Calculus: Analytic Geometry and Calculus, with Vectors

582 Partial derivatives

These formulas and the simple formula

(4)

1 a2u 1 a2u

p2

a4)_

r2 sin2 34)2

enable us to transform important expressions from cylindrical coordinates p, 4, z

to spherical coordinates r, ¢, B. To put fundamental consequences of our results

in the compact form which is very often used, we define V2u (read del squared u),

the Laplacian of u, by the first of the formulas

a2u a2u a2u

(5) 02u- ax2 + ay2 + az2

a2u I all 1 a2u a2u

(6) 211 = a p2 +Pa p +p2 a02+ az2

a2u 2 au I a2u cos 0 au^1 a2u

(7)

02u

= art + r ar + r2 a6_+ r2 sin 6 TO + r2 sin2 9

Then (5) gives V2u in terms of rectangular coordinates x, y, z. As Problem 4

showed, (6) gives V2u in terms of the cylindrical coordinates p, ¢, z of Figure

10.11. As we see by adding (2), (3), and (4), the formula (7) gives V2u in terms

of the spherical coordinates r, 4), 0 of Figure 10.12.

6 The results of Problems 3, 4, and 5 are important. Instead of proposing

that similar but less important problems be solved, the author suggests that

these problems be solved again and again.

7 This long problem involves a theorem which is called an implicit function

theorem. The equation

(1) x3+xy+y3-31=0,

which happens to be satisfied when x = 3 and y = 1, provides an introduction

to the subject. If we know that y is a differentiable function of x for which (1)

holds, then we can differentiate with respect to x to obtain

(2) 3x2+xdx+y+3y2dx

=0

or

(3) (3x2 + y) + (x + 3y2) dx = 0

and hence

(4)

dy _3x2 + y

dx x + 3y2

provided x + 3y2 0 0. We can be pleased by our abilities to calculate deriva-

tives of differentiable functions, but we can also be irked and frustrated when we

realize that we do not know whether there is a differentiable function f for which

(5) x3 + xf(x) + [f(x)]3 - 31 = 0.

With the aid of partial derivatives, we can obtain very satisfying information

about hordes of problems of which the one considered above is a special case.