ln(E)
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
ln("t)
From the two plots we can see that as the step size, t, gets smaller and smaller, the Euler
method for approximation gets more an
The equilibrium points occur at:
=
When n is an even integer, the point is unstable, when it is an odd integer it is stable, and
when it is zero the point is unstable.
2.7.6
Plot the potential function V(x) and identify all of the equilibrium
To account for the direction of the flow, a negative sign needs to be added to show that the
height of the water in the bucket is decreasing and therefore:
= 2
If:
= 2
Then:
=
2.7.3
Plot the potential function V(x) and identify all
= %
E=^(1)-x(1)
x
Evaluating at x(1) we get that x(t) = 1/e
(b) Using the Euler method with t = 1 estimate x(1) numerically:
'() = ' + (' )
'() = ' '
'() = ' (1 )
1 '
'() = ' 1
Plugging values for n=100, 101, 102, 103, 104 into t
When r < -0.4, there is one stable equilibrium point.
When -0.4 < r < 0.4 there are three equilibrium points: two stable points with an unstable
point between them.
When r > 0.4, there is one stable equilibrium point which can be seen below:
8
"
= "
This unit conversion demonstrates total volumetric flow-rate. The above equation represents
the total water currently in the bucket at any time t (A*h(t) and the total amount of water
leaving the bucket (a*v(t). This is also known as the Cons
2.5.2
Looking at = 1 + % instead of &' we have:
=
1 + %
tan-& = +
= tan
0
0
The solutions for this function only exist for < < and we can see from the plot below
0
%
%
x(t)
that as that () in a finite amount of time as can be see
x(t)
For a = 0, there is one fixed point at =
For a > 0, there are three fixed points at = and =
2.4.9
= 1
= 1
1
= 1
1
1
6 = +
2
=
+
From this solution we can see that the function decays not exponentially, but instead
2.1.1
Fixed points occur where = 0 therefore:
= sin
0 = sin
=
2.1.2
The greatest velocity occurs when = . To find the maximum values, differentiate and set
equal to zero:
= sin ()
= cos ()
0 = cos ()
= +
2.2.1
= 4 4 16
0
2.2.10
(a) Every number is a fixed point
=
(b) Every integer is a fixed point, and there are no others
= ()
(c) There are precisely three fixed points, and all of them are stable
This case is not possible because between any two stable fixed