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Exercise Answer1.2
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(2) The unit vector with the direction of
is
Exercise Answer1.3
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(1) The area of parallelogram is
Exercise Answer1.5
1.
.Thus they are not coplanar.
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Exercise Answer2.1
1. The components of
are
. Then
and the trace of
,
is on the paraboloid
.
2. Differentiate eachcomponents, we have
3.
.
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4. Let
. Then
.
Thus, by the derivative of the vector function, we have
Exercise Answer2.2
1. The straight line to be found has a starting point of and a direction of
. Let the arbitrary point on the line as
. Then,
3. The straight line to be found has a starting point of and a direction of
. Let the arbitrary point on the line as
. Then,
4. Since
,we have
. Thus,
is a smooth curve that spirally rotates around a cylinder with a radius of 1.
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Other way is
Exercise Answer3.1
1. Let the level surface through the point
be
Next, find the directional unit vector to find the directional derivative in the direction at the point
.
.Then the directional derivative is
Also, the equation of the tangent plane is
3. Let
be a streamline equation. Then
expresses the normal vector of
.
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7. Note that
of
is orthogonal to
.Therefore the unit normal vector
is
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Exercise Answer3.2
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3. Use
4. Note that
of
is orthogonal to
.Therefore the unit normal vector
is
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6.
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Exercise Answer3.3
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2.
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3. Note that if satisfies
,then
has a scalar potential and
. Thus we find
so that,
.
4. The force field
has a potential
. Then
.This suggest that the equation of motion of this mass point is
5.
If the origin O is centered on this plane and the circle with radius
is
, the equation of motion of the mass point is parametrized by
.Thus,
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8. The directional derivative of at P
in the direction of
is,
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Exercise Answer3.4
1.
(1) When the curved surface is projected onto the
plane,
maps to
. Also,from the surface
, if the corresponding
is the position vector,
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2. Since the surface is region on the
plane,the unit normal vector is
.
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3. By projecting onto
plane, then
aps to
.
Next by letting
be a position vector
.
Thus,the normal vector of the surface
is
.
Therefore,,
4. By projecting the surface onto
plane,
maps to
.
Next,let
be a postion vector
.
Thus,the normal vector of
is
.
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5. The position vector is
. Then
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6. Let
be the position vector corresponds to
. Then
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8. Since
,for
,the position vector is
. Thus,
For the plane ABCO: since , the position vector is
. Here,the positive direction is the direction of the back of plane ABCO to the front.Thus,the unit normal vector is
For the plane ABEF: since , the position vector is
. Here the positive direction is the direction of the back of the plane ABEF to the front. Thus, the unit normla vector is
. Thus,
.
and
.Therefore,
.
Exercise Answer3.5
Basic formula Let
. Then (1)
1.
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2.
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4.
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(3) (2) implies,
. By symmetry
. Therefore,
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8. Differentiation of composite function implies
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(2) By the differentiation of composite functions,,
Exercise Answer3.6
1.
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Note that there exits so that
implies
.
8. Since
is conservative, there exists
so that
.Now we find
. Then
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Exercise Answer4.1
1.
(1) Gauss's divergence theorem implies
(2) Tranform
into the surface integral.Using any constant vetor
and the triple scalar product,
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(3) Gauss's dievergence theorem implies
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(4) Transfrom
into surface integral.Using a constant vector
,
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(5) Transform
into surface integral.Using a constant vector
and the triple scalar product,
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(6) Transform
into surface integral.Using a constant vector
, we have
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2. Using
,we write into surfacee integral.
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3. Let be the boundary of the curved surface
, and let
and
be the curved surfaces separated by the boundary. Also,let the unit normal vector of
be
,the unt normal vector of
be
.Then let the unit normal vector of the surface
be
. We have
or
.Here, ,
4.
(1) Gauss's divergence theorem implies
(2) Gauss's divergence theorem implies
(3) Gauss's divergence theorem implies
(4) Gauss's divergence theorem implies
5.
(1)
labelenshu:4-1-5-1
is the directional derivative of the direction of unit normal vector
implies
.Thus,
(2)
is the directional derivative of the direction of unit normal vector implies
.Thus,
(3) The result of (1) subtracts the result of (2). Then,
(4) is harmonic function means that
.Therefore,using (2),
(5)
are harmonic functions means that
.Therefore,using (3),
(6) is harmonic function. Then (4)implies
6. If
,then Gauss's divergence theorem implies
7.
.Here, using a constant vector
, rewrite into surface integral.
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Since
,projection onto
plane,
implies
. Here,we consier
and
.
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Next projection onto plane. Then
implies
.Thus,for
and
,
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Adding these,
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Here using the polar coorinates,we have
. Thus
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(1) Using the arbitrary constant, we express by the surface integral and apply Gauss's divergence theorem. Then
(2) By Gauss's divergence theorem,
(3) Using a constant vector,we express by the surface integral and apply the triple scalar product and Gauss's divergance theorem,
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(2) is harmonic function. Then
. Thus,(1) implies
Exercise Answer4.2
1.
2.
(1) Rewrite a line integral into
.To do so, using a constant vector
and the triple scalar product,we have
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(3) Let be a constant vector. Then using Stokes' theorem,
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4. Rewrite the line integral into
.Then using a constant vector
and the triple scalar product,and Stokes' theorem,we have
Here using the triple scalar product, we have
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6 The boundary of ,
is a circle
.Thus the position vector is
. Then we find the line integral
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Next we find the surface integral..
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