Thus
f
(
x
+
h
)
−
f
(
x
) =
f
(
h
)(1 +
f
(
x
)
2
)
1
−
f
(
x
)
f
(
h
)
,
so the Newtonian quotient for
f
is given by
f
(
x
+
h
)
−
f
(
x
)
h
=
f
(
h
)
h
braceleftBig
1 +
f
(
x
)
2
1
−
f
(
x
)
f
(
h
)
bracerightBig
.
Hence by the properties of limits,
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
= (1 +
f
(
x
)
2
)
braceleftBig
lim
h
→
0
f
(
h
)
h
bracerightBig
×
×
1
1
−
lim
h
→
0
f
(
x
)
f
(
h
)
.
But then by the second set of properties for
f
,
lim
h
→
0
f
(
h
)
h
= 6
,
lim
h
→
0
f
(
h
) = 0
.
Consequently,
f
′
(
x
) = 6(1 +
f
(
x
)
2
).
028(part2of2)10.0points
(ii) By solving the differential equation in Part
(i) for
f
, find the value of
f
(
π
18
).
1.
f
parenleftBig
π
18
parenrightBig
=
1
√
3
2.
f
parenleftBig
π
18
parenrightBig
=
√
3
correct
3.
f
parenleftBig
π
18
parenrightBig
=
√
2
4.
f
parenleftBig
π
18
parenrightBig
=
1
√
2
5.
f
parenleftBig
π
18
parenrightBig
=
√
3
2
Explanation:
The differential equation for
f
can be writ
ten as
dy
dx
= 6(1 +
y
2
)
,
setting
y
=
f
(
x
). This is a separable variables
equation which in turn can be written as
integraldisplay
1
1 +
y
2
dy
=
integraldisplay
6
dx.
After integration, therefore,
arctan
y
= 6
x
+
C
with
C
an arbitrary constant. Thus
f
(
x
) = tan(6
x
+
C
)
.
Now
f
(0) = 0, so
f
(
x
) = tan 6
x
.
Hence at
x
=
π
18
,
f
(
π
18
) = tan(
π
3
) =
√
3.
029
10.0points
Solve
dy
dx
=
7
x
13
y
1 +
x
14
for
y.
pacheco (jnp926) – Homework 5 – staron – (52840)
15
1.
arcsin(1 +
x
14
) +
C
2.
arccos(1 +
x
14
) +
C
3.
arctan(1 +
x
14
) +
C
4.
None of these
correct
Explanation:
y
′
=
7
x
13
y
1 +
x
14
y
′
y
=
7
x
13
1 +
x
14
integraldisplay
y
′
y
dx
=
integraldisplay
7
x
13
dx
1 +
x
14
ln

y

=
1
2
integraldisplay
14
x
13
dx
1 +
x
14
=
1
2
ln

1 +
x
14

+
C
0
y
=
e
1
2
ln

1+
x
14

+
C
0
=
e
ln
√
1+
x
14
e
C
0
y
=
C
radicalbig
1 +
x
14
030
10.0points
Newton’s Law of Cooling asserts that the
rate of change of temperature of an object
is proportional to the difference between its
temperature and the ambient temperature.
A hot poker is immersed in water whose
temperature is held constant at 70
◦
F
. After
6 minutes the temperature of the poker has
dropped to 94
◦
F
, while after 9 minutes it has
dropped to 86
◦
F
. What was the temperature
of the poker initially?
5.
initial temp = 109
◦
F
Explanation:
If
T
=
T
(
t
) is the temperature of the poker
t
minutes after it is immersed in the water,
Newton’s Law of Cooling ensures that
dT
dt
=
−
k
(
T
−
70)
.
This is a separable variables differential equa
tion which becomes
integraldisplay
1
T
−
70
dT
=
−
integraldisplay
k dt
after
separating variables
and
integrating.
Consequently,
ln(
T
−
70) =
−
kt
+
C,
i
.
e
.,
T
−
70 =
e
−
kt
+
C
.
Thus
(
‡
)
T
(
t
) = 70 +
Be
−
kt
,
and so the inital temperature of the poker is
given by
T
(0) = 70 +
B
.
To determine
B
we use the conditions
031
10.0points
pacheco (jnp926) – Homework 5 – staron – (52840)
16
Diana is sick in hospital with a severe bac
terial infection.
She is to be fed antibiotics
intravenously at a constant infusion rate, and
the doctor knows that the antibiotic will be
eliminated from the bloodstream at a rate
proportional to the amount
y
(
t
) present in
the bloodstream at time
t