The binomial series 65
Hence new volume
≈ πr
2 h(1 − 0.08)(1 + 0.02)
≈ πr
2 h(1 − 0.08 + 0.02), neglecting
products of small terms
≈ πr
2 h(1 − 0.06) or 0.94πr
2 h, i.e. 94%
of the original volume
Hence the volume is reduced by approximately 6%.
(b) Curved surface area of cylinder =2πrh.
New surface area
= 2π[(1 − 0.04)r][(1 + 0.02)h]
= 2πrh(1 − 0.04)(1 + 0.02)
≈ 2πrh(1 − 0.04 + 0.02), neglecting
products of small terms
≈ 2πrh(1 − 0.02) or 0.98(2πrh),
i.e. 98% of the original surface area
Hence the curved surface area is reduced by
approximately 2%.
Problem 18. The second moment of area of a
rectangle through its centroid is given by
bl 3
12
.
Determine the approximate change in the second
moment of area if b is increased by 3.5% and l is
reduced by 2.5%.
New values of b and l are (1 + 0.035)b and (1 − 0.025)l
respectively.
New second moment of area
=
1
12
[(1 + 0.035)b][(1 − 0.025)l]
3
=
bl 3
12
(1 + 0.035)(1 − 0.025)
3
≈
bl 3
12
(1 + 0.035)(1 − 0.075), neglecting
powers of small terms
≈
bl 3
12
(1 + 0.035 − 0.075), neglecting
products of small terms
≈
bl
3
12
(1 − 0.040) or (0.96)
bl
3
12
, i.e. 96%
of the original second moment of area
Hence the second moment of area is reduced by
approximately 4%.
Problem 19. The resonant frequency of a
vibrating shaft is given by: f =
1
2π
k
I
, where k is
the stiffness and I is the inertia of the shaft. Use the
binomial theorem to determine the approximate
percentage error in determining the frequency using
the measured values of k and I when the measured
value of k is 4% too large and the measured value
of I is 2% too small.
Let f , k and I be the true values of frequency, stiffness
and inertia respectively. Since the measured value of
stiffness, k 1 , is 4% too large, then
k 1 =
104
100
k = (1 + 0.04)k
The measured value of inertia, I 1 , is 2% too small, hence
I 1 =
98
100
I = (1 − 0.02)I
The measured value of frequency,
f 1 =
1
2π
k 1
I 1
=
1
2π
k
1
2
1 I
−
1
2
1
=
1
2π
[(1 + 0.04)k]
1
2 [(1 − 0.02)I ]
−
1
2
=
1
2π
(1 + 0.04)
1
2 k
1
2 (1 − 0.02)
−
1
2 I
−
1
2
=
1
2π
k
1
2 I
−
1
2 (1 + 0.04)
1
2 (1 − 0.02)
−
1
2
i.e. f 1 = f (1 + 0.04)
1
2 (1 − 0.02)
−
1
2
≈ f
1 +
1
2
(0.04)
1 +
−
1
2
(−0.02)
≈ f (1 + 0.02)(1 + 0.01)
Neglecting the products of small terms,
f 1 ≈ (1 + 0.02 + 0.01) f ≈ 1.03 f
Thus the percentage error in f based on the measured
values of k and I is approximately [(1.03)(100) − 100],
i.e. 3% too large.
Hence new volume
≈ πr
2 h(1 − 0.08)(1 + 0.02)
≈ πr
2 h(1 − 0.08 + 0.02), neglecting
products of small terms
≈ πr
2 h(1 − 0.06) or 0.94πr
2 h, i.e. 94%
of the original volume
Hence the volume is reduced by approximately 6%.
(b) Curved surface area of cylinder =2πrh.
New surface area
= 2π[(1 − 0.04)r][(1 + 0.02)h]
= 2πrh(1 − 0.04)(1 + 0.02)
≈ 2πrh(1 − 0.04 + 0.02), neglecting
products of small terms
≈ 2πrh(1 − 0.02) or 0.98(2πrh),
i.e. 98% of the original surface area
Hence the curved surface area is reduced by
approximately 2%.
Problem 18. The second moment of area of a
rectangle through its centroid is given by
bl 3
12
.
Determine the approximate change in the second
moment of area if b is increased by 3.5% and l is
reduced by 2.5%.
New values of b and l are (1 + 0.035)b and (1 − 0.025)l
respectively.
New second moment of area
=
1
12
[(1 + 0.035)b][(1 − 0.025)l]
3
=
bl 3
12
(1 + 0.035)(1 − 0.025)
3
≈
bl 3
12
(1 + 0.035)(1 − 0.075), neglecting
powers of small terms
≈
bl 3
12
(1 + 0.035 − 0.075), neglecting
products of small terms
≈
bl
3
12
(1 − 0.040) or (0.96)
bl
3
12
, i.e. 96%
of the original second moment of area
Hence the second moment of area is reduced by
approximately 4%.
Problem 19. The resonant frequency of a
vibrating shaft is given by: f =
1
2π
k
I
, where k is
the stiffness and I is the inertia of the shaft. Use the
binomial theorem to determine the approximate
percentage error in determining the frequency using
the measured values of k and I when the measured
value of k is 4% too large and the measured value
of I is 2% too small.
Let f , k and I be the true values of frequency, stiffness
and inertia respectively. Since the measured value of
stiffness, k 1 , is 4% too large, then
k 1 =
104
100
k = (1 + 0.04)k
The measured value of inertia, I 1 , is 2% too small, hence
I 1 =
98
100
I = (1 − 0.02)I
The measured value of frequency,
f 1 =
1
2π
k 1
I 1
=
1
2π
k
1
2
1 I
−
1
2
1
=
1
2π
[(1 + 0.04)k]
1
2 [(1 − 0.02)I ]
−
1
2
=
1
2π
(1 + 0.04)
1
2 k
1
2 (1 − 0.02)
−
1
2 I
−
1
2
=
1
2π
k
1
2 I
−
1
2 (1 + 0.04)
1
2 (1 − 0.02)
−
1
2
i.e. f 1 = f (1 + 0.04)
1
2 (1 − 0.02)
−
1
2
≈ f
1 +
1
2
(0.04)
1 +
−
1
2
(−0.02)
≈ f (1 + 0.02)(1 + 0.01)
Neglecting the products of small terms,
f 1 ≈ (1 + 0.02 + 0.01) f ≈ 1.03 f
Thus the percentage error in f based on the measured
values of k and I is approximately [(1.03)(100) − 100],
i.e. 3% too large.