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
3πr^2 hUsing equation (2), the rate of change of volume,
dV
dt=∂V
∂rdr
dt+∂V
∂hdh
dt∂V
∂r=2
3πrhand∂V
∂h=1
3πr^2Since the height is increasing at 3mm/s,
i.e. 0.3cm/s, then
dh
dt
=+ 0. 3and since the radius is decreasing at 2mm/s,
i.e. 0.2cm/s, then
dr
dt=− 0. 2Hence
dV
dt=(
2
3πrh)
(− 0. 2 )+(
1
3πr^2)
(+ 0. 3 )=− 0. 4
3πrh+ 0. 1 πr^2However, 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 /sThus 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
∂ada
dt+∂A
∂cdc
dt+∂A
∂BdB
dtSince A=1
2acsinB,∂A
∂a=1
2csinB,∂A
∂c=1
2asinBand∂A
∂B=1
2accosBda
dt= 0 .4 units/s,dc
dt=− 0 .8 units/sanddB
dt= 0 .2 units/sHencedA
dt=(
1
2csinB)
( 0. 4 )+(
1
2asinB)
(− 0. 8 )+(
1
2accosB)
( 0. 2 )Whena= 3 ,c=4andB=π
6then:dA
dt=(
1
2( 4 )sinπ
6)
( 0. 4 )+(
1
2( 3 )sinπ
6)
(− 0. 8 )+(
1
2( 3 )( 4 )cosπ
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.
Cb
B z 5 3cmx^5
y 5 5cm
4cm
AFigure 35.1DiagonalAB=√
(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 )