Total differential, rates of change and small changes 353
Hence the rate of change of z,
dz
dt
=75.14units/s,
correct to 4 significant figures.
Problem 5. The height of a right circular cone is
increasing at 3mm/s and its radius is decreasing at
2mm/s. Determine, correct to 3 significant figures,
the rate at which the volume is changing (in cm^3 /s)
when the height is 3.2cm and the radius is 1.5cm.
Volume of a right circular cone,V=
1
3
πr^2 h
Using equation (2), the rate of change of volume,
dV
dt
=
∂V
∂r
dr
dt
+
∂V
∂h
dh
dt
∂V
∂r
=
2
3
πrhand
∂V
∂h
=
1
3
πr^2
Since the height is increasing at 3mm/s,
i.e. 0.3cm/s, then
dh
dt
=+ 0. 3
and since the radius is decreasing at 2mm/s,
i.e. 0.2cm/s, then
dr
dt
=− 0. 2
Hence
dV
dt
=
(
2
3
πrh
)
(− 0. 2 )+
(
1
3
πr^2
)
(+ 0. 3 )
=
− 0. 4
3
πrh+ 0. 1 πr^2
However, h= 3 .2cmandr= 1 .5cm.
Hence
dV
dt
=
− 0. 4
3
π( 1. 5 )( 3. 2 )+( 0. 1 )π( 1. 5 )^2
=− 2. 011 + 0. 707 =− 1 .304cm^3 /s
Thus the rate of change of volume is 1.30 cm^3 /s
decreasing.
Problem 6. The areaAof a triangle is given by
A=^12 acsinB,whereBis the angle between sidesa
andc.Ifais increasing at 0.4 units/s,cis
decreasing at 0.8 units/s andBis increasing at 0.2
units/s, find the rate of change of the area of the
triangle, correct to 3 significant figures, whenais 3
units,cis 4 units andBisπ/6radians.
Using equation (2), the rate of change of area,
dA
dt
=
∂A
∂a
da
dt
+
∂A
∂c
dc
dt
+
∂A
∂B
dB
dt
Since A=
1
2
acsinB,
∂A
∂a
=
1
2
csinB,
∂A
∂c
=
1
2
asinBand
∂A
∂B
=
1
2
accosB
da
dt
= 0 .4 units/s,
dc
dt
=− 0 .8 units/s
and
dB
dt
= 0 .2 units/s
Hence
dA
dt
=
(
1
2
csinB
)
( 0. 4 )+
(
1
2
asinB
)
(− 0. 8 )
+
(
1
2
accosB
)
( 0. 2 )
Whena= 3 ,c=4andB=
π
6
then:
dA
dt
=
(
1
2
( 4 )sin
π
6
)
( 0. 4 )+
(
1
2
( 3 )sin
π
6
)
(− 0. 8 )
+
(
1
2
( 3 )( 4 )cos
π
6
)
( 0. 2 )
= 0. 4 − 0. 6 + 1. 039 = 0 .839units^2 /s,correct
to 3 significant figures.
Problem 7. Determine the rate of increase of
diagonalACof the rectangular solid, shown in
Fig. 35.1, correct to 2 significant figures, if the sides
x,yandzincrease at 6mm/s, 5mm/s and 4mm/s
when these three sides are 5cm, 4cm and 3cm
respectively.
C
b
B z 5 3cm
x^5
y 5 5cm
4cm
A
Figure 35.1
DiagonalAB=
√
(x^2 +y^2 )
DiagonalAC=
√
(BC^2 +AB^2 )
=
√
[z^2 +{
√
(x^2 +y^2 )}^2
=
√
(z^2 +x^2 +y^2 )
LetAC=b,thenb=
√
(x^2 +y^2 +z^2 )