Solutions

Problems from Anton, x10.1
27. Solve dy
dx
; xy = x, y(0) = 3.
Math 120 Fall 2000|HW #8
First, we rewrite the dierential equation as:
dy
dx
Then we separate and integrate:
Z
dy
y +1 = xdx =)
Evaluating the integrals:
= xy + x = x(y +1):
Z
dy
y +1 =
xdx:
ln jy +1j = x2
2 + C
where C is any real number. Taking e to both sides:
So,
jy +1j = e x2
2 +C = e C e x2
2 :
y +1=e C e x2
2 =) y = e C e x2
2 ; 1:
We have missed one possible solution in our derivation so far we divided by
y + 1 in the second step, and thus lost the constant solution y = ;1, which
does indeed satisfy the original dierential equation. So, the solution to the
dierential equation is:
y = e C e x2 2 ; 1 or y = ;1:
Since C is any real number, e C is any non-zero number, and thus we may
encapsulate all of the above solutions into the simpler form:
y = C 1 e x2
2 ; 1
where C 1 is any real number. Now, since y(0) = 3,
or C 1 = 4. So,
3 = C 1 e 0 ; 1
y = 4e x2
2 ; 1:
1

43. At time t = 0, a tank contains 25 ounces of salt dissolved in 50
gallons of water. Then brine containing 4 ounces of salt per gallon of
brine is allowed to enter the tank at a rate of 2 gallons per minute
and the mixed solution is drained from the tank at the same rate.
Let s(t) be the amount of salt (in oz) present in the tank at time t. The rate of
inow of salt is (2 gal/min)(4 oz/gal), or 8 oz/min. At timet, the tank contains
s(t) oz of salt in 50 gal, or s oz/gal. Thus, the outow ofsalt is (2 gal/min)( s 50 50
oz/gal), or s oz/min. So, the dierential equation is:
25
We separate and integrate:
ds
8 ; s
25
ds
= 8 ; s
dt 25 :
= dt =)
Z
ds
8 ; s
25
We perform the integration (using a u = 8 ; s
25
ln j8 ; s
25 j
;1=25
=
Z
dt:
substitution not shown here):
= t + C =) ln j8 ; s
25 j = ; t
25 ; C 25
where C is any real number. Taking e to both sides:
Thus,
Now, at t = 0, s = 25, so
so e ;C=25 = 7. Thus,
Thus,
j8 ; s
25 j = e; t
25 ; C 25 = e
;C=25 e
;t=25 :
8 ; s
25 = e;C=25 e ;t=25 :
8 ; 25
25 = e;C=25 e 0
8 ; s
25 = 7e;t=25 :
8 ; 7e ;t=25 = s
25 =) 200 ; 175e;t=25 = s:
At t = 25, s = 200 ; 175e ;1 = 135:6 ounces.
2

55. Suppose that a particle moving along the x-axis encounters a
resisting force that results in an acceleration of a = dv=dt = ;0:04v 2 .
Given that x = 0 cm and v = 50 cm/s at time t = 0, nd the velocity
v and position x as a function of t for t 0.
We know that
Separating and integrating:
dv
dt = ;0:04v2 :
Z
;25 dv
= v dt =) ;25 dv
2 v 2
= Z
dt =) 25
where C is any real number. Since at t = 0, v = 50, we have:
So,
Taking one over each side:
v
25 = 1
t + 1 2
25
50 = 0+C =) C = 1 2 :
25
To nd x, we note that v = dx
dt . So,
Separating and integrating:
v = t + 1 2 :
= 2
2t +1
dx
= 50
dt 2t +1 :
dx = 50dt
2t +1 =) Z
v
=) v =
50
2t +1 :
dx =
Z
50dt
2t +1 :
= t + C
Performing the integrals (using a substitution u = 2t +1not shown):
x =
50 ln j2t +1j
2
+ C = 25 ln(2t +1)+C:
(we drop the absolute values since 2t +1 > 0 for t 0). Since x = 0 when
t = 0,
0 = 25 ln(1) + C =) C = 0:
3

Thus,
x = 25 ln(2t +1):
Problems from Anton, x10.3
29. A cup of water with a temperature of 95 C is placed in a room
with a constant temperature 21 C. After one minute, the water is
85 C. When will it be 51 C?
The basic law of cooling is:
dT
dt
= k(T ; 21):
Separating and integrating:
Z
dT
T ; 21 = kdt =)
Evaluating the integrals:
Z
dT
T ; 21 =
kdt:
ln jT ; 21j = kt + C
where C is any real number. We know that when t = 0, T = 95, so
So,
ln 74 = 0+C =) C = ln 74:
ln jT ; 21j = kt +ln74:
We may drop the absolute values, since clearly the water temperature will
always be greater than 21 C:
Taking e to both sides:
So,
ln(T ; 21) = kt +ln74:
T ; 21 = e kt+ln 74 = e ln 74 e kt = 74e kt :
T = 21 + 74e kt :
Now, we know that when t = 1, T = 85, so
85 = 21 + 74e k =) 64
74 = ek =) k = ln 64
74 = ;0:1452:
4

So,
T = 21 + 74e ;0:1452t :
Thus, to nd out when T = 51, we solve:
51 = 21 + 74e ;0:1452t =) 30
74 = e;0:1452t =) t =
Problems from Anton, x11.3
3. Determine if P 1
k=1
ln
30
74
;0:1452
= 6:22 min
;
;
3
4 k;1
converges, and if so, nd its sum.
This is a geometric series, with ratio r = ;3=4, and rst term a = ; ; 3 4 1;1
= 1.
So, it converges (jrj < 1), and its sum is
a
1;r = 1
1+ 3 4
= 4 7 .
23. A ball is dropped from a height of 10 m. Each time it strikes
the ground it bounces vertically to a height that is 3 of the preceding
4
height. Find the total distance the ball will travel if it is assumed to
bounce innitely often.
In the rst fall, the ; ball travels 10 m. In the next up-and-down, it travels 3(10)
4
3 2
twice in the next 4
(10) twice etc. Thus, the total distance is:
10 + (2) 3 3
2
3
3
4 4 (10) + (2) (10) + (2) (10) + ::::
4
The terms excluding the rst term form a geometric series, with ratio r = 3 4
and initial term a = (2) 3 a
(10) = 15. Their sum is thus
4 1;r = 15 = 60. So, the
1; 3 4
total distance is 10 + 60 = 70 m.
27. In each part, nd all values of x for which the series converges,
and nd the sum of the series for those values of x.
a. x ; x 3 + x 5 ; x 7 + x 9 ; :::
This is geometric with ratio r = ;x 2 and initial term a = x. It thus converges
when jrj = j;x 2 j = x 2 < 1, i.e. when ;1 < x < 1. In that case, it converges
to
a
1;r = x
1;(;x = x
2 ) 1+x
. 2
b.
1
x 2 + 2 x 3 + 4 x 4 + 8 x 5 + :::
This is geometric with ratio r = 2 x and initial term a = 1 x 2 . It thus converges
5

when jrj = j 2 x j = 2
jxj
case, it converges to
< 1, i.e. when 2 < jxj, i.e. when x > 2 or x < ;2. In that
a
1;r = 1=x2
1;2=x = 1
x 2 ;2x .
c. e ;x + e ;2x + e ;3x + e ;4x + :::
This is geometric with ratio r = e ;x and initial term a = e ;x . It thus converges
when jrj = je ;x j = e ;x < 1, i.e. when ;x < 0, i.e., when x > 0. In that case,
it converges to
a
1;r = e;x
1;e
= 1
;x e . x ;1
29. Show that P 1
; 1
k=1 k ; 1
k+2 =
3
. 2
The partial sum s n simplies quite a bit by cancellation:
s n =
1
1 ; 1 3
1
+
+
+
1
2 ; 1 4
n ; 3 ; 1
n ; 1
1
n ; 1 ; 1
n +1
= 1+ 1 2 ; 1
n +1 ; 1
n +2 :
1
3 ; 1 5
+ +
1
+
n ; 2 ; 1 n
1
+ ; 1
n n +2
By denition, the series' sum is lim n!1 s n = 1+ 1 2 = 3 2 .
Problems from Anton, x11.4
9. Determine whether P 1 1
converges.
k=1 k+6
1
1
4 ; 1 6
+
Consider the continuous, positive, decreasing function R f(x) = 1 . By the
x+6
1 1
Integral Test, our series converges if and only if dx converges. So, we
1 x+6
compute:
Z 1
Z
1
L
1
dx = lim
x +6 L!1
1 x +6 dx
= lim
ln jx +6j j L 1
L!1
= lim (ln jL +6j;ln 7)
L!1
Since this limit does not exist, the improper integral diverges, and thus the
innite series diverges.
14. Determine whether P 1 ln k
k=3 k
converges.
6

Consider the continuous, positive, decreasing function f(x) = ln x
x
for x 3 (to
check this is decreasing, observe that f 0 x(1=x);(ln x)(1)
(x) = = 1;ln x is
x 2
x R negative
2
1 ln x
for x > e). By the Integral Test, our series converges if and only if
3 x dx
converges. So, we compute (using the substitution u = ln x):
Z 1
3
ln x
x
dx = lim
L!1
Z L
3
Z ln L
= lim
L!1
ln 3
u 2
= lim
L!1 2
ln x
x dx
udu
ln L
ln 3
= lim
L!1
(ln L)
2
2
!
;
(ln 3)2
2
Since this limit does not exist, the improper integral diverges, and thus the
innite series diverges.
18. Determine whether P 1 k 2 +1
k=1 k 2 +3
converges.
We compute:
k 2 +1
lim
k!1 k 2 +3 = lim
2k
=
k!1 2k 1
where the rst step is L'H^opital's Rule. Since the limit of the terms of the series
is not zero, the series diverges by the Divergence Test.
32a. Find the n = 10 approximation to of P 1
S =
k=1
corresponding error bounds.
1
(2k+1) 2
and the
R 1 dx
We know that if we approximate the series by s 10 + , then the error
11 (2x+1) 2
is greater than 0 and less than a 11 = 1 = 1 = 0:0019. Now we evaluate the
(23) 2 529
integral (using the substitution u = 2x + 1):
Z 1
Z L
dx
dx
= lim
11 (2x +1) 2 L!1
11 (2x +1)
Z 2 2L
du=2
= lim
L!1
23 u 2
2L
!
= lim ; 1
L!1 2u
23
= lim ; 1 + 1
L!1 4L 46
7
= 1
44

Next, we evaluate the tenth partial sum:
s 10 = 1 3 2 + 1 5 2 + 1 7 2 + 1 9 2 + 1
(11) 2 + 1
(13) 2 + 1
(15) 2 + 1
(17) 2 + 1
(19) 2 + 1
(21) 2 = 0:2110:
Our approximation is therefore 0:2110 + 1 = 0:2327. Thus, the innite series
46
is between 0:2327 and 0:2327 + 0:0019 = 0:2346.
33. Approximate P 1 1
k=1 k
to within 0.01.
3
R 1 dx
We know that if we approximate the series by s n +
n+1 x
, then the error is
3
greater than 0 and less than a n+1 = 1 . We want to choose
(n+1) 3 n great enough
so the width of this error bound is less than
p
0.01. In other words, we want
1
(n+1) 3 < 0:01, or (n +1) 3 > 100, or n +1> 3 100 = 4:64, or n > 3:64. So, we
choose n = 4. Thus, our approximation is:
1X
k=1
1
k 3 s 4 +
Z 1
dx
5
x 3
= 1 1 + 1 2 3 + 1 3 3 + 1 4 3 + x;2
;2
= 1:1777 ; 1
2x 2
1
1
5
= 1:1777 ; 0+ 1
2(25) = 1:1977
So, the sum of the innite series is between 1:1977 and 1:1977 + a 5 = 1:1977 +
1
5 3 = 1:2057.
5
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