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3 Newton’s Law of Cooling

# 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:= step1 capts ekt

t0 350, capts

Out= 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:= step2

Out= 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:= 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:= FindRoot step2

Out=

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:= Table t, step2//N , t, 60, 100, 10

TableForm

60 76.5214

70 75.6398

Out= 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:= sol1

Πt

1

70 10 Cos

4

12

60 , u t , t //Simplify

DSolve

u t

u 0

Out=

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:= 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:= FindRoot Evaluate

t, 15

Out=

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:= sol2

Out=

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:= 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:= FindRoot Evaluate

t, 15

Out=

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:= Clear v, t

step1 DSolve v t

v t ,t

Out=

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:= step2 DSolve s t

s t ,t

Out=

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:= 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:= t00 FindRoot

t, 35

Out=

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:= 32

Out= 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:= Clear v, t, s

sol

v t

DSolve

g

cv t

,v 0

m

v0 ,

v t ,t

ct

m

Out=

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:= 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:= velocity

Out=

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:= pos

Out=

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:= 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:= Plot

velocity

Dashing

0.01

A more accurate approximation, t

together with N

In:= root

0.985572, is obtained using Solve

N Solve

t position

Out=

t

1

1

,

, 32, 48, 0, t

128 160

0, t

0.985572

or with FindRoot.

In:= FindRoot velocity

t, 1

Out=

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:= 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

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3 Newton’s Law of Cooling

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