346 Higher Engineering Mathematics
(b) To find
∂z
∂ y
, x is kept constant.
Since z =(5x 4 ) + (2x 3 )y 2 − 3y
then,
∂z
∂ y
= (5x
4
)
d
d y
(1) + (2x
3
)
d
d y
(y
2
) − 3
d
d y
( y)
= 0 + (2x
3
)(2y) − 3
Hence
∂z
∂y
= 4x 3 y − 3.
Problem 2. Given y = 4 sin3x cos 2t , find
∂ y
∂x
and
∂ y
∂t
To find
∂ y
∂x
, t is kept constant.
Hence
∂ y
∂x
= (4 cos 2t )
d
dx
(sin 3x)
= (4 cos 2t )(3 cos3x)
i.e.
∂y
∂x
= 12 cos 3x cos 2t
To find
∂ y
∂t
, x is kept constant.
Hence
∂ y
∂t
= (4 sin 3x)
d
dt
(cos 2t )
= (4 sin 3x)(−2 sin2t )
i.e.
∂y
∂t
= −8 sin 3x sin 2t
Problem 3. If z =sin x y show that
1
y
∂z
∂x
=
1
x
∂z
∂ y
∂z
∂x
= y cos x y, since y is kept constant.
∂z
∂ y
= x cos x y, since x is kept constant.
1
y
∂z
∂x
=
1
y
( y cos x y) = cos x y
and
1
x
∂z
∂ y
=
1
x
(x cos x y) = cos x y.
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 (keeping y constant)
=
−x
(x 2 + y 2 )
3
2
=
−x
(x 2 + y 2 )
3
∂z
∂ y
= −
1
2
(x
2
+ y
2
)
−3
2 (2y), (keeping x constant)
=
−y
(x 2 + y 2 )
3
Problem 5. Pressure p of a mass of gas is given
by pV = mRT, where m and R are constants, V is
the volume and T the temperature. Find expressions
for
∂ p
∂T
and
∂ p
∂V
.
Since pV = mRT then p =
mRT
V
To find
∂ p
∂T
, V is kept constant.
Hence
∂ p
∂T
=
mR
V
d
dT
(T ) =
mR
V
To find
∂ p
∂V
, T is kept constant.
Hence
∂ p
∂V
= (mRT)
d
dV
1
V
= (m RT )(−V
−2
) =
−mRT
V 2
Problem 6. The time of oscillation, t , of
a pendulum is given by t = 2π
l
g
where l is the
length of the pendulum and g the free fall
acceleration due to gravity. Determine
∂t
∂l
and
∂t
∂g