Partial differentiation 347
To find
∂t
∂l
, g is kept constant.
t = 2π
l
g
=
2π
√ g
√
l =
2π
√ g
l
1
2
Hence
∂t
∂l
=
2π
√ g
d
dl
(l
1
2 ) =
2π
√ g
1
2
l
−1
2
=
2π
√ g
1
2
√
l
=
π
lg
To find
∂t
∂g
, l is kept constant.
t = 2π
l
g
= (2π
√
l)
1
√ g
= (2π
√
l)g
−1
2
Hence
∂t
∂g
= (2π
√
l)
−
1
2
g
−3
2
= (2π
√
l)
−1
2
g 3
=
−π
√
l
g 3
= −π
l
g 3
Now try the following exercise
Exercise 138 Further problems on first
order partial derivatives
In Problems 1 to 6, find
∂z
∂x
and
∂z
∂ y
1. z =2x y
∂z
∂x
= 2y
∂z
∂ y
= 2x
2. z = x 3 − 2x y + y 2
⎡
⎢
⎢
⎣
∂z
∂x
= 3x 2 − 2y
∂z
∂ y
= −2x + 2y
⎤
⎥
⎥
⎦
3. z =
x
y
⎡
⎢
⎣
∂z
∂x
=
1
y
∂z
∂ y
=
−x
y 2
⎤
⎥
⎦
4. z =sin(4x + 3y) ⎡
⎢
⎢
⎣
∂z
∂x
= 4 cos(4x + 3y)
∂z
∂ y
= 3 cos(4x + 3y)
⎤
⎥
⎥
⎦
5. z = x
3 y
2
−
y
x 2 +
1
y
⎡
⎢
⎢
⎣
∂z
∂x
= 3x 2 y 2 +
2y
x 3
∂z
∂ y
= 2x 3 y −
1
x 2 −
1
y 2
⎤
⎥
⎥
⎦
6. z =cos 3x sin 4y
⎡
⎢
⎢
⎣
∂z
∂x
= −3 sin3x sin 4y
∂z
∂ y
= 4 cos3x cos 4y
⎤
⎥
⎥
⎦
7. The volume of a cone of height h and base
radius r is given by V =
1
3 πr 2 h. Determine
∂V
∂h
and
∂V
∂r ∂V
∂h
=
1
3
πr 2 ∂V
∂r
=
2
3
πrh
8. The resonant frequency f r in a series electrical circuit is given by f r =
1
2π
√
LC
. Show
that
∂ f r
∂ L
=
−1
4π
√
CL 3
9. An equation resulting from plucking a
string is:
y = sin
nπ
L
x
k cos
nπb
L
t + c sin
nπb
L
t
Determine
∂ y
∂t
and
∂ y
∂x
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
∂ y
∂t
=
nπb
L
sin
nπ
L
x
c cos
nπb
L
t
− k sin
nπb
L
t
∂ y
∂x
=
nπ
L
cos
nπ
L
x
k cos
nπb
L
t
+ c sin
nπb
L
t
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦