PARTIAL DIFFERENTIATION 345G
t= 2 π√
l
g=(
2 π
√
g)√
l=(
2 π
√
g)
l1
2Hence
∂t
∂l=(
2 π
√
g)
d
dl(l1(^2) )=
(
2 π
√
g
)(
1
2
l
− 1
2
)
(
2 π
√
g
)(
1
2
√
l
)
π
√
lg
To find
∂t
∂g
,lis kept constant
t= 2 π
√
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 139 Further problems on first
order partial derivatives
In Problems 1 to 6, find
∂z
∂x
and
∂z
∂y
1.z= 2 xy
[
∂z
∂x
= 2 y
∂z
∂y
= 2 x
]
2.z=x^3 − 2 xy+y^2
⎡
⎢
⎣
∂z
∂x
= 3 x^2 − 2 y
∂z
∂y
=− 2 x+ 2 y
⎤
⎥
⎦
3.z=
x
y
⎡
⎢
⎣
∂z
∂x
1
y
∂z
∂y
−x
y^2
⎤
⎥
⎦
4.z=sin (4x+ 3 y)
⎡
⎢
⎣
∂z
∂x
=4 cos (4x+ 3 y)
∂z
∂y
=3 cos (4x+ 3 y)
⎤
⎥
⎦
5.z=x^3 y^2 −
y
x^2
- 1 y ⎡ ⎢ ⎢ ⎣
∂z
∂x
= 3 x^2 y^2 +
2 y
x^3
∂z
∂y
= 2 x^3 y−
1
x^2
−
1
y^2
⎤
⎥
⎥
⎦
6.z=cos 3xsin 4y
⎡
⎢
⎢
⎣
∂z
∂x
=−3 sin 3xsin 4y
∂z
∂y
=4 cos 3xcos 4y
⎤
⎥
⎥
⎦
- The volume of a cone of heighthand base
radiusris given byV=^13 πr^2 h. Determine
∂V
∂h
and∂V
∂r [
∂V
∂h=1
3πr^2∂V
∂r=2
3πrh]- The resonant frequencyfrin a series electrical
circuit is given byfr=1
2 π√
LC. Show that
∂fr
∂L=− 1
4 π√
CL^3- An equation resulting from plucking a
string is:
y=sin(nπL)
x{
kcos(
nπb
L)
t+csin(
nπb
L)
t}Determine∂y
∂tand∂y
∂x
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
∂y
∂t=nπb
Lsin(nπL)
x{
ccos(
nπb
L)
t−ksin(
nπb
L)
t}∂y
∂x=nπ
Lcos(nπL)
x{
kcos(
nπb
L)
t+csin(
nπb
L)
t}⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦- In a thermodynamic system,k=Ae
TS−H
RT ,
whereR,kandAare constants.Find (a)∂k
∂T(b)∂A
∂T(c)∂(S)
∂T(d)∂(H)
∂T