Calculus: Analytic Geometry and Calculus, with Vectors

10.3 Areas and integrals involving polar coordinates 551

in such a way that 0 is a function of t having two derivatives, then the displace-

ment -vector and velocity of P are

(2) r(t) =

(3) v(t) =

a

1 - e cos fi(t)[cos 0(t)i + sin 0(t)j]

-ea4'(t) sin q5(t)

[1 - e cos 0(t)12[cos 0(t)i sin ¢(t)j]

+

aO'(t)

sin 4'(t)i + cos O(t)j].

1 - e cos q5(t)

For the case in s hich

(4) '[p(t)]201(t) = C,

so that the radius vector from 0 to P sweeps over equal areas in time intervals

having equal lengths, show that

(5) 0'(t) = [p(t)]2=a2 [1- e cos 0(t)]2

(6) v(t) _ -Zeesin 0(t)[cos q5(t)i + sin 4(t)j]

+ a [1 - e cos 4(t)][- sin 4(t)i + cos 4(t)j]

z

(7) a(t) _ -

a[p(t)]2

[cos ¢(t)i + sin q5(t)j].

Remark: This shows that if a particle P moves along a conic in such a way that
the radius vector from the focus to P sweeps over equal areas in time intervals
having equal lengths, then the particle is always accelerated toward the focus
and the magnitude of the acceleration is inversely proportional to the square
of the distance from the focus to the particle. Kepler discovered that, except
for minor perturbations, the planets move in ellipses with the sun at a focus
and that the "equal areas" property holds. As we did in this problem, Newton
used these laws to derive his famous inverse square law of gravitation.
13 Let S be a bounded convex set in the xy plane and let the origin 0 be an
inner point of S. (Basic definitions are given in a remark at the end of Problems
5.19.) Prove that to each q5 there corresponds exactly one positive number
/(0) such that the point P having polar coordinates (f(45), 0) lies on the boundary
B of S. Solution: Choose positive numbers S and R such that the circular disk
with center at 0 and radius 26 is a subset of S and S is a subset of the circular
disk with center at 0 and radius R. Let 0 be a given angle. Let f(4,) be the
least upper bound of numbers po such that S contains each point having polar
coordinates (p,4) for which 0 < p < po. Then 2S S f(4,) 5 R. Let P be the
point having polar coordinates (f(4,), ¢) and, as in Figure 10.392, let Q' and QZ
be the points of tangency of the lines through P tangent to the circle of radius
S having its center at the origin. With the possible exception of the point P
itself, each point of the line segment OP is a point of S. Since Ql and Q. are also
points of S and since S is convex, it follows that each point inside the triangle