312 DIFFERENTIAL CALCULUS

[Obviously, in this case, the exact value of dy

may be obtained by evaluatingywhenx= 1 .02,

i.e. y=4(1.02)^2 − 1. 02 = 3 .1416 and then sub-

tracting from it the value ofywhenx=1, i.e.

y=4(1)^2 − 1 =3, givingδy= 3. 1416 − 3 =0.1416.

Usingδy=

dy

dx

·δxabove gave 0.14, which shows

that the formula gives the approximate change iny

for a small change inx.]

Problem 25. The time of swingTof a pendu-

lum is given byT=k

√

l, wherekis a constant.

Determine the percentage change in the time of

swing if the length of the pendulumlchanges

from 32.1 cm to 32.0 cm.

IfT=k

√

l=kl

1

(^2) , then
dT
dl
=k
(
1
2
l
− 1
2
)

k
2
√
l
Approximate change inT,
δt≈
dT
dl
δl≈
(
k
2
√
l
)
δl
≈
(
k
2
√
l
)
(− 0 .1)
(negative sinceldecreases)
Percentage error

(
approximate change inT
original value ofT
)
100%

(
k
2
√
l
)
(− 0 .1)
k
√
l
×100%

(
− 0. 1
2 l
)
100%=
(
− 0. 1
2(32.1)
)
100%
=− 0 .156%
Hence the change in the time of swing is a decrease
of 0.156%.
Problem 26. A circular template has a radius
of 10 cm (±0.02). Determine the possible error
in calculating the area of the template. Find also
the percentage error.
Area of circular template,A=πr^2 , hence
dA
dr
= 2 πr
Approximate change in area,
δA≈
dA
dr
·δr≈(2πr)δr
Whenr=10 cm andδr= 0 .02,
δA=(2π10)(0.02)≈ 0. 4 πcm^2
i.e.the possible error in calculating the template
area is approximately 1.257 cm^2.
Percentage error≈
(
0. 4 π
π(10)^2
)
100%
=0.40%
Now try the following exercise.
Exercise 127 Further problems on small
changes

1. Determine the change inyifxchanges from
   2.50 to 2.51 when

(a)y= 2 x−x^2 (b)y=

5

x

[(a)−0.03 (b)−0.008]

2. The pressurepand volumevof a mass of
   gas are related by the equationpv=50. If the
   pressure increases from 25.0 to 25.4, deter-
   mine the approximate change in the volume
   of the gas. Find also the percentage change
   in the volume of the gas. [−0.032,−1.6%]


3. Determine the approximate increase in (a) the
   volume, and (b) the surface area of a cube
   of sidexcm ifxincreases from 20.0 cm to
   20.05 cm. [(a) 60 cm^3 (b) 12 cm^2 ]


4. The radius of a sphere decreases from 6.0 cm
   to 5.96 cm. Determine the approximate
   change in (a) the surface area, and (b) the
   volume. [(a)−6.03 cm^2 (b)−18.10 cm^3 ]


5. The rate of flow of a liquid through a
   tube is given by Poiseuilles’s equation as:

Q=

pπr^4

8 ηL

whereQis the rate of flow,p