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1 sections June 11, 2009

2 Population growth/decay When we model population growth, the simplest model is the exponential (or Malthusian) model. Basic ideas: P = P(t) = population size as a function of time. (1) Rate of population growth is proportional to the difference between the birth rates (inflow) and death rates (outflow) (2) Birth rate and death rate are each proportional to the population size P(t) at time t. Equation: dp = bp dp. dt Here we assume that b and d are constants.in fact, we let k = b d. So the equation becomes dp dt = kp.

3 Population growth/decay, cont d k > 0: exponential growth k < 0: exponential decay How to solve dp dt = kp subject to P(0) = P 0 (initial condition) Answer: Separation of variables. So Therefore dp P = kdt dp P = kdt lnp = kt + C

4 Population growth/decay, cont d P = Ce kt Since P(0) = P 0, one gets C = P 0, or P(t) = P 0 e kt So all we need for this model are the values of k and P 0. Example 1: An island (think Australia (*)) was populated by a population of rabbits. We know that after 2 years, there were 200 rabbits. After 4 years there were 800. (a) Find the initial population of the rabbits P 0 and the growth rate k? (b) How many rabbits will inhabit this island after 100 years? (*) Read the Wikipedia entry for Rabbits in Australia.

5 Example 1 (a) We have 200 = P 0 e 2k and 800 = P 0 e 4k. Dividing these two equations leads to: = P 0e 4k P 0 e 2k = e2k. Therefore k = ln = 1 ln4 = ln2. 2 To get P 0, we substitute back: 200 = P 0 e 2 ln2 = 4P 0 So P 0 = 50.

6 Example 1, cont d (b) What happens in 100 years? P(100) = 50e 100 ln2 = rabbits Conclusion: Exponential model is bad at predicting long range behavior. Malthus (population analyst from the 18th century): This model does not take into account availability of food (land), resources, and conflicts over them. This leads to the logistic model.

7 Logistic Model ) dp (1 dt = k P PP1 Can we analyze it before we solve? (Hint: Phase plane analysis; What is the role of the orbit P = P 1 ) How to solve it? Separation of variables dp ) = k dt P (1 P P1 So dp ) = kt + C P (1 P P1

8 Logistic Model, cont d We can finish this problem using partial fractions, or observing that 1 ( ) = 1 P P 1 + P P ) 1 = P 1 P P 1 P (1 1 1 P P + P 1 1 P P1 P 1 Remember our integral: dp ) = kt + C P (1 P P1 This becomes: So: ) lnp ln (1 PP1 = kt + C ln P (1 P P1 ) = kt + C

9 Logistic Model, cont d In summary P (1 P P1 ) = C e kt How to find C? From the initial condition P(0) = P 0. What is the role of P 1? Note that P = P 1 is the equation of the stable orbit. P 1 is called the carrying capacity. When we run through the algebra (not shown here), we are led to the following equation P(t) = P 0 P 1 P 0 + (P 1 P 0 ) e kt

10 Logistic Model, cont d In order to solve here, one needs to find P 0,P 1, and k. Example 2: Predict what happens after 100 years, if the island s population started with P 0 = 20 rabbits, with the additional facts: P(2) = 200 and P(4) = 800. Answer: (After working out the algebra) with P 0 = 20, one finds P 1 = 1100, and k = So after 100 years, P(100) 1100 rabbits (why?).

11 Newton s Law of Cooling Basic Thermodynamic Principle: Heat flows from higher temperature towards the lower one. Heat flow is proportional to the difference between the two temperatures Model: T(t): temperature of an object placed in an environment at temperature M(t). Basic model: dt = k (T M) dt Note: k > 0. For T < M, heat flow dt dt > 0 For T > M, heat flow dt dt < 0

12 Newton s law of cooling; basic model with added heat source Basic model with added Heat source H(t) (such as a body in a building with a furnace): dt = k (T M) + H(t) dt Let s solve this. dt + kt = km + H dt Multiplier: µ(t) = exp kdt = e kt. So: ( e T) kt = e kt (km + H) Or e kt T = T = e kt e kt (km + H) dt + C e kt (km + H) dt + Ce kt

13 Newton s law of cooling: 2 special cases Question 1: Find the equilibrium temperature (also called steady state), when M = M 0 = constant and H = constant Claim: T equilibirum = M 0 H/k. Just set dt dt = 0 and solve. Question 2: Find the maximum and minimum temperatures when M = M 0 B cos ω (t t 0 ); H = 0. Claim: We need to solve for T = M 0 + Ce kt B cos ω(t t 0)+(ω/k) sin ω(t t 0 ) 1+(ω/k) 2. Here I used the basic integral in the front of the book: e au au a cos nu + n sinnu cos nudu = e a 2 + n 2 So: the maximum and minimum temperature are given by: M 0 ± B 1 + (ω/k) 2

14 Newtonian Mechanics What is Mechanics? It is the study of motion of objects under the effect of forces acting on those objects This is the foundation of several branches of physics and engineering Newton s law of motion: (1) When a body is subject to no external force, it moves at constant velocity (2) When a body is subject to one or more extrenal forces, the time rate of change of the body s momentum is equal to the vector sum of external forces acting on it. (3) For every action, there is an equal and opposite reaction.

15 What is momentum? We will only deal with one dimension motion m: mass of an object position of the object (with respect to some frame): x = x(t) velocity: v = dx dt acceleration: a = dv dt = d2 x dt 2 momentum p = m v resultant (external) force exerted on the body of mass m: F = F(t,x,v) Newton s Law of Motion: In terms of the velocity v F = m a m dv dt = F(t,x,v)

16 Newton s law of motion, cont d In terms of position m d2 x dx = F(t,x, dt2 dt ) So, in general, we are dealing with a second order differential equation. In terms of the velocity v (once more) Suppose F = F(t,v) (the force is independent of position). The equation reduces to: m dv dt = F(t,v) Which is a first order DFQ in v. To get position, we must first solve for v = v(t), then integrate x(t) = v(t)dt.

17 Example 1 Example 1: An object of mass m is given an initial downward velocity v 0 and allowed to fall under the influence of gravity. Assuming the gravitational force is constant and assuming the force due to air resistance is proportional to the velocity of the object, find its equation of motion? (That means find x(t)). Solution Force of gravity: F 1 = m g Force due to resistance of air: F 2 = b v (why? what is the significance of the - sign?) Resultant force: F = F 1 + F 2 = mg bv

18 Example 1, cont d Newton s law of motion: m dv dt = m g b v Subject to v(0) = v 0 (this is given). Separate: Integrate both sides: dv dt = g b m v dv g b m v = dt dv g b m v = dt

19 Example 1, cont d m b ln(m g b v) = t + C Or m g b v = C e b m t Or v = m g b + C e b m t At t = 0, v = v 0, so: C = v 0 m g b Putting these together, one gets: v = m g ( b + v 0 m g ) b e b m t

20 What is the equation of motion? So x(t) = x(t) = v(t)dt (m g ( b + v 0 m g b That is: x(t) = m g b t m b Now use x(0) = x 0 to get Or c m b c = m b ( ( ( v 0 m g b v 0 m g b v 0 m g b ) e b t) m dt ) e b m t + c ) = x 0 ) + x 0

21 What happens when t? The speed reaches the terminal velocity of mg b (why?) Initially (small t), x(t) behaves like the linear function x(t) x 0 + v 0 t Note that the exponential piece is just a transient state. When t is large, x(t) is given by: x(t) m g b t + m ( b v0 m g ) b + x0 What if you wanted to find the terminal velocity when the falling object is modeled by Argue the answer: v = m dv dt = m g b v2? m g b.

22 Problem 25, p. 124: A projectile escaping from Earth According to Newton s law of grativation, the attractive force between two objects (with masses m and M) varies inversely as the square of the distance r between them. That is F g = G m M r 2 (what is the significance of the - sign?) (a) Find the differential equation that describes the motion of a projectile m, under the influence of the Earth s gravitational force (use M for the mass of Earth, R for the radius of Earth, assumed to be roughly spherical, and g = GM/R 2 )

23 Prob 25, cont d (b) Use the fact that dr dt can be put in the form = v to show that the equation in part (a) v dv dr = gr2 r 2

24 Prob 25, cont d (c) If the projectile leaves Earth at velocity v 0, show that v 2 = 2gR2 r + v 2 0 2gR

25 Problem 25, p. 124: A projectile escaping from Earth According to Newton s law of grativation, the attractive force between two objects (with masses m and M) varies inversely as the square of the distance r between them. That is F g = G m M r 2 (what is the significance of the - sign?) (a) Find the differential equation that describes the motion of a projectile m, under the influence of the Earth s gravitational force (use M for the mass of Earth, R for the radius of Earth, assumed to be roughly spherical, and g = GM/R 2 )

26 Prob 25, cont d (b) Use the fact that dr dt can be put in the form = v to show that the equation in part (a) v dv dr = gr2 r 2

27 Prob 25, cont d (c) If the projectile leaves Earth at velocity v 0, show that v 2 = 2gR2 r + v 2 0 2gR

28 Prob 25, cont d (b) Use the fact that dr dt can be put in the form = v to show that the equation in part (a) v dv dr = gr2 r 2

29 Electrical Circuits Goal: Application of first-order DFQ s to simple electrical circuits consisting of a voltage source, a resistor, and either an inductor or a capacitor. Physical Principles: Kirchhoff s Current Law: The algebraic sum of the currents flowing into any junction must be zero. Kirchhoff s Voltage Law: The algebraic sum of the instantaneous change in potential (voltage drop) across a device around any closed loop must be zero.

30 Devices/More Laws Resistor (R, measured in Ohms): Ohm s Law: The voltage drop across a resistor is proportional to the current I passing through it: V R = RI. The proportionality constant is called the resistance R. Inductance (L, measured in Henrys): Faraday s Law: The voltage drop across an inductor is proportional to the instantaneous rate of change of the current I: V L = L di dt. The proportionality constant is called inductance L.

31 Devices/More Laws Capacitor (C, measured in Farads): The voltage drop E C across a capacitor is proportional to the charge q on the capacitor. V C = 1 C q. The charge q is measured in coulombs What we have to remember is that the current I = dq dt (measured in Amperes).

32 Basic Differential Equations: RL Circuit: Solve for I = I(t) (see schematics p. 125) E L + E R = E So: Solution: I(t) = I L L di dt + RI = E t 0 E(τ)e R L (t τ) dτ.

33 Basic Differential Equations: RC Circuit: Solve for q = q(t) E R + E C = E So: RI + 1 C q = E Solution: q(t) = e t RC R dq dt + 1 C q = E (K + ) E(τ)e RC τ R dτ.

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