Higher Engineering Mathematics

64 NUMBER AND ALGEBRA

5.

1

√

1 + 3 x

⎡

⎢

⎢

⎣

(

1 −

3

2

x+

27

8

x^2 −

135

16

x^3 +···

)

|x|<

1

3

⎤

⎥

⎥

⎦

6. Expand (2+ 3 x)−^6 to three terms. For what
   values ofxis the expansion valid?
   ⎡

⎢

⎢

⎣

1

64

(

1 − 9 x+

189

4

x^2

)

|x|<

2

3

⎤

⎥

⎥

⎦

7. Whenxis very small show that:

(a)

1

(1−x)^2

√

(1−x)

≈ 1 +

5

2

x

(b)

(1− 2 x)

(1− 3 x)^4

≈ 1 + 10 x

(c)

√

1 + 5 x

√ (^31) − 2 x≈ 1 +
19
6
x

8. Ifxis very small such thatx^2 and higher pow-
   ers may be neglected, determine the power

series for

√

x+ 43

√

8 −x

√ (^5) (1+x) 3
[
4 −
31
15
x
]

9. Express the following as power series in
   ascending powers ofxas far as the term in
   x^2. State in each case the range ofxfor which
   the series is valid.

(a)

√(

1 −x

1 +x

)

(b)

(1+x)^3

√

(1− 3 x)^2

√

(1+x^2 )

⎡

⎢

⎢

⎣

(a) 1−x+

1

2

x^2 ,|x|< 1

(b) 1−x−

7

2

x^2 ,|x|<

1

3

⎤

⎥

⎥

⎦

7.5 Practical problems involving the

binomial theorem

Binomial expansions may be used for numerical

approximations, for calculations with small vari-

ations and in probability theory (see Chapter 57).

Problem 16. The radius of a cylinder is

reduced by 4% and its height is increased by 2%.

Determine the approximate percentage change

in (a) its volume and (b) its curved surface area,

(neglecting the products of small quantities).

Volume of cylinder=πr^2 h.

Letrandhbe the original values of radius and

height.

The new values are 0.96ror (1− 0 .04)rand 1.02h

or (1+ 0 .02)h.

(a) New volume=π[(1− 0 .04)r]^2 [(1+ 0 .02)h]

=πr^2 h(1− 0 .04)^2 (1+ 0 .02)

Now (1− 0 .04)^2 = 1 −2(0.04)+(0.04)^2

=(1− 0 .08),

neglecting powers of small terms.

Hence new volume

≈πr^2 h(1− 0 .08)(1+ 0 .02)

≈πr^2 h(1− 0. 08 + 0 .02), neglecting

products of small terms

≈πr^2 h(1− 0 .06) or 0. 94 πr^2 h, i.e. 94%

of the original volume

Hence the volume is reduced by approxi-

mately 6%.

(b) Curved surface area of cylinder= 2 πrh.

New surface area

= 2 π[(1− 0 .04)r][(1+ 0 .02)h]

= 2 πrh(1− 0 .04)(1+ 0 .02)

≈ 2 πrh(1− 0. 04 + 0 .02), neglecting

products of small terms

≈ 2 πrh(1− 0 .02) or 0.98(2πrh),

i.e. 98% of the original surface area

Hence the curved surface area is reduced by

approximately 2%.

Problem 17. The second moment of area of

a rectangle through its centroid is given by

bl^3

12

.

Determine the approximate change in the second

moment of area ifbis increased by 3.5% and

lis reduced by 2.5%.