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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

⋅ + ⋅ + = =

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

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

0 1

1 1 ( ) ( ) ( ) ( )

2 2

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

( ) ( ) ( ) ( )

. 2

n 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

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

20 1 0.2146018366

x dx

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

− , 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

x I dx

+= = − +∫

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

Average lifetime fraction that fail in [ , ]

fraction that fail in [ , ]

fraction that fail in [ , ]n n n

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:

( ) ( ) 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

dL cL dHL

= −

= − +

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

1 1

( ) 2 (cos cos )

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 2 2 20 2 2

4 4 (sin ( ) sin ( ))

T dθ

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 20

( )

1 sin

K k

d T

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 ( / )

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

( / )

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 /

1 1 ( )

1 1

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 / ( )

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

( ) ( ) 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

+ ′ ′′ = (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

1 1 .

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?