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.
- z=x^3 +y^2 [3x^2 dx+ 2 ydy]
- z= 2 xy−cosx [( 2 y+sinx)dx+ 2 xdy]
- z=
x−y
x+y
[
2 y
(x+y)^2
dx−
2 x
(x+y)^2
dy
]
- z=xlny
[
lnydx+
x
y
dy
]
- z=xy+
√
x
y
− 4
[(
y+
1
2 y
√
x
)
dx+
(
x−
√
x
y^2
)
dy
]
- 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
]
- 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