Mathematical Physics Vol 1

4.6 Examples

123

The unit vector in the direction 2 i − j − 2 k is n = v | v | = 2 i − j − 2 k p 2 2 +( − 1 ) 2 +( − 2 ) 2 = Then the required increment in the direction of v (see p.84) is ∇ φ · n =( 8 i − j − 10 k ) · 2 3 i − 1 3 j − 2 3 k = 16 3 + 2 3 i − 1 3

2 3

j −

k .

1 3

20 3

37 3

+

=

.

Note that the solution is positive, namely d φ d s

> 0, which means that φ is growing in the

R

given direction.

Problem39 Prove that the gradient of a complex function ϕ ( f ) is given in the form

d ϕ d f

∇ ϕ ( f )=

∇ f ,

where f = f ( x , y , z ) .

Proof

∂ϕ ∂ x

∂ϕ ∂ y

∂ϕ ∂ z

d ϕ d f

∂ f ∂ x

d ϕ d f

∂ f ∂ y

d ϕ d f

∂ f ∂ z

∇ ϕ =

i +

j +

k =

i +

j +

k =

d ϕ d f

k =

∂ f ∂ x⃗

∂ f ∂ y⃗

∂ f ∂ z⃗

d ϕ d f

∇ f .

i +

j +

=

Let us apply this formula to the function ϕ = 1

r (see example 33c).

d ϕ d r

1 r 2

∇ ϕ =

∇ r = −

∇ r .

Given that (see example 33a, p. 119),

r r

∇ r =

= r 0

we finally obtain

r r 3

1 r 2

∇ ϕ = − r 0 , which is the same result as in the aforementioned example. = −