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12.4 Independence of Path and Potential Theory 387
to think of F as a vector field in 3-space. Now compute
i j k
∂g ∂ f
∇× G = ∂/∂x ∂/∂y ∂/∂z = − k.
∂x ∂y
f (x, y) g(x, y) 0
The 3-space condition ∇× G=O therefore reduces to equation (12.8) if the vector field is in the
plane.
Theorem 12.7 can be proved when Stokes’s theorem is available to us.
SECTION 12.4 PROBLEMS
In each of Problems 1 through 10, determine whether F 21. Prove the law of conservation of energy, which states
is conservative in the given region D.If D is not defined that the sum of the kinetic and potential energies of an
explicitly, it is understood to be the entire plane or 3-space. object acted on by a conservative force is a constant.
2
If the vector field is conservative, find a potential. Hint: The kinetic energy is (m/2) R (t) ,where m
is the mass and R(t) describes the trajectory of the
3
2
1. F = y i + (3xy − 4)j particle. The potential energy is −ϕ(x, y, z),where
xy
xy
2. F = (6y + e )i + (6x + xe )j F =∇ϕ.
2
3. F = 16xi + (2 − y )j 22. Complete the proof of Theorem 12.5 by filling in the
details of the following argument. By differentiation,
4. F = 2xy cos(x )i + sin(x )j
2
2
it has already been shown that, if F has a potential
2x 2y function, then
5. F = i + j D is the plane with
2
2
x + y 2 x + y 2 ∂g ∂ f
the origin removed. ∂x = ∂y .
6. F = 2xi − 2yj + 2zk To prove the converse, assume equality of these par-
tial derivatives for (x, y) in D. We must produce a
7. F = i − 2j + k
potential function ϕ for F.
8. F = yz cos(x)i + (z sin(x) + 1)j + y sin(x)k First use Green’s theorem to show that F ·
C
dR = 0for anyclosedpathin D. Thus conclude that
2
2
9. F = (x − 2)i + xyzj − yz k
F · dR is independent of path in D. Choose a point
C
2
2
10. F = e xyz (1 + xyz)i + x zj + x yk P 0 = (a,b) in D. Then, for any (x, y),define
(x,y)
In each of Problems 11 through 20, determine a potential ϕ(x, y) = F · dR.
function to evaluate F · dR for C anypathfromthefirst P 0
C
point to the second. This is a function because the integral is independent
of path, hence depends only on (x, y). To show that
3
2
3
2
11. F = 3x (y − 4y)i + (2x y − 4x )j;(−1,1),(2,3) ∂ϕ/∂x = f (x, y), first show that
x
x
12. F = e cos(y)i − e sin(y)j;(0,0),(2,π/4) ϕ(x + x, y) − ϕ(x, y)
13. F=2xyi+(x −1/y)j;(1,3),(2,2) (The path cannot (x+ x,y)
2
cross the x -axis). = f (ξ,η)dξ + g(ξ,η)dη.
(x,y)
14. F = i + (6y + sin(y))j;(0,0),(1,3) Parametrize the horizontal line segment from (x, y) to
(x + x, y) by ξ = x + t x for 0 ≤ t ≤ 1toshowthat
2
3
3
2
2
15. F = (3x y − 6y )i + (2x y − 18xy )j;(0,0),(1,1)
1
2
16. F=(y cos(xz)− xyz sin(xz))i+ x cos(xz)j− x sin(xz)k; ϕ(x + x, y) − ϕ(x, y) = x f (x + t x, y)dt.
(1,0,π),(1,1,7) 0
Use this to show that
3
2
17. F = i − 9y zj − 3y k;(1,1,1),(0,3,5) ϕ(x + x, y) − ϕ(x, y)
= f (x + t 0 x, y)
2
18. F =−8y i − (16xy + 4z)j − 4yk;(−2,1,1),(1,3,2) x
for some t 0 in (0,1). Now take the limit as x → 0to
3
yz
2 yz
yz
3
19. F=6x e i+2x ze j+2x ye k;(0,0,0),(1,2,−1)
show that ∂ϕ/∂x = f (x, y). A similar argument shows
2
2
20. F = (y − 4xz)i + xj + (3z − 2x )k;(1,1,1),(3,1,4) that ∂ϕ/∂y = g(x, y).
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