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Calculus 2nd Edition Rogawski

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SECTION 17.1 Vector Fields 949<br />

THEOREM 2 Uniqueness of Potential Functions If F is conservative on an open<br />

connected domain, then any two potential functions of F differ by a constant.<br />

Proof If both V 1 and V 2 are potential functions of F, then<br />

∇(V 1 − V 2 ) = ∇V 1 −∇V 2 = F − F = 0<br />

However, a function whose gradient is zero on an open connected domain is a constant<br />

function (this generalizes the fact from single-variable calculus that a function on an<br />

interval with zero derivative is a constant function—see Exercise 36). Thus V 1 − V 2 = C<br />

for some constant C, and hence V 1 = V 2 + C.<br />

The next two examples consider two important radial vector fields.<br />

The result of Example 5 is valid in R 2 : The<br />

function<br />

V (x, y) = √ x 2 + y 2 = r<br />

is a potential function for e r .<br />

where<br />

In R 2 ,<br />

REMINDER<br />

〈 x<br />

e r =<br />

r , y r , z 〉<br />

r<br />

r = (x 2 + y 2 + z 2 ) 1/2<br />

〈 x<br />

e r =<br />

r , y 〉<br />

r<br />

where r = (x 2 + y 2 ) 1/2 .<br />

z<br />

EXAMPLE 5 Unit Radial Vector Fields Show that<br />

√<br />

V (x, y, z) = r = x 2 + y 2 + z 2<br />

is a potential function for e r . That is, e r = ∇r.<br />

Solution We have<br />

Similarly, ∂r<br />

∂y = y r<br />

∂r<br />

∂x = ∂<br />

∂x<br />

√<br />

x 2 + y 2 + z 2 =<br />

x<br />

√<br />

x 2 + y 2 + z 2 = x r<br />

∂r<br />

and<br />

∂z = z 〈 x<br />

r . Therefore, ∇r = r , y r , z 〉<br />

= e r .<br />

r<br />

The gravitational force exerted by a point mass m is described by an inverse-square<br />

force field (Figure 9). A point mass located at the origin exerts a gravitational force F on<br />

a unit mass located at (x, y, z) equal to<br />

F = − Gm<br />

r 2 e r = −Gm<br />

〈 x<br />

r 3 , y r 3 , z<br />

r 3 〉<br />

where G is the universal gravitation constant. The minus sign indicates that the force is<br />

attractive (it pulls in the direction of the origin). The electrostatic force field due to a<br />

charged particle is also an inverse-square vector field. The next example shows that these<br />

vector fields are conservative.<br />

EXAMPLE 6 Inverse-Square Vector Field Show that<br />

( )<br />

e r −1<br />

r 2 = ∇ r<br />

Solution Use the Chain Rule for gradients (Theorem 1 in Section 15.5) and Example 5:<br />

∇(−r −1 ) = r −2 ∇r = r −2 e r<br />

x<br />

y<br />

17.1 SUMMARY<br />

FIGURE 9 The vector field − Gme r<br />

r 2<br />

represents the force of gravitational<br />

attraction due to a point mass located at the<br />

origin.<br />

• Avector field assigns a vector to each point in a domain.Avector field in R 3 is represented<br />

by a triple of functions<br />

F = ⟨F 1 ,F 2 ,F 3 ⟩<br />

A vector field in R 2 is represented by a pair of functions F = ⟨F 1 ,F 2 ⟩. We always assume<br />

that the components F j are smooth functions on their domains.

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