Page 134 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 6] PARTIAL DERIVATIVES 125

Under the transformation (10)a closed region r of the xy plane is, in general, mapped into a closed
region r of the uv plane. Then if A xy and A uv denote respectively the areas of these regions, we can
0
show that

A xy @ðx; yÞ
lim ¼ ð11Þ

A uv @ðu; vÞ
where lim denotes the limit as A xy (or A uv ) approaches zero. The Jacobian on the right of (11)is
often called the Jacobian of the transformation (10).
If we solve (10) for u and v in terms of x and y,we obtain the transformation u ¼ f ðx; yÞ, v ¼ gðx; yÞ
often called the inverse transformation corresponding to (10). The Jacobians @ðu; vÞ and @ðx; yÞ of these
@ðx; yÞ @ðu; vÞ
transformations are reciprocals of each other (see Problem 6.43). Hence, if one Jacobian is different
from zero in a region, so also is the other.
The above ideas can be extended to transformations in three or higher dimensions. We shall deal
further with these topics in Chapter 7, where use is made of the simplicity of vector notation and
interpretation.

CURVILINEAR COORDINATES
If ðx; yÞ are the rectangular coordinates of a point in the xy plane, we can think of ðu; vÞ as also
specifying coordinates of the same point, since by knowing ðu; vÞ we can determine ðx; yÞ from (10). The
coordinates ðu; vÞ are called curvilinear coordinates of the point.

EXAMPLE. The polar coordinates ð ; Þ of a point correspond to the case u ¼ , v ¼ . In this case the
transformation equations (10)are x ¼ cos , y ¼ sin .
For curvilinear coordinates in higher dimensional spaces, see Chapter 7.

MEAN VALUE THEOREM
If f ðx; yÞ is continuous in a closed region and if the first partial derivatives exist in the open region
(i.e., excluding boundary points), then
0 < < 1
f ðx 0 þ h; y 0 þ kÞ f ðx 0 ; y 0 Þ¼ hf x ðx 0 þ h; y 0 þ kÞþ kf y ðx 0 þ h; y 0 þ kÞ ð12Þ
This is sometimes written in a form in which h ¼ x ¼ x x 0 and k ¼ y ¼ y y 0 .

Solved Problems

FUNCTIONS AND GRAPHS
1 2
2
3
6.1. If f ðx; yÞ¼ x 2xy þ 3y , find: (a) f ð 2; 3Þ; ðbÞ f ; ; ðcÞ f ðx; y þ kÞ f ðx; yÞ ;
x y k
k 6¼ 0.
2
3
f ð 2; 3Þ¼ð 2Þ 2ð 2Þð3Þþ 3ð3Þ ¼ 8 þ 12 þ 27 ¼ 31
ðaÞ
3 2
1 2 1 1 2 2 1 4 12
f ; 2 þ 3
x y x x y y x xy y
ðbÞ ¼ ¼ 3 þ 2