Assignment 6 for Mathematical Modeling | MATH 142, Assignments of Mathematics

Material Type: Assignment; Class: Mathematical Modeling; Subject: Mathematics; University: University of California - Los Angeles; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/30/2009

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Math 142 Homework #6
1. 37.4
Answer:
2. The infected population should grow in proportion to the uninfected population.
2. 38.2
Answer:
The growth is governed by
dN
dt =R(N)N
Equilibrium occurs when the time derivative is zero. This can happen when N= 0 and
when R= 0. Since Ris zero at two non-zero Nvalues, there are three equilibrium positions.
At N=๎˜N1, the growth rate is negative, and the population returns to zero. So N= 0 is
stable.
At the left most root of R(N), the growth rate is positive for values slightly larger than Neq.
This is unstable.
Finally, at the other root of R(N), the growth rate is negative to the right, and positive to
the left. This is stable.
3. Phase plane of linear systems
Sketch the phase plane solutions of the following systems
a.
dx
dt =x+y
dy
dt =x
Answer:
The phase plane equation is
dy
dx =x
x+y
This gives a zero isocline at x= 0 and an infinity isocline at y=โˆ’x. When x= 0,
dx
dt =ywhich means that xis increasing when yis positive, and vice versa. When y=โˆ’x,
dy
dt =โˆ’ywhich means yis decreasing when yis positive.
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Math 142 Homework #

Answer:

  1. The infected population should grow in proportion to the uninfected population.

Answer: The growth is governed by

dN dt =^ R(N^ )N

Equilibrium occurs when the time derivative is zero. This can happen when N = 0 and when R = 0. Since R is zero at two non-zero N values, there are three equilibrium positions. At N = N 1 , the growth rate is negative, and the population returns to zero. So N = 0 is stable. At the left most root of R(N ), the growth rate is positive for values slightly larger than Neq. This is unstable. Finally, at the other root of R(N ), the growth rate is negative to the right, and positive to the left. This is stable.

  1. Phase plane of linear systems

Sketch the phase plane solutions of the following systems

a. dx dt

= x + y dy dt =^ x Answer: The phase plane equation is dy dx =^

x x + y This gives a zero isocline at x = 0 and an infinity isocline at y = โˆ’x. When x = 0, dx dydt^ =^ y^ which means that^ x^ is increasing when^ y^ is positive, and vice versa. When^ y^ =^ โˆ’x, dt =^ โˆ’y^ which means^ y^ is decreasing when^ y^ is positive.

Straight-line trajectories are given by

m = x x + mx m = 1 1 + m m + m^2 โˆ’ 1 = 0 m = โˆ’^1 2

m โ‰ˆ 0. 62 , โˆ’ 1. 62

x

y

b.

dx dt =^2 x^ +^ y dy dt

= โˆ’x + y

Answer: The phase plane equation is

dy dx

= โˆ’x^ +^ y 2 x + y

This gives a zero isocline at y = x and an infinity isocline at y = โˆ’ 2 x. When y = x, dx dt = 3x^ which means that^ x^ is increasing when^ x^ is positive, and vice versa.^ When y = โˆ’ 2 x, dydt = โˆ’ 3 x which means y is decreasing when x is positive.