Higher Engineering Mathematics, Sixth Edition

Total differential, rates of change and small changes 355

Problem 8. Pressurepand volumeVof a gas are

connected by the equationpV^1.^4 =k. Determine

the approximate percentage error inkwhen the

pressure is increased by 4% and the volume is

decreased by 1.5%.

Using equation (3), the approximate error ink,

δk≈

∂k

∂p

δp+

∂k

∂V

δV

Letp,Vandkrefer to the initial values.

Since k=pV^1.^4 then

∂k

∂p

=V^1.^4

and

∂k

∂V

= 1. 4 pV^0.^4

Since the pressure is increased by 4%, the change in

pressureδp=

4

100

×p= 0. 04 p.

Since the volume is decreased by 1.5%, the change in

volumeδV=

− 1. 5

100

×V=− 0. 015 V.

Hence the approximate error ink,

δk≈(V)^1.^4 ( 0. 04 p)+( 1. 4 pV^0.^4 )(− 0. 015 V)

≈pV^1.^4 [0. 04 − 1. 4 ( 0. 015 )]

≈pV^1.^4 [0.019]≈

1. 9

100

pV^1.^4 ≈

1. 9

100

k

i.e.the approximate error inkis a 1.9% increase.

Problem 9. Modulus of rigidityG=(R^4 θ)/L,

whereRis the radius,θthe angle of twist andLthe

length. Determine the approximate percentage error

inGwhenRis increased by 2%,θis reduced by

5% andLis increased by 4%.

Using δG≈

∂G

∂R

δR+

∂G

∂θ

δθ+

∂G

∂L

δL

Since G=

R^4 θ

L

,

∂G

∂R

=

4 R^3 θ

L

,

∂G

∂θ

=

R^4

L

and

∂G

∂L

=

−R^4 θ

L^2

SinceRis increased by 2%,δR=

2

100

R= 0. 02 R

Similarly,δθ=− 0. 05 θandδL= 0. 04 L

HenceδG≈

(

4 R^3 θ

L

)

( 0. 02 R)+

(

R^4

L

)

(− 0. 05 θ)

+

(

−

R^4 θ

L^2

)

( 0. 04 L)

≈

R^4 θ

L

[0. 08 − 0. 05 − 0 .04]≈− 0. 01

R^4 θ

L

,

i.e. δG≈−

1

100

G

Hence the approximate percentage error inGis a

1% decrease.

Problem 10. The second moment of area of a

rectangle is given byI=(bl^3 )/3. Ifbandlare

measured as 40mm and 90mm respectively and the

measurement errors are−5mm inband+8mmin

l, find the approximate error in the calculated value

ofI.

Using equation (3), the approximate error inI,

δI≈

∂I

∂b

δb+

∂I

∂l

δl

∂I

∂b

=

l^3

3

and

∂I

∂l

=

3 bl^2

3

=bl^2

δb=−5mmandδl=+8mm

HenceδI≈

(

l^3

3

)

(− 5 )+(bl^2 )(+ 8 )

Sinceb=40mm andl=90mm then

δI≈

(

903

3

)

(− 5 )+ 40 ( 90 )^2 ( 8 )

≈− 1215000 + 2592000

≈1377000mm^4 ≈ 137 .7cm^4

Hence the approximate error in the calculated value

ofIis a 137.7cm^4 increase.

Problem 11. The time of oscillationtof a

pendulum is given byt= 2 π

√

l

g

. Determine the
approximate percentage error intwhenlhas an
errorof0.2%toolargeandg0.1% too small.