# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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You might think that the path of a sky diver in free fall looks like the downward path of the ball in the simplest juggler problem of vertical motion. However, as the sky diver’s velocity becomes large the effects of air resistance (or drag) become noticeable and must be included in the model. A revised model (starting with Step 3) follows:

3. In this case, Newton’s second law says that mass times acceleration is equal to the force due to gravity plus that due to air resistance. Experience has shown that the force of air resistance can be modeled fairly well by a term that is proportional to velocity and opposite in direction.

4. We have mh" = mv' = —mg — kv, where k is a constant coefficient of air resistance. The initial velocity of the sky diver is vo = 0 ft/sec; the initial height when the sky diver jumps from the plane is ho ft.

5. We solve the second-order ODE for h in two steps, first for v (by separating the variables) and then for h (by integrating the expression we

32

Chapter 2

C is an arbitrary constant.

So the sky diver ’s free fall

ODE is hi’ = —32 — (k/tyv.

find for v, since v = h'). Here are the steps:

k

v' = — g---------v, v(0) = 0

m

1 dv

g + kv/m dt

f dv = - f dt

J g + kv/m J

(m/k) ln(g + kv/m) = —t + C ln(g + kv/m) = (k/ m)( — t + C)

Exponentiating and setting K = exp(kC/m) we obtain g + kv/m = Ke—(k/ m)t Since v = 0 when t = 0, we find that K = g. Solving for v we obtain

v = ^g+rnge_(k/m)t

kk

That means that h(t) solves the IVP

h' = v = -^- + ^-e-<k/m)t, h(0)=ho

kk

We find the formula for h( t) by integration and the fact that h = h0 at

t = 0:

k k2 k2

In our example of free fall (Screen 4.3), these equations become

h" = v' = —32 — if m = 5 slugs

, —160 160 (k/5)t

H = V =-----------1 e-W5T

kk

—160 800 (k/5)t 800 1ornn

h= —— t- —re~{k/5)t+ —r + 13500 (5)

k k2 k2

See Figure 2.4 for some time-height curves.

Since the mass m of the sky diver doesn't drop out of the ODE when damping is added, we have to use appropriate units for the mass. In English

units (which the English have been wise enough to discard) we have

force weight lbs

mass =-------------=-----------= 5- = slugs

acceleration gravity ft/sec2

Opening the Parachute

If we wish to model what happens when the parachute opens, we'll need to alter the model slightly to account for the sudden change in drag—that is, for how the value of k suddenly changes.

The Sky Diver

33

Opening the chute changes k from kff to kp.

References

t-h

Time (seconds)

Figure 2.4: Six sky divers in free fall from 13,500 ft: viscous damping constants range from 0.5 to 1.5 slug/sec. Which sky diver has the smallest damping constant?

4. We can use experimental values for the drag coefficients: in free fall kff = 0.86, and, after the parachute opens, kp = 6.71, both in slugs/sec. The parachute opens at time tp, when h is 2500 feet. It’s hard to calculate tp from formula (5), so we can approximate it by reading the graph of h vs. t (use the Explore feature on graphs of Screen 4.4).

We noticed on Screen 4.5 of Module 2 that an instantaneous opening of the parachute would exert an enormous force on the sky diver, so the model was further revised to allow the chute to open over a few seconds (a more realistic model), and we let k grow gradually, in a linear way, as it goes from kff to kp. Take a look at Exploration 2.4, Problem 3.

Borrelli, R. L., and Coleman, C. S., Differential Equations: A Modeling Perspective, (1998: John Wiley & Sons, Inc.)

Boyce, W. E., and DiPrima, R. C., Elementary Differential Equations and Boundary Value Problems, 6th ed. (1997: John Wiley & Sons, Inc.)

Hale, M., and Skidmore, A., A Guided Tour of Differential Equations, (1997: Prentice-Hall)

GODE-ENewsletter, http://www.math.hmc.edu/codee, for articles on modeling with ODEs

IDEA (Internet Differential Equations Activities), created by Thomas LoFaro and Kevin Cooper, offers an interactive virtual lab book with models. http://www.sci.wsu.edu/idea

34 Chapter 2

Answer questions in the space provided, or on

attached sheets with carefully labeled graphs. A

notepad report using the Architect is OK, too.

Name/Date______

Course/Section

Exploration 2.1. ODEs and Their Solutions

1. Where is that constant?

Solution formulas for first-order ODEs often involve an arbitrary constant C, and it can show up in all sorts of strange places in the formulas. Solve each of the following ODEs for y in terms of t and C.

(a) y = 1 + sin t (b) y = —y/3 (c) y = t/y (d) y = 2t2y/lny

2. Let’s check out the ODE Architect.

You can see how good the ODE Architect solver is by creating initial value problems for the ODEs of Problem 1 and using the Architect to solve them and graph the solutions. Then compare the solver graphs with those obtained using the solution formula. For example, use ODE Architect to solve and plot the solution of the IVP / = —y/3, y(0) = 1. Then graph the solution y = e—t/3 and compare. To do this, enter the following two equations on the editor screen:

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