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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT 256 15 "volterra.mws   " }{TEXT -1 22 " Predator-prey mod
els\n" }{TEXT 261 43 "Slightly modified 10/31/2003 by Matt Miller" }
{TEXT 262 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 122 "We use Maple's DEtool
s to study solutions of the Lotka-Volterra system and its refinements \+
as described in section 4.1 of " }{TEXT 257 19 "Calculus in Context" }
{TEXT -1 82 " and chapter 6 of E-K.  As you play with the models, keep
 these questions in mind:" }}{PARA 0 "" 0 "" {TEXT -1 822 "\n1.  What \+
is the long term behavior of the system?\n2.  In the case of oscillati
ons, what is the period (time interval from peak to peak or trough to \+
trough), and what is the amplitude?\n3.  How does changing the initial
 conditions affect your answers to qustions 1 and 2?  \n4.  Does the s
ystem have any steady states (equilibria)?  Do these appear to be stab
le or unstable?\n5.  If there are steady states, are they in any way r
elated to the long term behavior?\n\nWe first consider problem 6 on pa
ge 190 of CIC; see also E-K pp 218-220, equations (7ab).  Here h repre
sents the hare population and u represents 60 times the lynx populatio
n (since the lynx population is numerically much smaller than the hare
 population, we scale it up to fit on the same graph).  We are going t
o work with three different initial conditions. " }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 37 "restart: with(plots):  with(DEtools):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "rate_eqn1:= diff(h(t),t)=(0
.1)*h(t)-(0.005)*h(t)*(1/60)*u(t); rate_eqn2:=diff(u(t),t)=(0.00004)*h
(t)*u(t)-(0.04)*u(t);\nvars:= [h(t), u(t)];  " }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 110 "init1:=[h(0)=2000,u(0)=600]; init2:=[h(0)=2000,
u(0)=1200]; init3:=[h(0)=2000, u(0)=3000]; \ndomain := 0 .. 320;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 368 "\n We plot the hare and lynx popu
lations jointly against time using the first of the given initial cond
itions.   You should repeat this with the other initial conditions.  G
et a feeling for the accuracy of the computations by changing the step
 size, and for the long term behavior by changing the time interval.  \+
Keep a record of the results with questions 1-5 in mind!" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 207 "Lplot:= DEplot(\{rate_eqn1, rate_e
qn2\}, vars, domain,\{init1 \}, stepsize=0.5, scene=[t, u], arrows=NON
E):\nHplot:= DEplot(\{rate_eqn1, rate_eqn2\}, vars, domain,\{init1 \},
 stepsize=0.5, scene=[t, h], arrows=NONE):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 69 "display( \{Lplot, Hplot\} , title = `Hares and 60 *
 Lynxes vs. time` );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "\nWhich graph is which?  You may
 want to inset options such as linecolor= or thickness= to distinguish
 them.  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 341 "Next we plot the hare and lynx populations against one a
nother in what is called a PHASE PORTRAIT.  We do this for three diffe
rent initial conditions.  ***Can you identify which curve goes with wh
ich initial condition?  How is the independent variable t showing up i
n these pictures?  (Hint: try it again with time interval t = 0 .. 20.
) ***" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 161 "DEplot(\{rate_eqn
1, rate_eqn2\}, vars, t= 0 .. 160, \{init1, init2, init3\}, stepsize=0
.5, scene=[h,u], title=`Hares vs. 60 * Lynxes for t = 0 .. 160`, arrow
s=NONE);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 115 "What is the significance of the next calculation?  (Hint
: try using these values of h and u as initial conditions.)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "equil:= solve( \{(0.1)*H-(0.
005)*H*(1/60)*U = 0, (0.00004)*H*U-(0.04)*U = 0\}, \{H, U\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "CQfunction:= (H, U) -> 0.000
04*H + (0.005/60)*U - 0.04*ln(H)- 0.1*ln(U);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 97 "Fill in (H, U) coordinates of several points from one o
f the \"orbits\" above. What do you observe?" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 109 "CQvalue:= CQfunction(  %? , %? ); # use the TAB
 key to get to the %? and replace them with actual coordinates" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 137 "implicitplot( 0.00004*H + (
0.005/60)*U - 0.04*ln(H)- 0.1*ln(U) = CQvalue, H = 0 .. 4200, U = 0 ..
 3500, thickness = 2, numpoints = 1000);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 178 "Does this la
st plot help convince you that the \"orbits\" really do close up?  (Hi
nt: along each orbit there appears to be a conserved value, CQ. What i
s the significance of this?)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "\n***Discuss the answers
 to questions 1-5 above in light of your examination of the model.***
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
489 " Next we study solutions of the Lotka-Volterra system of CIC pp 1
84-186; see also E-K pp 220-222 equations (8ab).  In this model the pr
ey is assumed to grow logistically in the absence of any predators.  C
an you see how the rate equations have been changed from the original \+
L-V model to incorporate this assumption?  This time h represents the \+
hare (rabbit) population and u represents 100 times the lynx (fox) pop
ulation.   We are going to work with three different initial condition
s.  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "restart: with(plots
):  with(DEtools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "rate
_eq1:= diff(h(t),t) = (0.1)*h(t)-.00001*(h(t))^2-(0.005)*h(t)*(1/100)*
u(t) ;\nrate_eq2:= diff(u(t),t) = (0.00004)*h(t)*u(t)-(0.04)*u(t) ;\nv
ars:= [h(t), u(t)];  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "i
nit1:=[h(0)=2000,u(0)=500]; init2:=[h(0)=2000,u(0)=1000]; init3:=[h(0)
=2000,u(0)=4000]; \ndomain:= 0 .. 380;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 237 "\n First we plot the hare and lynx populations jointly a
gainst time using the first of the given initial conditions.   As abov
e you should repeat this with the other initial conditions, different \+
time intervals, different step sizes, etc." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 204 "Lplot:= DEplot([rate_eq1, rate_eq2], vars, domain,\{
init1 \},stepsize=0.5, scene=[t, u], arrows=NONE):\nHplot:= DEplot([ r
ate_eq1, rate_eq2], vars, domain, \{init1 \}, stepsize=0.5, scene=[t, \+
h], arrows=NONE):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "displa
y( \{Lplot, Hplot\}, title = `Hares and 100 * Lynxes vs. time` );" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 139 "Repeat these plots with the other initial conditions. Wh
at do you observe, especially as regards the frequency and amplitude o
f the cycles?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 366 "\nNext we plot \+
the hare and lynx populations against one another in what is called a \+
PHASE PORTRAIT.  We do this for two different initial conditions.  Can
 you identify which curve goes with which initial condition?  How is t
he independent variable t showing up in these pictures?  (Hint: try it
 again with time interval t = 0 .. 20, or use the option arrows = SLIM
.)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 202 "DEplot([ rate_eq1, rate_eq2 \+
], vars, 0 .. 480, [init1, init2, init3], linecolor=[blue, black, gree
n], arrows = SLIM, stepsize= 0.5, title =\"Hares vs. 100 * Lynxes for \+
t = 0 .. 480\", dirgrid = [15, 15]);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 89 "equil:= solve( \{(0.1)*H-.00001*H^2-(0.005)*H*(1/100)
*U, (0.00004)*H*U-(0.04)*U\}, \{H, U\});" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 95 "\nWhat is the significance of this last calculation?  ***
Answer questions 1-5 for this model.***" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 494 "Finally we study solutio
ns of the May system (problem 5 on page 189 of CIC; see also E-K page \+
265 problem 32 where the predators are whales and the prey are krill).
   Here x is measured in units of \"hectohares\" (i.e., the number of \+
hares in units of 100) and y is the number of lynxes.  Choose a variet
y of initial conditions, time intervals, stepsizes, and so forth.  ***
Are there any QUALITATIVE similarities and/or differences that you not
ice between the May model and the two L-V models?***" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 82 "eq1:=diff( x(t), t) = 0.6 * x(t) *(1 -(x(
t) / 10)) - 0.5 * x(t) * y(t) /(x(t)+ 1);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 61 "eq2:=diff( y(t), t) = 0.1 * y(t) * ( 1 - y(t) / (2 * \+
x(t)) );" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "vars:= [x(t), y(t)];" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "init1:=[x(0)=8, y(0)=10];
 init2:=[x(0)=8, y(0)=15]; init3:=[x(0)=8, y(0)=5]; init4:=[x(0)=1.4, \+
y(0)=2.5]; domain:= 0 .. 150;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 94 "Xplot:= DEplot([eq1, eq2], vars, domain, \{init1 \}, stepsize=0.
5, scene=[t,x], linecolor=blue):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 95 
"Yplot:= DEplot([eq1, eq2], vars, domain, \{init1 \}, stepsize=0.5, sc
ene=[t,y], linecolor=green):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 79 "display( \{Xplot, Yplot\}, title = `May model: Rabbits/100 and F
oxes vs. time` );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "Repeat these plots with the other
 initial conditions. What do you observe, especially as regards the fr
equency and amplitude of the cycles?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 188 "DEplot([eq1, eq2], vars, 0 .. 280, [init1, init2, in
it3, init4], linecolor=[blue, black, aquamarine, plum],  stepsize=0.25
, scene=[x,y], title = `May model phase portrait`, arrows = SLIM);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 102 "equil:= solve( \{0.6 * X *(1 -(X / 10)) - 0.5 * X \+
* Y /(X + 1), 0.1 * Y * ( 1 - Y / (2 * X))\}, \{X, Y\});" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 397 "\nWhat is the significance of the last c
alculation?  *** In Biology 301 you may have heard the term \"limit cy
cle\".  Can you explain where this term comes from?  Predict what will
 happen if you use initial conditions close to, but not at, the equili
brium values.  Test your prediction!***  Suggest two very different ex
amples of global trajectories with the same local behavior near the eq
uilibrium." }}{PARA 0 "" 0 "" {TEXT -1 816 "\n***At last it's time to \+
judge the models.  What sort of field measurements would you want to h
ave in order to choose one over another?  Are there any purely mathema
tical features of the predictions of the models that might help?  (One
 can critique the models on the basis of unreasonable assumptions that
 might go into their construction--this is a separate matter.)  For ex
ample, you have surely observed that each of the three models (L-V wit
h unbounded prey growth, L-V with bounded prey growth, and May) exhibi
ts a different possibility for long term behavior.  Why is this consid
ered to be such an important aspect of the model?  What about sensitiv
ity to initial conditions--how does long term behaviour depend on init
ial conditions, and what does this mean in terms of actual observation
 of populations?***  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{SECT 1 {PARA 3 "" 0 "" {TEXT 259 57 "Lotka-Volterra type models wit
h local eigenvalue analysis" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 300 "\n
We first consider problem 6 on page 190 of CIC; see also E-K pp 218-22
0, equations (7ab).  Here h represents the hare population and u repre
sents 60 times the lynx population (since the lynx population is numer
ically much smaller than the hare population, we scale it up to fit on
 the same graph).  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 38 "restart: with(linalg):  with(plots):  " }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "rate_eqn1:= diff(h(t),t)=
(0.1)*h-(0.005)*h*(1/60)*u; rate_eqn2:= diff(u(t),t)=(0.00004)*h*u-(0.
04)*u;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "equil:= solve( \{
rhs(rate_eqn1), rhs(rate_eqn2)\}, \{h , u \});" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 40 "F:= rhs(rate_eqn1);  G:= rhs(rate_eqn2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "J:= jacobian([F, G], [h, u] \+
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "sel_equil:=  ; # sele
ct the desired equilibrium point (copy and paste from output above)" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "A:= evalf(subs(sel_equil, \+
evalm(J)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eigenvals( A
 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 98 "What is this calculation telling you about the nat
ure of the equilibrium point in phase space for " }{TEXT 260 24 "the l
inearized system?  " }{TEXT -1 73 "What conclusion can you draw, if an
y, for the original non-linear system?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 426 " Next we study solutions of the Lotka-Volterra sy
stem of CIC pp 184-186; see also E-K pp 220-222 equations (8ab).  In t
his model the prey is assumed to grow logistically in the absence of a
ny predators.  Can you see how the rate equations have been changed fr
om the original L-V model to incorporate this assumption?  This time h
 represents the hare (rabbit) population and u represents 100 times th
e lynx (fox) population.   " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 118 "rate_eq1:= diff(h(t),t) = (
0.1)*h-.00001*h^2-(0.005)*h*(1/100)*u ;\nrate_eq2:= diff(u(t),t) = (0.
00004)*h*u-(0.04)*u  ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "e
quil:= solve( \{rhs(rate_eq1), rhs(rate_eq2)\}, \{h , u\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "F:= rhs(rate_eq1);  G:= rhs(
rate_eq2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "J:= jacobian(
[F, G], [h, u] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "sel_e
quil:=  ; # select the desired equilibrium point (copy and paste from \+
output above, complete with the curly braces)" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 37 "A:= evalf(subs(sel_equil, evalm(J)));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eigenvals( A );" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 53 "\nWhat is the significance of this last c
alculation?  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 494 "F
inally we study solutions of the May system (problem 5 on page 189 of \+
CIC; see also E-K page 265 problem 32 where the predators are whales a
nd the prey are krill).   Here x is measured in units of \"hectohares
\" (i.e., the number of hares in units of 100) and y is the number of \+
lynxes.  Choose a variety of initial conditions, time intervals, steps
izes, and so forth.  ***Are there any QUALITATIVE similarities and/or \+
differences that you notice between the May model and the two L-V mode
ls?***" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 68 "eq1:=diff( x(t), t) = 0.6 * x *(1 -(x / 10)) -
 0.5 * x * y /(x + 1);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "eq2:=diff
( y(t), t) = 0.1 * y * ( 1 - y / (2 * x) );" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 45 "equil:= solve( \{rhs(eq1), rhs(eq2)\}, \{x, y\});
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "F:= rhs(eq1);  G:= rhs(
eq2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "J:= jacobian([F, G
], [x, y] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "sel_equil:=
   ; # select the desired equilibrium point (copy and paste from outpu
t above)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "A:= evalf(subs(
sel_equil, evalm(J)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "e
igenvals( A );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 330 "\nWhat is the \+
significance of the last calculation?  *** In Biology 301 you may have
 heard the term \"limit cycle\".  Can you explain where this term come
s from?  Predict what will happen if you use initial conditions close \+
to, but not at, the equilibrium values.  Test your prediction, using t
he first section of this worksheet. ***" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "0 0 0" 0 
}{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }