9.6 Some differential properties 269
The inverse relationships are obtained in the same way from (9.32):
(9.34a)
(9.34b)
EXAMPLE 9.17From cartesian to polar coordinates
Letz 1 = 1 f(x, y)be a function of the cartesian coordinates of a point in the xy-plane. The
position of the point is specified equally well in terms of the polar coordinates rand θ,
wherex 1 = 1 x(r, θ) 1 = 1 r 1 cos 1 θandy 1 = 1 y(r, θ) 1 = 1 r 1 sin 1 θ. Then, replacing uby rand vby θ
in equations (9.33),
(9.35)
The first of these is identical to equation (9.5) and has the same graphical interpretation.
The second is identical to the result obtained in Example 9.13 for motion around a
circle. The inverse relationships are, by equations (9.34),
(9.36)
0 Exercises 47, 48
Laplace’s equation in two dimensions
The Laplace equation in two dimensions is
(9.37)
∂
∂
∂
∂
=
22220
f
x
f
y
xxr
z
y
z
r
r
y
∂ z
∂
=
∂
∂
∂
∂
∂
∂
θ
θ
∂
∂
=
∂
∂
∂
∂
xr
y
z
rr
θ z
θ
θ
θ
θ
sin
cos
yyr
z
x
z
r
r
x
∂ z
∂
=
∂
∂
∂
∂
∂
∂
θ
θ
∂
∂
=
∂
∂
−
∂
∂
θ
θ
θ
θ
θ
x
z
rr
z
yr
cos
sin
ryrx
zz
x
xz
y
∂
∂
=
∂
∂
∂
∂
∂
∂
θθ
∂
∂
=−
∂
∂
∂
∂
ryx
y
y
z
x
x
z
θ y
θθ
∂
∂
=
∂
∂
∂
∂
∂
∂
z
r
z
x
x
r
z
y
yx
∂
∂
=
∂
∂
∂
∂
θ
θ
y
r
z
x
z
y
yx
cos ssinθ
xxu
z
y
z
u
u
y
∂ z
∂
=
∂
∂
∂
∂
+
∂
∂
v
v
∂
∂
x
y
v
yyu
z
x
z
u
u
x
∂ z
∂
=
∂
∂
∂
∂
∂
∂
v
v
∂
∂
y
x
v