Calculus: Analytic Geometry and Calculus, with Vectors

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.