Higher Engineering Mathematics

G

Differential calculus

34

Partial differentiation

34.1 Introduction to partial derivatives

In engineering, it sometimes happens that the varia-
tion of one quantity depends on changes taking place
in two, or more, other quantities. For example, the
volumeVof a cylinder is given byV=πr^2 h. The
volume will change if either radiusror heighth
is changed. The formula for volume may be stated
mathematically asV=f(r,h) which means ‘V is
some function ofrandh’. Some other practical
examples include:

(i) time of oscillation,t= 2 π

√

l

g

i.e.t=f(l,g).

(ii) torqueT=Iα, i.e.T=f(I,α).

(iii) pressure of an ideal gasp=

mRT

V

i.e.p=f(T,V).

(iv) resonant frequencyfr=

1

2 π

√

LC

i.e.fr=f(L,C), and so on.

When differentiating a function having two vari-
ables, one variable is kept constant and the dif-
ferential coefficient of the other variable is found
with respect to that variable. The differential coef-
ficient obtained is called apartial derivativeof
the function.

34.2 First order partial derivatives

A ‘curly dee’,∂, is used to denote a differential coef-
ficient in an expression containing more than one
variable.

Hence ifV=πr^2 hthen

∂V

∂r

means ‘the partial

derivative ofVwith respect tor, withhremaining
constant’. Thus,

∂V

∂r

=(πh)

d

dr

(r^2 )=(πh)(2r)= 2 πrh.

Similarly,

∂V

∂h

means ‘the partial derivative ofVwith

respect toh, withrremaining constant’. Thus,

∂V

∂h

=(πr^2 )

d

dh

(h)=(πr^2 )(1)=πr^2.

∂V

∂r

and

∂V

∂h

are examples offirst order partial

derivatives, sincen=1 when written in the form

∂nV

∂rn

.

First order partial derivatives are used when finding

the total differential, rates of change and errors for

functions of two or more variables (see Chapter 35),

when finding maxima, minima and saddle points for

functions of two variables (see Chapter 36), and with

partial differential equations (see Chapter 53).

Problem 1. Ifz= 5 x^4 + 2 x^3 y^2 − 3 yfind

(a)

∂z

∂x

and (b)

∂z

∂y

(a) To find

∂z

∂x

,yis kept constant.

Since z= 5 x^4 +(2y^2 )x^3 −(3y)

then,

∂z

∂x

=

d

dx

(5x^4 )+(2y^2 )

d

dx

(x^3 )−(3y)

d

dx

(1)

= 20 x^3 +(2y^2 )(3x^2 )− 0.

Hence

∂z

∂x

= 20 x^3 + 6 x^2 y^2.

(b) To find

∂z

∂y

,xis kept constant.

Sincez=(5x^4 )+(2x^3 )y^2 − 3 y

then,

∂z

∂y

=(5x^4 )

d

dy

(1)+(2x^3 )

d

dy

(y^2 )− 3

d

dy

(y)

= 0 +(2x^3 )(2y)− 3