten. Next come Mason and Winawer, with a score of five- 

 and-a-half each, and two draws to play off. 



The play itself is excellent, and continues to interest a large 

 number of daily visitors. Some of the dailies also give considerable 

 attention to the Tournament, notably the Evenin'i Standard and 



