PROBLEMS ON ABSOLUTE ERROR RELATIVE ERROR AND PERCENTAGE ERROR

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 n^th root of a number is approximately 1/n times the percentage error in the number.

Solution :

f(x)  =  x^1/n

y  =  x^1/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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