PRACTICE PROBLEMS: SET 5
MATH 100: PROF. DRAGOS GHIOCA
1. Problems
Problem 1. Find
dy
dx
if
arctan(xy) =
ey−x
x+y
Problem 2. A model boat is put into the water at rest and its engines are turned
on. Let v(t) denote the velocity of the boat in m/s, with t in seconds after the
engines are turned on. It is known that v(t) solves the differential equation
dv
1
= − (v − 10) .
dt
10
(a) Let y(t) = v(t) − 10, and determine the differential equation satisfied by y.
(b) Determine the velocity v(t) explicitly as a function of t.
(c) Determine
lim v(t).
t→+∞
Problem 3. A freshly brewed cup of coffee has temperature 95 degrees Celsius in
a 20 degrees Celsius room. When its temperature is 70 degrees Celsius, it is cooling
at a rate of 1 degree Celsius per minute. After how many minutes does this occur?
Problem 4. A turkey is taken out of a hot oven at 2PM and placed on the table in
the dining room. At 2:30PM the temperature of the turkey is 250F, while at 3PM
the temperature of the turkey is 200F. Knowing that the dining room temperature
is constant equal to 70F, what was the temperature of the turkey at 2PM?
2. Solutions.
Problem 1. We differentiate both sides and obtain:
1
ey−x (y 0 − 1) · (x + y) − ey−x · (1 + y 0 )
0
·
(y
+
xy
)
=
1 + (xy)2
(x + y)2
We solve then for y 0 :
xy 0
ey−x y 0 · (x + y) − ey−x y 0
y
−ey−x (x + y) − ey−x
−
=
−
+
.
1 + x2 y 2
(x + y)2
1 + x2 y 2
(x + y)2
We obtain
0
y ·
x
ey−x (x + y − 1)
−
1 + x2 y 2
(x + y)2
=−
1
y
ey−x (x + y + 1)
−
1 + x2 y 2
(x + y)2
2
MATH 100: PROF. DRAGOS GHIOCA
and therefore
0
y =−
y
1+x2 y 2
+
x
1+x2 y 2
−
ey−x (x+y+1)
(x+y)2
ey−x (x+y−1)
(x+y)2
.
Problem 2.
(a) The approach is same as in the Newton’s Law of Cooling. So,
1
1
y 0 (t) = v 0 (t) = − (v(t) − 10) = − y(t).
10
10
So, the differential equation satisfied by y is
1
y 0 = − y.
10
(b) The solution of the above differential equation satisfied by y is
1
y(t) = y(0) · e− 10 t .
Moreover,
y(0) = v(0) − 10 = 0 − 10 = −10,
since the boat is intially at rest (i.e., v(0) = 0). So,
1
y(t) = −10e− 10 t
and therefore because v(t) = 10 + y(t) we get
t
v(t) = 10 − 10e− 10 .
t
(c) As t → +∞ we have that − 10
→ −∞ and thus
t
lim e− 10 = 0.
t→+∞
In conclusion,
lim v(t) = 10 m/s.
t→+∞
Problem 3. We denote by T (t) the temperature of the coffee at time t, where
T (0) = 95. Also, we let
f (t) = T (t) − 20.
So, f (0) = 75. We know – by the Newton’s Law of Cooling – that the differential
equation satisfied by the temperature function of the coffee is of the following form:
dT
= k(T − 20).
dt
So,
f 0 (t) = kf (t),
which yields that
f (t) = f (0)ekt .
Since f (0) = 75, we know that f (t) = 75ekt .
We know that when f (t) = 70 − 20 = 50, then f 0 (t) = −1. Note that since we
are dealing with a cooling process, the rate of change f 0 is indeed negative. This
allows us to compute the constant k since then
−1 = k · 50 (because f 0 (t) = k · f (t)),
PRACTICE PROBLEMS: SET 5
3
and so,
1
.
50
Finally, we are asked for what t is it that f (t) = 50? So, we solve:
k=−
1
50 = 75e− 50 ·t ,
and then
which yields
t
50
= e− 50 ,
75
75
t
− ln
=− .
50
50
In conclusion,
t = 50 ln(1.5) minutes.
Problem 4. We let time t = 0 be 2PM; at that time the unknown temperature of
the turkey (when taken out of the oven) is T (0). We know that T (30) = 250 while
T (60) = 200 (time being measured in minutes). We have the differential equation:
dT
= k(T − 70),
dt
since the room temperature is constant equal to 70F. We let y(t) be the function which measures the difference between the turkey temperature and the room
temperature at any given time t. So,
(1)
y = T − 70, which yields
dy
dT
=
, and so
dt
dt
dy
= ky.
dt
Hence y(t) = y(0) · ekt , i.e.
y(t) = y(0)ekt and thus
T (t) = 70 + y(0)ekt .
We know that T (30) = 250, i.e.
70 + y(0)ek·30 = 250 and so,
(2)
y(0)e30k = 180.
Similarly, using that T (60) = 200, we get
70 + y(0)ek·60 = 200, and so
(3)
y(0)e60k = 130.
We divide side-by-side equations (2) and (3) and obtain
y(0)e60k
130
=
, i.e.
30k
y(0)e
180
e30k =
13
, and so
18
4
MATH 100: PROF. DRAGOS GHIOCA
13
ln 18
13
and thus k =
.
18
30
Using the value of k computed above, and moreover the fact that e30k =
compute y(0) from (2):
180
180
3240
y(0) = 30k = 13 =
.
e
13
18
30k = ln
13
18 ,
we
Now, the temperature of the turkey is T (0) which is related to y(0) according to
(1), and so the temperature of the turkey at 2PM is
4150
3240
=
degrees F.
T (0) = 70 +
13
13