James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

1094 Chapter 16 Vector Calculus
Comparing this equation with Equation 16, we see that
PsAd 1 KsAd − PsBd 1 KsBd
which says that if an object moves from one point A to another point B under the influence
of a conservative force field, then the sum of its potential energy and its kinetic
energy remains constant. This is called the Law of Conservation of Energy and it is the
reason the vector field is called conservative.
1. The figure shows a curve C and a contour map of a function f
whose gradient is continuous. Find y C
=f dr.
y
9. Fsx, yd − sy 2 cos x 1 cos yd i 1 s2y sin x 2 x sin yd j
10. Fsx, yd − sln y 1 yyxd i 1 sln x 1 xyyd j
C
10
20
30
60
50
40
11. The figure shows the vector field Fsx, yd − k2xy, x 2 l and
three curves that start at (1, 2) and end at (3, 2).
(a) Explain why y C
F dr has the same value for all three
curves.
(b) What is this common value?
0
2. A table of values of a function f with continuous gradient is
given. Find y C
=f dr, where C has parametric equations
x − t 2 1 1 y − t 3 1 t 0 < t < 1
x
y
3
2
x
0
y
0 1 2
1
1 6 4
1
2
3
8
3–10 Determine whether or not F is a conservative vector field.
If it is, find a function f such that F − =f .
3. Fsx, yd − sxy 1 y 2 d i 1 sx 2 1 2xyd j
4. Fsx, yd − sy 2 2 2xd i 1 2xy j
5. Fsx, yd − y 2 e xy i 1 s1 1 xyde xy j
6. Fsx, yd − ye x i 1 se x 1 e y d j
7. Fsx, yd − sye x 1 sin yd i 1 se x 1 x cos yd j
8. Fsx, yd − s2xy 1 y 22 d i 1 sx 2 2 2xy 23 d j, y . 0
5
2
7
9
0
1
12–18 (a) Find a function f such that F − =f and (b) use
part (a) to evaluate y C
F dr along the given curve C.
12. Fsx, yd − s3 1 2xy 2 d i 1 2x 2 y j,
C is the arc of the hyperbola y − 1yx from s1, 1d to (4, 1 4)
13. Fsx, yd − x 2 y 3 i 1 x 3 y 2 j,
C: rstd − kt 3 2 2t, t 3 1 2tl, 0 < t < 1
14. Fsx, yd − s1 1 xyde xy i 1 x 2 e xy j,
C: rstd − cos t i 1 2 sin t j, 0 < t < y2
15. Fsx, y, zd − yz i 1 xz j 1 sxy 1 2zd k,
C is the line segment from s1, 0, 22d to s4, 6, 3d
2
3
x
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