Total differential, rates of change and small changes 355
Problem 8. Pressure p and volume V of a gas are
connected by the equation pV 1.4 = k. Determine
the approximate percentage error in k when the
pressure is increased by 4% and the volume is
decreased by 1.5%.
Using equation (3), the approximate error in k,
δk ≈
∂k
∂ p
δp +
∂k
∂V
δV
Let p, V and k refer 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
pressure δp =
4
100
× p = 0.04 p.
Since the volume is decreased by 1.5%, the change in
volume δV =
−1.5
100
× V =−0.015V .
Hence the approximate error in k,
δk ≈ (V )
1.4
(0.04 p) + (1.4 pV
0.4
)(−0.015V )
≈ 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 in k is a 1.9% increase.
Problem 9. Modulus of rigidity G = (R 4 θ)/L,
where R is the radius, θ the angle of twist and L the
length. Determine the approximate percentage error
in G when R is increased by 2%, θ is reduced by
5% and L is increased by 4%.
Using δG ≈
∂G
∂ R
δ R +
∂G
∂θ
δθ +
∂G
∂ L
δL
Since
G =
R 4 θ
L
,
∂G
∂ R
=
4R 3 θ
L
,
∂G
∂θ
=
R 4
L
and
∂G
∂ L
=
−R 4 θ
L 2
Since R is increased by 2%, δ R =
2
100
R = 0.02R
Similarly, δθ =−0.05θ and δL =0.04L
Hence δG ≈
4R 3 θ
L
(0.02R) +
R 4
L
(−0.05θ)
+
−
R 4 θ
L 2
(0.04L)
≈
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 in G is a
1% decrease.
Problem 10. The second moment of area of a
rectangle is given by I = (bl 3 )/3. If b and l are
measured as 40 mm and 90 mm respectively and the
measurement errors are −5 mm in b and +8 mm in
l, find the approximate error in the calculated value
of I .
Using equation (3), the approximate error in I ,
δ I ≈
∂ I
∂b
δb +
∂ I
∂l
δl
∂ I
∂b
=
l 3
3
and
∂ I
∂l
=
3bl 2
3
= bl
2
δb = −5 mm and δl = +8 mm
Hence δ I ≈
l 3
3
(−5) + (bl 2 )(+8)
Since b = 40 mm and l = 90 mm then
δ I ≈
90 3
3
(−5) + 40(90)
2
(8)
≈ −1215000 + 2592000
≈ 1377000 mm
4
≈ 137.7 cm
4
Hence the approximate error in the calculated value
of I is a 137.7 cm 4 increase.
Problem 11. The time of oscillation t of a
pendulum is given by t = 2π
l
g
. Determine the
approximate percentage error in t when l has an
error of 0.2% too large and g 0.1% too small.