Calculus: Analytic Geometry and Calculus, with Vectors

10.2 Polar curves, tangents, and lengths 545

are orthonormal vectors because they are unit vectors and their scalar (or dot)
product is zero. Henceforth we consider only time intervalsover which p(t) > 0.
When motion of the particle P is being considered, the first vector in (4) is said
to be radial because it has the direction of the "radius vector" from the origin
to P, and the second vector is transverse because it is orthogonal to the radius
vector. Thus the right member of (3) displays, in order, the radial and the
transverse vector components of the acceleration of P. This acceleration is said
to be radial (the kind produced by a "central force field" having its center at
the origin) when its transverse component is zero, that is,

(5) P(t)O"(t) + 2p'(t)4,'(t) = 0.

This is another one of those derivative equations that is called a differential
equation and from which information can be extracted. Since p(t) > 0, the
left member of (5) is zero if and only if the product of it and p(t) is zero, that is,

(6) [P(t)]20"(i) + 2P(t)P'(t)O'(t) = 0.

The virtue of (6) lies in the fact that it can be put in the form

(7) dt{{P(t)]20'(t)) = 0,

and this fact should be carefully checked. The virtue of (7) lies in the fact that

it holds over an interval of values oft if and only if there is a constant c such that

(8) .{PUT 1(t) = c
for each t in the interval. The physical significance of (8) will be revealed in
Section 10.3; it is an important fact that (8) holds if and only if the radius vector
from the origin to the particle P sweeps over regions of equal area in time intervals
of equal lengths.
9 A circular race track has cylindrical equations p = a and z = 0, and it
has rectangular equations x2 + y2 = a2 and z = 0. A wheel of radius b rolls,
without slipping, around the track. The center of the wheel is always above the
track, it travels with constant speed, and it makes a circuit of the track in T
minutes. At time t = 0 the center of the wheel is going in the direction of the
positive y axis, and a pink tack P in the tire on the wheel is at the point having
rectangular coordinates (a,0,0). Letting r,(t) denote the vector running from
the origin 0 to the center Q of the wheel at time t, show that

(1) rl(t) = a(cos2Tti-+ - sin2Ttj) +bk.

Letting x(t), y(t), z(t) denote the rectangular coordinates of the pink tack P at
time t, show that
2aat
(2) z(t) = b (1 - cos bT).

Using the fact that the line PQ (a spoke of the wheel) is perpendicular to the

horizontal line from Q to the z axis, obtain the formula

(3) x(t) cos T + y(t) sin 2T = a 2