Higher Engineering Mathematics, Sixth Edition

352 Higher Engineering Mathematics

Since pV=kT,k=

pV

T

Hence dp=

(

pV

T

)

V

dT−

(

pV

T

)

T

V^2

dV

i.e. dp=

p

T

dT−

p

V

dV

(b) Total differential dT=

∂T

∂p

dp+

∂T

∂V

dV

Since pV=kT,T=

pV

k

hence

∂T

∂p

=

V

k

and

∂T

∂V

=

p

k

Thus dT=

V

k

dp+

p

k

dV and substituting k=

pV

T

gives:

dT=

V

(

pV

T

)dp+

p

(

pV

T

)dV

i.e. dT=

T

p

dp+

T

V

dV

Now try the following exercise

Exercise 140 Further problems on the

total differential

In Problems 1 to 5, find the total differential dz.

1. z=x^3 +y^2 [3x^2 dx+ 2 ydy]


2. z= 2 xy−cosx [( 2 y+sinx)dx+ 2 xdy]


3. z=

x−y

x+y

[

2 y

(x+y)^2

dx−

2 x

(x+y)^2

dy

]

4. z=xlny

[

lnydx+

x

y

dy

]

5. z=xy+

√

x

y

− 4

[(

y+

1

2 y

√

x

)

dx+

(

x−

√

x

y^2

)

dy

]

6. If z=f(a,b,c) and z= 2 ab− 3 b^2 c+abc,
   find the total differential, dz.
   [
   b( 2 +c)da+( 2 a− 6 bc+ac)db
   +b(a− 3 b)dc

]

7. Givenu=lnsin(xy)show that
   du=cot(xy)(ydx+xdy).

35.2 Rates of change

Sometimes it is necessary to solve problems in which

different quantitieshave different rates of change. From

equation (1), the rate of change ofz,

dz

dt

is given by:

dz

dt

=

∂z

∂u

du

dt

+

∂z

∂v

dv

dt

+

∂z

∂w

dw

dt

+··· (2)

Problem 4. Ifz=f(x,y)andz= 2 x^3 sin2yfind

the rate of change ofz, correct to 4 significant

figures, whenxis 2 units andyisπ/6 radians and

whenxis increasing at 4 units/s andyis decreasing

at 0.5 units/s.

Using equation (2), the rate of change ofz,

dz

dt

=

∂z

∂x

dx

dt

+

∂z

∂y

dy

dt

Sincez= 2 x^3 sin2y,then

∂z

∂x

= 6 x^2 sin2yand

∂z

∂y

= 4 x^3 cos2y

Sincexis increasing at 4 units/s,

dx

dt

=+ 4

and sinceyis decreasing at 0.5 units/s,

dy

dt

=− 0. 5

Hence

dz

dt

=( 6 x^2 sin2y)(+ 4 )+( 4 x^3 cos2y)(− 0. 5 )

= 24 x^2 sin2y− 2 x^3 cos2y

Whenx=2 units andy=

π

6

radians, then

dz

dt

= 24 ( 2 )^2 sin[2(π/ 6 )]− 2 ( 2 )^3 cos[2(π/ 6 )]

= 83. 138 − 8. 0