128 Functions of Several Variables(c) Let r = r(u,v) be a parametric representation of the surface. As-
sume that the partial derivatives ru and rv exist, are continuous, and are
not equal to zero at a point (UQ,VO). Let Po be the point on the sur-
face corresponding to the position vector T{UQ,VQ), andC, a smooth curve
on the surface passing through the point PQ. (i) Verify that the vector
ru{uo,vo) x rv(uo,vo) is perpendicular to the unit tangent vector to C at
Po- Hint: C is represented by a vector function r(u(t),v(t)) for suitable functions
u,v; at the point Po this curve has a tangent vector ruv! + rvv'. (ii) Write the
equation of the tangent plane to the surface at the point Po. Hint: the
plane has ru(uo,vo) x rv(uo,vo) as the normal vector.
The same surface can have different representations. The graph z =
f(x,y) is a level set F(x,y,z) = 0 with F(x,y,z) — f(x,y) — z or
F(x,y,z) = z — f(x,y). The same graph can also be written in the
parametric form r(u,v) — ui + vk + f(u,v)k. Similarly, the vector
r — x(u, v) i + y(u, v)j+ z(u, v) k defines a parametric representation of
the level set F(x, y,z) = c if and only if F(x(u, v), y(u, v), z(u, v)) = c for
all (u,v). In particular, one can write the level set in parametric form by
solving the equation F(x,y,z) = c for one of the variables x,y,z. FOR
EXAMPLE, the hemisphere x^2 + y^2 + z^2 = 4, x > 0, can be written as
x = y/A — y^2 — z^2. Taking u = y and v = z, we find the equivalent para-
metric representation r(u, v) = %/4 — u^2 — v^2 i+uj+v K, where u^2 +v^2 < 4.
The most convenient representation of the given surface depends on the
particular problem.
EXERCISE 3.1.10.c Verify that the surface x = ^/4 — y^2 — z^2 can be rep-
resented as r(u,v) = 2cosvz + 2cosusinvj + 2sinusint;K, 0 < u < 2TT,
0<v< IT/2.
EXAMPLE. Consider the function f(x, y) — 2x^2 +3y^2. This function has
continuous partial derivatives of every order everywhere in K^2 : fx = Ax,
fxx = 4, fy = 6y, fyy = 6, and all other derivatives are equal to zero. Thus,
Vf(x,y) = Axi + 6yj. Consider the point A = (1,2). Then V/(A) =
Al + 12 j, so that the most rapid increase of / at A is in the direction of
i + 3J, with the rate ||Vf(A)\ — Ay/16. Similarly, the most rapid decrease
of / at A is in the direction of -i — 3j, with the rate -|| V/(J4)|| = —4y/I6.
The rate of change of / at A in the direction toward the origin is -28/\/5
and is computed by (3.1.6) as follows: B — (0,0), AB = — i — 2j, UB =
-(i+2j)/y/5, Vf(A) • uB = -28/V5.
Next, consider the surface z = 2x^2 + 3y^2. Writing 2x^2 + 3y^2 — z = 0,