Mecflu - matrizes, vetores e c?lculos de vetor

Mecflu - matrizes, vetores e c?lculos de vetor

Matrices, Vectors, and Vector Calculus

1-1.

So the transformation matrix is:

1-2. a) βγ O E x

From this diagram, we have cosOEOAα= cosOEODγ= Taking the square of each equation in (1) and adding, we find

But and

Therefore,

Thus, b) x

First, we have the following trigonometric relation:

MATRICES, VECTORS, AND VECTOR CALCULUS 3

But,

2 cos cos cos cos

cos cos or, cos cos cos cos cos cos 2 cos cos cos cos cos cos

2 cos cos cos cos cos cos cos cos cos cos cos cos cosθ αα β β γ γ=+ +′ ′ ′ (10)

1-3.

e e e A e ′

Denote the original axes by , , , and the corresponding unit vectors by e,, . Denote the new axes by , , and the corresponding unit vectors by 1x 2x 3x 1 2e 3e

e e e e e e e e e e e e

1-4. a) Let C = AB where A, B, and C are matrices. Then,

( )t ji jkk i ki jkij

Identifying ()t ki ik

B=B and ()t jk kj

1-5. Take λ to be a two-dimensional matrix:

Then,

2 2 λ λλ λλ λ λ λ λ λλ λ λ λλ λ λ λ λ λ λ λ λ λλ λλ λ λ λ λ

But since λ is an orthogonal transformation matrix, ijkjikj λλδ=∑.

Thus,

1-6. The lengths of line segments in the jx and jx′ systems are

MATRICES, VECTORS, AND VECTOR CALCULUS 5

If , then L=′

The transformation is

Then, ji k k ji k

ki k i ki x x λλ λλ i x

But this can be true only if

which is the desired result.

1-7.

There are 4 diagonals:

1D, from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D; 1 2D, from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = ; 2D 3D, from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = ; and 3D

4D, from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D. 4 The magnitudes of the diagonal vectors are

so that

Similarly,

1-8. Let θ be the angle between A and r. Then, 2A⋅=Ar can be written as 2cosArAθ=

This implies

Therefore, the end point of r must be on a plane perpendicular to A and passing through P.

1-92=+−Aijk23=−++Bijk

b) component of B along A

The length of the component of B along A is B cos θ. cosABθ⋅=AB

The direction is, of course, along A. A unit vector in the A direction is

MATRICES, VECTORS, AND VECTOR CALCULUS 7

So the component of B along A is

+−ijk

c) 3 cos

ij k ABijk

e)32−=−−ABijk5+=−+ABij

ij k ABAB2

1-10. 2sincosbtbtωω=+rij a) 2

2c os sin

2s in cos bt b t bt b t ωω ω ω 2ω ωω ω ω == −

4c os sin bt b bt t ωω ω ωωωω

So, at this time, bω=−vj, 22bω=−ai

1-1.

a) Since ()ijkjkijk ABε×=∑AB, we have

() ijk j k i ij k

AB C× (1)

We can also write

We notice from this result that an even number of permutations leaves the determinant unchanged.

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

()()3area of the base =altitude area of the base

=volume of the parallelepiped ⋅× = ⋅ × ×

CA B C e

1-12. O

C ha c a – c c – b b – a

The distance h from the origin O to the plane defined by A, B, C is ab a c b ba c b

ab c a c a b bc a c a b ab c

The area of the triangle ABC is:

1-13. Using the Eq. (1.82) in the text, we have

a)

Expand by the first row.

d) ?t−=ABBA

1-15. If A is an orthogonal matrix, then

MATRICES, VECTORS, AND VECTOR CALCULUS 1

1-16.

x P r θ xa r θ a r cos θ constant⋅=ra cosconstantraθ=

It is given that a is constant, so we know that cosconstantrθ=

But cosrθ is the magnitude of the component of r along a.

The set of vectors that satisfy all have the same component along a; however, the component perpendicular to a is arbitrary. constant⋅=ra

Th is us the surface represented by constant a plane perpendicular to . ⋅=ra a

1-17. a θ bBc

Consider the triangle a, b, c which is formed by the vectors A, B, C. Since

or,

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

γ β bc

where e is the unit vector perpendicular to the triangle abc. Therefore, 3 sinsinBCABαγ= (5) or, sin sinCA γα=

Similarly, sin sin sin CA B γαβ== (6) which is the sine law of plane trigonometry.

1-19.

x a α x a b a b b β a) We begin by noting that

cos cos sin sin sin cos sin cos 2 cos cos sin sin 2c os cos sin sin ab a b ab a b ab ab ab ab β (2)

MATRICES, VECTORS, AND VECTOR CALCULUS 13

Thus, comparing (1) and (2), we conclude that ()coscoscossinsinαβαβα−=+β (3)

sin 1 cos 1 cos cos sin sin 2cos sin cos sin

1 cos 1 sin sin 1 cos 2cos sin cos sin

sin cos 2sin sin cos cos cos sin sin cos cos sin so that

()sin sin cos cos sinα βα β α−= − βj

1-20. a) Consider the following two cases:

When i≠0ijδ= but 0ijkε≠.

When i=j0ijδ≠ but 0ijkε=. Therefore,

b) We proceed in the following way: When j = k, 0ijkijjεε==.

ε ε ε εε ε ε εε εε ε ε ε= + +++ +∑ A A A A A= Now, suppose i, then, 1==A

, 2ijk jk ijk c) ijk ijk ijk

or,

1-21. ()ijkjkijk ABε×=∑AB (1)

( ) ijk j k i ij k ABCε×⋅=∑∑ABC (2)

By an even permutation, we find ijki j k ijk ABCε=∑ABC (3)

1-2. To evaluate ijkmkk εε∑A we consider the following cases:

a) :0 for all ,,ijkmkiikmkkk ij i mεε εε== =∑∑ A A b) :1 for

0 for ijk mk ijk imkkk ij jm εε εε== = ==≠

∑∑A and,mkij≠

i j i c) :0 for

1 for and , ijk mk ijk ikkk im jjk

A d) :0 for

1 for and , ijk mk ijk jmkkk jm mi k i j

MATRICES, VECTORS, AND VECTOR CALCULUS 15 e) :0 for

1 for and , ijk mk ijk jkkk jm iik j km f) :0 for all ,,ijkmkijkkkk mi εε εε== =∑∑ A A g) : This implies that i = k or i = j or m = k. or i≠A

h) for all or :0ijkmkk jm εε≠= ∑ A , , ,ij mA

Now, consider ijmimjδδδδ−A and examine it under the same conditions. If this quantity behaves in the same way as the sum above, we have verified the equation ijkm k i jmi m jk εεδδδδ=−∑A a) :0 for all ,,iimimiijimδδδδ=−=A b) :1 if , ,

0 if i jm im jij jm m i j mδ δδ δ=− = = ≠=≠

0 if ij i i jim j i j δ δδ δ=− = − ==≠ A AA mi d) :1 if ,

e) :1 if ,

0 if im m im mjm i mi δ δδ δ=− = ==≠ A AA all,,j f) :0 for ijiljmiδδδδ=−=A g) ,:0 for all ,,,ijmimjimijmδδδδ≠−=A h) ,:0 for all ,,,ijmimijmijmδδδδ≠−=A Therefore, ijkm k i jmi m jk εεδδδδ=−∑A (1)

First ()ijkjkijk BCε×=∑BC. Then, mn m mn m njk j kn mn mn jk

mn njk m j k mn jkn m j k jkmn jkmn lmn jkn m j k jkm n jl km k jm m j k jkm

m m m m m m m m m m

1-23. Write

( ) j mmj m A Bε×= ∑AB A

Then,

ijk j m m krs r si jk m rs ijk j m krs m r s jk mrs j m ijk rsk m r s jm rs k

j m ir js is jr m r s jm rs jm m i j m i j jm jm j m i j m jm j j mi m

Therefore,

But,

e(A · e) is the component of A in the e direction, while e × (A × e) is the component of A perpendicular to e.

1-25.

e e

The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by ( ) cos cos , cos sin , sin sin , cos , 0

sin cos , sin sin , cosr θφ

sinrφθφθθ=+e e (4) Now, let any position vector be x. Then,

( )sinsinrr r r r r r r φθφθ

=+ + xe e e e e

sin cos sin sin

2 sin 2 cos sin sin

2s in cos r r r r r r r r r r r r r r r xe e e e e

(7) or,

MATRICES, VECTORS, AND VECTOR CALCULUS 19

1 sin sin cos

sin r d r r r r rd t d r rd t θφ

(8)

cos sinrkrk θθ θ θθ θ

Now, the velocity vector in polar coordinates is [see Eq. (1.97)]

2 cos vr r k

Using (1), we find

Differentiating (5) and using the expression for r, we obtain sin sin

The acceleration vector is [see Eq. (1.98)]

cos sin 1 cos sin cos 1 cos

21 cos

1c os 2cos 1

21 cos

3 1c os r r k k k

=− (9) or,

In a similar way, we find

From (10) and (1), we have

or,

we have

d dt dt rv r r r v r v r r a r v v r v v v v r ra r

arvvrra (1) Thus,

MATRICES, VECTORS, AND VECTOR CALCULUS 21

where

Therefore, i i i x x x

so that

or,

1-29. Let describe the surface S and 29r=121xyz++= describe the surface S. The angle 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

Sr x y z grad grad grad e e e e

Sx y z grad grad

Therefore, cos grad grad grad grad

or,

from which

1 i i i

i ixx x ix φψφ

Thus, () φψφψψφ=+gradgradgrad

1-31. a) n n i j ij i

i j ij i j ij n i i r rx n x xn x xn r grad e e

MATRICES, VECTORS, AND VECTOR CALCULUS 23 i i i ij ij i i j ij i i fr f r r fr fr x fr x fx rd r grad e e e f r fr c) ln ln ln j ij i

ij j i i j j ij ij i i i j j ij i ji j i r rx x x x x x x x x xr r r r

1-32. Note that the integrand is a perfect differential:

Clearly,

1-3. Since

we have r d dt dt

r dt r from which

where C is the integration constant (a vector).

1-34. First, we note that

But the first term on the right-hand side vanishes. Thus,

so that

MATRICES, VECTORS, AND VECTOR CALCULUS 25 1-35.

z y

We compute the volume of the intersection of the two cylinders by dividing the intersection volume into two parts. Part of the common volume is that of one of the cylinders, for example, the one along the y axis, between y = –a and y = a:

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 2yax=−2, z = a to

8s in 32 a x a x a

Vd x dy dz dx ax ax a xa x ax

26 CHAPTER 1 1-36.

d z x y c = x + y

The form of the integral suggests the use of the divergence theorem.

Since ∇⋅, we only need to evaluate the total volume. Our cylinder has radius c and height d, and so the answer is 1=A

1-37. z y xR

To do the integral directly, note that A, on the surface, and that . 3rR=e5 rdd a=ae

To use the divergence theorem, we need to calculate ∇⋅A. This is best done in spherical coordinates, where A. Using Appendix F, we see that 3rr=e

Therefore,

Alternatively, one may simply set dv in this case. 24rdrπ=

MATRICES, VECTORS, AND VECTOR CALCULUS 27 1-38.

C x + y = 1 z = 1 – x – y

By Stoke’s theorem, we have

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 dθθ=se and sincosθθ=+Aik on the curve. Since sinθθ⋅=−ei and 0θ⋅=ek, we have

1-39.

1-40. a) At the top of the hill, z is maximum;

so x = –2 ; y = 3, and the hill’s height is max[z]= 72 m. Actually, this is the max value of z, because the given equation of z implies that, for each given value of x (or y), z describes an upside down parabola in term of y ( or x) variable.

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

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 then dx dy dx dz dxα

The hill is steepest when tanβ is minimum, and this happens when α = –45 degrees with respect to W-E axis. (note that α = 135 does not give a physical answer).

then if only a = 1 or a = 0. 0⋅=AB

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