Some applications of differentiation 313
Problem 24. Given y = 4x
2
− x, determine the
approximate change in y if x changes from 1 to
1.02.
Since y = 4x 2 − x, then
d y
dx
= 8x − 1
Approximate change in y,
δy ≈
d y
dx
· δx ≈ (8x − 1)δx
When x = 1 and δx = 0.02, δy ≈ [8(1) − 1](0.02)
≈ 0.14
[Obviously, in this case, the exact value of dy
may be obtained by evaluating y when x = 1.02, i.e.
y = 4(1.02) 2 − 1.02 = 3.1416 and then subtracting from
it the value of y when x = 1, i.e. y = 4(1) 2 − 1 = 3, giving
δy = 3.1416 −3 =0.1416.
Using δy =
d y
dx
· δx above gave 0.14, which shows that
the formula gives the approximate change in y for a
small change in x.]
Problem 25. The time of swing T of a pendulum
is given by T = k
√
l, where k is a constant.
Determine the percentage change in the time of
swing if the length of the pendulum l changes from
32.1 cm to 32.0 cm.
If T = k
√
l = kl
1
2 , then
dT
dl
= k
1
2
l
−1
2
=
k
2
√
l
Approximate change in T ,
δt ≈
dT
dl
δl ≈
k
2
√
l
δl
≈
k
2
√
l
(−0.1)
(negative since l decreases)
Percentage error
=
approximate change in T
original value of T
100%
=
k
2
√
l
(−0.1)
k
√
l
× 100%
=
−0.1
2l
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
d A
dr
= 2πr
Approximate change in area,
δ A ≈
d A
dr
· δr ≈ (2πr)δr
When r = 10 cm and δr = 0.02,
δ A = (2π10)(0.02) ≈ 0.4π cm 2
i.e. the possible error in calculating the template area
is approximately 1.257 cm 2 .
Percentage error ≈
0.4π
π(10) 2
100%
= 0.40%
Now try the following exercise
Exercise 125 Further problems on small
changes
1. Determine the change in y if x changes from
2.50 to 2.51 when
(a) y = 2x − x 2 (b) y =
5
x
[(a) −0.03 (b) −0.008]