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数学方法物理工程解题指南

Mathematical

Methods

Physics

Engineering

Riley

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"Student Solutions Manual for Mathematical Methods for Physics and Engineering, Third Edition by K. F. Riley and M. P. Hobson" 这本《物理与工程数学方法学生解题手册》是针对K. F. Riley和M. P. Hobson编写的第三版《物理与工程数学方法》教科书的配套指南。这本教科书在物理学和工程学的本科教育中享有很高的声誉，它全面地教授了所有在自然科学领域本科课程中必需的数学知识。书中的讲解清晰，包含大量实例，以及超过800个练习题，旨在帮助学生巩固理解。新版本的特色在于增加了独立的章节，专门讨论物理学中的"特殊函数"，扩大了复变函数在实际应用中的范围，并引入了量子算符的基础知识。这样的更新使教材内容更加丰富，更加符合现代科学教育的需求。解题手册与主教材同步，提供了超过400个练习题的完整解答，这些解答对应的是主教材中提供提示和答案的奇数编号练习题。偶数编号的练习题则没有提示、答案或完整的解题过程，旨在作为无辅助的家庭作业问题，以鼓励学生的独立思考和解决问题的能力。教师可以在受密码保护的网站www.cambridge.org/978052167上获取这些偶数编号练习题的完整解决方案，以供教学参考。此解题手册的价值在于它为学生提供了自我检查和深入学习的工具，有助于他们更好地掌握教科书中涵盖的数学概念，并将其应用于物理学和工程学的实际场景。通过解决这些习题，学生可以提升自己的数学技能，为他们在未来的科学研究和工程实践中奠定坚实的基础。

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PRELIMINARY ALGEBRA

we have

sin

+sin

=2sin

cos

√

cos

√

3+1

√

(b) Using, successively, the identities

sin(A + B)=sinA cos B +cosA sin B,

sin(π −θ)=sinθ

and cos(

π −θ)=sinθ,

we obtain

sin





=sin

cos

+cos

sin

7π

√

sin

5π

√

3+1

√

cos

√

3+1

√

1.9 Find the real solutions of

(a) 3 sin θ −4cosθ =2,

(b) 4 sin θ +3cosθ =6,

We use the result that if

a sin θ + b cos θ = k

then

θ =sin

−1





−φ,

where

= a

+ b

and φ =tan

−1

PRELIMINARY ALGEBRA

Recalling that the inverse sine yields two values and that the individual signs of

a and b have to be taken into account, we have

(a) k =2,K =

√

=5,φ =tan

−1

(−4/3) and so

θ =sin

−1

−tan

−1

−4

=1.339 or − 2.626.

(b) k =6,K =

√

=5.Sincek>Kthere is no solution for a real angle θ.

√

= 13, φ =tan

−1

(−5/12) and so

θ =sin

−1

−6

−tan

−1

−5

= −0.0849 or −2.267.

1.11 Find all the solutions of

sin θ + sin 4θ = sin 2θ + sin 3θ

that lie in the range −π<θ≤ π. What is the multiplicity of the solution θ =0?

Using

sin(A + B)=sinA cos B +cosA sin B,

and cos A − cos B = −2sin



A + B



sin



A − B



and recalling that cos(−φ)=cos(φ), the equation can be written successively as

2sin

5θ

cos



−

3θ



=2sin

5θ

cos



−



sin

5θ



cos

3θ

−cos



=0,

−2sin

5θ

sin θ sin

=0.

The ﬁrst factor gives solutions for θ of −4π/5, −2π/5, 0, 2π/5and4π/5. The

second factor gives rise to solutions 0 and π, whilst the only value making the

third factor zero is θ =0.Thesolutionθ = 0 appears in each of the above sets

and so has multiplicity 3.

PRELIMINARY ALGEBRA

Coordinate geometry

1.13 Determine the forms of the conic sections described by the following equa-

tions:

(a) x

+ y

+6x +8y =0;

(b) 9x

−4y

−54x − 16y +29=0;

+2y

+5xy − 4x + y − 6=0;

(d) x

+ y

+2xy − 8x +8y =0.

(a) x

+ y

+6x +8y =0. The coeﬃcients of x

and y

are equal and there is

no xy term; it follows that this must represent a circle. Rewriting the equation in

standard circle form by ‘completing the squares’ in the terms that involve x and

y, each variable treated separately, we obtain

(x +3)

+(y +4)

−(3

)=0.

The equation is therefore that of a circle of radius

√

=5centredon

(−3, −4).

(b) 9x

− 4y

− 54x − 16y + 29 = 0. This equation contains no xy term and so

the centre of the curve will be at ( 54/(2 × 9), 16/[2 × (−4)] ) = (3, −2), and in

standardised form the equation is

9(x −3)

−4(y +2)

+29−81 + 16 = 0,

(x − 3)

−

(y +2)

=1.

The minus sign between the terms on the LHS implies that this conic section is a

hyperbola with asymptotes (the form for large x and y and obtained by ignoring

the constant on the RHS) given by 3(x − 3) = ±2(y + 2), i.e. lines of slope ±

passing through its ‘centre’ at (3, −2).

+2y

+5xy − 4x + y − 6=0. As an xy term is present the equation

cannot represent an ellipse or hyperbola in standard form. Whether it represents

two straight lines can be most easily investigated by taking the lines in the form

x+b

y+1 = 0, (i =1, 2) and comparing the product (a

x+b

y+1)(a

x+b

y+1)

with −

(2x

+2y

+5xy − 4x + y −6). The comparison produces ﬁve equations

which the four constants a

, b

,(i =1, 2) must satisfy:

−6

+ a

−4

−6

+ b

−6

and

+ b

−6

PRELIMINARY ALGEBRA

Combining the ﬁrst and third equations gives 3a

−2a

−1 = 0 leading to a

and

having the values 1 and −

, in either order. Similarly, combining the second

and fourth equations gives 6b

+ b

−2 = 0 leading to b

and b

having the values

and −

, again in either order.

Either of the two combinations (a

= −

, b

= −

, a

=1,b

)and(a

=1,

, a

= −

, b

= −

) also satisﬁes the ﬁfth equation [note that the two

alternative pairings do not do so]. That a consistent set can be found shows that

the equation does indeed represent a pair of straight lines, x +2y − 3=0and

2x + y +2=0.

(d) x

+ y

+2xy −8x +8y =0. We note that the ﬁrst three terms can be written

as a perfect square and so the equation can be rewritten as

(x + y)

=8(x − y).

The two lines given by x + y =0andx − y = 0 are orthogonal and so the

equation is of the form u

=4av,which,forCartesiancoordinatesu, v,represents

a parabola passing through the origin, symmetric about the v-axis (u =0)and

deﬁned for v ≥ 0. Thus the original equation is that of a parabola, symmetric

about the line x + y = 0, passing through the origin and deﬁned in the region

x ≥ y.

Partial fractions

1.15 Resolve

(a)

2x +1

+3x −10

, (b)

−3x

into partial fractions using each of the following three methods:

(i) Expressing the supposed expansion in a form in which all terms have the

same denominator and then equating coeﬃcients of the various powers of x.

(ii) Substituting speciﬁc numerical values for x and solving the resulting simul-

taneous equations.

(iii) Evaluation of the fraction at each of the roots of its denominator, imagining

a factored denominator with the factor corresponding to the root omitted –

often known as the ‘cover-up’ method.

Verify that the decomposition obtained is independent of the method used.

(a) As the denominator factorises as (x +5)(x −2), the partial fraction expansion

must have the form

2x +1

+3x −10

x +5

x − 2

PRELIMINARY ALGEBRA

(i)

x +5

x −2

x(A + B)+(5B − 2A)

(x +5)(x − 2)

Solving A + B =2and−2A +5B = 1 gives A =

and B =

(ii) Setting x equal to 0 and 1, say, gives the pair of equations

−10

−2

;

−6

−1

−1=2A −5B; −3=A − 6B,

with solution A =

and B =

(iii)

A =

2(−5) + 1

−5 − 2

; B =

2(2) + 1

2+5

All three methods give the same decomposition.

(b) Here the factorisation of the denominator is simply x(x−3) or, more formally,

(x − 0)(x − 3), and the expansion takes the form

−3x

x −3

(i)

x −3

x(A + B) −3A

(x − 0)(x − 3)

Solving A + B =0and−3A = 4 gives A = −

and B =

(ii) Setting x equal to 1 and 2, say, gives the pair of equations

−2

;

−2

−1

−4=2A −B; −4=A − 2B,

with solution A = −

and B =

(iii)

A =

0 − 3

= −

; B =

3 − 0

Again, all three methods give the same decomposition.

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