The binomial series 65

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 ( 2 πrh),

i.e. 98% of the original surface area

Hence the curved surface area is reduced by

approximately 2%.

Problem 18. 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% andlis

reduced by 2.5%.

New values ofbandlare( 1 + 0. 035 )band( 1 − 0. 025 )l
respectively.

New second moment of area

=

1

12

[( 1 + 0. 035 )b][( 1 − 0. 025 )l]^3

=

bl^3

12

( 1 + 0. 035 )( 1 − 0. 025 )^3

≈

bl^3

12

( 1 + 0. 035 )( 1 − 0. 075 ),neglecting

powers of small terms

≈

bl^3

12

( 1 + 0. 035 − 0. 075 ),neglecting

products of small terms

≈

bl^3

12

( 1 − 0. 040 )or( 0. 96 )

bl^3

12

,i.e. 96%

of the original second moment of area

Hence the second moment of area is reduced by

approximately 4%.

Problem 19. The resonant frequency of a

vibrating shaft is given by:f=

1

2 π

√

k

I

,wherekis

the stiffness andIis the inertia of the shaft. Use the

binomial theorem to determine the approximate

percentage error in determining the frequency using

the measured values ofkandIwhen the measured

value ofkis 4% too large and the measured value

ofIis 2% too small.

Letf,kandIbe the true values of frequency, stiffness

and inertia respectively. Since the measured value of

stiffness,k 1 , is 4% too large, then

k 1 =

104

100

k=( 1 + 0. 04 )k

Themeasured valueof inertia,I 1 ,is2%toosmall,hence

I 1 =

98

100

I=( 1 − 0. 02 )I

The measured value of frequency,

f 1 =

1

2 π

√

k 1

I 1

=

1

2 π

k

1

2

1 I

−^12

1

=

1

2 π

[( 1 + 0. 04 )k]

1

(^2) [( 1 − 0. 02 )I]−
1
2

1
2 π
( 1 + 0. 04 )
1
(^2) k
1
(^2) ( 1 − 0. 02 )−
1
(^2) I−
1
2

1
2 π
k
1
(^2) I−
1
(^2) ( 1 + 0. 04 )
1
(^2) ( 1 − 0. 02 )−
1
2
i.e. f 1 = f( 1 + 0. 04 )
1
(^2) ( 1 − 0. 02 )−
1
2
≈f
[
1 +
(
1
2
)
( 0. 04 )
][
1 +
(
−
1
2
)
(− 0. 02 )
]
≈f( 1 + 0. 02 )( 1 + 0. 01 )
Neglecting the products of small terms,
f 1 ≈( 1 + 0. 02 + 0. 01 )f≈ 1. 03 f
Thus the percentage error infbased on the measured
values ofkandIis approximately [( 1. 03 )( 100 )−100],
i.e.3% too large.