is Mr. WHEWELL on the 
dx = (a'r — aq) dt. 
Similarly, dy = (y'r_ — y''9) 4t, 
and dz = (sr — 2"q) dt. 
Hence, dx = (adr - a"dq + rda' — qdz") dt, 
d'y = (y'dr — y'dq + rdy' — qdy") dt, 
@z = (dr — 2"dq + rdz — qdz") dt; 
And multiplying these equations by 2’, y', #, and adding, 
observing the simplifications which result from (a), (5), and from 
the equations 
vdx +y'dy + dz =0 
adx +ydy +2dz =0 
(8); 
ada" + y"dy" fe "de!" = re) 
which arise from differentiating (0); we shall find 
o, abe +y dy +x%@2= {[dr—q(a/da" +y/dy" + 2'dz") dt} = (dr+pqdt) dt. 
Similarly, we should find 
ald?e +y"d?y +2"d*2 =(dp+qrdt) dt, 
ede" +ydy’+2d°3" =(dq+prdt) dt. 
Also, 
ada +ydy +2d'x' =(-dr+pqdt) dt, 
ada" +7 Cy" +2d'3" =(—dp+qrdt) dt, 
x'dx +y'd’y +2"d*x =(-dq+prdt) dt. 
Hence, equations (d) become 
m (dr+pqdt) =m (-—dr+pqdt), 
m (dp + qrdt) = m'(—dp + qrdt); 
m (dq +prdt) = m'(-dq + prat), 
(m + m’) dp + (m — m’) qrdt=0 
“. (m + m’)dr + (m — m) pqdt = 0 
(i). 
(m' +m) dq +(m" — m) prdt =0 
