
4-56 Calculus – Differentiation and Integration
Multiplying Eq. (1) by y, and Eq. (2) by x and adding, we get
y
Z
u
x
Z
v
y x
Z
y
e v v
Z
y
u
∂
∂
+
∂
∂
= +
∂
∂
= +
∂
∂
( )
(sin cos )
2 2
2 2 2
∴
∂
∂
+
∂
∂
=
∂
∂
y
Z
u
x
Z
v
e
Z
y
u2
.
Example 4.52 If u x y= −
−
sin ( ),
1
where x t y t= =3 4
3
and , then show that
du
dt
t
=
−
3
1
2
.
Solution:
du
dt
u
x
dx
dt
u
y
dy
dt
x y x y
t
=
∂
∂
+
∂
∂
=
− −
⋅ ⋅ +
− −
⋅ − ⋅
1
1
1 3
1
1
1 12
2 2
( ) . ( ) .
( ) (
22
2
2
2
2 2
2
2 4
3 12
1
3 12
1 2
3 1 4
1 9 24 16
)
( )
( )
=
−
− −
=
−
− + −
=
−
− + −
t
x y
t
x xy y
t
t t
tt
t
t t t
t
t t
6
2
2 2 4
2
2 2 2
3 1 4
1 1 8 16
3 1 4
1 1 4
=
−
− − +
=
−
− −
( )
( )( )
( )
( )( )
∴ =
−
du
dt
t
3
1
2
.
4.14 RATES AND THEIR APPLICATIONS
We have studied derivative as the rate of change and seen that the position of ...