354 Higher Engineering Mathematics
Using equation (2), the rate of change of diagonal b is
given by:
db
dt
=
∂b
∂x
dx
dt
+
∂b
∂ y
dy
dt
+
∂b
∂z
dz
dt
Since b =
(x 2 + y 2 + z 2 )
∂b
∂x
=
1
2
(x 2 + y 2 + z 2 )
−1
2 (2x) =
x
(x 2 + y 2 + z 2 )
Similarly,
∂b
∂ y
=
y
(x 2 + y 2 + z 2 )
and
∂b
∂z
=
z
(x 2 + y 2 + z 2 )
dx
dt
= 6 mm/s = 0.6 cm/s,
dy
dt
= 5 mm/s = 0.5 cm/s,
and
dz
dt
= 4 mm/s = 0.4 cm/s
Hence
db
dt
=
x
(x 2 + y 2 + z 2 )
(0.6)
+
y
(x 2 + y 2 + z 2 )
(0.5)
+
z
(x 2 + y 2 + z 2 )
(0.4)
When x = 5 cm, y = 4 cm and z = 3 cm, then:
db
dt
=
5
(5 2 + 4 2 + 3 2 )
(0.6)
+
4
(5 2 + 4 2 + 3 2 )
(0.5)
+
3
(5 2 + 4 2 + 3 2 )
(0.4)
= 0.4243 + 0.2828 + 0.1697 = 0.8768 cm/s
Hence the rate of increase of diagonal AC is
0.88 cm/s or 8.8 mm/s, correct to 2 significant figures.
Now try the following exercise
Exercise 141 Further problems on rates of
change
1. The radius of a right cylinder is increasing at
a rate of 8 mm/s and the height is decreasing
at a rate of 15 mm/s. Find the rate at which the
volume is changing in cm 3 /s when the radius
is 40 mm and the height is 150 mm.
[+226.2 cm 3 /s]
2. If z = f (x, y) and z = 3x 2 y 5 , find the rate of
change of z when x is 3 units and y is 2 units
when x is decreasing at 5 units/s and y is
increasing at 2.5 units/s.
[2520 units/s]
3. Find the rate of change of k, correct to 4
significant figures, given the following data:
k = f (a, b, c); k = 2b ln a + c
2 e
a
; a is increasing at 2 cm/s; b is decreasing at 3 cm/s; c is
decreasing at 1 cm/s; a = 1.5 cm, b = 6 cm and
c = 8 cm.
[515.5 cm/s]
4. A rectangular box has sides of length x cm,
y cm and z cm. Sides x and z are expanding at
rates of 3 mm/s and 5 mm/s respectively and
side y is contracting at a rate of 2 mm/s. Determine the rate of change of volume when x is
3 cm, y is 1.5 cm and z is 6 cm.
[1.35 cm 3 /s]
5. Find the rate of change of the total surface area
of a right circular cone at the instant when the
base radius is 5 cm and the height is 12 cm if the
radius is increasing at 5 mm/s and the height
is decreasing at 15 mm/s.
[17.4 cm 2 /s]
35.3 Small changes
It is often useful to find an approximate value for
the change (or error) of a quantity caused by small
changes (or errors) in the variables associated with the
quantity. If z = f (u, v, w, ...) and δu, δv, δw,... denote
small changes in u, v, w, ... respectively, then the corresponding approximate change δz in z is obtained from
equation (1) by replacing the differentials by the small
changes.
Thus δz ≈
∂z
∂u
δu +
∂z
∂v
δv +
∂z
∂w
δw + · · ·
(3)
Using equation (2), the rate of change of diagonal b is
given by:
db
dt
=
∂b
∂x
dx
dt
+
∂b
∂ y
dy
dt
+
∂b
∂z
dz
dt
Since b =
(x 2 + y 2 + z 2 )
∂b
∂x
=
1
2
(x 2 + y 2 + z 2 )
−1
2 (2x) =
x
(x 2 + y 2 + z 2 )
Similarly,
∂b
∂ y
=
y
(x 2 + y 2 + z 2 )
and
∂b
∂z
=
z
(x 2 + y 2 + z 2 )
dx
dt
= 6 mm/s = 0.6 cm/s,
dy
dt
= 5 mm/s = 0.5 cm/s,
and
dz
dt
= 4 mm/s = 0.4 cm/s
Hence
db
dt
=
x
(x 2 + y 2 + z 2 )
(0.6)
+
y
(x 2 + y 2 + z 2 )
(0.5)
+
z
(x 2 + y 2 + z 2 )
(0.4)
When x = 5 cm, y = 4 cm and z = 3 cm, then:
db
dt
=
5
(5 2 + 4 2 + 3 2 )
(0.6)
+
4
(5 2 + 4 2 + 3 2 )
(0.5)
+
3
(5 2 + 4 2 + 3 2 )
(0.4)
= 0.4243 + 0.2828 + 0.1697 = 0.8768 cm/s
Hence the rate of increase of diagonal AC is
0.88 cm/s or 8.8 mm/s, correct to 2 significant figures.
Now try the following exercise
Exercise 141 Further problems on rates of
change
1. The radius of a right cylinder is increasing at
a rate of 8 mm/s and the height is decreasing
at a rate of 15 mm/s. Find the rate at which the
volume is changing in cm 3 /s when the radius
is 40 mm and the height is 150 mm.
[+226.2 cm 3 /s]
2. If z = f (x, y) and z = 3x 2 y 5 , find the rate of
change of z when x is 3 units and y is 2 units
when x is decreasing at 5 units/s and y is
increasing at 2.5 units/s.
[2520 units/s]
3. Find the rate of change of k, correct to 4
significant figures, given the following data:
k = f (a, b, c); k = 2b ln a + c
2 e
a
; a is increasing at 2 cm/s; b is decreasing at 3 cm/s; c is
decreasing at 1 cm/s; a = 1.5 cm, b = 6 cm and
c = 8 cm.
[515.5 cm/s]
4. A rectangular box has sides of length x cm,
y cm and z cm. Sides x and z are expanding at
rates of 3 mm/s and 5 mm/s respectively and
side y is contracting at a rate of 2 mm/s. Determine the rate of change of volume when x is
3 cm, y is 1.5 cm and z is 6 cm.
[1.35 cm 3 /s]
5. Find the rate of change of the total surface area
of a right circular cone at the instant when the
base radius is 5 cm and the height is 12 cm if the
radius is increasing at 5 mm/s and the height
is decreasing at 15 mm/s.
[17.4 cm 2 /s]
35.3 Small changes
It is often useful to find an approximate value for
the change (or error) of a quantity caused by small
changes (or errors) in the variables associated with the
quantity. If z = f (u, v, w, ...) and δu, δv, δw,... denote
small changes in u, v, w, ... respectively, then the corresponding approximate change δz in z is obtained from
equation (1) by replacing the differentials by the small
changes.
Thus δz ≈
∂z
∂u
δu +
∂z
∂v
δv +
∂z
∂w
δw + · · ·
(3)