Multivariate Calculus - Bruce E. Shapiro

172 LECTURE 21. VECTOR FIELDS
Not all vector fields are gradient field. In the next lecture we will derive a test
for determining if a vector field is a gradient field, and a method for determining
the potential function that gives rise to the gradient field.
Example 21.3 Find the gradient field for the scalar function (given in polar coordinates)
f(r, θ) = a/r (21.5)
where a is a constant.
Solution. We begin by converting to rectangular coordinates, using the substitution
r = √ x 2 + y 2 , to give
a
f(x, y) = √
x 2 + y 2
Differentiating,
( ∂
F(x, y) =
∂x , ∂ )
a
√
∂y x 2 + y
(
2
)
∂ a
= √
∂x x 2 + y , ∂ a
√ 2 ∂y x 2 + y 2
(
)
ax
= −
(x 2 + y 2 ) 3/2 , − ay
(x 2 + y 2 ) 3/2
a
= −
(x 2 + y 2 (x, y)
) 3/2
Returning to polar coordinates, we observe that since r = (x, y) and r 2 = x 2 + y 2 ,
we have
F(r, θ) = − ar
r 3 (21.6)
Equations 21.5 and 21.6 are the formulas for the Newtonian Gravitational potential
and force, respectively, between two bodies of mass m 1 and m 2 if we make the
substitution a = Gm 1 m 2 , where G ≈ 6.6742 × 10 −11 meters 3 / (second 2 kilogram)
is the universal constant of gravity.
We can treat the gradient operator
grad = ∇ =
( ∂
∂x , ∂ ∂y , ∂ )
∂z
in many ways just like a vector. We can use it, for example, in a dot product or
cross product to define new vector operators.
Definition 21.6 The divergence of a vector field F = (F 1 , F 2 , F 3 ) is given by
the dot product
( ∂F1
div F = ∇ · F =
∂x , ∂F 2
∂y , ∂F )
3
(21.7)
∂z
The divergence of a vector field is a scalar field.
Revised December 6, 2006. Math 250, Fall 2006