VECTOR DIFFERENTIATION. 
85 
If, instead of applying the third and fourth operators di¬ 
rectly, we apply the parts into which they resolve, we obtain 
(rV) 2_ = 8 _ d (r 2 p 2 ) 2 
drdp rp 
dp r 2 p 
2 _ Q <?ov) 2 
rp dr rp 2 
4. 
4. 
In the second last differentiation the multiplier is 2 — 2, 
which reduces the term to zero. The total is 18, as before. 
When the function depends on the spherical variables 
r, 0, <p, then 
< 7 = W Fr + j0r° + 
V9- 
To determine pr, pO, p <p apply the operator to the radius 
vector R. 
r> dR , dR . dR 
v R=- wV r + — V 0 + V9 . 
But 
R = rp — r | cos Oi 4- sin 6 cos <pj -j- sin 6 sin <pk J. 
Therefore 
Hence 
dR _ t dR dp dR dp 
~JJ — p ’ -Jo ~ 1 le’ — 
vR = p pr + r pO + r p<f. 
But we know that pR = 3 absolutely, and in order that the 
above three Terms ma}^ yield 3, it is necessary that 
pr = —; pO 
P 
*P_ 
do 
V9 = 
As 
dp_ 
d(f 
P — cos 0 . i - j- sin 0 cos <p . j + sin 0 sin <p . k ; 
— w n 0 • i cos 0 cos <p . j -\- cos 0 sin <p . k ; 
~~ == sin 0 | — sin <p . j -j- cos <p . & j. 
