9.6 Some differential properties 263
EXAMPLE 9.12Givenz 1 = 1 x
21 + 1 y
3, wherex 1 = 1 e
tandy 1 = 1 e
−t, finddz 2 dt.
(i) By substitution:
(ii) By the chain rule (9.21), we have
Therefore,
and this is identical to the result obtained by substitution.
0 Exercises 37–39
EXAMPLE 9.13Walking on a circle
The equation of a circle in the xy-plane with centre at the origin and radius ais
x
21 + 1 y
21 = 1 a
2. A displacement on the circle is most easily described when the equation
of the circle is expressed in terms of the polar coordinates rand θ:
x 1 = 1 a 1 cos 1 θ, y 1 = 1 a 1 sin 1 θ
These have the form of the pair of equations (9.19), with treplaced by θ. In general,
such equations are called the parametric equationsof a curve. Letz 1 = 1 f(x, y). Then,
by equation (9.21), sincedx 2 dθ 1 = 1 −yanddy 2 dθ 1 = 1 x,
For example, ifz 1 = 1 xythen and
0 Exercise 40
dz
d
xy
θ
=−
22∂
∂
=
∂
∂
=
z
x
y
z
y
, ,x
dz
d
z
x
dx
d
z
y
dy
d
y
z
yx y
θθθ
=
∂
∂
+
∂
∂
=−
∂
∂xx
x
z
y
x
+
∂
∂
dz
dt
z
x
dx
dt
z
y
dy
dt
= xx y y
∂
∂
∂
∂
=×+ ×−=()()( )()23
2223
23xy−
∂
∂
=,
∂
∂
= = = , =− =−
−z
x
x
z
y
y
dx
dt
ex
dy
dt
ey
tt23
2and
ze e
dz
dt
ee xy
tt t t=+ , = − = −
23 −−2 3 2323 23