344 DIFFERENTIAL CALCULUS

Hence

∂z

∂y

= 4 x^3 y− 3.

Problem 2. Giveny=4 sin 3xcos 2t, find

∂y

∂x

and

∂y

∂t

.

To find

∂y

∂x

,tis kept constant

Hence

∂y

∂x

=(4 cos 2t)

d

dx

(sin 3x)

=(4 cos 2t)(3 cos 3x)

i.e.

∂y

∂x

=12 cos 3xcos 2t

To find

∂y

∂t

,xis kept constant.

Hence

∂y

∂t

=(4 sin 3x)

d

dt

(cos 2t)

=(4 sin 3x)(−2 sin 2t)

i.e.

∂y

∂t

=−8 sin 3xsin 2t

Problem 3. Ifz=sinxyshow that

1

y

∂z

∂x

=

1

x

∂z

∂y

∂z

∂x

=ycosxy, sinceyis kept constant.

∂z

∂y

=xcosxy, sincexis kept constant.

1

y

∂z

∂x

=

(

1

y

)

(ycosxy)=cosxy

and

1

x

∂z

∂y

=

(

1

x

)

(xcosxy)=cosxy.

Hence

1

y

∂z

∂x

=

1

x

∂z

∂y

Problem 4. Determine

∂z

∂x

and

∂z

∂y

when

z=

1

√

(x^2 +y^2 )

.

z=

1

√

(x^2 +y^2 )

=(x^2 +y^2 )

− 1

2

∂z

∂x

=−

1

2

(x^2 +y^2 )

− 3

(^2) (2x), by the function of a
function rule (keepingyconstant)

−x
(x^2 +y^2 )
3
2

−x
√
(x^2 +y^2 )^3
∂z
∂y
=−
1
2
(x^2 +y^2 )
− 3
(^2) (2y), (keepingxconstant)

−y
√
(x^2 +y^2 )^3
Problem 5. Pressurepof a mass of gas is given
bypV=mRT, wheremandRare constants,
Vis the volume andTthe temperature. Find
expressions for
∂p
∂T
and
∂p
∂V
SincepV=mRTthenp=
mRT
V
To find
∂p
∂T
,Vis kept constant.
Hence
∂p
∂T

(
mR
V
)
d
dT
(T)=
mR
V
To find
∂p
∂V
,Tis kept constant.
Hence
∂p
∂V
=(mRT)
d
dV
(
1
V
)
=(mRT)(−V−^2 )=
−mRT
V^2
Problem 6. The time of oscillation,t, of a pen-
dulum is given byt= 2 π
√
l
g
wherelis the length
of the pendulum andgthe free fall acceleration
due to gravity. Determine
∂t
∂l
and
∂t
∂g
To find
∂t
∂l
,gis kept constant.