1
Solutions to Section 1.1
TrueFalse Review:
1.
FALSE.
A derivative must involve
some
derivative of the function
y
=
f
(
x
), not necessarily the first
derivative.
2. TRUE.
The initial conditions accompanying a differential equation consist of the values of
y, y , . . .
at
t
= 0.
3. TRUE.
If we define positive velocity to be oriented downward, then
dv
dt
=
g,
where
g
is the acceleration due to gravity.
4.
TRUE.
We can justify this mathematically by starting from
a
(
t
) =
g
, and integrating twice to get
v
(
t
) =
gt
+
c
, and then
s
(
t
) =
1
2
gt
2
+
ct
+
d
, which is a quadratic equation.
5. FALSE.
The restoring force is directed in the direction
opposite
to the displacement from the equilibrium
position.
6.
TRUE.
According to Newton’s Law of Cooling, the rate of cooling is proportional to the
difference
between the object’s temperature and the medium’s temperature.
Since that difference is greater for the
object at 100
◦
F
than the object at 90
◦
F
, the object whose temperature is 100
◦
F
has a greater rate of
cooling.
7.
FALSE.
The temperature of the object is given by
T
(
t
) =
T
m
+
ce

kt
, where
T
m
is the temperature
of the medium, and
c
and
k
are constants.
Since
e

kt
= 0, we see that
T
(
t
) =
T
m
for all times
t
.
The
temperature of the object
approaches
the temperature of the surrounding medium, but never equals it.
8. TRUE.
Since the temperature of the coffee is falling, the temperature
difference
between the coffee and
the room is higher initially, during the first hour, than it is later, when the temperature of the coffee has
already decreased.
9. FALSE.
The slopes of the two curves are
negative
reciprocals of each other.
10. TRUE.
If the original family of parallel lines have slopes
k
for
k
= 0, then the family of orthogonal tra
jectories are parallel lines with slope

1
k
. If the original family of parallel lines are vertical (resp. horizontal),
then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines.
11. FALSE.
The family of orthogonal trajectories for a family of circles centered at the origin is the family
of lines passing through the origin.
Problems:
1.
d
2
y
dt
2
=
g
=
⇒
dy
dt
=
gt
+
c
1
=
⇒
y
(
t
) =
gt
2
2
+
c
1
t
+
c
2
.
Now impose the initial conditions.
y
(0) = 0 =
⇒
c
2
= 0
.
dy
dt
(0) =
⇒
c
1
= 0
.
Hence, the solution to the initialvalue problem is:
y
(
t
) =
gt
2
2
.
The object hits the
ground at time,
t
0
, when
y
(
t
0
) = 100. Hence 100 =
gt
2
0
2
, so that
t
0
=
200
g
≈
4
.
52 s, where we have taken
g
= 9
.
8 ms

2
.
2.
From
d
2
y
dt
2
=
g
, we integrate twice to obtain the general equations for the velocity and the position of the