If s¢ is proportional to s, there is a constant
k such that s¢(t) = k s(t). For s(t) ¹ 0,
s¢(t)/s(t) is constant.
If S(t) = c t2, S¢(t) = 2 c t. For t ¹ 0, S(t) ¹ 0, and S¢(t)/S(t) = 2/t, which is not constant.
(b)
If s(t) = (a/2) t2,
s¢(t)
= a t.
s¢¢(t)
= a.
Note that
(s¢(t))2
= (a t)2
= 2 a s(t).
12.
Speed limit at position x is L(x). Position of
A at time t is denoted by a(t).
(a)
A travels at the speed limit means: For all t,
a¢(t) = L(a(t)).
(b)
Suppose A travels at the speed limit and b(t) = a(t -1). Then b¢(t) = a¢(t-1) = L(a(t - 1)) = L(b(t)), and B travels at the speed limit.
(c)
If b(t) = a(t) - k, b¢(t) = a¢(t) = L(a(t)).
Then b¢(t) = L(b(t)) for all t, if and only if L(b(t)) = L(a(t) - k) = L(a(t)), or L(x) is periodic with period k.
18.
f is the oneoverq function. If r is a rational
number, f is not continuous at r. Thus f is not differentiable at r.
If a is an irrational number, f(a) = 0. If h is rational,
the difference quotient is 0. Thus if f¢(a) exists,
f¢(a) = 0.
Let a have the nonrepeating
decimal expansion m.a1 a2 ¼an ¼. Define the irrational number hn = -0.00¼0 an an+ 1¼, so that a + hn = m.a1 a2¼ an-1.
Now |hn| £ 101-n, |1/hn| ³ 10n-1, and f(a + hn) = 1/q with
q £ 10n-1 so that |f(a + hn)| = 1/q ³ 101-n.
It follows that
ê ê
f(a+hn)- f(a)
hn
ê ê
=
|f(a +hn)|
|hn|
=
1/q
|hn|
³ 101 -n 10n-1 = 1.
Conclude that f¢(a) does not exist.
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