Some applications of differentiation 313

Problem 24. Giveny= 4 x^2 −x, determine the

approximate change inyifxchanges from 1 to

1.02.

Sincey= 4 x^2 −x,then

dy

dx

= 8 x− 1

Approximate change iny,

δy≈

dy

dx

·δx≈( 8 x− 1 )δx

Whenx=1andδx= 0. 02 ,δy≈[8( 1 )−1]( 0. 02 )

≈ 0. 14

[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 subtracting from

itthevalueofywhenx=1,i.e.y= 4 ( 1 )^2 − 1 =3,giving

δy= 3. 1416 − 3 =0.1416.

Usingδy=

dy

dx

·δxabovegave0.14,whichshowsthat

the formula gives the approximate change inyfor a

small change inx.]

Problem 25. The time of swingTof a pendulum

is given byT=k

√

l,wherekis a constant.

Determine the percentage change in the time of

swing if the length of the pendulumlchanges from

32.1cm to 32.0cm.

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
10cm (±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≈( 2 πr)δr
Whenr=10cm andδr= 0 .02,
δA=( 2 π 10 )( 0. 02 )≈ 0. 4 πcm^2
i.e.thepossibleerrorin calculating thetemplatearea
is approximately 1.257cm^2.
Percentage error≈
(
0. 4 π
π( 10 )^2
)
100%
=0.40%
Now try the following exercise
Exercise 125 Further problemson 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]