6.5-3. First order chemical reactions.
Sol :
In some chemical reactions, the rate at which the amount of a substance change with
time is proportional to the amount present. For the change of -gluconolactone into
gluconic acid, for example,
dy = ;0:6y
dt
when t is measured in hours. If there are 100 grams of -gluconolactone present
when t = 0, how many grams will be left after the rst hour?
y = y e; : t . When t = 0, y = 100, so y = 100: When t = 1, y = 100e; : 54:88:
0
06
0
06
6.5-13. Continuously compounded interest.
You have just placed A dollars in a bank account that pays 4% interest, conpounded
continuously.
(a) How much money will you have in this account in 5 years?
(b) How long will it take your money to double? to triple?
(a) A(t) = A ert
A(5) = A e : = A e : = 1:22A :
(b) A(x) = A e : x = 2A :
e : x = 2:
x = 17:33: A(y) = A e : y = 3A :
e : y = 3:
x = 27:47:
0
Sol :
0
0 04 5
0
0 04
0
0 04
02
0
0
0
0
0 04
0 04
0
6.5-18. Polonium-210.
Sol :
The half-time of polonium is 139 days, but your sample will not be useful to you after
95% of the radioactive nuclei present on the day the sample arrives has disintegrated.
For how many days after the sample arrives will you be able to use polonium?
y = y;3 e; ;3 t = 0:95y
e; t = 0:95
t 10:26:
5 10
0
5 10
0
6.5-21. Cooling soup.
Sol :
Suppose that a cup of soup cooled from 90oC to 60oC after 10 minutes in a room
whose temperature was 20oC . Use Newton's law of cooling to answer the following
questions:
(a) How much longer will it take to the soup to cool to 35oC ?
(b) Instead of being left to stand in the room, the cup of 90oC soup is put in a
freezer whose temperature is ;15o C . How long will it take the soup to cool
from 90oC to 35oC ?
T ; Ts = (T ; Ts)e;kt
T = 20 + (90 ; 20)e;kt = 20 + 70e;ktT = 60 when t = 10 Hence, 60 = 20 +
70e; k e; k = k = 0:0566:
(a) 35 = 20 + 70e;ktt 27:53
(b) 35 = ;15 + 105e;ktt 13:26
1
10
10
4
7
6.5-25. The age of Crater Lake.
The charcoal from a tree killed in the volcanic eruption that formed Crater Lake
in Oregon contained 44:5% of carbon-14 found in living matter. About how old is
Crater Lake?
k=
e;ktlog =2 0:445
e; 5700 t = 0:445
t 6658:3.
6.6-21. Use L'H^opital rule to nd the limit: lim 3 ; 1
log 2
5700
Sol :
sin
!0
3 ; 1 = lim log 3 3 cos = log 3
: lim
!
!
1
sin
Sol
0
0
1 ; 1
6.6-35. Use L'H^opital rule to nd the limit: xlim
! + x ; 1 log x
1 ; 1
: xlim
! + x ; 1 log x
1
Sol
1
log x ; x + 1
= xlim
! + (x ; 1) log x
1
= xlim
!+
;1
log x + 1 ; x
= xlim
!+
+ x2
x
1
1
= ;21
1
x
1
;21
1
x
1
Zx
1
6.6-38. Use L'H^opital rule to nd the limit: xlim
[log t]dt
!1 x log x
Sol
1
Zx
1
[log t]dt
: xlim
!1 x log x
x
(
t
log
t
;
t
)
j
= xlim
!1 x log x
x log x ; x + 1
= xlim
!1
x log x
log x + 1 ; 1
= xlim
!1 log x + 1
1
1
1
x
= xlim
!1
1
=1
x
6.6-51. Find the limit limx! xx
0
2
Sol :
Let y = xx ) log y = x log x ) y = ex
log x
lim
xx
!
x
0
= xlim
ex
!
log x
0
= exp(lim
x log x) (since exponential is continuous)
x!
= exp(lim log x )
0
!0
1
x
x
1
x
) (by L'H^opital rule)
= exp(lim
x! ;
0
1
x2
= exp(lim
;x)
x!
=e
=1
0
0
6.6-61. Fnd a value c that makes the function f (x) =
Sol :
at x = 0. Explain why your value of c works.
(a) limx! f (x)
9x ; 3 sin 3x
= xlim
!
5x
9 ; 9 cos 3x
= xlim
!
15x
27 sin 3x
= xlim
! 30x
81 cos 3x
= xlim
!
30
81
= 30
= 27
10
=c
(b) limx! + f (x)
0
0
3
0
3
0
0
0
= xlim
f (x)
!;
0
= 27
10
=c
= f (0)
Hence f is continuous at x = 0.
3
( 9x;3 sin 3x
;
3
5x
c;
=0
continuous
x = 0
x
6
6.6-63. The continuous compound interest formula.
In deriving the formula A(t) = A ert in Section 6.5, we claimed that
0
r )kt = A ert :
lim
A
(1
+
k !1
k
0
The equation will hold if
0
r )kt = ert:
lim
(1
+
k !1
k
and this, in turn, will hold if
r )k = er :
lim
(1
+
k !1
k
As you can see, the limit leads to the indeterminate form 11. Verify the limit using
L'Hopital rule. item [ ]We want to verify that limx!1 (1 + xr )x = er :
Let y = (1 + xr )x ) log y = x log(1 + xr ) ) y = ex
. Then
Sol :
r
log (1+ x
)
r )x
lim
(1
+
x!1
x
r ))
= xlim
exp(
x
log(1
+
!1
x
= exp(xlim
x log(1 + xr )) (since exponential is continuous)
!1
= exp(xlim
!1
log xx r
+
)
1
x
log(x + r) ; log x )
= exp(xlim
!1
1
;x
1
+
1
x r
= exp(xlim
!1 ;2
x
1
x
) (by L'H^opital rule)
x ; (x + r) (;x ))
= exp(xlim
!1 x(x + r)
2
rx )
= exp(xlim
!1 x(x + r)
r
=e
2
6.7-7. Order the following functions from the slowest growing to the fastest growing as
x ! 1.
ex = lim e 2 = 1; ex grows faster than e 2 :
(a) Since xlim
!1 e 2 x!1
x
Sol :
x
x
x = lim 1 = 1; xx grows faster than (log x)x:
(b) Since xlim
!1 log x x!1 x
(c) Since lim log x = 1; (log x)x grows faster than ex:
1
e
Hence d < a < c < b.
!1
x
4
6.7-10. True or false? As x ! 1.
(a) x = O( x )
(c) x ; x2 = o( x )
(e) ex + x = O(ex)
(g) ln ln x = O(ln x)
1
Sol :
1
+3
(b) x + x2 = O( x )
(d) 2 + cos x = O(2)
(f) x ln x = o(x )
(h) ln x = o(ln(x + 1))
1
1
1
1
(a) True. xlim
!1
1
x+3
1
x
1
x
(b) True. xlim
!1
1
1
2
2
=1
+ x2
1
1
x
=1
; x2 = 1
(c) False. xlim
!1
1
1
x
1
x
2 + cos x 2
(d) True. xlim
!1
2
x
(e) True. lim e + x = 1
!1
x
ex
x log x = 0
(f) True. xlim
!1 x
log log x = 0
(g) True. xlim
!1 log x
log x = 1
(h) False. xlim
!1 log(x + 1) 2
2
2
6.7-21. (a) Show that1 log x grows slower as x ! 1 than x 1 for any postive integer n,
n
even x 1 000 000 .
;
;
(b) CALCULATOR
1
Although the values of x 1 000 000 eventually overtake the values of log x, you
have to go way out on 1the x-axis before this happens. Find a value of x greater
than 1 for which x 1 000 000 > log 1x. You might start by observing that when
x > 1 the equation log x = x 1 000 000 is equivalent to the equation log log x =
log x=1; 000; 000.
log x = lim x x 1 ; = lim 1 x 1 = 0 for any positive integer n:
: (a) xlim
x!1
!1 1
x!1
;
;
;
;
;
;
1
Sol
x
n
1
n
1
n
1
n
n
Therefore, log x grows slower than x 1 as x ! 1.
6
(b) ln(e ; ; ) = 17; 000; 000 < (e 6 1 10 = e 24; 154; 952:75
6.7-23. (a) Suppose you have three dierent algorithms for solving the same problem and
each algorithm takes a number of steps that is of the order of one of the
functions listed here:
n log n; n = ; n(log n) :
Which of the algorithms is the most ecient in the long run? Give reasons for
your answer.
5
n
17 000 000
17
2
10 )
=
2 3
17
2
2
(b) Grapher Graph the functions in part (a) together to get a sense of how rapidly
each one grows.
Sol
: (a)
n log n = lim 1 = 0
lim
n!1 log n
n!1 n(log n)
2
2
2
2
) n log n grows slower than n(log n) ;
2
2
2
( n) 1
(n)
n
log
n
2 lim 1 = 0
lim
=
lim
=
lim
=
=
=
;
=
n!1
n
!1
n
!1
n
n
ln 2
ln 2 n!1 n =
( )n
ln
ln 2
1 2
2
3 2
1
1
2
1 2
1 2
) n log n grows slower than n = . Therefore, n log n grows at the slowest rate
) the algorithm that takes O(n log n) steps is the most ecient in the long
3 2
2
2
2
run.
(b) Graph of the functions y = n log n; y = n = ; y = n(log n) :
2 3
2
2
2
y
4500
4000
3500
2
y = n(log2n)
3000
2500
2000
1500
2/3
y=n
1000
500
y = nlog n
2
n
0
O
−500
0
10
20
30
40
50
6
60
70
80
90
100