3 Newton’s Law of Cooling
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158
Chapter 3 Applications of First-Order Equations
SOLUTION: In this example, T0
350 and Ts
these values into equation (3.11), we obtain T t
75 275ekt 75.
In[378]:= step1 capts ekt
t0 350, capts
Out[378]= 75 275 k t
75. Substituting
350 75 ekt
t0 /.
capts
75
To solve the problem we must ﬁnd k or ek . Because we also know that
75. Solving this equation for k or ek
T 15
150, T 15
275e15k
gives us:
275e15k
75
3
11
e15k
3
ln
11
3
ln
11
3
ln 11
15
ln 11
3
15
ln e15k
15k
k
k
Thus, T t
275e
275e15k
t ln 11/ 3 / 15
In[379]:= step2
Out[379]= 75
75
3
11
3
11
11
3
e15k
or
ek
ek
75
275
t
15
To ﬁnd the value of t for which T t
11 t/ 15
75 80 for t:
275
3
11 t/ 15
3
11 t/ 15
3
11 t/ 15
ln
3
11
t
ln
15
3
275
t
.
t/ 15
75.
80, we solve the equation
5
1
55
ln
1/ 15
3
1
Log
15
11
step1/.k
25 3t/15 111
11
3
1/ 15
1
55
ln 55
ln 55
15 ln 55
11
ln
3
46.264.
3.3 Newton’s Law of Cooling
159
350
300
250
200
150
100
50
20
Figure 3-26 The value of t where y
40
60
T t and y
80 is the solution to the problem
80
Alternatively, we can graph the solution together with the line y
as shown in Figure 3-26
In[380]:= Plot step2, 80 , t, 0, 90 ,
PlotStyle
GrayLevel 0 , Dashing
PlotRange
0, 350
0.01
80
,
and then use FindRoot to approximate the time at which the temperature of the pie reaches 80o F.
In[381]:= FindRoot step2
Out[381]=
t
80, t, 45
46.264
Thus, the pie will be ready to eat after approximately 46 minutes.
An interesting question associated with cooling problems is to
determine if the pie reaches room temperature. From the formula, T t
11 t/ 15
11 t/ 15
75, we see that the component 275
> 0, so T t
275
3
3
t/ 15
11
275
75 > 75. Therefore, the pie never actually reaches room
3
temperature according to our model. However, we see from the graph
and from the values in the following table that its temperature
approaches 75o F as t increases.
160
Chapter 3 Applications of First-Order Equations
In[382]:= Table t, step2//N , t, 60, 100, 10
TableForm
60 76.5214
70 75.6398
Out[382]= 80 75.2691
90 75.1132
100 75.0476
//
If the temperature of the surroundings, Ts , varies the situation is more complicated.
For example, consider the problem of heating and cooling a building. Over the
span of a 24-hour day, the outside temperature, Ts , varies so the problem of determining the temperature inside the building becomes more complicated. Assuming that the building has no heating or air conditioning system, the differential
equation that needs to be solved to ﬁnd the temperature u t at time t inside the
building is
du
k Ct ut ,
(3.11)
dt
where C t is a function that describes the outside temperature and k > 0 is a constant that depends on the insulation of the building. According to this equation,
if C t > u t , then du/ dt > 0, which implies that u increases. On the other hand, if
C t < u t , then du/ dt < 0 which means that u decreases.
The ﬁrst choice of C t has
average value of 70o F; the
second choice has an average
value of 80o F.
EXAMPLE 3.3.2: (a) Suppose that during the month of April in Atlanta,
Georgia, the outside temperature in degrees F is given by C t
70
10 cos Πt/ 12 , 0
t
24. Determine the temperature in a building
that has an initial temperature of 60o F if k 1/ 4. (b) Compare this to
the temperature in June when the outside temperature is C t
80
10 cos Πt/ 12 and the initial temperature is 70o F.
SOLUTION: (a) The initial-value problem that we must solve is
du
dt
u0
k 70
10 cos
Π
t
12
u
60.
The differential equation can be solved if we write it as du/ dt ku
k 70 10 cos Πt/ 12 and then use an integrating factor. This gives us
d kt
e u
dt
kekt 70
10 cos Πt/ 12 ,
3.3 Newton’s Law of Cooling
161
75
72.5
70
67.5
65
62.5
5
10
15
20
Figure 3-27 The temperature in a hypothetical building over a period of 24 hours
so we must integrate both sides of the equation. Of course, solving the
equation is most easily carried out through the use of DSolve.
In[383]:= sol1
Πt
1
70 10 Cos
4
12
60 , u t , t //Simplify
DSolve
u t
u 0
Out[383]=
1
10 63
Π2
Πt
t/ 4 2
Π
9 Cos
12
Πt
3 Π Sin
12
u t
9
u t
,
7 Π2
We then use Plot to graph the solution for 0
t
24 in Figure 3-27.
In[384]:= Plot u t /.sol1, t, 0, 24
Note that the temperature reaches its maximum (approximately 77 F)
near t
15.5 hours which corresponds to 3:30 p.m. A more accurate
estimate is obtained with FindRoot by setting the ﬁrst derivative of
the solution equal to zero and solving for t.
In[385]:= FindRoot Evaluate
t, 15
Out[385]=
t
15.1506
t
sol1 1, 1, 2
0 ,
162
Chapter 3 Applications of First-Order Equations
85
82.5
80
77.5
75
72.5
5
10
15
20
Figure 3-28 The plot is almost identical to the plot obtained in (a)
(b) This problem is solved in the same manner as the previous case.
In[386]:= sol2
Out[386]=
DSolve u t
Πt
1
80 10 Cos
u t ,u 0
4
12
70 , u t , t //Simplify
1
10 72
Π2
Πt
t/ 4 2
Π 9 Cos
12
Πt
3 Π Sin
12
u t
9
8 Π2
The solution is graphed with Plot in Figure 3-28. From the graph, we
see that the maximum temperature appears to occur near t 15 hours.
In[387]:= Plot u t /.sol2, t, 0, 24
Again, a more accurate value is obtained with FindRoot by setting
the ﬁrst derivative of the solution equal to zero and solving for t. This
calculation yields 15.15 hours, the same as that in (a).
In[388]:= FindRoot Evaluate
t, 15
Out[388]=
t
15.1506
t
sol2 1, 1, 2
0 ,
3.4 Free-Falling Bodies
163
3.4 Free-Falling Bodies
The motion of objects can be determined through the solution of ﬁrst-order initialvalue problems. We begin by explaining some of the theory that is needed to set
up the differential equation that models the situation.
Newton’s Second Law of Motion: The rate at which the momentum of a body changes with respect to time is equal to the resultant force acting on the body.
Because the body’s momentum is deﬁned as the product of its mass and velocity,
this statement is modeled as
d
mv
F,
dt
where m and v represent the body’s mass and velocity, respectively, and F is the
sum of the forces (the resultant force) acting on the body. Because m is constant,
differentiation leads to the well-known equation
m
dv
dt
F.
If the body is subjected only to the force due to gravity, then its velocity is determined by solving the differential equation
m
dv
dt
mg
or
dv
dt
g,
where g 32 ft/ s2 (English system) and g 9.8 m/ s2 (international system). This
differential equation is applicable only when the resistive force due to the medium
(such as air resistance) is ignored. If this offsetting resistance is considered, we
must discuss all of the forces acting on the object. Mathematically, we write the
equation as
dv
forces acting on the object
m
dt
where the direction of motion is taken to be the positive direction. Because air
resistance acts against the object as it falls and g acts in the same direction of the
motion, we state the differential equation in the form
m
dv
dt
mg
FR
or
m
dv
dt
mg
FR ,
where FR represents this resistive force. Note that down is assumed to be the positive direction. The resistive force is typically proportional to the body’s velocity, v,
164
Chapter 3 Applications of First-Order Equations
or the square of its velocity, v2 . Hence, the differential equation is linear or nonlinear based on the resistance of the medium taken into account.
EXAMPLE 3.4.1: Determine the velocity and displacement functions
of an object with m
1 slug where 1 slug
lb s2 / ft, that is thrown
downward with an initial velocity of 2 ft/ s from a height of 1000 feet.
Assume that the object is subjected to air resistance that is equivalent
to the instantaneous velocity of the object. Also, determine the time at
which the object strikes the ground and its velocity when it strikes the
ground.
SOLUTION: First, we set up the initial-value problem to determine
the velocity of the object. Because the air resistance is equivalent to the
instantaneous velocity, we have FR v. The formula m dv/ dt mg FR
then gives us dv/ dt
32 v. Of course, we must impose the initial
velocity v 0
2. Therefore, the initial-value problem is
dv/ dt
v0
32
v
2
which is both separable and ﬁrst-order linear. We solve it as a linear
ﬁrst-order equation and so we multiply both sides of the equation by
32t. Integrating
the integrating factor et , which results in d/ dt et v
both sides gives us et v 32et C, so v 32 Ce t . Applying the initial
velocity, we have v 0
32 C 2. Therefore, the velocity of the object
is v 32 30e t . We obtain the same result with DSolve, naming the
resulting output step1.
In[389]:= Clear v, t
step1 DSolve v t
v t ,t
Out[389]=
v t
2 t 15 16
32
v t ,v 0
2 ,
t
To determine the position, or distance traveled at time t, s t , we solve
the ﬁrst-order equation ds/ dt
32 30e t with initial displacement
s0
0. Notice that we use the initial displacement as a reference and
let s s t represent the distance traveled from this reference point.
In[390]:= step2 DSolve s t
s t ,t
Out[390]=
s t
2 t 15 15
32
t
16
30e t , s 0
t
t
0 ,
3.4 Free-Falling Bodies
165
2000
1500
1000
500
10
Figure 3-29
20
30
Plots of s
40
50
60
s t and the line s
70
1000
Thus, the displacement of the object at time t is given by s
32t
t
30.
30e
Because we are taking s 0
0 as our starting point, the object strikes
the ground when s t
1000. Therefore, we must solve s 32t 30e t
30 1000. The roots of this equation can be approximated with Find
Root. We begin by graphing the function s s t and the line s 1000
with Plot in Figure 3-29.
In[391]:= Plot
30 30e t 32t, 1000 , t, 0, 70 ,
PlotStyle
GrayLevel 0 , GrayLevel 0.5
From the graph of this function, we see that s t
1000 near t
obtain a better approximation, we use FindRoot
In[392]:= t00 FindRoot
t, 35
Out[392]=
t
30
30e
t
32t
35. To
1000,
32.1875
so the object strikes the ground after approximately 32.1875 seconds.
The velocity at the point of impact is found to be 32.0 ft/ s by evalvt
32 30e t , at the time at which the
uating the derivative, s t
object strikes the ground, t 32.1875.
In[393]:= 32
Out[393]= 32.
30e t /. t00 1
166
Chapter 3 Applications of First-Order Equations
EXAMPLE 3.4.2: Determine a solution (for the velocity and the displacement) of the differential equation that models the motion of an
object of mass m when directed upward with an initial velocity of v0
from an initial displacement y0 assuming that the air resistance equals
cv, where c is constant.
SOLUTION: Because the motion of the object is upward, mg and FR act
against the upward motion of the object; mg and FR are in the negative
direction. Therefore, the initial-value problem that must be solved in
this case is the linear problem,
dv
dv
v0
c
v
m
g
v0
which we solve with DSolve, naming the resulting output sol.
In[394]:= Clear v, t, s
sol
v t
DSolve
g
cv t
,v 0
m
v0 ,
v t ,t
ct
m
Out[394]=
ct
m
gm
v t
gm
c v0
c
Next, we use sol to deﬁne velocity. This function can be used to
investigate numerous situations without re-solving the differential
equation each time.
In[395]:= velocity m , c , g , v0 , t
gm
ce
ct
m
gm
c
v0
c
For example, the velocity function for the case with m 128 slugs, c
88e 4t/ 5 40.
1/ 160, g 32 ft / s2 , and v0 48 ft / s is v t
In[396]:= velocity
Out[396]=
40
88
1
1
,
, 32, 48, t //Expand
128 160
4 t/ 5
The displacement function s t that represents the distance above the
ground at time t is determined by integrating the velocity function. This
is accomplished here with DSolve using the initial displacement y0 . As
3.4 Free-Falling Bodies
167
with the previous case, the output is named pos so that the displacement formula may be used to deﬁne the function position.
In[397]:= pos
Out[397]=
DSolve y t
velocity m, c, g, v0, t ,
y 0
y0 , y t , t
1
ct
ct
m
m g m2
y t
g m2
c2
ct
c m g m t c m v0
ct
m
c
m v0
c2
ct
m
y0
In[398]:= position m , c , g , v0 , y0 , t
1
ct
ct
e m gm2 cgmt ce m mv0
c2
c2
gm2
c2
mv0
c
y0
The displacement and velocity functions are plotted in the following
using the parameters m 128 slugs, c 1/ 160, g 32 ft/ s2 , and v0
48 ft / s as well as y0 0.
The time at which the object reaches its maximum height occurs
when the derivative of the displacement is equal to zero. From the
vt
0 when t 1.
graph in Figure 3-30 we see that s t
1
1
,
, 32, 48, t ,
128 160
1
1
position
,
, 32, 48, 0, t , t, 0, 2 ,
128 160
PlotStyle
GrayLevel 0 ,
In[399]:= Plot
velocity
Dashing
0.01
A more accurate approximation, t
together with N
In[400]:= root
0.985572, is obtained using Solve
N Solve
t position
Out[400]=
t
1
1
,
, 32, 48, 0, t
128 160
0, t
0.985572
or with FindRoot.
In[401]:= FindRoot velocity
t, 1
Out[401]=
t
0.985572
1
1
,
, 32, 48, t
128 160
0,
168
Chapter 3 Applications of First-Order Equations
40
30
20
10
0.5
1
1.5
2
-10
-20
Figure 3-30
The maximum height of the object occurs when its velocity is 0
We now compare the effect that varying the initial velocity and displacement has on the displacement function. Suppose that we use the
same values used earlier for m, c, and g. However, we let v0 48 in one
function and v0 36 in the other. We also let y0 0 and y0 6 in these
two functions, respectively. See Figure 3-31.
1
1
,
, 32, 48, 0, t ,
128 160
1
1
position
,
, 32, 36, 6, t ,
128 160
t, 0, 2 , PlotStyle
GrayLevel 0 ,
In[402]:= Plot
position
Dashing
0.01
Figure 3-32 demonstrates the effect that varying the initial velocity only
has on the displacement function. The values of v0 used are 48, 64,
48. Notice that as the
and 80. The darkest curve corresponds to v0
20
15
10
5
0.5
1
1.5
Figure 3-31 Varying v0 and y0
2