Mathematical Physics Vol 1

Chapter 4. Field theory

124

Problem40 A scalar function φ = x 2 yz 3 is given. a) Find the direction, through point A ( 2 , 1 , − 1 ) , in which the function is growing the fastest. b) Compute the maximum increment of the function.

Solution a) Bearing in mind the statement on page 84, that "...the gradient determines the direction in which the scalar field changes the fastest", we need to find the gradient of the function φ in point A ( 2 , 1 , − 1 ) ∇ φ = ∇ ( x 2 yz 3 )= 2 xyz 3 i + x 2 z 3 j + 3 x 2 yz 2 k = = − 4 i − 4 j + 12 k . Thus the direction, which passes through point A , and in which the function s growing the fastest is ∇ φ = − 4 i − 4 j + 12 k . b) The maximum increment is | ∇ φ | = q ( − 4 ) 2 +( − 4 ) 2 +( 12 ) 2 = √ 176 = 4 √ 11 .

Problem41 Find the angle between the surfaces φ 1 ≡ x

2 + y 2 + z 2 = 9 and φ

2 + y 2 − 3, at

2 ≡ z = x

point A ( 2 , − 1 , 2 ) .

Solution The angle between two surfaces at a given point is the angle between their normals, i.e, their gradients, at that point. The gradient (normal) of the surface x 2 + y 2 + z 2 = 9 at point ( 2 , − 1 , 2 ) is (see Example 39) ∇ φ 1 = ∇ ( x 2 + y 2 + z 2 )= 2 x i + 2 y j + 2 z k = 4 i − 2 j + 4 k , and the gradient (normal) of the surface z = x 2 + y 2 − 3 at point ( 2 , − 1 , 2 ) is ∇ φ 2 = ∇ ( x 2 + y 2 − z )= 2 x i + 2 y j − k = 4 i − 2 j − k . The scalar product of these vectors is ( ∇ φ 1 ) · ( ∇ φ 2 )= | ∇ φ 1 || ∇ φ 2 | cos θ , and thus the