Understanding Engineering Mathematics

We have (treatingdx,dy, etc. as small increments here)

dz=αxα−^1 yβdx+βxαyβ−^1 dy

so the relative error inzdue to ‘errors’dx,dyinx,yrespectively is

dz

z

=

αxα−^1 yβ

z

dx+

βxαyβ−^1

z

dy

=α

xα−^1 yβ

xαyβ

dx+

βxαyβ−^1

xαyβ

dy

=α

dx

x

+β

dy

y

So the percentage error inzis

dz

z

× 100 =

(

α

dx

x

+β

dy

y

)

100

=α

(

dx

x

× 100

)

+b

(

dy

y

× 100

)

=αpercentage error inx+βpercentage error iny

Notice the pattern in which the exponents in the original expression become the linear
weightings in the expression for the percentage error.
An easier derivation is to first take logs

lnz=ln(xαyβ)=αlnx+βlny

Now differentiate through

dz

z

=α

dx

x

+β

dy

y

and then multiplying through by 100 gives the required relation between percentage errors.
Extension to functions of more than two variables is obvious.

Problem 16.5

Find the percentage error in the volume of a rectangular box in terms of

the percentage errors in the sides of the box.

With sidesa,b,cthe volume of the box is

V=abc=a^1 b^1 c^1

So the percentage error inV is

dV

V

× 100 = 1.

da

a

× 100 + 1.

db

b

× 100 + 1.

dc

c

× 100

=percentage error ina+percentage error inb

+percentage error inc