352 Higher Engineering Mathematics
Since pV = kT, k =
pV
T
Hence d p =
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
d p +
∂T
∂V
dV
Since pV = kT, T =
pV
k
hence
∂T
∂ p
=
V
k
and
∂T
∂V
=
p
k
Thus dT =
V
k
d p +
p
k
dV and substituting k =
pV
T
gives:
dT =
V
pV
T
d p +
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 + 2y dy]
2. z = 2x y − cos x
[(2y + sin x) dx + 2x dy]
3. z =
x − y
x + y
2y
(x + y) 2 dx −
2x
(x + y) 2 dy
4. z = x ln y
ln y d x +
x
y
dy
5. z = x y +
√
x
y
− 4
y +
1
2y
√
x
dx +
x −
√
x
y 2
dy
6. If z = f (a, b, c) and z =2ab − 3b 2 c + abc,
find the total differential, dz.
b(2 + c) da + (2a − 6bc + ac) db
+ b(a − 3b) dc
7. Given u = ln sin(x y) show that
du = cot(x y)(y dx + x dy).
35.2 Rates of change
Sometimes it is necessary to solve problems in which
different quantities have different rates of change. From
equation (1), the rate of change of z,
dz
dt
is given by:
dz
dt
=
∂z
∂u
du
dt
+
∂z
∂v
dv
dt
+
∂z
∂w
dw
dt
+ · · ·
(2)
Problem 4. If z = f (x, y) and z = 2x 3 sin 2y find
the rate of change of z, correct to 4 significant
figures, when x is 2 units and y is π/6 radians and
when x is increasing at 4 units/s and y is decreasing
at 0.5 units/s.
Using equation (2), the rate of change of z,
dz
dt
=
∂z
∂x
dx
dt
+
∂z
∂ y
dy
dt
Since z =2x 3 sin 2y, then
∂z
∂x
= 6x
2 sin 2y and
∂z
∂ y
= 4x
3 cos 2y
Since x is increasing at 4 units/s,
dx
dt
=+4
and since y is decreasing at 0.5 units/s,
dy
dt
=−0.5
Hence
dz
dt
= (6x 2 sin 2y)(+4) + (4x 3 cos 2y)(−0.5)
= 24x 2 sin 2y − 2x 3 cos 2y
When x = 2 units and y =
π
6
radians, then
dz
dt
= 24(2)
2 sin[2(π/6)] − 2(2)
3 cos[2(π/6)]
= 83.138 − 8.0