Homework for Lecture 24 of Dr. Z.’s Dynamical Models in Biology class
Email the answers (either as .pdf file and/or .txt file) to ShaloshBEkhad@gmail.com by 8:00pm Monday, Nov. 29,, 2021. Subject: hw24 with an attachment hw24FirstLast.pdf and/or hw24FirstLast.txt
Also please indicate (EITHER way) whether it is OK to post
Comment: The goal of this homework and all the remainig ones is to prepare you to the ‘basic concepts/skills’ qualifying exam on Dec. 7. As I hope you know, this is a necessary condition for passing.
1.	Carefully read, and understand, the posted solution to the attendance quiz for Lecture 24 https://sites.math.rutgers.edu/~zeilberg/Bio21/att24S.pdf .
If you get any of it wrong, explain what you did wrong.

I was absent from class because I was sick, but I read and understand the solutions for each of the problems. 

2.	For a certain particle, moving in a straight line, the rate of change of the rate of change of the rate of change of the acceleration is 120 meters/sec3. At the very beginning its velocity,acceleration , rate of change of acceleration, and rate of change of rate of change of the acceleration are 0 meters/sec, 0 meters/sec2, 0 meters/sec3, and 0 meters/sec4,respectively. How far from the starting point is the particle after 2 seconds?
X’’’(t) = 120
X’’(t) = 120t
X’(t) = 60t^2
X(t) = 20t^3
X(2) = 160
The particle is 160 meters from the starting point at t=2. 

3.	(i) A ball is dropped vertically from a building of height 100 meters. How long will it take until it hits the ground. Ignore air resistance
X(0) = -100, X’(0) = 0
X”(t) = -10
X’(t) = -10t
X(t) = -5t^2
-100 = -5t^2
t = 4.47 s

(ii) A ball is dropped vertically from a building of height 100 meters. The air-resistance is twice times the mass times the velocity. Set up a differential equation with the appropriate initial conditions, and then use Maple’s dsolve to find how long it will take to hit the ground.
X(0) = 100
Air resistance = 2m*x’(t)
ma = mg – 2m*v
x”(t) = g – 2*x’(t)
x”(t) + 2*x’(t) + 10 = 0
t = 20.5 s

4. Define
(a)	A discrete-time dynamical system with one-quantity (variable). Write down its format.
X(n) = f(x(n-1))
(b)	The orbit of a discrete-time dynamical system with one-quantity (variable) starting at x(n) = x0 up to n = K.
The orbit represents the set of points that a particular function approaches (also known as trajectory), and can be used to find stable fixed points. 
(c)	The notion of an equilibrium solution (Hint: it is a solution of the difference equation that is always constant)
An equilibrium solution is a solution that results in a derivative of 0 everywhere. 
(d)	The notion of a stable equilibrium solution (Hint: it is a solution of the difference equation that is always constant and if you start x(0) not too far from it, you also wind-up, in the long run very close to it.)
A stable equilibrium point results in a derivative of 0. The system always approaches this point in the trajectory. 
5.
(a)	Describe how to numerically locate (at least approximately, but very reliably) the stable fixed points (aka stable discrete-equilibria), using orbits (implemented by Orb)
To find the stable fixed points, plug in each value into the equation to find the next value, and continue this process until the numbers being repeating. The first repeated value is the stable fixed point. The set of the values between the repeated values is the orbit. 

(b)	In terms of the underlying function, call it f(x) describe how to use algebra to find the set of fixed points (aka discrete-equilibrium-solutions)
Set x = f(x) and solve for x. 
(c)	In terms of the underlying function, call it f(x) describe how to use calculus to find the subset of the above set, the set of stable fixed points (aka discrete-stable equilibrium-solutions)
Find the derivative of f(x), called f’(x). For each fixed point, solve f’(x) by plugging each fixed point into x. If the absolute value is less than 1, the fixed point is stable. 
(d)	Apply (a) (using Orb in DMB.txt with the appropriate inputs) , (b), (c) to the following discrete time first-order dynamical system

(i)
	 	,
F(x) = x + 1 / x + 2
Fixed points at {0.618, -1.618}. 
-1.618 is a stable fixed point. 

(ii)
	 	,
F(x) = 5x/2(1 – x) = (5x/2 – (5x^2)/2)
Fixed points at {0, 0.6}. 
0.6 is a stable fixed point. 

(iii)
	 	,
F(x) = 7x/2(1 – x)  = (7x/2 – (7x^2)/2)
Fixed points at {0, 1). 
There are no stable fixed points.