Find the absolute, percentage and relative errors if x is rounded-off to three decimal digits.. If 0.333 is the approximate value of 1 3 , then find its absolute, relative and percentage
Trang 1Using the principle of equal effects, which states 1 2
∂ ∂ = = n
n
X x x
∂ δ
∂ this implies that
1 1
∂
δ = δ
∂
X
X n x
x or 1
1
X x
X n x
δ
δ = ∂
∂
Similarly we get 2 3
,
n
n
X
x
X
n
x
δ
δ = ∂
∂
and so on
This form is useful where error in dependent variable is given and also we are to find errors
in both independent variables
Remark: The Error 1 10
2
n
−
= × , if a number is correct to n decimal places Also Relative
error is less than 1 1
10n
l× − , if number is correct to n significant digits and l is the first significant digit of a number
1.4.6 Error in Evaluating x k
Let x k be the function having k is an integer or fraction then Relative Error for this function is
given
Relative Error = k x or X k x
Example 2 Find the absolute, percentage and relative errors if x is rounded-off to three decimal
digits Given x = 0.005998.
Sol If x is rounded-off to three decimal places we get x = 0.006 Therefore
Error = True value – Approximate value Error = 005998 – 006 = – 000002
Absolute Error = E a = Error = 0.000002
0.005998 0.005998
r
E True value
Percentage Error = E p = E r× 100 = 0.33344
Example 3 Find the number of trustworthy figure in (0.491) 3 assuming that the number 0.491 is correct to last figure.
Sol We know that Relative Error, E r = δX ≤ δx
k
Here δx = 0.0005 because 1× −3=
10 0.0005 2
Trang 2or 3 0.00053 3 0.0005 0.01267
0.118371 (0.491)
x
k
x
×
δ
Therefore, Absolute Error = E r X
or Absolute Error < 0.01267 × (0.491)3
= 0.01267 × 0.118371
= 0.0015 The error affects the third decimal place, therefore, (0.491)3 = 0.1183 is correct to second decimal places
Example 4 If 0.333 is the approximate value of 1
3 , then find its absolute, relative and percentage errors.
Sol Given that True value ( ) 1
3
x = , and its Approximate value (x′) = 0.333
Therefore, Absolute Error, 1 0.333 0.333333 0.333 0.000333
3
a
Relative Error, E r = 0.000333 0.000999
0.333333
a
E
Percentage Error, E p = E r×100=0.000999 100× =0.099%
Example 5 Round-off the number 75462 to four significant digits and then calculate its absolute
error, relative error and percentage error.
Sol After rounded-off the number to four significant digits we get 75460
Therefore Absolute Error E a= 75462 75460− =2
75462 75462
Percentage Error E p = E r × 100 = 0.00265
Example 6 Find the relative error of the number 8.6 if both of its digits are correct.
Sol Since = ×1 −1=
10 0.05 2
a
E therefore, Relative Error = =0.5=0.0058
8.6
r
Example 7 Three approximate values of number 1
3 are given as 0.30, 0.33 and 0.34 Which of these three is the best approximation?
Sol The number, which has least absolute error, gives the best approximation
3=
When approximate value x′ is 0.30 the Absolute Error is given by:
E a = x x− ′ = 0.33333 0.30− =0.03333
Trang 3When approximate value x′ is 0.33 the Absolute Error is given by:
E a = x x− ′= 0.33333 0.33− =0.00333
When approximate value x′ is 0.34 the Absolute Error is given by:
E a = x x− ′= 0.33333 0.34− =0.00667 Here absolute error is least when approximate value is 0.33 Hence 0.33 is the best approximation
Example 8 Calculate the sum of 3, 5 and 7 to four significant digits and find its absolute and relative errors.
Sol Here 3=1.732, 5=2.236, 7 =2.646
Hence Sum = 6.614 and
Absolute Error = E a = 0.0005 + 0.0005 + 0.0005 = 0.0015 (Because 1× −3
10 = 0.0005)
2 Also the total absolute error shows that the sum is correct up
to 3 significant figures Therefore S = 6.61 and
Relative Error, E r = 0.0015 =0.0002
Example 9 Approximate values of 7 1 and 1
11 , correct to 4 decimal places are 0.1429 and 0.0909 respectively Find the possible relative error and absolute error in the sum of 0.1429 and 0.0909.
Sol The maximum error in each case =1× −4 = ×1 =
10 0.0001 0.00005
1 Relative Error, E r = δ < 0.00005 + 0.00005
0.2338 0.2338
X
Therefore, δ < 0.0001=0.00043
0.2338
X X
2 Absolute Error, E a = 0.0001 0.2338 0.0001
0.2338
X X X
Example 10 Find the number of trustworthy figures in (367) 1/5 where 367 is correct to three significant figures.
Sol Relative Error <1δ ;
5
r
x E x
Therefore, 1δ = ×1 0.5 =0.0003
5 5 367
x x
Similarly, Absolute Error E a<(367)1/5×0.0003=3.258 0.0003× =0.001
Hence Absolute Error <0.001
Thus error effects fourth significant figure and hence (367)1/5 ≈ 3.26 correct to the three figures
Trang 4Example 11 Find the relative error in calculation of 7.342 0.241 Where numbers 7.342 and 0.241 are correct to three decimal places Determine the smallest interval in which true result lies.
Sol Relative Error ≤ δ 1 + δ 2
Here δx1 = δx2 = 0.0005, x1 = 7.342, x2 = 0.241
Therefore, Relative Error ≤ 0.0005 + 0.0005
7.342 0.241
×
×
1 1 0.0005 7.583
7.342 0.241 7.342 0.241 Similarly, Absolute Error ≤ × = × =
1 2
0.0021 7.342
0.241
x x
2
x
x = 7.342=30.4647 0.241
Hence true value of 7.342
0.241 lies between 30.4647 – 0.0639 = 30 4008 and 30.4647 + 0.0639 = 30.5286
Example 12 Find the product of 346.1 and 865.2 and state how many figures of the result are
trustworthy, given that the numbers are correct to four significant figures.
Sol For given numbers 346.1 and 865.2,
δx1 = 0.05 = δx2 Because Error = 1×10−
2
n
Also, X = 346.1 × 865.2 = 299446 (correct to six significant figures)
Therefore Relative Error ≤ δ 1 + δ 2 = +
0.05 0.05 346.1 865.2
r
E
= 0.000144 + 0.000058 = 0.000202
Similarly, Absolute Error E a = E r X ≤ 0.000202 × 299446 ≈ 60
So, true value of the product of the given numbers lies between
299446 – 60 = 299386 And 299446 + 60 = 299506
Hence the mean of these values is 299386 299506+ =
299446
2 which is written as 299.4 × 103 This is correct to four significant figures
Example 13 Find the relative error in the calculation of 3.724 × 4.312 and determine the interval
in which true result lies Given that the numbers 3.724 and 4.312 are correct to last digit?
Sol For product of numbers, Relative Error = δ ≤ δ 1 + δ 2 + + δ
n
n
x
X
Trang 5Therefore, Relative Error, E r = 0.0005 + 0.0005 =0.0002501
3.724 4.312
(Because Error = ×1 − = ×1 −3 =
10 10 0.0005)
n
Absolute Error, E a = E r X = 0.002501 × 3.724 × 4.312
= 0.0040157
Product x1x2 = 3.724 × 4.312 = 16.057888
Lower limit is given by 16.057888 – 0.004016 = 16.053872
Upper limit is given by 16.057888 + 0.004016=16.061904
Hence true value lies between 16.0539 and 16.0619
Example 14 Find the absolute error in calculating (768) 1/5 and determine the interval in which true value lies 768 is correct its last digit.
Sol Relative Error, E r = δx
k
x = ×1 0.5
5 768
= 0.1 =0.0001302 768
Absolute Error, E a = E r × (768)1/5
= 0.0001302 × 3.77636
= 0.0004916 Therefore, lower limit = 3.77636 – 0.00049 = 3.77587 and
Upper limit = 3.77636 + 0.00049 = 3.77685
Hence value of (768)1/5 lies between 3.77587 and 3.77685
Example 15 Find the number of correct figure in the quotient 65.3
7 , assuming that the numerator
is correct to last figure.
Sol Since Relative Error, 1 2
r
E
x x
Here δx1 = 0.05 = δx2, x1 = 65.7 and x2 = 2.6
Therefore, Relative Error, 0.05 0.05 0.05 68.3 0.01999
65.7 2.6 65.7 2.6
r
×
Also, Absolute Error, 0.01999 65.3 0.502
2.6
a
E ≤ × = (since the error affects the first decimal place)
Example 16 Find the percentage error if 625.483 is approximated to three significant figures Sol Here x = 625.483 and x′ = 625.0 therefore,
Absolute Error, E a= 625.483 625− =0.483,
Trang 6Relative Error, 0.483 0.000772
625.483 625.483
a r
E
Percentage Error, E p= ×E r 100 0.077%.=
Example 17 Find the relative error in taking the difference of numbers 5.5 = 2.345 and 6.1 = 2.470 Numbers should be correct to four significant figures.
Sol Relative Error 1 2
r
E
Here δx1 = 0.0005 = δx2
Therefore, Relative Error = 1 0.0005
2.470 2.345
x X
δ =
−
= 20.0005 0.001 0.008 0.125 =0.125=
Example 18 If X = x + e prove that X− x ≈ e
2 X .
Sol L.H.S X− x =
1 2
1 e
x
− − = − −
2
e
X
− −
=
− + ≈ R.H.S proved
Example 19 If u = 4x y 2 3 4
z and errors in x, y, z be 0.001, compute the relative maximum error in
u when x = y = z = 1.
Sol We know u u x u y u z
δ = δ + δ + δ
Also the errors δx, δy, δz may be positive or negative, therefore absolute values of terms on
R.H.S is,
( )δu max = 8xy43δ +x 12x y2 24 δ +y 12x y52 3δz
= 8(0.001) + 12(0.001) + 16(0.001) = 0.036 Also, Max Relative Error = 0.036=0.009
4 (Because E r(max) =
δ
;
u
u u = 4 at x = y = z = 1).
Trang 7Example 20 It is required to obtain the roots of X 2 – 2X + log 10 2 = 0 to four decimal places To what accuracy should log 10 2 be given?
Sol Roots of the equation X2 – 2X + log10 2 = 0 are given by,
= ± − 10
10
2 4 4 log 2
1 1 log 2 2
Therefore, δX = 1 (log 2) 0.5 10 4
2 1 log 2
− δ
< ×
−
or δ(log 2) 2 0.5 10 (1 log 2)< × × −4 − 1/2<0.83604 10× −4
≈ 8.3604 × 10–5
Example 21 If r = 3h(h 6 – 2), find the percentage error in r at h = 1, if the percentage error in
h is 5.
Sol We know δx n = δ
∂
∂ n
X X n x
where X = f(x1, x2, , x n)
∂ (21 6 6)
r
h
δ ×r 100
− δ ×
−
6 7
21 6
100
h
h
= − δ × = × = −
h h
Percentage Error = = δ ×100 =25%
p
r E
Example 22 Find the relative error in the function 1 2
1m 2m m n
n
Sol Given function 1 2
1m 2m m n
n
On taking log both the sides, we get
log y = log a + m1 log x1 + m2 log x2 + + m n log x n
Therefore, ∂ = ∂ = ∂ =
3
Hence Relative Error, = ∂ δ + ∂ δ + + ∂ δ
∂ 1 1 ∂ 2 2 ∂ n
r
n
E
n n
n
x
Trang 8Since errors δx1, δx2 may be positive or negative, therefore absolute values of terms on R.H.S give,
max
n
n
x
Remark: If y = x1x2x3 x n, then relative error is given by
δ
n
r
n
x
E
Therefore relative error of n product of n numbers is approximately equal to the algebraic
sum of their relative errors
Example 23 The discharge Q over a notch for head H is calculated by the formula
Q = kH 5/2 , where k is a given constant If the head is 75 cm and an error of 0.15 cm is possible in its measurement, estimate the percentage error in computing the discharge.
Sol Given that Q = kH5/2
On taking Log both the sides of the equation, we have
log Q = log +5
2
k log H
On differentiating, we get
5 2
δ = δ
Q Q
Example 24 Compute the percentage error in the time period T=2 1
g
π for l = 1m if the error
in the measurement of l is 0.01.
g
On taking log both the sides, we get
log T = log 2π + 1log −1log
2 l 2 g
δ 1 2
l l
×
0.01
100 100 0.5%
l
Example 25 If u = 2V 6 – 5V, find the percentage error in u at V = 1, if error in V is 0.05.
Sol Given u = 2V6 – 5V
δu = ∂ δ = − δ
∂ (12 5 5)
u
V
Trang 9∂ ×u 100
δ ×
5 6
12 5
100
V
V
= − × × = − × = −
−
Hence maximum percentage error (E p)max = 11.667%
Example 26 How accurately should the length and time of vibration of a pendulum should be
measured in order that the computed value of g is correct to 0.01%.
Sol Period of vibration T is given by T = 2π l
g , where l is the length of pendulum.
∂
4 l g 4
l
∂
2 3
4 2
δl = and
T
δ =
(1)
But the percentage error in g is
δ × 100
g
2
100 0.01 4
g l T
δ
(a) Percentage Error in l = δ ×
100
l l
l
l
∂
=
100 100 2
= 1×0.01 0.005=
(b) Percentage Error in T = T 100
T × δ
2
g
δ
∂ ∂
=
2 3
100 4
4
g l T
δ
×
π
= 1×100=0.0025
Trang 10Example 27 Calculate the value of x – x cos θ correct to three significant figures if x = 10.2 cm, and θ = 5° Find permissible errors also in x and θ.
Sol Given that θ = 5° = 5π = 11
radian
180 126
2 4
1 1
2! 4!
= 0.0038107 – 0.0000024 ≈ 0.0038083
Therefore X = x(1 – cos θ) = 10.2(0.0038083) = 0.0388446 ≈ 0.0388
∂
0.0005
0.0656
2 0.0038083 2
X X x
∂θ
0.0005 0.0005
2 sin 2 10.2 0.0871907 2
X
3
0.0871907 3! 5! 126 6 126
×
0.0005
0.0002809 0.00028
Example 28 The percentage error in R which is given by R= r 2 +h
2h 2 , is not allowed to exceed 0.2%, find allowable error in r and h when r = 4.5 cm and h = 5.5 cm.
Sol The percentage error in R = δR×100=0.2
R
Therefore δR = 0.2 0.2 ( )4.52 5.5
100 R 100 2 5.5 2
2 2
h
= 0.2×50.50= 0.002 50.50×