Kunci Jawaban Bab 10 n 11 Print

7/25/2019 Kunci Jawaban Bab 10 n 11 Print

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Chapter 10

4.4 I = 2

15

π −1 0−1 π 0

0 0 π

Principal moments:

2

15(π − 1, π , π + 1); principal

axes along the vectors: (1, 1, 0), (0, 0, 1), (1,−1, 0).

4.5 I =

4 0 00 4 −2

0 −2 4

Principal moments: (2, 4, 6); principal axes along the

vectors: (0, 1, 1), (1, 0, 0), (0, 1,−1).

4.6 I =

9 0 −3

0 6 0−3 0 9

Principal moments: (6, 6, 12); principal axes along the

vectors: (1, 0,−1) and any two orthogonal vectors in the plane z = x, say(0, 1, 0) and (1, 0, 1).

4.7 I = 1

120

4 −1 −1−1 4 −1−1 −1 4

Principal moments:

1

60 ,

1

24 ,

1

24

; principal axes

along the vectors: (1, 1, 1) and any two orthogonal vectors in the planex + y + z = 0, say (1,

−1, 0) and (1, 1,

−2).

5.5 1 if j k = m n (6 cases); −1 if j k = n m (6 cases); 0 otherwise5.6 (a) 3 (b) 0 (c) 2 (d) −2 (e) −1 (f) −1

5.7 (a) δ kqδ ip − δ kpδ iq (b) δ apδ bq − δ aqδ bp

6.9 to 6.14 r, v, F, E are vectors; ω, τ , L, B are pseudovectors; T is a scalar.6.15 (a) vector (b) pseudovector (c) vector

6.16 vector (if V is a vector); pseudovector (if V is a pseudovector)

8.1 hr = 1, hθ = r, hφ = r sin θds = erdr + eθrdθ + eφr sin θdφdV = r2 sin θdrdθdφar = i sin θ cos φ + j sin θ sin φ + k cos θ = er

aθ = ir cos θ cos φ + jr cos θ sin φ − kr sin θ = reθaφ = −ir sin θ sin φ + jr sin θ cos φ = r sin θeφ

8.2 d2s/dt2 = er(r − r θ2) + eθ(rθ + 2r θ) + ez z

8.3 ds/dt = er r + eθr θ + eφr sin θ φ

d2s/dt2 = er(r − r θ2 − r sin2 θ φ2)

+eθ(rθ + 2r θ−r sin θ cos θ φ2)

+eφ(r sin θ φ + 2r cos θ θ φ + 2 sin θ r φ)

8.4 V = −reθ + k

49



Chapter 10 50

8.5 V = er cos θ − eθ sin θ − eφr sin θ

8.6 hu = hv = (u2 + v2)1/2, hz = 1ds = (u2 + v2)1/2(eudu + evdv) + ezdzdV = (u2 + v2) du dv dzau = iu + jv = (u2 + v2)1/2euav =

−iv + ju = (u2 + v2)1/2ev

az = k = ez

8.7 hu = hv = a(sinh2 u + sin2 v)1/2, hz = 1ds = a(sinh2 u + sin2 v)1/2(eudu + evdv) + ezdz

dV = a2(sinh2 u + sin2 v) du dv dzau = ia sinh u cos v + ja cosh u sin v = hueuav = −ia cosh u sin v + ja sinh u cos v = hvevaz = k = ez

8.8 hu = hv = (u2 + v2)1/2, hφ = uvds = (u2 + v2)1/2(eudu + evdv) + uveφdφdV = uv (u2 + v2) du dv dφau = iv cos φ + jv sin φ + ku = hueuav = iu cos φ + ju sin φ

−kv = hvev

aφ = −iuv sin φ + juv cos φ = hφeφ

8.9 hu = hv = a(cosh u + cos v)−1

ds = a(cosh u + cos v)−1(eudu + evdv)dA = a2(cosh u + cos v)−2 du dvau = (h2

u/a)[i(1 + cos v cosh u) − j sin v sinh u] = hueuav = (h2

v/a)[i sinh u sin v + j(1 + cos v cosh u)] = hvev

8.11 deu/dt = (u2 + v2)−1(uv − vu)evdev/dt = (u2 + v2)−1(v u − uv)euds/dt = (u2 + v2)1/2(eu u + ev v) + ez z

d2s/dt2 = eu(u2 + v2)−1/2[(u2 + v2)u + u(u2 − v2) + 2v uv]+ev(u2 + v2)−1/2[(u2 + v2)v + v(v2 − u2) + 2uuv] + ez z

8.12 deu/dt = (sinh2 u + sin2 v)−1(v sinh u cosh u

− u sin v cos v)ev

dev/dt = (sinh2

u + sin2

v)−1

(u sin v cos v − v sinh u cosh u)euds/dt = a(sinh2 u + sin2 v)1/2(eu u + ev v) + ez z

d2s/dt2 = eua(sinh2 u + sin2 v)−1/2

×[(sinh2 u + sin2 v)u + (u2 − v2) sinh u cosh u + 2uv sin v cos v]

+eva(sinh2 u + sin2 v)−1/2[(sinh2 u + sin2 v)v+(v2 − u2)sin v cos v + 2uv sinh u cosh u] + ez z

8.13 deu/dt = (u2 + v2)−1(uv − v u)ev + (u2 + v2)−1/2v φeφdev/dt = (u2 + v2)−1(v u − uv)eu + (u2 + v2)−1/2u φeφdeφ/dt = −(u2 + v2)−1/2(veu + uev) φ

ds/dt = (u2 + v2)1/2(eu u + ev v) + eφuv φ

d2s/dt2 = eu(u2 + v2)−1/2[(u2 + v2)u + u(u2 − v2) + 2v uv − uv2 φ2]

+ev(u2 + v2)−1/2[(u2 + v2)v + v(v2

− u2) + 2uuv

−u2v φ2]

+eφ(uvφ + 2vu φ + 2uv φ)8.14 deu/dt = −(cosh u + cos v)−1(u sin v + v sinh u)ev

dev/dt = (cosh u + cos v)−1(u sin v + v sinh u)euds/dt = a(cosh u + cos v)−1(eu u + ev v)d2s/dt2 = eua(cosh u + cos v)−2[(cosh u + cos v)u + (v2 − u2) sinh u + 2uv sin v]

+eva(cosh u + cos v)−2[(cosh u + cos v)v + (v2 − u2)sin v − 2uv sinh u]



Chapter 10 52

9.18 −r−1eθ, r−1er, 3

9.19 2eφ, er cos θ − eθ sin θ, 3

9.20 r−1, r−3, 0

9.21 2r−1, 6, 2r−4, −k2eikr cos θ

11.4 Vector

11.5 ds2 = du2 + h2vdv2, hu = 1, hv = u(2v − v2)−1/2,

dA = u(2v − v2)−1/2dudv, ds = eu du + hvevdv,eu = i(1− v) + j(2v − v2)1/2, ev = −i(2v − v2)1/2 + j(1 − v)au = eu = au, av = hvev, av = ev/hv

11.6 m

u − uv2

v(2 − v)

= −∂V

∂u = F u

m

uv + 2uv

[v(2 − v)]1/2 +

uv2(v − 1)

[v(2 − v)]3/2

= −u−1[v(2− v)]1/2

∂V

∂v = F v

11.7 ∇U = eu∂U/∂u + evu−1

v(2 − v) ∂U/∂v

∇ ·V = u−1∂ (uV u)/∂u + u−1

v(2− v) ∂V v/∂v

∇2U =

1

u

∂

∂uu

∂U

∂u+

1

u2 v(2

−v)

∂

∂V v(2

−v)

∂U

∂V

11.8 u−1, u−1k, 0



Chapter 11

3.2 3/2 3.3 9/10 3.4 25/143.5 32/35 3.6 72 3.7 83.8 Γ(5/3) 3.9 Γ(5/4) 3.10 Γ(3/5)3.11 1 3.12 Γ(2/3)/3 3.13 3−4Γ(4) = 2/273.14 −Γ(4/3) 3.15 Γ(2/3)/4 3.17 Γ( p)

7.1 12

B(5/2, 1/2) = 3π/16 7.2 12

B(5/4, 3/4) = π√ 2/8

7.3 1

3B(1/3, 1/2) 7.4 1

2B(3/2, 5/2) = π/32

7.5 B(3, 3) = 1/30 7.6 1

3B(2/3, 4/3) = 2π

√ 3/27

7.7 1

2B(1/4, 1/2) 7.8 4

√ 2B(3, 1/2) = 64

√ 2/15

7.10 4

3B(1/3, 4/3) 7.11 2B(2/3, 4/3)/B(1/3, 4/3)

7.12 (8π/3)B(5/3, 1/3) = 32π2√

3/277.13 I y/M = 8B(4/3, 4/3)/B(5/3, 1/3)

8.1 B(1/2, 1/4)

2l/g = 7.4163

l/g

Compare 2π

l/g.

8.2 1

4

35/11 B(1/2, 1/4) = 2.34 sec

8.3 t = π a/g

10.2 Γ( p, x) ∼ x p−1e−x[1 + ( p − 1)x−1 + ( p − 1)( p − 2)x−2 + · · · ]10.3 erfc (x) = Γ

1/2, x2

/√

π10.5 (a) E 1(x) = Γ(0, x)

(b) Γ(0, x) ∼ x−1e−x[1− x−1 + 2x−2 − 3! x−3 + · · · ]10.6 (a) Ei(ln x) (b) Ei(x) (c) −Ei(ln x)

11.4 1/√

π 11.5 1 11.10 e−1

12.1 K = F (π/2, k) = (π/2)

1 +1

2

2k2 +

1·32·4

2k4 + · · ·

E = E (π/2, k) = (π/2)

1 −1

2

2k2 − 1

2·4

2 · 3k4 − 1·32·4·6

2 · 5k6 · · ·

Caution : For the following answers, see the text warning about elliptic integralnotation just after equations (12.3) and in Example 1.

12.4 K (1/2) ∼= 1.686 12.5 E (1/3) ∼= 1.52612.6 1

3F π3

, 13

∼= 0.355 12.7 5E 5π4

, 15

∼= 19.46

12.8 7E π3

, 27

∼= 7.242 12.9 F π6

,√ 3

2

∼= 0.542

12.10 1

2F π4

, 12

∼= 0.402 12.11 F 3π8

, 3√ 10

+ K

3√ 10

∼= 4.097

12.12 10E π6

, 110

∼= 5.234 12.13 3E π6

, 23

+ 3E

arc sin 3

4, 23

∼= 3.96

53

Kunci Jawaban Bab 10 n 11 Print

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