Consider the equations describing the interactions of robins r and worms w:
dw/dt = w - wr, and dr/dt = -r + rw.
A nullcline for this system is a curve in the w, r phase plane such that either = dw/dt = 0, or dr/dt = 0.
Critical points occur where nullclines intersect. Another application of nullclines is to divide the phase plane into regions where dw/dt and dr/dt are each either positive or negative.
(a) What are the (non-zero) nullclines for this system?
(b) Your nullclines divide the phase plane into four regions. Give a sample point in each region, and indicate for that point whether each of the populations is increasing or decreasing (by entering the word increasing or decreasing appropriate blank):

i) (w, r) = ( , ) is in one region, where the population of worms, w is and the population of robins, r is .
ii) (w, r) = ( , ) is in a second region, where the population of worms, w is and the population of robins, r is .
iii) (w, r) = ( , ) is in a third region, where the population of worms, w is and the population of robins, r is .
iv) (w, r) = ( , ) is in the fourth region, where the population of worms, w is and the population of robins, r is .
Notice what your conclusions about these four regions say about how the populations change with time.

a) See Picture attached

b)

i) (w, r) = ( 1.5 , 0.5 ) is in one region, where the population of worms, w is increasing and the population of robins, r is increasing.

ii) (w, r) = ( 1.5 , 1.5 ) is in one region, where the population of worms, w is decreasing and the population of robins, r is increasing.

iii) (w, r) = ( 0.5 , 1.5 ) is in one region, where the population of worms, w is decreasing and the population of robins, r is decreasing.

iv) (w, r) = ( 0.5 , 0.5 ) is in one region, where the population of worms, w is increasing and the population of robins, r is increasing.

answer: hey there,

the answer you are looking for is a..

hope this you!

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.