188 
MR. TALBOT’S RESEARCHES IN THE INTEGRAL CALCULUS. 
— Arc (#) — Arc (z) = — 1"267 
+ Arc (y) = + 1*044 
Also 
Sum = — 0223 
r' — r = — 0-297 
f (V - r) = — 0-223 
in accordance with the theorem. 
I had at first some difficulty to perceive the reason why some of the arcs were to 
be considered negative rather than the others. This question was one of a novel 
nature, which had not hitherto occurred to analysts, and therefore no solution of it 
was to be met with in books. On the other hand, to leave so essential a point with- 
out any demonstration was unsatisfactory. But the following considerations appeared 
to afford an explanation of this fact. 
In the first example, since x -j- y -j- z = 0, and both x and y are positive quantities, 
z must be negative. Therefore xy is positive, but x z and y z are negative. 
Now we have 
y * _ a/ i + x *- i 
which therefore must be negative: .-. also ^/l — 1 must be negative. But this 
quantity would necessarily be positive if the radical had a positive sign. Therefore 
the radical must have a negative sign. 
Reasoning in the same manner, because 
x z 
O 
v' ] + ?/ 4 
this radical has neces- 
sarily a negative sign ; and because ~ = 
ccy_ _ a/] + z 4 _ i 
, this radical has a positive sign. 
Attributing, therefore, these signs to the radicals, the three hyperbolic arcs are re- 
spectively, 
’dx , f*dv . f'dz 
/l 4 - r* — / , 
y 
+«*■*> j 1 -f +yv + * 4 - 
On the other hand, during the change of x, y, z to x', y', z' respectively, we have seen 
that x and z increase while y diminishes (see equations [4.]) ; .'. dx and dz are to be 
accounted positive, and dy negative. This consideration renders it necessary to write 
[e-] -fird rr^, +f^J t+¥, +fir 
And thus the assertion that arc (<r) is to be considered negative in this example is 
justified. 
Second example . — Here the reasoning remains the same as far as regards the signs 
of the different radicals ; but it appears from the equations [5.], that while x increases, 
both //and 2 diminish, :.dx is positive, and dy and dz arc negative. Therefore the 
only difference between this example and the former respects the sign of dz. There- 
fore equation [6.] must be written 
