Homework 1 Solutions - University of Massachusetts Amherst 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2

Homework 1 Solutions - University of Massachusetts Amherst 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2
Homework 1 Solutions - University of Massachusetts Amherst 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2
Homework 1 Solutions - University of Massachusetts Amherst 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2
Homework 1 Solutions - University of Massachusetts Amherst 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2
Homework 1 Solutions - University of Massachusetts Amherst 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2
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Transcript of Homework 1 Solutions - University of Massachusetts Amherst 26 CHAPTER 1 1-36. d z x y c2 = x 2 + y 2

  • 1

    Homework 1 Solutions

    1.

    2 CHAPTER 1

    1-2.

    a)

    x1 A

    B

    C

    D

    α

    β γ

    O

    E

    x2

    x3

    From this diagram, we have

    cosOE OAα =

    cosOE OBβ = (1)

    cosOE ODγ =

    Taking the square of each equation in (1) and adding, we find

    2 2 2 2cos cos cos OA OB ODα β γ+ + = + +

    2 2 2 OE (2)

    But

    2 2

    OA OB OC+ = 2 (3)

    and

    2 2

    OC OD OE+ = 2

    (4)

    Therefore,

    2 2 2

    OA OB OD OE+ + = 2 (5)

    Thus,

    2 2 2cos cos cos 1α β γ+ + = (6)

    b)

    x3

    A A′ x1

    x2O

    E D

    C

    C′

    B′ E′D′

    First, we have the following trigonometric relation:

    2 2

    2 cosOE OE OE OE EEθ 2′ ′+ − = ′ (7)

  • 2MATRICES, VECTORS, AND VECTOR CALCULUS 3

    But,

    2 2 2 2

    2 2

    2

    cos cos cos cos

    cos cos

    EE OB OB OA OA OD OD

    OE OE OE OE

    OE OE

    β β α

    γ γ

    ′ ′ ′ ′= − + − + −

    ′ ′= − + −′ ′

    ′+ −′

    α

    (8)

    or,

    2 2 22 2 2 2 2 2

    2 2

    cos cos cos cos cos cos

    2 cos cos cos cos cos cos

    2 cos cos cos cos cos cos

    EE OE OE

    OE OE

    OE OE OE OE

    α β γ α β

    α α β β γ γ

    γ

    α α β β γ

    ′ ′= + + + + +′ ′ ′

    ′− + +′ ′ ′

    ′= + − + +′ ′ ′ γ ′ (9)

    Comparing (9) with (7), we find

    cos cos cos cos cos cos cosθ α α β β γ γ= + +′ ′ ′ (10)

    1-3.

    x1 e3′

    x2

    x3

    O

    e1 e2

    e3 A e2′

    e1′ e2

    e1

    e3

    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

    1x′ 2x′ 3x′ 1′e , 2′e , e . The effect of the rotation is e e , , e . Therefore, the transformation matrix is written as:

    3′

    1 3′→ 2 1′e e→ 3 → 2′e

    ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

    1 1 1 2 1 3

    2 1 2 2 2 3

    3 1 3 2 3 3

    cos , cos , cos , 0 1 0 cos , cos , cos , 0 0 1

    1 0 0cos , cos , cos ,

    ′ ′ ′

    λ ′ ′ ′= =

    ′ ′ ′

    e e e e e e

    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,

    ij ik kj k

    C A= B∑ (1)

    ( )t ji jk ki ki jkij k k

    C C A B B A= = =∑ ∑

    ..

  • 3

    2.

  • 4

    3. 26 CHAPTER 1

    1-36.

    d

    z

    x

    y

    c2 = x2 + y2 The form of the integral suggests the use of the divergence theorem.

    (1) S V

    d⋅ = ∇ ⋅∫ ∫A a A dv

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

    1=A

    (2) 2 V

    dv c dπ=∫

    1-37.

    z

    y

    x

    R

    To do the integral directly, note that A , on the surface, and that . 3 rR= e

    5

    rd da=a e

    3 3 24 4 S S

    d R da R R Rπ π⋅ = = × =∫ ∫A a (1)

    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 3 rr= e

    ( )221 5rrr r 2r

    ∂ ∇ ⋅ = =

    ∂ A A (2)

    Therefore,

    ( )2 2 2 0 0 0

    sin 5 4 R

    V dv d d r r dr R

    π π 5θ θ φ π∇ ⋅ = =∫ ∫A∫ ∫ (3)

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

    4. ..

    24 CHAPTER 1

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

    ( ) (2 2 d da b a b dt dt

    ⋅ + ⋅ = ⋅ + ⋅r r r r r r r r) (1)

    Clearly,

    ( ) 2 22 2 cona b dt ar br⋅ + ⋅ = + +∫ r r r r st. (2)

    1-33. Since

    2 d r r dt r r r 2

    r r

    − = = −

    r r r r r (1)

    we have

    2 r d

    dt dt r r dt r − =∫ ∫

    r r r (2)

    from which

    2 r

    dt r r r − = +∫

    r r r C (3)

    where C is the integration constant (a vector).

    1-34. First, we note that

    ( )ddt × = × + ×A A A A A A (1) But the first term on the right-hand side vanishes. Thus,

    ( ) ( )ddt dtdt× = ×∫ ∫A A A A (2) so that

    ( )dt× = × +∫ A A A A C (3) where C is a constant vector.

  • 1

    5.

    24 CHAPTER 1

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

    ( ) (2 2 d da b a b dt dt

    ⋅ + ⋅ = ⋅ + ⋅r r r r r r r r) (1)

    Clearly,

    ( ) 2 22 2 cona b dt ar br⋅ + ⋅ = + +∫ r r r r st. (2)

    1-33. Since

    2 d r r dt r r r 2

    r r

    − = = −

    r r r r r (1)

    we have

    2 r d

    dt dt r r dt r − =∫ ∫

    r r r (2)

    from which

    2 r

    dt r r r − = +∫

    r r r C (3)

    where C is the integration constant (a vector).

    1-34. First, we note that

    ( )ddt × = × + ×A A A A A A (1) But the first term on the right-hand side vanishes. Thus,

    ( ) ( )ddt dtdt× = ×∫ ∫A A A A (2) so that

    ( )dt× = × +∫ A A A A C (3) where C is a constant vector.