390 Guided Projects Guided Project 31 Cooling cof.docx

390 Guided Projects
Guided Project 31: Cooling coffee
Topics and skills: Derivatives, exponential functions
Imagine pouring a cup of hot coffee and letting it cool at room
temperature. How does the temperature of the
coffee decrease in time? How long must you wait until the
coffee is cool enough to drink? When should you
add an ounce of cold milk to the coffee to accelerate the cooling
as much as possible?
A fairly accurate model to describe the temperature changes in a
conducting object is Newton’s Law of
Cooling. Suppose that at time t ≥ 0 an object has a temperature
of T(t). The Law of Cooling says that the rate at
which the temperature of the object increases or decreases is
given by
( ( ) ) , (1)
dT
k T t A
dt
= − −
where A is the ambient (surrounding) temperature and k > 0 is a

constant called the conductivity (which is a
property of the cooling object). Newton’s Law of Cooling
assumes that the cooling body has a uniform
temperature throughout its interior. This is not strictly accurate,
as a cooling body loses heat through its surface.
1. Explain in words what equation (1) means. Specifically, in
terms of T and A, when is 0
dT
dt
> and when is
0
dT
dt
< ? For the case of hot coffee cooling to room temperature,
which case do you expect to see?
2. Verify by substitution that the solution to equation (1)
subject to the initial condition T(0) = T0 is
0( ) ( ) . (2)
ktT t A T A e−= + −
3. Before graphing the temperature function, use equation (2)
to evaluate T(0) and limt→∞ T(t). Are these the
values you expect?
4. Consider the case of a cup of hot coffee cooling with an

ambient room temperature of A = 60◦ F and the
initial temperature of the coffee is T0 = 200
◦ F. Use a graphing utility to plot the temperature function for
k = 0.3, 0.2, 0.1, and 0.05. Comment on how the curves change
with k. Do larger values of k produce faster
or slower rates of temperature change?
5. For the values of A and T0 in Step 4, estimate the value of k
that describes the case in which the coffee
cools to 100 degrees in 10 minutes.
Here is an interesting question. Suppose you want to cool your
hot coffee to 100◦ F as quickly as possible.
Suppose also that you have one ounce of cold milk with a
temperature of 40◦ F that you can add to the
cooling coffee at any time. When should you add the milk to
cool the coffee to 100◦ F as quickly as
possible?
6. We need to make an assumption about the effect of cold milk
on the temperature of the coffee. A
reasonable assumption is that when milk is added to coffee, the
temperature of the coffee immediately
decreases to the average of the coffee temperature and the milk
temperature, where the average is weighted
by the volumes. So if we add 1 ounce of milk with temperature
Tm to 8 ounces of coffee with temperature
T, the temperature of the mixture will be
1 8 8
. (3)

1 8 9
m m
new
T T T T
T
⋅ + ⋅ +
= =
+
Copyright © 2011 Pearson Education, Inc. All rights reserved.
Cooling coffee 391
(For example, if we add one ounce of 40◦ F milk to eight ounces
of 150◦ F coffee, the coffee temperature is
lowered immediately to about 138◦ F.) Assume the coffee is
allowed to cool for t1 minutes, where t1 will be
determined. Use equation (2) and the value of k found in Step 5
to determine the temperature of the coffee
at t = t1; call it T
*.
7. Use assumption (3) to determine the temperature of the
coffee when one ounce of 40◦ F milk is added to
eight ounces of coffee, which has a temperature of T*. Call this
new coffee temperature Tnew.

8. Use equation (2) to show that the temperature of the coffee
for t > t1 using Tnew as the initial temperature is
( ) 60 ( 60) ktnewT t T e
−= + − ,
where t is now measured in minutes after t1.
9. We now solve for the value of t such that T(t) = 100 and call
this value t2. Remember that t2 measures time
after the milk is added. Show that
12
1 18
ln
56 1kt
t
k e−
⎛ ⎞= − ⎜ ⎟⎝ ⎠−
.
10. Notice that as t1 increases, t2 decreases. In fact, show that
if t1 > 8.65 (approximately), then t2 < 0 and the
solutions are not meaningful. Therefore, we consider t1 in the
interval 10 8.65t≤ ≤ .
11. The total time needed to cool the temperature to 100◦ F is τ
= t1 + t2. Notice that the total time τ depends on
t1, the time at which the milk is added. Therefore, the last step

is to graph the total time τ = t1 + t2 as a
function of t1 for 10 8.65t≤ ≤ . Where does this function have a
minimum on this interval? This is the time
at which the milk should be added to give the quickest cooling.
12. Interpret the solution. What is the value of t2 in the optimal
solution? What is the temperature of the
coffee/milk mixture immediately after the milk is added?
392 Guided Projects
Guided Project 32: Simpson’s rule
Topics and skills: Integration, calculator
This project assumes that you are familiar with the Trapezoid
Rule and the Midpoint Rule, which are numerical
integration methods used to approximate definite integrals. As a
review, recall that to approximate ( )
b
a
f x dx∫ ,
we subdivide the interval [a, b] into n subintervals of equal
width Δx = (b – a)/n (Figure 1) and identify the grid
points
x0 = a, x1 = a + Δx, …, xk = a + kΔx, …, xn = b.

x
x0 � a x1 x2 x3 xk xn � b
�x � b � an
Figure 1
The Trapezoid Rule (Figure 2) approximation to the integral
with n subintervals, which we denote T(n), is
1
0
1
1 1
( ) ( ) ( ) ( )
2 2
n
k n
k
T n f x f x f x x
−
=
⎛ ⎞

= + + ∆⎜ ⎟
⎝ ⎠
∑ .
Figure 2
The Midpoint Rule approximation to the integral with n
subintervals, which we denote M(n), is
1 2
1
1
( ) ( ) ( ) ( )
.
2
n
n
k k
k
M n f m x f m x f m x
x x
f x−

=
= ∆ + ∆ + + ∆
+⎛ ⎞= ∆⎜ ⎟⎝ ⎠∑
where mk is the midpoint of the subinterval [xk−1, xk], for k =
0, 1, …, n.
Simpson’s Rule takes groups of three consecutive points on the
curve y = f(x), say (x0, f(x0)), (x1, f(x1)), and
(x2, f(x2)), and approximates the curve with a segment of a
parabola. The net area bounded by this parabola can
be computed exactly. When this idea is applied to every group
of three consecutive points along the interval of
Simpson’s rule 393
integration, the result is Simpson’s Rule. This method generally
gives more accurate approximations to definite
integrals than either the Midpoint or Trapezoid Rules. In one
form, Simpson’s Rule with n subintervals, which
we denote S(n), appears as
0 1 2 3 2 1
( ) ( )

( ( ) 4 ( ) 2 ( ) 4 ( ) 2 ( ) 4 ( ) ( )) .
3
b
a
n n n
f x dx S n
x
f x f x f x f x f x f x f x− −
≈
∆= + + + + + + +
∫
Notice that apart from the first and last terms, the coefficients
alternate between 4 and 2; n must be an even
integer for this rule to work.
1. You can use the formula for Simpson’s Rule given above;
but here is a better way. If you already have the
Trapezoid Rule approximations T(2n) and T(n), the next
Simpson’s Rule approximation follows
immediately with a simple calculation:
4 (2 ) ( )
(2 )

3
T n T n
S n
−= .
Verify that for n = 8, the two forms of Simpson’s Rule are the
same.
2. Consider the integral
21
20
1 0.2146018366
41
x
dx
x
π= − ≈
+∫ .
Compute the Trapezoid Rule and Simpson’s Rule
approximations to the integral for n = 4, 8, 16, 32, and
64, and fill in the columns labeled T(n) and S(n) in the table
below.
Table 1

n T(n) S(n) Error T(n) Error S(n)
4
8
16
32
64
3. If the exact value of a number is x and a computed
approximation to x is c, then the absolute error in c as an
approximation to x is |c – x|. The relative error in c as an
approximation to x is
c x
x
−
, provided x ≠ 0. Compute
the absolute errors for the approximations in
Table 1 and fill in the corresponding columns in
Table 1.
4. Which rule appears to give a better approximation to the
integral? Explain.
5. Notice that each successive row in
Table 1 corresponds to increasing n by a factor of 2 or a
reducing Δx by a factor of 2. Complete the following
sentences:

• If Δx is reduced by a factor of 2, the error in T(n) is reduced
approximately by a factor of ________.
• If Δx is reduced by a factor of 2, the error in S(n) is reduced
approximately by a factor of ________.
394 Guided Projects
6. Approximate the following integrals using Simpson’s Rule.
Experiment with values of n to insure that the
error is less than 10−6. It is best to compute Simpson’s Rule
approximations and Trapezoid Rule
approximations in tandem. Exact values of the integrals are
given so the error may be computed.
1
0
ln(2 cos ) ln(1 3 / 2)I x dx
π
π= + = +∫
22
2
0
1
ln 32 2
2

x
I dx
x
+= = −
+∫
7. In practice, you would apply numerical integration to a
definite integral whose exact value is not known to
you. Use Simpson’s Rule to obtain your best approximation to
the following integrals.
/2
7
3
0
cosI x dx
π
= ∫
1
2 3
4
0
lnI x x dx= ∫

Euler’s method for differential equations 395
Guided Project 33: Euler’s method for differential equations
Topics and skills: Derivatives, computing
You have seen how direction fields make it possible to visualize
the solution of a differential equation. In this
project we combine computation with direction fields to develop
a method that gives numerical approximations
to solutions of differential equations.
1. Let’s review direction fields by considering the differential
equation ( ) 2 4y t y′ = − + . For what values of y
is ( ) 0y t′ = ? How are these values of y reflected in the
direction field?
2. For what values of y is ( ) 0y t′ > ? For these values of y
what is the appearance of the direction field? For
what values of y is ( ) 0y t′ < ? For these values of y what is the
appearance of the direction field?
3. The direction field for this differential equation is shown in
Figure 1. Is it consistent with your conclusions
in Steps 1 and 2?

Figure 1
4. On the direction field in Figure 1 sketch the solution curve
that results from the initial condition y(0) = –2.
Sketch the solution curve that results from the initial condition
y(–2) = 3.
5. Now suppose that you want numerical values of points on a
solution curve. Here is a way, known as
Euler’s method, for approximating these numerical values.
Consider the initial value problem
0( ) ( , ), (0)y t f t y y y′ = = , where f is a given function
involving t and y, and y0 is a given initial value. We
first select a time step Δt, which is generally small in value (say
Δt < 0.5). Then we define the following
grid points on the t-axis:
t0 = 0, t1 = Δt, t2 = 2Δt, t3 = 3Δt, …, tn = n Δt.
396 Guided Projects
Figure 2
The grid points cover the interval of interest on the t-axis. Now

imagine starting at the initial condition
(0, y(0)) = (t0, y0). If we step forward in time one time step,
what is the solution at the point t = t1 = Δt? The
slope of the solution curve at (t0, y0) is given by the direction
field; the slope is f(t0, y0), which is a number
we can compute. Suppose we draw a line segment in the
positive t-direction at (t0, y0) with slope f(t0, y0)
(Figure 2). Explain why the change in y along this line segment
over a horizontal distance Δt is
Δy = f(t0, y0) Δt.
6. As shown in Figure 2, an approximation to the solution at t =
t1 is
y1 = y0 + Δy = y0 + f(t0, y0) Δt;
that is, y(Δt) ≈ y1. Explain why y1 is an approximation to the
solution y(Δt) and is not generally exact.
7. Now we repeat the same argument at the point (t1, y1). The
slope of the solution curve at (t1, y1) is f(t1, y1).
We draw a line segment in the positive t-direction at (t1, y1)
with slope f(t1, y1) (Figure 2). Explain why the
change in y along this line segment over a horizontal distance
Δt is Δy = f(t1, y1) Δt. Show that an
approximation to the exact solution y(t2) is
y2 = y1 + f(t1, y1) Δt.
8. In general, assuming we are at the point (tk, yk), an

approximation to the solution at t = tk+1 is
y(tk+1) ≈ yk+1 = yk + f(tk, yk) Δt , for k = 0, 1, 2, 3, …
This procedure is best carried out in a small computer program
that has the following structure:
1. Given f and y0, choose a time step Δt and a maximum number
of time steps N.
2. For k = 0 to N - 1, compute yk+1 = yk + f(tk, yk) Δt
Carry out this procedure for the initial value problem ( )y t′ = –
2y + 4, y(0) = 1. Fill in the following table
assuming Δt = 0.2 and N = 5.
k tk f(tk, yk) yk
0 0 1
1 0.2
2 0.4
3 0.6
4 0.8
5 1.0
9. The exact solution of the differential equation in Step 8 is
y(t) = 2 – e–2t. Compare the approximation to y(1)
obtained in Step 8 with the exact value of y(1). What is the
percent error in the approximation?
Euler’s method for differential equations 397

10. How would you change the time step to improve the
approximations obtained in Step 8? Explain your
answer.
11. Repeat the calculations of Step 8 with Δt = 0.1 and N = 10.
k tk f(tk, yk) yk
0 0 1
1 0.1
2 0.2
3 0.3
4 0.4
5 0.5
6 0.6
7 0.7
8 0.8
9 0.9
10 1.0
12. What is the error in the approximation to y(1) obtained in
Step 11? Did the error increase or decrease when
Δt was decreased from 0.2 to 0.1?
13. Write a short program to carry out Euler’s method for any
positive time step Δt. Use the program to
generate approximate solutions to the initial value problem ( )y
t′ = t + 3y/t, y(1) = 2 on the interval [1, 3].
Compare your approximations to the exact solution y(t) = –t2 +
3t3. Experiment with your choice of Δt.
What value of Δt is required to make the error in the

approximation to y(3) less than 10–2?
398 Guided Projects
Guided Project 34: How long will your iPod last?
Topics and skills: Integration, graphing, improper integrals
None of the many gadgets we use – laptops, dishwashers, iPods,
skateboard wheels – last forever; they all have
finite lifetimes. And yet predicting those lifetimes is generally
difficult. A digital camera battery may last two
years or two months depending on the conditions in which it is
used (hot, cold, wet, dry), the number of times it
is recharged, or many other factors. Because of the variability
and randomness involved in the lifetime of a
gadget, the best we can do is talk about the average lifetime.
For example, the average lifetime of a 60-watt
compact fluorescent light bulb is rated as 6000 hours; some
bulbs don’t last that long and others last longer. In
this project we investigate a few of the basic questions in
computing lifetimes.
1. In order to study the lifetime of a
particular iPod model, suppose you
collect data from 1000 owners who
carefully keep a record of how many
hours their iPod operated before it failed.
Now imagine compiling and graphing
the data as shown in Figure 1. The

horizontal axis measures time in hours
and the vertical axis is the fraction of
devices that have failed before a given
number of hours. Explain why the
vertical scale on the graph extends from
0 to 1. Explain why the data values form
an increasing sequence as time increases.
Figure 1
2. The graph in Step 1 is one way to show the distribution of
failure times. It’s easier to work with the data if
they are approximated by a continuous function. The
distribution of failure times for many devices is well
approximated by functions of the form F(t) = 1 – e–λt, where λ
(“lambda”) is a positive constant; this
function is called a failure distribution function (in probability
it is called a cumulative distribution
function). Graph F(t) = 1 – e–λt with λ = 0.2, 0.5, 1, 1.5, and 2,
and 0 10t≤ ≤ . Do the curves have a shape
similar to the data graph in Step 1?
3. Describe how the graphs in Step 2 change with λ. Do larger
values of λ correspond to devices that have
shorter or longer lifetimes? If device A has an average lifetime
of 2000 hours and device B has an average
lifetime of 200 hours, which device has a larger value of λ in its
failure distribution function?

4. For the failure distribution function in Step 2, evaluate and
interpret F(0).
5. For the failure distribution function in Step 2, evaluate
limt→∞ F(t). Interpret this limit.
6. Compute ( )F t′ to show that F is an increasing function.
7. Consider the failure distribution function F(t) = 1 – e–t/100
(λ = 0.01). Compute the following quantities:
a. the fraction of devices that fail in the first 10 hours
b. the fraction of devices that still work after 5 hours
c. the fraction of devices that fail between 10 and 15 hours.
d. the fraction of devices that fail between a and b hours, where
b > a.
8. The goal now is to calculate the average lifetime of a device
when we are given its failure distribution
function F. For the next few calculations we will assume that all
failure times lie in an interval [0, T]. Let’s
follow a familiar process that leads to an integral. We subdivide
the interval [0, T] into n subintervals of
equal width Δt = T/n. Define the grid points tk = kΔt, for k = 0,
1, …, n, and then focus on the kth
How long will your iPod last? 399
subinterval [tk–1, tk], for k = 1, …, n. On this subinterval,
explain why the fraction of devices that fail is

F(tk) – F(tk–1) (see Step 7).
9. Remember that we are aiming to compute the average
lifetime of all devices. Explain why the lifetime of
the devices that fail in the interval [tk–1, tk] is approximately
tk.
10. To find the average of all the lifetimes we add up the
individual lifetimes multiplied by the fraction of
devices with that lifetime:
0 1 1
1 2 2
1
Average lifetime fraction that fail in [ , ]
fraction that fail in [ , ]
fraction that fail in [ , ]n n n
t t t
t t t
t t t−
≈ ×
+ ×
+ + ×

Now recall that the fraction of devices that fail in the interval
[tk–1, tk] is F(tk) – F(tk–1) (Step 8). Explain why
( )1
1 lifetimefraction that fail
Average lifetime ( ) ( )
n
k k k
k
F t F t t −
=
≈ −∑ .
11. We would like to identify this sum as a Riemann sum and
convert it to an integral. As it stands, we do not
have the necessary Δt term. We introduce it by multiplying and
dividing by Δt:
1
1
( ) ( )
Average lifetime
n

k k
k
k
F t F t
t t
t
−
=
−≈ ∆
∆∑
Assuming that F ′ is a continuous function, as Δt → 0 and n →
∞, two things happen:
• the quotient 1
( ) ( )k kF t F t
t
−−
∆
approaches the derivative ( )F t′
• the sum approaches an integral.
Explain in your own words why the expected lifetime is given
by
0
Average lifetime ( )

T
tF t dt= ′∫ .
Average lifetime calculations are easier if we return to the
assumption that lifetimes are theoretically
infinite. Then we have
0
Average lifetime ( )tF t dt
∞
= ′∫ .
12. The hard work is complete. Return to the failure
distribution function F(t) = 1 – e–λt and compute ( )F t′ (or
see Step 6), which is called the failure density function (in
probability, it is a probability density function).
13. Evaluate the average lifetime integral (using integration by
parts and handling an improper integral). Show
that the average lifetime for any λ > 0 is 1/λ.
14. Step 13 says that the larger the value of λ, the shorter the
average lifetime. Go back to your answers to Step
3 and check that they are consistent with this conclusion.
15. Suppose your iPod data is well approximated by a failure
distribution function with λ = 1.4 × 10–4. What is
the average lifetime of an iPod?

400 Guided Projects
Guided Project 35: Mercator projections
Topics and skills: Integration, geometry
Fourteenth century mariners, using only a compass for
navigation, could not easily determine longitude.
However they could hold a compass bearing (for example, 45
degrees east of north) for long distances.
Unfortunately such fixed-bearing courses (called rhumb lines or
loxodromes) had a disadvantage: On a globe,
a fixed-bearing course is a spiral and on the fixed-grid maps of
the day, on which one degree of latitude equals
one degree of longitude, these courses are curves (Figure 1).
This fact complicated navigation and led to route-
finding errors.
Figure 1
The solution to this problem was a map devised by
a Flemish geographer named Gerardus Mercator
(1512-1594). On the Mercator map the distance
corresponding to one degree of latitude increases
as one moves away from the equator. When done
according to Mercator’s plan, this stretching of

latitude has the desired effect of making fixed-
bearing courses straight lines on a flat map (Figure
2). How should the stretching of the map be done?
Figure 2
Mercator projections 401
1. To transfer the spherical Earth to a flat piece of
paper, stretching must be done in two
directions. First the longitude lines that
converge at the poles on a sphere become
parallel lines on a flat map (Figure 2). This
stretching is in the east-west direction. Once we
determine the stretching factor for the longitude
lines, the same stretching factor is applied to
the latitude circles (in the north-south
direction). Figure 3 shows a slice from a ball of
radius R with two longitude lines separated by
an angle φ. It also shows two arcs on circles of
latitude, one at the Equator and one at a latitude
of 0 / 2θ π< < . Explain why these two arcs on
circles of latitude have length s Rϕ= and
( cos )s R θ ϕ′ = , respectively.

Figure 3
2. Compute /s s′ for the arc lengths in Step 1.
3. We see that an arc of a latitude circle at
latitude θ is shorter than the
corresponding arc at the equator by a
factor of cos θ. Explain why, to make the
longitude lines parallel on a flat map,
horizontal distances at latitude θ must be
stretched by a factor of sec θ (Figure 4).
Explain why the stretching factor
increases with latitude. Notice that at a
fixed latitude, the stretching of longitude
lines is the same at all longitudes.
Figure 4
4. We now come to the key step in the Mercator projection. The
loxodromes (paths with a fixed bearing) cut
the longitude lines on a sphere at the same angle. A flat
Mercator map must also have this property; that is,
a loxodrome, which is a straight line on the flat map, must cut
the parallel longitude lines at the same angle
(said differently, angles must be preserved when going from the
sphere to the flat map). Explain why the
same stretching factor sec θ that is used in the east-west
direction must also be used in the north-south

direction.
5. The stretching in the north-south direction must be done
carefully, because the amount of stretching of
latitude circles itself varies with latitude. Let L(θ) be the
distance on the map along a longitude line from
the Equator (θ = 0) to the latitude θ. Refer to Figure 3 and
explain why, on a sphere of radius R, if the
latitude θ increases from θ to θ + Δθ, the corresponding change
in distance on the longitude line is RΔθ.
6. Recall that north-south distances on the map are stretched by
a factor sec θ compared to similar distances
on the sphere, where again we consider latitudes with 0 ≤ θ <
π/2. Therefore, if the latitude θ increases
from θ to θ + Δθ on the flat map, the distance L changes by
approximately ΔL = RΔθ sec θ. The total
change in L between the Equator (θ = 0) and a latitude circle θ
is found by adding up the changes ΔL at
each latitude; that is we integrate. Therefore, because L(0) = 0,
0
( ) secL θ R u du
θ
= ∫ .

402 Guided Projects
Notice that a dummy variable u has been used. At this point, we
drop the factor of R as it determines only
the final size of the flat map. Evaluate the integral for L(θ) and
show that, for 0 ≤ θ < π/2, the map distance
function is
( )( ) ln sec tanL θ θ θ= + .
Because Mercator and his contemporaries who worked on this
problem pre-dated calculus, L(θ) was not
originally expressed as an integral. In fact, the integral wasn’t
evaluated in its “modern” form until about
1650 (by Sir Isaac Newton’s teacher, Isaac Barrow).
7. Show that the map distance function may also be written
( ) ln tan
2 4
θ π
L θ ⎡ ⎤ ⎛ ⎞= +⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦
.
8. Graph L as a function of θ, for 0 ≤ θ < π/2.
9. Evaluate
/2
lim ( )

θ π
L θ
−→
. Explain why the Mercator projection is claimed to be valid
only between the latitudes
of 70o S and 70o N.
Predator-prey models 403
Guided Project 36: Predator-prey models
Topics and skills: Integration
The remarkable graph in Figure 1 shows nearly 100 years of
data of hare and lynx populations (collected south
of Hudson Bay in Canada). Both populations show distinct
cycles with periods of approximately 12 years that
are slightly out of phase (the peaks of the two curves are offset
by a few years). Wildlife ecologists theorize that
the hare and lynx populations displayed in this graph must have
interacted in a special way to produce these
regular cycles. Specifically, lynx (the predator) fed on hares
(the prey) until the hare population was reduced to
low levels. With a shortage of food, the lynx began to die.
However, the absence of lynx allowed the hare
population to recover, which in turn increased the food supply
for lynx. Such predator-prey cycles are exhibited

clearly in these data. Our goal is to devise a simple
mathematical model that shows the same kind of oscillation.
(Adapted from Odum, Fundamentals of Ecology, Saunders,
1953)
Figure 1
1. The model is based on two assumptions:
A1: In the absence of lynx, the hare population increases
exponentially. However, the hare population
decreases in proportion to encounters between hares and lynx.
A2: In the absence of hares, the lynx population decreases
exponentially. However, the lynx population
increases in proportion to encounters between hares and lynx.
We let H(t) and L(t) denote the lynx and hare populations,
respectively, at time t ≥ 0. Consider the
following differential equations that give the rates of change of
H and L.
natural hare-lynx
growth rate interactions
natural hare-lynx
decay rate interactions
,
dH
aH bHL
dt

dL
cL dHL
dt
= −
= − +
where a, b, c, and d are positive constants. Explain how the
terms aH and –bHL in the first differential
equation reflect assumption A1.
2. Explain how the terms –cL and dHL in the second
differential equation reflect assumption A2.
404 Guided Projects
3. The differential equations given above involve
two unknown function, H and L. So we need a
useful way to display both solutions at once.
Think of H(t) and L(t) as functions of a parameter
t, which together describe a parametric curve in
the HL-plane (this plane is called the phase
plane). The initial populations (H(0), L(0))
correspond to a point in the plane. As t increases,
a curve is generated consisting of the points
(H(t), L(t)) (Figure 2). Question: If H and L vary

cyclically, as in Figure 1, what would be the
general shape of the solution curve in the
HL-plane?
Figure 2
4. Let’s now use specific values of the coefficients; consider
the equations
0.4 0.02
0.3 0.005
dH
H HL
dt
dL
L HL
dt
= −
= − +
(1)
It turns out that these equations cannot be solved
explicitly for H and L. Instead we will do some
graphical analysis that tells us a lot about the
solutions. Because H and L are populations, they
are positive and we consider only the first quadrant

of the HL-plane.
An equilibrium in the system occurs when
0
dH dL
dt dt
= = (neither population changes in
time). Solve two algebraic equations to show that
the equilibrium states of the system are
H = L = 0, which is not very interesting, and H =
60, L = 20. The equilibrium point is shown in
Figure 3 as the intersection of the lines H = 60 and
L = 20. Figure 3
5. We now work in several steps to produce a direction field in
the HL-plane that shows the general shape of
the solution curves. Let’s start with the first equation in (1).
Show that ( ) 0H t′ = when L = 20. This means
that if a solution curve crosses the line L = 20, the hare
population is not changing, so the solution curve is
vertical along L = 20. We indicate this fact by putting small
vertical line segments on L = 20 (Figure 3).
6. Use the first equation in (1) to show that
• 0 if 0 20
dH
L

dt
> < <
• 0 if 20
dH
L
dt
< >
Predator-prey models 405
7. Explain why the observations of Step 6 are reflected in the
HL-plane using right-arrows (→) and left-
arrows (←) in Figure 4.
8. Using the second equation in (1), argue as in Step 5 and
show that if H = 60, then 0
dL
dt
= . Explain why we
put small horizontal line segments along the line H = 60.
9. Use the second equation in (1) show that 0
dL
dt

> if 60H > and 0dL
dt
< if 0 60H< < .
10. Explain why the observations of Step 9 are reflected in the
HL-plane using up-arrows (↑) and down-arrows
(↓) in Figure 4.
Figure 4
11. In each of the four regions in Figure 4, look at the
combined effect of the arrows to find the overall
direction of the solution curves. For example, in the region 0 <
H < 60, 0 < L < 20, the solutions move in
the negative L-direction and the positive H-direction. Do a
similar analysis in the other three regions.
Conclude that the solutions move in a counterclockwise
direction around the equilibrium point (Figure 5).
Figure 5
406 Guided Projects
12. Assuming that the solution curves are actually closed
curves, explain why the solution curves in Figure 5

correspond to periodic functions for the hare and lynx
populations.
13. As mentioned earlier, it is not possible to solve equations
(1) for H and L as functions of t. But it’s fun to
see how close we can get. Divide the second equation of (1) by
the first equation to eliminate t. Then write
the resulting equation in separable form:
0.4 0.02 0.3 0.005L dL H
L dH H
− − +
= .
14. Now integrate both sides with respect to H and rearrange to
show that 0.4 0.3ln 0.005 0.02L H H L C= + + ,
where C is an arbitrary constant.
15. If you have a graphing utility that plots implicit functions,
choose several values of C (C must be chosen
carefully within a fairly small interval), and graph the resulting
solution curves. Verify that you obtain
closed curves as in Figure 5.

Period of the pendulum 407
Guided Project 37: Period of the pendulum
Topics and skills: Integration
The common pendulum may be idealized as an object of mass m
connected to a frictionless pivot by a light
massless wire or rod of length ℓ. We assume that the object
swings in a plane, so its motion may be described
by the angular displacement θ(t) (Figure 1), where θ is positive
when the displacement is to the right of vertical.
The pendulum is set in motion by giving it an initial angular
displacement θ(0) = θ0 and an initial angular
velocity of (0) 0θ ′ = .
The motion of the pendulum is described by Newton’s
second law of motion (mass · acceleration = applied
force). In this case the only force acting on the
pendulum is the gravitational force and the governing
equation (derived in elementary physics books) is
mass acceleration graviatational force
( ) sin ( )m t mg tθ θ
⋅
′′ = − .
An important fact is that the acceleration is in the
negative direction when θ(t) > 0 and in the positive
direction when θ(t) < 0; therefore, gravity acts as a
restoring force that returns the pendulum to the vertical

position.
Canceling m and dividing through by ℓ, the equation of the
pendulum is
2( ) sin ( ) 0, where
g
t tθ ω θ ω+ = =′′ . (1)
The surprising fact is that we cannot find an explicit formula for
the solution of this differential equation.
However, it is possible to determine the period of the pendulum,
which is the time required to make one
complete swing.
1. The first step in finding the period of the pendulum is to
multiply both sides of equation (1) by ( )tθ ′ . Then
show that equation (1) can be written as 2 2
1
( ( )) cos ( ) 0
2
d
t t
dt
θ ω θ⎛ ⎞′ − =⎜ ⎟⎝ ⎠ .
2. Both sides of this equation may now be integrated with
respect to t. Show that the result is

2 2( ( )) 2 cos ( )t t Cθ ω θ− =′ ,
where C is an arbitrary constant.
3. The arbitrary constant is determined by using the initial
conditions 0(0)θ θ= and (0) 0θ ′ = . Determine C
and show that
2 2
0( ( )) 2 (cos ( ) cos )t tθ ω θ θ= −′ . (2)
Figure 1
408 Guided Projects
4. Let the period of the pendulum be T. When the pendulum
swings from θ = θ0 to θ = 0 (from its initial
position to the vertical position), it swings through one-quarter
of a period. Solving equation (2) for ( )tθ ′ ,
we have
2
0( ) 2 (cos ( ) cos )t tθ ω θ θ= ± −′ .
For this part of the period, explain why ( ) 0tθ ′ < . Therefore,
the negative sign (rather than the positive
sign) in front of the square root must be used.

5. Rewrite the result of Step 4 in the form
2
0
1 1
( ) 2 (cos cos )
dt
d tθ θ ω θ θ
= = −
′ −
.
Now integrate this expression with respect to θ from (t = 0, θ =
θ0) to (t = T/4, θ = 0). Show that
0
20
0
4 2 (cos cos )
T dθ θ
ω θ θ
=
−∫ .
6. We now have an expression for the period of the pendulum.
The goal is to express this integral in a more
manageable form. First use the identify 2cos 1 2sin ( / 2)θ θ= −

(a double angle formula) to obtain
0
0 2 2 20
2 2
4 4 (sin ( ) sin ( ))
T dθ
θ θ
θ
ω
=
−
∫ .
7. While this expression looks worse than before, it allows us
to make a significant change of variables.
Define a new variable of integration φ by 0
2 2
sin( ) sin( )sinθθ ϕ= . Note that when θ = 0, then φ = 0 and
when
θ = θ0, then φ = π/2. Show that the new integral with respect to
φ is
/2

2 20
( )
4
1 sin
K k
d
T
k
π ϕ
ω ϕ
=
−∫ ,
where k2 = sin2 (θ0/2). This new integral, which depends on the
value of k (and hence on θ0), is called a
complete elliptic integral of the first kind, and it is denoted K.
So we can write the period in the form
4
( )T K k
ω
= .

8. First notice that if the initial displacement θ0 (measured in
radians) is very small, then k = sin (θ0/2) is also
very small; for example, if θ0 = 0.1 radian (about 6 degrees),
then k = 0.05. If we set k = 0 in the integral,
show that T = 2π /ω, which is the approximate period for small
amplitudes.
Period of the pendulum 409
9. The elliptic integral cannot be evaluated analytically.
However, it is a well-known function and its values
are available. Table 2 shows a few values of the integral.
Table 2
θ0 (radians) K θ0 (radians) K
π/6 1.5981 2π/3 2.1565
π/4 1.6327 3π/4 2.4003
π/3 1.6858 5π/6 2.7681
π/2 1.8541 π ∞
Consider a pendulum with m = 1 kg and ℓ = 1 m. Using the
values in Table 2, find and plot the period of
the pendulum with initial displacements of π/6, π/4, π/3, π/2,
2π/3, 3π/4, and 5π/6 radians. Comment on
how the period of the pendulum varies with the initial
displacement. Is the pattern consistent with your

intuition?
10. If the mass of the pendulum bob in Step 9 is doubled, how
do the periods change? If the length of the
pendulum in Step 9 is halved, how do the periods change?
11. Give a physical interpretation of the case 0 ,θ π= which
implies that ( ) .K k = ∞
410 Guided Projects
Guided Project 38: Terminal velocity
Topics and skills: Integration, graphing
When an object falls in Earth’s gravitational field (think of a
skydiver jumping from an airplane or a marble
falling in a tank of oil), it accelerates due to the force of
gravity. If gravity were the only force acting on the
object, then all objects—elephants and feathers alike—would
fall at the same rate. But gravity is not the only
force present. Moving objects also experience resistance or
friction from the surrounding medium; it would be
air resistance for a skydiver and fluid resistance for a marble
falling in oil. The strength of the resistance
depends on several factors, among them the shape of the object
and the thickness (or viscosity) of the
surrounding medium. The effect of resistance is that a falling
object does not accelerate forever, as it would
without resistance. Eventually the gravitational force acting

downward and the resistance force acting upward
balance each other. As this balance is reached, the object
approaches a constant terminal velocity.
The motion of moving objects is described by Newton’s second
law of motion, which says that
Mass × acceleration = sum of external forces.
For a falling object there are two significant external forces:
gravity and resistance. We let x(t) be the position of
the falling object where x = 0 is the point at which the object is
released and the positive direction is downward
(Figure 1).
The velocity of the object is ( )
dx
v t
dt
= and its
acceleration is ( )
dv
a t
dt
= . The force due to gravity is
mg, where m is the mass of the object and g ≈ 9.8 m/s2
is the acceleration due to gravity; it acts in the positive
(downward) direction. We denote the resistance force R;

it acts in the negative (upward) direction.
The equation of motion now takes the form
dv
ma m mg R
dt
= = − .
Figure 1
The goal is to solve this equation for the velocity of the object.
All of the terms have been specified except the
resistance force. Air or fluid resistance is usually modeled in
one of two ways. For small velocities, often in a
heavy medium such as water or oil, it is common to assume that
R = kv, where k > 0 is a coefficient of
resistance. This says that the resistance force increases linearly
with the velocity. For large velocities, often in
air, a better assumption is R = Kv2, where K > 0 is a (different)
coefficient of resistance. With this model, the
resistance force increases with the square of the velocity. Our
immediate task is to solve the equation of motion
for the velocity using both types of resistance.
1. Let’s begin with the assumption that R = kv. What are the
units (dimensions) of the coefficient k in terms of
kilograms, meters, and/or seconds?

2. The equation of motion becomes
dv
m mg kv
dt
= − .
This equation is said to be separable because of the terms
involving the unknown v can be collected on one
side of the equation:
1
1
( / )
dv
dt g k m v
⎛ ⎞
=⎜ ⎟⎝ − ⎠
.
Terminal velocity 411
Now it is a matter of integrating both sides of the equation with
respect to t:
1

( / )
dv
dt dt
g k m v dt
⎛ ⎞
=⎜ ⎟⎝ − ⎠∫ ∫ .
A change of variables on the left side results in an integral with
respect to v:
( / )
dv
dt
g k m v
=
−∫ ∫ .
Evaluate the integrals on both sides of this equation. Then, use
the initial condition that v(0) = 0 to
determine the arbitrary constant of integration. Show that the
velocity function is given by
/( ) (1 )kt m
mg
v t e
k
−= − .

3. Assume m = 0.1 kg and graph the velocity function for k =
0.1, 0.5, 1.0. Describe the graph and check that
the initial condition v(0) = 0 is satisfied.
4. The terminal velocity is limt→∞ v(t). What is the terminal
velocity in each case in Step 3? Use Step 2 to find
an expression for the terminal velocity in terms of m, g, and k.
How does the terminal velocity vary with k?
5. We now turn to the assumption that R = Kv2. What are the
units (dimensions) of the coefficient K in terms
of kilograms, meters, and/or seconds? Note that k and K have
different units, which means we cannot
compare numerical values of the two parameters.
6. The equation of motion now becomes 2
dv
m mg Kv
dt
= − .
This equation is also separable because all of the terms
involving v can be brought to the left side of the
equation:
2
1
1

( / )
dv
dt g K m v
⎛ ⎞
=⎜ ⎟⎝ − ⎠
.
As before, we integrate both sides of the equation with respect
to t and use a change of variables. The
resulting equation is
2( / )
dv
dt
g K m v
=
−∫ ∫ .
It’s easiest to write the equation as 2
2 2
, where
dv K mg
dt a
m Ka v
= =

−∫ ∫ .
Evaluate the integrals on both sides of this equation and use the
initial condition v(0) = 0 to determine the
arbitrary constant. Show that the velocity function is given in
either of the two forms
2 / 2 /
2 / 2 /
1 1
( )
1 1
Kg mt Kg mt
Kg mt Kg mt
mg e mg e
v t
K Ke e
−
−
⎛ ⎞ ⎛ ⎞− −
= =⎜ ⎟ ⎜ ⎟
+ +⎝ ⎠ ⎝ ⎠
.

7. Assume m = 0.1 kg and graph the velocity function for K =
0.1, 0.5, 1.0. Describe the graph and check that
the initial condition v(0) = 0 is satisfied.
8. What is the terminal velocity in each case in Step 7? Find an
expression for the terminal velocity in terms
of m, g, and K. How does the terminal velocity vary with K?
9. Consider the following velocity measurements of a marble
falling in light oil.
Time (sec) 0.2 0.4 0.6 0.8 1.0 1.2 1.4
Velocity (m/sec) 1.5 2.3 2.8 3.0 3.1 3.2 3.3
Graph the data and graph the velocity functions in Steps 2 and 6
with m = 0.1 for various values of k and K.
Which model and which value of k or K give the best fit to the
data?
412 Guided Projects
10. Integrate the velocity function found in Step 2 to find the
position function x(t) where x(0) = 0. Graph the
position function for m = 0.1 kg and k = 0.1, 0.5, and 1.0.
Discuss how the graphs vary with k. How is the
terminal velocity reflected in these graphs?

11. Integrate the velocity function found in Exercise 6 to find
the position functions x(t) where x(0) = 0. Graph
the position for m = 0.1 kg and K = 0.1, 0.5, and 1.0. Discuss
how the graphs vary with K. How is the
terminal velocity reflected in these graphs?
Logistic growth 413
Guided Project 39: Logistic growth
Topics and skills: Integration, graphing
One of the most common models of population growth is the
exponential model. These models use functions of
the form p(t) = p0e
rt, where p0 is the initial population and r > 0 is the rate
constant. Because exponential models
describe unbounded growth, they are unrealistic over long
periods of time. Due to shortages of space and
resources, all populations must eventually have decreasing
growth rates. Logistic growth models allow for
exponential growth when the population is small. However, as
the population approaches a critical size called
the carrying capacity, the growth rate approaches zero. The
result is a self-regulating population.
We let p(t) be the population of a community or species at time
t ≥ 0, which means that ( )p t′

is the population
growth rate. The population growth rate for pure exponential
growth satisfies ( )p t rp′ = , where r > 0 is the rate
constant; that is, the growth rate is proportional to the
population size. By contrast, the growth rate for logistic
growth is given by ( ) 1
p
p t rp
K
⎛ ⎞′ = −⎜ ⎟⎝ ⎠
, where r is the rate constant and K is the carrying capacity.
1. Graph the exponential growth rate, ( )p t rp′ = as a function
of p with r = 0.1. As p increases, describe how
( )p t′
changes.
2. The growth rate for a logistic model, ( ) 1
p
p t rp
K
⎛ ⎞′ = −⎜ ⎟⎝ ⎠
, is graphed as a function of p in Figure 1 using
r = 0.1 and K = 500.

Figure 1
a. For what populations is the growth rate zero?
b. For what population is the growth rate a maximum?
c. Does the population ever decrease in size?
d. Assume that the initial population is p(0) = 10 individuals.
Using the growth rate in Figure 1, make a
rough sketch of the population as a function of time. The scale
on the horizontal (time) axis is not
important.
e. What value does the population approach as t →∞?
414 Guided Projects
3. The goal is to find the actual population function p that has a
growth rate given by ( ) 1
p
p t rp
K
⎛ ⎞′ = −⎜ ⎟⎝ ⎠
,
where we assume that r and K are specified constants. This task
requires solving a differential equation,

which in this case means evaluating integrals. We first separate
the dependent variable from the
independent variable. Dividing both sides of the equation by p(1
– p/K) gives
( )
( )
1 /
p t
r
p p K
′
=
−
.
We now integrate both sides of this equation with respect to t.
Show that on the left side a change of
variables leads to the integral
( )
1
( )
1 / ( )
dp

K
p t dt dp
p p K p K p
′ =
− −∫ ∫ .
4. Show that on the right side we have r dt rt C= +∫ , where C is
an arbitrary constant.
5. Use a partial fraction decomposition for the integral on the
left side to show that the population p satisfies
ln
p
rt C
K p
= +
−
.
6. Assume that if the initial population p(0) is between 0 and
K, then 0 < p(t) < K for all t > 0. Explain why
the absolute value may be removed to give
ln
p

rt C
K p
⎛ ⎞
= +⎜ ⎟−⎝ ⎠
.
7. Exponentiate both sides of the expression in Step 6 to obtain
rtp Ce
K p
=
−
,
where C is a new arbitrary constant.
8. Assume that the initial population p(0) = p0 is given.
Substitute t = 0 and p = p0 into the expression in Step
7 and solve for the arbitrary constant. Show that C = p0/(K –
p0).
9. Now solve for p in Step 7, use C from Step 8, and do several
steps of algebra to show that the population
function is

0
0 0
( )
( ) rt
Kp
p t
K p e p−
=
− +
.
10. Verify that p(0) = p0 as specified.
11. Evaluate limt→∞ p(t) and interpret this result.
12. Graph the population function for r = 0.1, p0 = 10, and K =
500. Is the graph similar to the sketch you made
in Step 2d?
Logistic growth 415
13. On the same set of axes, graph the population function for
p0 = 10, K = 500, and r = 0.05, 0.1, 0.2, and 0.5
(four curves). Describe the effect of changing the rate constant

r.
14. On the same set of axes graph the population function for
p0 = 10, r = 0.1, and K = 100, 500, 1000, and
1500 (four curves). Describe the effect of changing the carrying
capacity K.
15. Prove that for a logistic growth model, the maximum
growth rate occurs when the population reaches K/2.
416 Guided Projects
Guided Project 40: A pursuit problem
Topics and skills: Derivatives, integration, graphing
From photographers stalking wildlife to heat-seeking missiles
chasing incoming rockets, pursuit problems take
many forms with varying degrees of complexity. In this project
we solve a benign pursuit problem that
illustrates the essential features. Here is the problem.
At the moment a dog begins walking north
from a crossroads at 1 mi/hr, the dog’s
master begins walking from a point 1 mi
east of the crossroads (Figure 1). The
master walks at a constant speed of s > 1
mi/hr and at all times walks directly at the
dog. Find a description of the master’s

path and determine the point at which the
master overtakes the dog.
Figure 1
Let (xd(t), yd(t)) be the coordinates of the dog at time t ≥ 0,
where t is measured in hours. Let (x(t), y(t)) be the
coordinates of the master at time t ≥ 0, where (x(0), y(0)) = (1,
0). The goal is to find the function f such that the
path of the master is the curve y = f(x).
1. Why is the condition s > 1 imposed on the master’s speed?
2. Explain why the coordinates of the dog are (xd(t), yd(t)) =
(0, t), for all t ≥ 0.
3. Note that ( )x t′ is the velocity of the master in the x-
direction and ( )y t′ is the velocity of the master in the
y-direction. Give an argument for the fact that at all times the
slope of the line tangent to the master’s path
is ( ) ( ) / ( )y x y t x t′ = ′ ′ .
4. The speed of the master s is specified. Show that 2 2 2( ( )) (
( ))s x t y t= ′ + ′ .
5. Combine Steps 3 and 4 to conclude that
2
( ) .

1 ( ( ))
s
x t
y x
′ = −
+ ′
(1)
Explain why ( ) 0x t′ < .
6. Use Figure 1 to show that at a particular time t ≥ 0, the
slope of the line tangent to the master’s path is
( ) .
y t
y x
x
−′ = (2)
Is ( )y x′ positive or negative? Explain.
7. The immediate task is to eliminate t from equations (1) and
(2) leaving a single equation in x and y. Toward
this end, multiply both sides of (2) by x and then differentiate
carefully with respect to t to show that

( ) ( ) ( ) ( ) ( ) 1xy x x t y x x t y t′′ ′ + ′ ′ = ′ − .
A pursuit problem 417
8. Divide both sides of the equation in Step 7 by ( )x t′ and use
(1) to give the single equation in x and y
21 ( ( ))
( ) .
y x
xy x
s
+ ′
′′ = (3)
9. This equation, which involves ( )y x′ and ( )y x′′ , must now
be solved for y. It can be done by integrating
twice. Before doing so, explain why it is true that (1) (1) 0y y= ′
= .
10. Equation (3) is easiest to attack if we make the substitution
( ) ( )v x y x= ′ . Show that equation (3) becomes

2
1 1
.
1
dv
dx sxv
=
+
(4)
11. Integrate both sides of (4) with respect to x and use a
change of variables on the left side to show that
2 lnln( 1 )
x
v v C
s
+ + = + , (5)
where C is an arbitrary constant and we have assumed that x >
0.
12. Use properties of logarithms to write (5) in the form

2 1/1 sv v Cx+ + = , (6)
where C is another arbitrary constant.
13. Show that the condition (1) (1) 0v y= ′ = implies that C = 1
in (6).
14. Solve (6) for ( ) ( )v x y x= ′ and show that
( )1/ 1/1( )
2
s sv y x x x−= = −′ . (7)
15. Remembering that the goal is to find y(x), integrate (7)
with respect to x. Then use the condition y(1) = 0 to
show that the arbitrary constant is C = s/(s2 – 1). Conclude that
the path of the master is described by the
function
( 1) / ( 1)/
2
( )
2 1 1 1
s s s ss x x s
y f x
s s s

+ −⎛ ⎞
= = − +⎜ ⎟+ − −⎝ ⎠
.
16. In terms of s, at what point does the master overtake the
dog?
17. Graph the path function f for s = 1.1, 1.5, 2, and 3.
Describe how the master’s path depends on s. Does the
length of the path increase or decrease with s?

390 Guided Projects Guided Project 31 Cooling cof.docx

Recommended

Recommended

More Related Content

Similar to 390 Guided Projects Guided Project 31 Cooling cof.docx

Similar to 390 Guided Projects Guided Project 31 Cooling cof.docx (20)

More from gilbertkpeters11344

More from gilbertkpeters11344 (20)

Recently uploaded

Recently uploaded (20)

390 Guided Projects Guided Project 31 Cooling cof.docx