Example 7-16. Let r =xˆe 1 +yˆe 2 +zˆe 3 denote the position vector to a
general point (x, y, z )and let r=|r |. Find grad f(r)where f=f(r)is any continuous
differentiable function of r.
Solution By definition
grad f(r) = ∂f
∂x
ˆe 1 +∂f
∂y
eˆ 2 +∂f
∂z
ˆe 3
where
∂f
∂x
=df
dr
∂r
∂x
=f′(r)x
r
∂f
∂y
=df
dr
∂r
∂y
=f′(r)y
r
∂f
∂z
=df
dr
∂r
∂z
=f′(r)z
r
so that
grad f(r) = f′(r)^1
r
r =f′(r)ˆer
Compare this result with the result from the previous example.
Example 7-17. If φ=φ(x, y, z )is continuous and possess derivatives which
are also continuous, show that the curl of the gradient of φproduces the zero vector.
That is, show
curl(grad φ) = ∇× (∇φ) = 0
Solution The function φis differentiable so that
grad φ=∇φ=∂φ∂x ˆe 1 +∂φ∂y ˆe 2 +∂φ∂z ˆe 3
and the curl of this vector is represented
curl(grad φ) = ∇× (∇φ) =
∣∣
∣∣
∣∣
ˆe 1 ˆe 2 ˆe 3
∂
∂x
∂
∂y
∂
∂z
∂φ
∂x
∂φ
∂y
∂φ
∂z
∣∣
∣∣
∣∣
curl(grad φ) = ˆe 1
(
∂^2 φ
∂y ∂z −
∂^2 φ
∂z ∂y
)
−eˆ 2
(
∂^2 φ
∂x ∂z −
∂^2 φ
∂z ∂x
)
+ˆe 3
(
∂^2 φ
∂x ∂y −
∂^2 φ
∂y ∂x
)
= 0
because the mixed partial derivatives inside the parenthesis are equal to one another.
