Mathematical Physics Vol 1

4.6 Examples

125

cosine of the required angle is

( 4 i − 2 j + 4 k ) · ( 4 i − 2 j − k ) | 4 i − 2 j + 4 k || 4 i − 2 j − k | = = 16 + 4 − 4 p ( 4 ) 2 +( − 2 ) 2 +( 4 ) 2 p ( 4 ) 2 +( − 2 ) 2 +( − 1 ) 2 = = 16 6 √ 21 .

cos θ =

8 3 √ 21

From here we obtain θ = arccos

= arccos0 . 5819 = 54 . 41 ◦ .

Problem42 By taking samples from an open pit, determining their densities and interpolating the data, a density field ρ = x 2 y was obtained. Find a) gradient of this field at point A ( 3 , 2 ) , b) density increment at point A in the direction defined by point P ( 1 , 2 ) , c) density increment at point A in the direction AB , where point B is B ( 1 , 3 ) , d) density increment at point A in the direction defined by point Q ( − 4 , 3 ) .

Solution a) grad ρ = 12 i + 9 j , b) ∆ u P ρ = 30 c) ∆ AB ρ = − 13

√ 5, density increases in this direction,

√ 5, density decreases in this direction, d) ∆ u Q ρ = 0, density is constant in this direction, i.e. in this direction the field is homogenous.

Problem43 Let R be the distance between a fixed point A ( a , b , c ) and an arbitrary point P ( x , y , z ) ( R = AP ). Prove that ∇ R is the unit vector of vector −→ AP .

Solution If r A and r P are position vectors a i + b j + c k and x i + y j + z k of points A and P respectively, then R = r P − r A =( x − a ) i +( y − b ) j +( z − c ) k and the magnitude of this vector (the distance AP ) is R = q ( x − a ) 2 +( y − b ) 2 +( z − c ) 2 .