(xii) ∇(A
·B
) = (B
·∇ )A
+ (A
·∇ )B
+B
×(∇× A
) + A
×(∇× B
)

(xiii) ∇·(∇u) = ∇^2 u=

∂^2 u

∂x^2 +

∂^2 u

∂y^2 +

∂^2 u

∂z^2

The operator ∇^2 =

∂^2

∂x^2 +

∂^2

∂y^2 +

∂^2

∂z^2 is called the Laplacian operator.

(xiv) ∇× (∇× A
) = ∇(∇·A
)−∇^2 A

(xv) ∇× (∇u) = curl (grad u) =
0 The curl of the gradient of uis the zero vector.

(xvi) ∇·(∇× A
) = div (curlA
) = 0 The divergence of the curl of A
is the scalar zero.

(xvii) If a vector field F
(x, y, z )is solenoidal, then it is derivable from a vector function

A
=A
(x, y, z )by taking the curl. One can then write F
= curlA
and hence div F
= 0.

The vector function A
is called the vector potential from which F
is derivable.

(xviii) If f is a function of u 1 , u 2 ,... , u n where ui=ui(x, y, z )for i= 1, 2 ,.. ., n , then

grad f=∇f=

∂f

∂u 1 ∇u^1 +

∂f

∂u 2 ∇u^2 +···+

∂f

∂u n∇un

Many properties and physical interpretations associated with the operations of

gradient, divergence and curl are given in the next chapter.

Example 7-13. Let
r =xˆe 1 +yˆe 2 +zˆe 3 denote the position vector to a general

point (x, y, z ). Show that

grad(r) = grad |
r |=^1

r

r =eˆr

where ˆeris a unit vector in the direction of
r.

Solution Let r=|
r |=

√

x^2 +y^2 +z^2 , then

grad(r) = grad |
r |=

∂r

∂x ˆe^1 +

∂r

∂y ˆe^2 +

∂r

∂z ˆe^3

where

∂r

∂x =

1

2 (x

(^2) +y (^2) +z (^2) )− 1 / (^22) x=x
r
∂r
∂y

1
2
(x^2 +y^2 +z^2 )−^1 /^22 y=
y
r
∂r
∂z

1
2
(x^2 +y^2 +z^2 )−^1 /^22 z=
z
r

Substituting for the partial derivatives in the gradient gives

grad(r) = grad |
r |=^1

r

r =eˆr