Calculus: Analytic Geometry and Calculus, with Vectors

5.3 Second derivatives, convexity, and fiexpoints 311

17 Make a sketch showing the points (x,y) on the graph of the equations

x=a(4)-sin 0), y=a(1 -cos0)

for which ¢ = 0, 7r/2, ir, 3,t/2, and 2a. Show that the graph is convex downward
at each point for which y ; 0. Remark: The graph is a cycloid. Answers can
be simplified by use of trigonometric identities. Thus

dx

d 40

= a(l - cos 4)) = 2a sin2

2

d = a sin = 2a sin^2 cos 2

dy sin 4)

cos 2^0

dx 1 - cos ¢ = cotT1

sin 2

so dy/dx > 0 when 0 < 0 < ir and dy/dx < 0 when Tr < 4, < 27r. Moreover,
since

d dcos 2 1

dO

cot2=dt

when sin 2 0 0, we find that

sin 2 2 sin2 2

d dy d 0

d2y dO ax d¢ cot 2

dx2 dx

TO

2a sin2

2

4a sin"

2

when sin

2

54 0 and hence when y 0 0. The slope is therefore decreasing when
y00.
18 Verify that the hypotheses and conclusion of the following theorem are
satisfied when f(x) = sin x, g(x) = (sin x)/x, and a = 7r.
Theorem If a > 0, if f has two derivatives over the interval 0 5 x < a, if
f(0) > 0, if f"(x) < 0 when 0 < x < a, and if g(x) =
f(x)/x, then g is decreasing over the interval 0 < x < a.
Remark: This theorem has a very interesting geometric
interpretation. The hypotheses imply that, as in Fig-
ure 5.393, f(0) >-- 0 and the graph of f is convex down-
ward. The graph can make us feel that, as x increases,
the angle 0 must decrease and hence f(x)/x must
decrease because f(x)/x is tan 0. It is, however, neces-
sary to recognize that feelings and impressions obtained
by looking at one or a dozen figures do not constitute a proof of the theorem.
To prove the theorem, we begin by observing that, when 0 < x < a,

xf'(x) - f(x)x2 =h(x)x2,