38 1 Mathematical Physics
A unit vector normal to the surface is obtained by dividing the above
vector by its magnitude. Hence the unit vector is
−ˆi+ˆj+kˆ
[(−1)^2 + 12 + 12 ]^1 /^2=−
ˆi
√
3+
ˆj
√
3+
kˆ
√
3(b)∇Φ=(
iˆ∂
∂x+ˆj∂
∂y+kˆ∂
∂z)
(x^2 yz+ 2 xz^2 )=(2xyz+ 2 z^2 )iˆ+x^2 zˆj+^4 xzkˆ
=−ˆj− 4 kˆat (1, 1 ,−1)
The unit vector in the direction of 2ˆi− 2 ˆj+ˆk,isnˆ=2 ˆi− 2 ˆj+kˆ
[2^2 +(−2)^2 + 12 ]^1 /^2= 2 ˆi/ 3 − 2 ˆj/ 3 +ˆk/ 3The required directional derivative is∇Φ.n=(−ˆj− 4 kˆ).(
2 ˆi
3−
2 ˆj
3+
2 kˆ
3)
=
2
3
−
4
3
=−
2
3
Since this is negative, it decreases in this direction.1.15 The inverse square force can be written as
fαr
r^3
∇.f=∇.r−^3 r=r−^3 ∇.r+r.∇r−^3But∇.r=(
iˆ∂
∂x+ˆj∂
∂y+kˆ∂
∂z)
·(ˆix+ˆjy+kzˆ )=
∂x
∂x+
∂y
∂y+
∂z
∂z= 3
Now∇rn=nrn−^2 r
so that∇r−^3 =− 3 r−^5 r
∴∇.(r−^3 r)= 3 r−^3 − 3 r−^5 r.r= 3 r−^3 −r−^3 = 0
Thus, the divergence of an inverse square force is zero.1.16 The angle between the surfaces at the point is the angle between the normal to
the surfaces at the point.
The normal tox^2 +y^2 +z^2 =1at(1,+ 1 ,−1) is
∇Φ 1 =∇(x^2 +y^2 +z^2 )= 2 xˆi+ 2 yˆj+ 2 zkˆ= 2 ˆi+ 2 ˆj− 2 kˆ
The normal toz=x^2 +y^2 −1orx^2 +y^2 −z=1at(1, 1 ,−1) is
∇Φ 2 =∇(x^2 +y^2 −z)= 2 xˆi+ 2 yˆj−kˆ= 2 iˆ+ 2 ˆj−kˆ
(∇Φ 1 ).(∇Φ 2 )=|∇Φ 1 ||∇Φ 2 |cosθ
whereθis the required angle.
(2ˆi+ 2 ˆj− 2 kˆ).(2iˆ+ 2 ˆj−kˆ)=(12)^1 /^2 (9)^1 /^2 cosθ
∴cosθ=10
6
√
3
= 0. 9623
θ= 15. 780