8.10. THE BOUNDED CONVERGENCE THEOREM FOR THE NATURAL INTEGRAL 109
Although this simple argument can be used to defend the Riemann integral we
still remain concerned that the requirement that the limit function be assumed to
be Riemann integrable, which c
APPLIED PROJECT CALCULUS AND BASEBALL D 725
(b) From part (b)(i), 90 mi/h = 132 ft/s. Assume oo = 1) (so) = 0 and 01 = 1) (s1) 3 132 ft/s (note that $1 is the
point of release ofthe baseball). m = 3%, so the work done is
W = %mv]2 %mvg = t . T52 . (132)2
728 [3 CHAPTER) DIFFERENllALEGUATIONS
5. (a) We will assume that the difference in the birth and death rates is 20 million/y. Let t = 0 correspond to the year
1 dP
1990 and use a unit of 1 million for all calculations. k 8 ~13 E 2 5100 (
dP P l P
_ = 1 _
760 C] CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES
16. x :44t,y=21+5,0 g t 5 2.): =42(2t) =42(y5) =2y+ l4,sotheparticlemovesalongthe
liney = x + 7 from (4, 5) to (4, 9).
17. x2 + y2 : cos2 7n + sin2 7rt : l, l s t 5 2, so the particle moves coun
CHAPTER 8
SOLUTIONS FOR SELECTED EXERCISES
8.1. Chapter 1
8.1.1. Solution for Exercise 5. To evaluate
Z 1
1
p dx
|x|
0
it is enough to find explicitly an indefinite integral. (Not always possible but here
it is.)
p
1
The function F (x) = 2 x = 2x 2 has f
SECTION 10.5 THE LOGlSllC EQUATION E] 733
15. (a) dP/dt = kP cos (rt ) :> (dP)/P = keos(rt ~q)dt :> f (dP)/P = kfcos (rt (b) dt 2;
In P = (k/r) sin (rt 913) + C. (Since this is a growth modeL P > 0 and we can write In P instead ofln IPI.)
Since P (O) = P0
7.8. CHAPTER 5
85
inductively, starting with A0 = ;. Then check that
f (x) =
1
X
rk Ak (x)
k=1
at every x.
7.8.11. Proof of Theorem 6.21. Let cfw_fn be a sequence of continuous functions defined on the real line. Suppose that f is a function on R for whi
CHAPTER) REVIEW E 745
2 (a) We sketch the direction field and four solution
curves, as shown. Note that the slope y = x/y is
not dened on the line y = 0.
(b)y =x/y <2) ydy :xdx (DJ/2 =x2+C.
For C = 0, this is the pair oflines y = ix. For
_ C
746 D CHAPTER 10 DIFFERENTIAL EQUATIONS
7. Since its linear, 1 (x) = eI_2/" I" = x2 and multiplying by 1(x) gives x"2y 2y3 2 l :> (x y)/ = l
2) y(x)=x2 (fldxlC) =x2[x+C]=Cx2+x3.
d 3
:(l+x2)dx => lnl2+yl=x+i3+c1 => 2+y=kex+x3/3and
the solution is y (x) = k
732 CHAPTER 10 DIFFERENTIAL EQUATIONS
dP P m KP Pm k
(c)=kP(l k)(l$)=kP( K )( P ):E(KP)(Pm) <=>
dP k
(KP)(Pm):/Kdt'
. 1 A B
Bypartialfract1ons,(1(_P)~(-)=K_P+P_m,soA(Pm)+B(KP)=1_
l l l l
lfP: B: "fP=K,A: , p_ _
m, Kml Km SOKm/(K P+
8.8. LIMITS AND LIMITATIONS
105
and when that happens i and j0 are inside an interval of length smaller than 2 .
Consequently
f (i )
f (j0 ) < /(b
a).
The proof is easy enough then to write up. Compare it with the version for the
natural integral.
Theorem
SECTION 10.4 EXPONENTIAL GROWTH AND DECAY D 721
5. (a) Let the population (in millions) in the year t be P (1). Since the initial time is the year 1750, we substitute
t 1750 for t in Theorem 2, so the exponential model gives P (t) 2 P (1750) 314117501 The
SECTION 10.4 EXPONENTIAL GROWTH AND DECAY
dy
14 (a) Let y (t) _ temperature after! minutes Newtons Law of Cooling implies that E: k (y 5). Let
du
u(t)=y(t)5Then27=ku,sou(t)=u(0)ek'153]" => y(t)=5+152k =5
y(1) = 5+ 156 = 12 => 6" = cfw_5 => k =ln37,soy(t)
Essay: Why the Riemann integral is more difficult
Most instructors of the calculus are likely to hold the mistaken belief that the
natural integral on the real line (presented in these notes) is a more difficult choice
for a theoretical calculus course th
8.3. CHAPTER 3
93
so that k (k = 1, 2, 3, . . . ) just contains elements ([ai , bi ], i ) from that happen
also to belong to k . Most of these new subpartitions are empty: in fact since
contains only n elements, there are at most n of the k that are nonem
740 CHAPTER 10 DIFFERENTIAL EQUATIONS
2. (a) dx/dt = 0.12x 0.0006x2 + 0.00001xy. dy/dt = 0.08y + 0.00004xy.
The xy terms represent encounters between the two species x and y. An increase in y makes dx/dt (the growth
rate ofx) larger due to the positiv
SECTION10J PREDATORPHEYSYSTEMS U 739
34. Let y (t) denote the amount of chlorine in the tank at time t (in seconds). y (0) z (0.05 g/L) (400 L) = 20 g, The
amount of liquid in the tank at time t is (400 6t) L since 4 L of water enters the tank each second
The Quiz
This quiz is designed for graduate students who have just completed their study
of the Lebesgue integral and are expected to remember their freshman calculus
study of the Riemann integral. They will be unable to use the natural integral,
of cours
CHAPTER) REVIEW El 149
(C) x (insects)
60,000
(birdS) t y Both graphs have the same period and the bird
400
insects population peaks about a quarter-cycle after the insect
" 300 population.
*' 200
40.000
20,000
. r 100
>
0 x
25. (a) dx/dt = 0.4x
SECTION10.6 LINEAR EQUATIONS B 737
26. Here n = 3, P (x) = 1, Q (x) = x and setting u = y-Z, u satises u 2o = 2x. Then 1(x) : efHW = 2-21
~1/2
and u : e2 [fer"l dx +C] = 62X (xezx + ezx +C) 2x + % +Cezx. Soy = [x + i +Cezx] .
d1 d1 .
27. (a) 2E + 10] = 40
8.16. QUESTION #4
117
8.15.2. The Riemann integral. Split this into the two integrals
Z 1
Z 0
Z 1
1
1
1
p
p +
p .
=
|x|
|x|
|x|
1
1
0
Then worry. There dont seem to be any theorems about the improper Riemann
integral in our calculus book that allow this,
SECTION 10.6 LINEAR EQUATIONS C1 735
12. I (x) = ef de = elnlcosxl z cosx (since % < x < g). Multiplying the differential equation by I (x) gives
y cos .x ytanx cosx = x cosx sin 2x => (y cosx) = x cosx sin 2x. SO
I l '
y [/xcosxsiandx4C] : [/ 2xcos2xsinx
742 E1 CHAPTER! DlFFERENTIALEQUATIONS
9. (a) Letting W = 0 gives us dR/dt = 0.08R (1 0.0002R). dR/at 2 O <:> R = 0 or 5000. Since dR/dt > 0
for O < R < 5000, we would expect the rabbit population to increase to 5000 for these values of R. Since
dR/dt < 0
748
21.
22.
23.
24.
III CHAPTER10 DIFFERENTIAL EQUATIONS
Let P be the population and I be the number of infected people. The rate of spread dI/dt is jointly proportional to
hP
I and to P I, so for some constant k, dI/dt 2 H (P I) => I = W (from the
discus
722 1: CHAPTER 10 DIFFERENTIAL EQUATIONS
8. (a) The mass remaining aftert days is y (t) = y (0) e : 2003. Since (1) Y
11111111111 1101s 140 days _y(140)=20014k 100 :1 e140" 5 20
=> 140k 2 In 5 => k 2 ~ (ln2)/140. so
y (1) Z 2008-(1n2)!/140 = 200 . 2