Another way to visualize a vector field is to create a bundle of curves in space,
where each curve in the bundle has the property that at every point (x, y, z)on any
one curve, the direction of the tangent vector to the curve is the same as the direction
of the vector field F(x, y, z)at that point. Curves with this property are called field
lines associated with the vector field F(x, y, z).Let r =xˆe 1 +yˆe 2 +zˆe 3 be a position
vector to a point on a field line curve, then dr =dx ˆe 1 +dy ˆe 2 +dz ˆe 3 is in the direction
of the tangent to the curve. If the curve is a field line, then one can write dr =αF ,
where αis a proportionality constant. That is, if the curve is a field line, then the
vectors dr and F are colinear at all points along the curve and one can write
dr =dx ˆe 1 +dy ˆe 2 +dz ˆe 3 =αF 1 (x, y, z)ˆe 1 +αF 2 (x, y, z)ˆe 2 +αF 3 (x, y, z )ˆe 3
where αis some proportionality constant. By equating like components in the above
equation, one obtains
dx
F 1 (x, y, z )
= dy
F 2 (x, y, z)
= dz
F 3 (x, y, z)
=α (8 .1)
The equations (8.1) represent a system of differential equations to be solved to obtain
the representation of the field lines.
Example 8-1. Find and sketch the field lines associated with the vector field
V =V(x, y ) = (3 −x)(4 −y)ˆe 1 + (6 −x^2 )(4 + y^2 )ˆe 2
Solution If r =xˆe 1 +yˆe 2 describes a field line, then one can write
dr =dx ˆe 1 +dy ˆe 2 =αV(x, y ) = α(3 −x)(4 −y)ˆe 1 +α(6 −x^2 )(4 + y^2 )ˆe 2
where αis a proportionality constant. Equating like components one can show that
the field lines must satisfy
dx =α(3 −x)(4 −y), dy =α(6 −x^2 )(4 + y^2 )
or one could write
dx
(3 −x)(4 −y)=
dy
(6 −x^2 )(4 + y^2 )=α (8 .2)
Separate the variables in equation (8.2) to obtain
6 −x^2
3 −x dx =
4 −y
4 + y^2 dy
