11.2 Increments, chain rule, and gradients 575
Po is obtained by putting sf = 00 in (5), differentiating with respect to 0, and
putting 0 = Bo in the result. Thus,
(7) t2 = a(cos Oo cos Ooi + sin Oo cos Ooj- sin 0ok).
While the north and south poles of a sphereare as good as any other points on
the sphere, it often happens that methods involving spherical coordinates refuse
to give information about them. When 0 < 00 < r, the two vectors tj and t2
are nonzero nonparallel tangents to S at the point Po, and with the aid of the
fact that S has a tangent plane and a normal at Po we conclude that the vector
product of t2 and to must be a normal to S at Po. Letting N = t2 X ti, we find
that N = a2 sin Oro, where ro is the vector running from the origin to Po. This
shows that ro is normal to Sat Po. In working out this elementary fact we have
shown how, in at least one case, (3) can be used to obtain information about the
surface which it represents.
(^23) As in (5) of the preceding problem, let r have its tail at the origin and let
r = a(cos 0 sin Oi + sin 0 sin Oj + cos Ok).
Let 0 and 0 be differential functions of t so the tip of r traverses a curve C on
the sphere S as t increases. Find r'(t). 11ns.:
r'(t) = aO'(t)[cos (k cos 0i + sin 0 cos Oj - sin Ok]
- ao'(t) sin 0[- sin q5i + cos Oil.
24 Let r have its tail at the origin and let
(1) r = (b + a cos 0) cos 4i + (b + a cos 0) sin c&j + a sin Bk.
Let 0 and 0 be differentiable functions of t so that, as we can see with the aid of
Problem 22 at the end of Section 2.2, the tip of r traverses a curve C on a torus
T as t increases. Show that
(2) r'(t) = a0'(t)[- sin 0 cos cbi - sin 0 sin Oj + cos Ok] - (b + a cos 0),'(t)[- sin ¢i + cos q5j].
Show that the two vectors in brackets are orthogonal. Work out the formula
n = cos 0 cos 01 + sin 0 cos Oj + sin Ok
for the unit normal to the torus.
25 This problem illustrates the fact that there are situations in which the
elegant and useful chain formula
(1) dt u(x,Y) = v (x,Y)
dx+ uy(x,Y)
dt
cannot be applied with impunity. Let
(2) u(0,0) = 0, u(x,Y) =x2T2y2 (x2 + Y= s 0).
Let a and b be nonzero constants and let x = at and y = bt. Show that, when
t = 0, the left member of (1) is ab2/(a2 + b2) and hence is not 0 while the right
member is 0. Remark: Because (1) is invalid, Theorem 11.24 implies that uz
and uy cannot be continuous at the origin.