Solution manual for classical dynamics of particles and systems 5th edition by thornton

CHAPTER 1Matrices, Vectors, and Vector Calculus 1-1.. Consider the triangle a, b, c which is formed by the vectors A, B, C... The unit vectors in spherical coordinates are expressed in t

CHAPTER 1

Matrices, Vectors, and Vector Calculus

1-1

x2 = x2′

x1′ 45˚

x1

x3′

x3

45˚

Axes x′1 and x′3 lie in the x x1 3 plane

The transformation equations are:

CHAPTER 1

x1A

B C

D

α β

(4) Therefore,

(5) Thus,

E D

Comparing (9) with (7), we find

1

1

C

But,

i i

4

Identifying B k i =(B t)ik and A jk =(A t)kj, at./

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A A

A

i x

But this can be true only if

ik i k i

There are 4 diagonals:

If L=L′, then

component of B along A

B A

( )

12

b) Consider vectors A and B in the plane defined by e , Since the figure defined by A, B,

1 3 2

= altitude area of the base

= volume of the parallelepiped

( ) ( ) ( ) ( )

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But cosr θ is the magnitude of the component of r along a

component perpendicular to a is arbitrary

constant

⋅ =

r a

This

a plane perpendicular to

⋅ =

r a a

which is the cosine law of plane trigonometry

1-18 Consider the triangle a, b, c which is formed by the vectors A, B, C

A

α C B

γ β

b c

a

1-16.

( )

b) Using (3), we can find sin(α β− ):

a) Consider the following two cases:

When i ≠ δij = but 0 εijk ≠ 0When i = j δij ≠ but 0 εijk = 0Therefore,

0

ijk ij ij

ε δ =

When j = k, εijk =εijj= 0Terms such as ε εj11 A11=0 Then,

g) i ≠ A or : This implies that i = k or i = j or m = k

jl km k jm m j k jkm

The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by

d

r dt d

r

θ φ

(

2 2

v k

1 cos2

r

v k

θ

θθ

1-29 Let r =2 9 describe the surface S and 1 x y z+ + 2 = describe the surface S The angle 1 θ

between and at the point (2,–2,1) is the angle between the normals to these surfaces at the point The normal to is

2 1

12

grad grad grad grad

(3)

or,

4cos

2 1

1 2 2

1 2 2

2

22

r

x r

x x

1 2 2

f r

f r x

f r

r f

1 2 2

1 2

2

1

j j

i i

j

j i

=

b)

where C is the integration constant (a vector)

1-34 First, we note that

The rest of the common volume is formed by 8 equal parts from the other cylinder (the one

along the x-axis) One of these parts extends from x = 0 to x = a, y = 0 to y= a2−x2, z = a to

0

88

16

23

a a

3

163

a

1-35.

Since ∇ ⋅ , we only need to evaluate the total volume Our cylinder has radius c and height

d, and so the answer is

z

y C

The curve C that encloses our surface S is the unit circle that lies in the xy plane Since the

element of area on the surface da is chosen to be outward from the origin, the curve is directed

counterclockwise, as required by the right-hand rule Now change to polar coordinates, so that

we have ds=dθeθ and A=sinθi+cosθk on the curve Since eθ⋅ = −i sinθ and eθ⋅ =k 0, we have

−

= =

−Further, AH=(x−1, ,y z) is perpendicular to n so one has 6(x− +1) 3y+2z= 0Solving these 3 equations one finds

c) Suppose that in the α direction ( with respect to W-E axis), at point A = (1,1,13) the hill is

steepest Evidently, dy = (tanα)dx and

b) At point A:x=y=1,z= 13 At this point, two of the tangent vectors tothe surface of thehill are

a.u/

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