P K P P < K - University of Miamidscheib/teaching/fall...MAP 2302 In-Class Activity #1 - Fall 2011...
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MAP 2302 In-Class Activity #1 - Fall 2011 (1) Let P (t) represent the density of a population at time t. Suppose the population grows with an intrinsic growth rate of r, has an environmental carrying capacity of K , and an initial population density of P (0) = P 0 . Then the rate at which the population’s density changes with time can be modeled by the logistic equation dP dt = rP 1 - P K . (a) If P >K (i.e., the population density if greater than the environmental carrying capacity), what does the model say will happen to the population’s density? If P<K (i.e., the population density is less than the environmental carrying capacity, what does the model say will happen to the population’s density? (b) Determine what will happen to the population density in the limit as P → K . Explain why this makes sense ecologically. Before we solve the logistic equation, let’s make the following substitutions: P = Ky and t = 1 r x (1) Using the chain rule, we get dP dt = dP dy · dy dx · dx dt = K · dy dx · r Substituting this into the logistic equation, along with y and x from (1) gives Kr dy dx = r(Ky) 1 - Ky K . After simplifying we are left with dy dx = y(1 - y) (2) This equation is referred to as the non-dimensionalized logistic equation and there are multiple ways to solve it. (c) View equation (2) as a Bernoulli equation and solve it for y(x). Then use the substitutions in (1) to get P (t). Note: You need to switch y and x back to P and t before you use the initial condition. (d) Determine the eventual fate of the population; that is, determine lim t→∞ P (t).
Transcript of P K P P < K - University of Miamidscheib/teaching/fall...MAP 2302 In-Class Activity #1 - Fall 2011...
MAP 2302 In-Class Activity #1 - Fall 2011
(1) Let P (t) represent the density of a population at time t. Suppose the populationgrows with an intrinsic growth rate of r, has an environmental carrying capacityof K, and an initial population density of P (0) = P0. Then the rate at which thepopulation’s density changes with time can be modeled by the logistic equation
dP
dt= rP
(1− P
K
).
(a) If P > K (i.e., the population density if greater than the environmentalcarrying capacity), what does the model say will happen to the population’sdensity? If P < K (i.e., the population density is less than the environmentalcarrying capacity, what does the model say will happen to the population’sdensity?
(b) Determine what will happen to the population density in the limit as P → K.Explain why this makes sense ecologically.
Before we solve the logistic equation, let’s make the following substitutions:
P = Ky and t =1
rx (1)
Using the chain rule, we get
dP
dt=
dP
dy· dydx· dxdt
= K · dydx· r
Substituting this into the logistic equation, along with y and x from (1) gives
Krdy
dx= r(Ky)
(1− Ky
K
).
After simplifying we are left with
dy
dx= y(1− y) (2)
This equation is referred to as the non-dimensionalized logistic equation and thereare multiple ways to solve it.
(c) View equation (2) as a Bernoulli equation and solve it for y(x). Then use thesubstitutions in (1) to get P (t). Note: You need to switch y and x back to Pand t before you use the initial condition.
(d) Determine the eventual fate of the population; that is, determine
limt→∞
P (t).
(2) Carbon dating in a technique used to determine the age of an object or artifact.The idea is based on the radioactive decay of carbon-14 atoms. What happensis that cosmic radiation (such as from the sun) enters the Earth’s atmospherecausing a chemical reaction that leads to carbon-14. It is well-known that allliving organisms contain both carbon-12 (which is necessary for life) and carbon-14 (which is absorbed from the atmosphere). Scientists believe that the amountof carbon-14 in the atmosphere is constant, so the ratio of carbon-12 to carbon-14 in all living organisms remains constant constant while the organism is alive(while carbon-14 decays, the living organism is able to absorb more by meansof eating to replace what has been lost). When the organism dies, the amountof carbon-12 remains the same, but carbon-14 decays without being replenished.By comparing the ratio of carbon-12 to carbon-14 in a dead organism to that ofa living organism, scientists can determine how much carbon-14 has decayed andhence how long the organism has been dead for.
(a) Let A(t) represent the amount of carbon-14 which remains in an organismafter t years. Suppose the initial amount of carbon-14 is A(0) = A0 andthat it has a decay constant of k. The rate at which carbon-14 decays isproportional to the amount remaining, so radioactive decay of carbon-14 canbe modeled by the equation
dA
dt= kA.
Solve this equation as a first-order linear equation to determine A(t).
(b) The half-life of an atom is the amount of time it takes for the atom to decay tohalf of its original amount. In other words, the half-life is the time it takes forA(t) = 1
2A0. It is known that the half-life of carbon-14 is approximately 5700
years. Use this information to determine the decay constant for carbon-14.
(c) Suppose an archaeologist finds a bone believed to belong to an animal whichlived approximately 24,000 years ago. If this archaeologist is correct, thenwhat percent of the original carbon-14 should be found in the bone when hetakes it back for analysis?
(d) Suppose the FBI is investigating a 1000 year old painting which was stolenfrom a museum. They receive a tip that someone is selling the paintingonline. When they arrive at the seller’s house, he claims that the painting isa duplicate made in the 1900’s. When the FBI tests the wood in the frame,they discover that 88.5% of the carbon-14 in the wood is still present. Shouldthe FBI believe the seller’s story or should they suspect that the paintingmight be stolen?
(3) According to Newton’s Second Law of motion,
F = ma.
We know that velocity and acceleration are related by the equation a = dvdt
, so wecan rewrite Newton’s Second Law of motion as
F = mdv
dt.
Suppose a skydiver jumps out of an airplane with initial vertical velocity of v(0) =0. Then the two forces acting on him are the force of gravity, Fg, and air resistance,FR. Since we are concerned with a falling body, we can take the positive directionto be downward, so the force due to gravity is
Fg = mg.
The force due to air resistance is in the opposite direction of motion, so it willbe negative. Empirical evidence suggests that at low speeds, air resistance isproportional to velocity and for high speeds, air resistance is proportional tovelocity squared. In the case of a skydiver, this means that the force due to airresistance before opening the parachute is
FR = −kv2
and after opening the parachute is
FR = −mv.
Thus, a model for the velocity of a skydiver before opening his parachute is
mdv
dt= mg − kv2 (3)
and for a skydiver after opening his parachute is
mdv
dt= mg − kv. (4)
If we divide both sides of both equation by m and make the replacement D = km
(D is referred to as the drag coefficient) then equations (3) and (4) become
dv
dt= g −Dv2 and
dv
dt= g −Dv.
(a) Solve both equations (you don’t have to use the same method for both) toget v(t).
(b) Determine the terminal velocity of the skydiver in both cases; that is, deter-mine
limt→∞
v(t).
(c) Suppose you have been contracted to design parachutes for the U.S. Army’s82nd Airborne Division. They want you to design a parachute that will enabletheir soldiers to land at under 15 mph (22 ft/sec). What is the minimal dragcoefficient you can build your parachutes to have?
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