Thomas Calculus 13th [Solutions]

Section 16.2 Vector Fields and Line Integrals: Work, Circulation, and Flux 1169
39.
40.
y
x
2 2 2 2
2 2
F i j on x y 4;
x y x y
at (2, 0), F j ; at (0, 2), F i;
at ( 2, 0), F j ; at (0, 2), F i;
at
at
at
at
3
F i 1 j
2 2
2, 2 , ;
3
F i 1 j
2 2
2, 2 , ;
3
F i 1 j
2 2
2, 2 , ;
3
2, 2 , F i 1 j
2 2
2 2
F i j F i
x y on x y 1; at (1, 0), ;
at ( 1, 0), F i; at (0,1), F j;
1 3 1 3
F j F i j
2 2 2 2
at (0, 1), ; at , , ;
1 3 1 3
F i j
2 2 2 2
at , , ;
1 3 1 3
F i j
2 2 2 2
at , , ;
1 3 1 3
at , , F i j .
2 2 2 2
41. (a) G P( x, y) i Q( x, y)
j is to have a magnitude
(b)
counterclockwise direction. Thus
2 2 2 2
tangent line at any point on the circle y at ( a, b ). Let
2 2
a b and to be tangent to
2 2 2 2
x y a b in a
x y a b 2x 2yy 0 y
x
is the slope of the
y
a
b
2 2
v bi aj | v | a b , with v in a
counterclockwise direction and tangent to the circle. Then let P( x, y)
y and Q( x, y)
2 2 2 2 2 2
G i j G i j G
y x for ( a, b) on x y a b we have b a and | | a b .
2 2 2 2
G F F
x y a b
.
x
42. (a) From Exercise 41, part a, yi
xj is a vector tangent to the circle and pointing in a counterclockwise
direction yi
xj is a vector tangent to the circle pointing in a clockwise direction
unit vector tangent to the circle and pointing in a clockwise direction.
(b) G F
yi
xj
G is a
x
2 2
y
43. The slope of the line through ( x, y ) and the origin is y x
pointing away from the origin
xi
yj
x
2 2
y
v xi yj is a vector parallel to that line and
F is the unit vector pointing toward the origin.
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