Higher Engineering Mathematics

350 DIFFERENTIAL CALCULUS

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 141 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+ 2 ydy]


2. z= 2 xy−cosx [(2y+sinx)dx+ 2 xdy]


3. z=

x−y

x+y

[

2 y

(x+y)^2

dx−

2 x

(x+y)^2

dy

]

4. z=xlny

[

lnydx+

x

y

dy

]

5. z=xy+

√

x

y

− 4

[(

y+

1

2 y

√

x

)

dx+

(

x−

√

x

y^2

)

dy

]

6. If z=f(a,b,c) and z= 2 ab− 3 b^2 c+abc,
   find the total differential, dz.
   [
   b(2+c)da+(2a− 6 bc+ac)db
   +b(a− 3 b)dc

]

7. Givenu=ln sin (xy) show that
   du=cot (xy)(ydx+xdy)

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 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 sin 2yfind

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 sin 2y, then

∂z

∂x

= 6 x^2 sin 2yand

∂z

∂y

= 4 x^3 cos 2y

Sincexis increasing at 4 units/s,

dx

dt

=+ 4

and sinceyis decreasing at 0.5 units/s,

dy

dt

=− 0. 5

Hence

dz

dt

=(6x^2 sin 2y)(+4)+(4x^3 cos 2y)(− 0 .5)

= 24 x^2 sin 2y− 2 x^3 cos 2y

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

Hence the rate of change ofz,

dz

dt

=75.14 units/s,

correct to 4 significant figures.

Problem 5. The height of a right circular cone

is increasing at 3 mm/s and its radius is decreas-

ing at 2 mm/s. Determine, correct to 3 significant

figures, the rate at which the volume is chang-

ing (in cm^3 /s) when the height is 3.2 cm and the

radius is 1.5 cm.

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