PROBLEMS ON ABSOLUTE ERROR RELATIVE ERROR AND PERCENTAGE ERROR

Suppose that certain quantity is to be determined. Its exact value is called the actual value. Some times we obtain its approximate value through some approximation process.

Absolute error  =  Actual value - Approximate value

Approximate value  =  Rate of change x difference 

If the actual value is not zero, then

Relative error

=  (Actual value - Approximate value)/Actual value

Percentage error  =  Relative error × 100

Problem 1 :

The radius of a circular plate is measured as 12.65 cm instead of the actual length 12.5 cm. find the following in calculating the area of the circular plate:

(i) Absolute error (ii) Relative error (iii) Percentage error

Solution :

(i)  Absolute error :

Absolute error  =  Actual value - Approximate value

  Actual length  =  12.5 cm approximate value  =  12.65 cm

Area of circle A(r) =  πr2

Actual change in area  =  π(12.65)2 π(12.5)2

=  π[160.0225 - 156.25]

=  π(3.7725)

=  3.7725π  ---(1)

Approximate change  =  A'(12.5) x change in radius

  =  2π(12.5) x 0.15

=  25π x 0.15

=  3.75π  ---(2)

(1) - (2)

Absolute error  =  3.7725π - 3.75π

=  0.0225π cm2

(ii)  Relative error  =  (Actual value - Approximate value)/Actual value

Relative error  =  0.0225π / 3.7725π

=  0.0059 cm2

(iii)  Percentage error  =  Relative error x 100%

=  0.0059 x 100%

=  0.6%

Problem 2 :

A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9 8 . cm. Find approximations for the following :

(i) change in the volume (ii) change in the surface area

Solution :

Volume of sphere V(r)  =  (4/3)πr3

Surface area of sphere A(r) =  4πr2

(i)  Change in volume  =  Rate of change in volume x difference

V'(r)  =  (4/3)π(3r2) dr

V'(r)  =  (4πr2) dr

V'(r)  =  (4π(10)2) (0.2)

V'(r)  =  80π

Volume decreases by 80π cm3.

(ii)  Change in surface area :

A(r) =  4πr2

A'(r)  =  8πr(dr)

A'(r)  =  8π(10)(0.2)

A'(r)  =  16π cm2

Surface area decreases by 16π cm2.

Problem 3 :

The time T , taken for a complete oscillation of a single pendulum with length l , is given by the equation

T  =  2π√(l/g)

where g is constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of 1.

Solution :

T  =  2π√(l/g)

T  =  2π(l/g)1/2

log T  =  log 2π + (1/2) [log l - log g]

(1/T) dT/dl  =  0 + (1/2) [(1/l) - 0]

dT/T  =  (1/2) (1/l) dl

Percentage error  =  (1/2) (1/l) x 0.02 l x 100

=  1%

Problem 4 :

Show that the percentage error in the nth root of a number is approximately 1/n times the percentage error in the number.

Solution :

f(x)  =  x1/n

y  =  x1/n

Taking log on both sides, we get

log y  =  (1/n) log x

(1/y) (dy/dx)  =  (1/n) (1/x)

dy/y  =  (1/n) (1/x) dx

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