(e) Use Newton s method to find the x coordinate that satisfies this equation, and your graph in part (b) to show that this is an inflection point.

Transcription

1 Chapter 6 Review problems 6.1 A strange function Consider the function (x 2 ) x. (a) Show that this function can be expressed as f(x) = e x ln(x2). (b) Use the spreadsheet, and a fine subdivision of the x axis (e.g., use increments of x = 0.01 along the x axis) to graph the function over the domain 2 x 1.5. (c) On the same graph, and using the spreadsheet, graph the first and second derivatives of the function. You need not compute these derivatives manually. You can ask the spreadsheet to approximate them (dy/dx y/ x). (d) By computing the first and second derivatives, show that if this function has an inflection point, that point has to satisfy (2 + ln(x 2 )) x = 0 (e) Use Newton s method to find the x coordinate that satisfies this equation, and your graph in part (b) to show that this is an inflection point. (a) Let y = (x 2 ) x. Then ln(y) = x ln(x 2 ) and e ln(y) = e x ln(x2) Using the fact that the exponential of base e and the natural logarithm are inverse functions, it follows that e ln(y) = y. Thus y = f(x) = e x ln(x2). (c) y (x) = e x ln(x2 ) (1 ln(x 2 ) + x 1x ) 2(2x) = e ( x ln(x2 ) ln(x 2 ) + 2 ) Note that the derivative can be expressed in terms of the original function, i.e. y (x) = y ( ln(x 2 ) + 2 ). The second derivative is then (using the product rule). y (x) = y (x) ( ln(x 2 ) + 2 ) +y(x) ( ln(x 2 ) + 2 ) = [y(x) ( ln(x 2 ) + 2 ) ] ( ln(x 2 ) + 2 ) +y(x) ( ) 1 x 22x v November 26,

2 Science 1 Problems (set by L. Keshet) Chapter y=(x^2)^x y(x) y (x) y (x) (b) Figure 6.1: Figure for solution to problem 6.1. The black curve is the function, the red curve is the first derivative, and the blue curve is the second derivative. So ( y (x) = y(x) (2 + ln(x 2 )) ) x (d) If there is an inflection point somewhere then it satisfies y (x) = 0. Since y(x) > 0 for all x, this can only be satisfied if (2 + ln(x 2 )) x = 0. (e) The inflection point is seen to be somewhere at negative x values (See Fig. 6.1). We start Newton s method at x 0 = 1 and iterate a few times. (See Table 6.1.) We find that the values converge to x = , which is the only inflection point of this function. x 0 f(x 0 ) f (x 0 ) x 1 = x 0 f(x 0) f (x 0 ) e Table 6.1: Values for the iterates of Newton s method, starting from the initial guess x 0 = Dividing up your attention: Suppose that the probability of finding food type 1 is P 1 (x) = x 2, 0 x 1 and the probability of finding food type 2 is P 2 (x) = x 3, 0 x 1 where x stands for the fraction of attention that v November 26,

3 Science 1 Problems Chapter 6 is dedicated to the task. If an animal searches for these two types of food simultaneously, it must devote some fraction x 1 to searching for type 1 and some fraction x 2 to searching for type 2 food. (a) Which food is harder to find? Why? (b) What constraint to the values of x 1 and x 2 satisfy if the animal is paying full attention to finding food? (c) Suppose that food that is harder to find is twice as nutritious, and that the energy gained is a sum of the form E = C 1 P 1 (x 1 ) + C 2 P 2 (x 2 ). Find out how the animal should subdivide its time to gain the maximal energy. (a) Food 2 is harder to find since for any x in 0 x 1, the function P 2 (x) = x 3 has a lower value than the function P 1 (x) = x 2 (b) Since x 1 + x 2 = 1. (c) Let x = x 1. Then x 2 = 1 x and C 2 = 2C 1 so Using the given functions, Looking for critical points, we compute E = C 1 P 1 (x 1 ) + C 2 P 2 (x 2 ) = C 1 P 1 (x) + 2C 1 P 2 (1 x). E(x) = C 1 (x 2 + 2(1 x) 3 ). de dx = C 1(2x + 6(1 x) 2 ( 1)) = 0. Note use of the Chain Rule in the last term. Solving for x: leads to the quadratic whose roots are x 3(1 x) 2 = 0, 3x 2 7x + 3 = 0, x = 7 ± Only one of these is in the interval of interest, since 0 x 1, (the smaller root). Next, we check the type of critical point by the second derivative test: d 2 E dx 2 = C 1(2 + 12(1 x)). But this is clearly positive whenever x is in 0 x 1, so the point we found is a local minimum! not a maximum. Next we consider the endpoints. Plugging x = 0 into E leads to E(0) = C 1 (0 + 2(1) 3 ) = 2C 1, and for x = 1 we get E(1) = C 1 ( ) = C 1. Thus the animal would do best by spending full attention on getting the second type of food that is twice as nutritious. v November 26,

4 Science 1 Problems (set by L. Keshet) Chapter More muli-tasking attention problems In this question, we consider the same issue of whether it pays to divide up attention between searching for two food types, but this time, we consider foods for which the probability of success is given by P 1 (x) = x 1/3, P 2 (x) = x 1/2, where x is the fraction of attention paid, and 0 x 1. Repeat the calculation done in Problem 6.2, but using these two types of food choices. To find the critical point, you will have to solve an algebraic equation that is not easy to solve analytically. Use Newton s method to find the solution in that case. In this question the food type 2 is harder to find, since x 1/2 < x 1/3 on the interval 0 < x < 1, as shown in Fig 6.2. We assume that the harder food is twice as nutritious, so that the energy gain is then E(x) = P 1 (x) + 2P 2 (1 x) = x 1/3 + 2(1 x) 1/2. Then the critical point satisfies de dx = 1 3 x 2/ (1 x) 1/2 ( 1) = 1 3 x 2/3 (1 x) 1/2 = 0 We need to find a value of x in 0 < x < 1 for which this is satisfied. This is not easily done analytically, so we use Newton s Method. The function of interest is f(x) = 1 3 x 2/3 (1 x) 1/2 = 0. We need the derivative of this for Newton s method, i.e., f (x) = 2 9 x 5/3 1 2 (1 x) 3/2. Since the solution is in the interval 0 < x < 1, we use an initial guess of x 0 = 0.5 and iterate x 1 = x 0 f(x 1 0) f (x 0 ) = x 3 0 x 2/3 0 (1 x 0 ) 1/2 2 9 x 5/3 0 1(1 x 2 0) 3/2 As shown by Table 6.2, the solution is x = , so the individual should spend 16.77% of its attention on food type 1 and the rest on food type Logistic growth curves Consider the Logistic equation, dn = rn(1 N/K). The curves N(t) versus t that satisfy this equation are called the solution curves. Show that these curves have an inflection point whenever they pass through the point N = K/2. v November 26,

5 Science 1 Problems Chapter Energy: E(x)=p1(x)+2p2(1-x) Food 1: p1(x)=x^(1/3) Food 1: p1(x)=x^(1/3) Food 2: p2(x)=x^(1/2) Food 2: p2(x)=x^(1/2) Attention, x Attention, x Figure 6.2: Figure for Problem 6.3. The two functions for probability of finding food given the fraction of attention devoted to the task are shown here (on the left). We also show the total energy gained, E(x) = P 1 (x) + 2P 2 (1 x) (on the right, rescaled so that E as well as P 1, P 2 fit on the same axes). Unlike Problem 6.2, in this case, the optimal strategy is to divide attention between the two food types. Directly from the differential equation, we have the first derivative of N satisfying dn = rn(1 N/K) = rn (r/k)n2. we compute the second derivative (using the chain rule, since N depends on t): d 2 N = rdn 2 2(r/K)N dn = rdn (1 2N/K) When N = K/2 (plugging in) we have d 2 N = 0. 2 Moreover, it is clear that the second derivative changes sign at N = K/2, so that at this point there is an inflection point for these curves. 6.5 Logistic growth with constant yield harvesting Consider a fish population number in a certain lake at time t is N(t). Suppose that N satisfies a logistic growth equation, i.e. dn = R(N) = rn(1 N/K) where r and K are positive constants. v November 26,

6 Science 1 Problems (set by L. Keshet) Chapter 6 x0 f(x0) f (x0) x0-f(x0)/f (x0) Table 6.2: Decimal value for solution to the optimal fraction of time to devote to food type 1 in Problem 6.3. (a) Sketch R and a function of N. (b) For what density of fish is the growth rate maximal? What is the maximal growth rate? (c) Now suppose that the Fisheries Board sells licenses to harvest a constant number of fish per unit time, H > 0, so that dn = R(N) H = rn(1 N/K) H This type of harvesting is called Constant yield harvesting. What is the largest value of H for which the fish in the lake will not go extinct? (d) What will be the population of fish in the lake be if the harvesting rate is at the level you found in part (c)? (Hint: what is the steady state behaviour of N?) (a) See Figure 6.3 for a sketch of R(N). (b) Rate of growth of the population is where r, K are constants. R(N) = rn(1 N/K) = rn (r/k)n 2 We can find the maximal growth rate by finding where the function R(N) is largest. We do this by taking the derivative, dr dn = r 2(r/K)N. Setting the derivative equal to zero, we solve for N in r 2(r/K)N = 0. We find that N = K/2 satisfies this equation. Further we note that the derivative is positive if N < K/2 and negative if N > K/2 which assures that this value of N makes the growth rate maximal. The maximal growth rate is R(K/2) = rk/4. (c) In this problem, the fish population satisfy dn = f(n) = rn(1 N/K) H v November 26,

7 Science 1 Problems Chapter 6 The positive constant, H, is a vertical shift of the parabola downwards. The greatest shift that keeps some part of the parabola positive is a vertical shift by H = rk/4, so this would be the largest harvesting rate that keeps fish from going extinct in the lake. (d) The steady state level (which is barely stable) that survives at the maximal harvesting rate is N = K/2. See Figure 6.3. rk/4 dn/ dn/ dn/ K/2 K/2 N Figure 6.3: Figure for solution to problem 6.5. The harvesting rate, H is absent on the left. The middle figure is for some intermediate level of harvesting, and the panel on the right shows the maximal harvesting rate, H = rk/4 that leads to a steady state population of N = K/2. This population is unstable in the sense that a very small decrease in N will send fish towards extinction. 6.6 Logistic growth with constant effort harvesting Consider a fish population N in the lake, and suppose that it is growing logistically, as in Problem 6.5 Suppose that the fisheries department decides to allow harvesting at constant effort E, i.e. that the rate of harvesting (i.e. removal) of the population is h(n) = qen where E, the effort of the fishermen, and q, the catchability of this type of fish, are positive constants. (a) At what density of fish does the growth rate exactly balance the harvesting rate? (This density is called the maximal sustainable yield: MSY.) (b) What is the largest effort E, for which will the fish population will not go extinct? (a) The growth rate and the harvesting rate balance when R(N) = h(n), v November 26,

8 Science 1 Problems (set by L. Keshet) Chapter 6 i.e. rn(1 N/K) = qen. We need to solve this for N. Cancelling out a common factor and simplifying leads to: 1 N/K = qe/r = N/K = 1 (qe/r). Thus, the maximal sustainable yield occurs when N MSY = K(1 qe/r). This is the density at which the growth rate of the fish exactly balances with the harvesting rate. (b) For the fish to avoid going extinct, it should be true that the maximal sustainable yield is a positive quantity, i.e. N MSY = K(1 qe/r) > 0. Thus, the largest value of E for which this holds is E = r/q. We can also understand this result from Figure 6.4, where both growth, R(N) and harvesting, h(n) are shown on the same graph. If the slope of h is too high (left), the fish cannot outgrow the harvesting, and they go extinct. If the effort (slope of h) is not too great, then there is a steady state towards which the fish population will evolve. (The steady state is at the point for which growth rate =harvesting rate.) The critical level of fishing is such that the slope of h is the same as the slope of R at N = 0 (not shown on figure). R (N) = r 2(r/K)N, so R (0) = r.. But h (N) = qe. These slopes are equal at r = qe, or E = r/q. dn/ harvesting growth dn/ growth harvesting N Figure 6.4: Figure for solution to problem 6.6. We show the growth rate curve together with the harvesting rate curve. In the panel on the left, the harvesting rate is always greater than the growth rate, for any fish population, and the population will go extinct. In the panel on the right, the harvesting rate has a shallower slope (smaller value of E), so that there is a range of value of N for which the population increases. v November 26,

9 Science 1 Problems Chapter Maximizing the Yield Consider the results of Problem 6.6(a). What level of fishing effort should be used to lead to the greatest harvest when the fish population is kept at its maximal sustainable yield level? If we insist that the density of the fish will be at the level then the harvest is just qe times this density, i.e. h(n MSY ) = qen MSY = qek(1 qe Kq2 ) = qke r r E2 We now want to find how we can maximize h by adjusting the level of effort, E. So think of E as a variable now. To maximize h we find the derivative of h with respect to E and set it equal to zero. dh de = qk 2Kq2 r E = qk ( 1 2q r E ) = 0 Solving for the effort, E, we get E = r/(2q). We see that for smaller values of E, the derivative is positive, and for larger values, it is negative, which assures that this leads to a local maximum. Therefore the fishing effort that should be used to lead to the greatest harvest is E = r/2q. 6.8 Disease dynamics with no immunity Consider the disease dynamics shown in Figure 6.5, where S(t) = the number of (healthy) susceptibles and I(t) = the number of infected people at time t, and suppose that N(t) = S(t) + I(t). Suppose that the mean duration of the disease is 5 days. r S β I Figure 6.5: Figure for problem 6.8. (a) Given that the infection spreads through contact between individuals, explain each term in the model below. di = βsi ri, ds = βsi + ri. (b) Show that the population, N is constant. v November 26,

10 Science 1 Problems (set by L. Keshet) Chapter 6 (c) What is the value of r for this disease? (d) Suppose that N = 100, β = 0.01 (per person per day). Treating a random fraction f of the population with anti-viral drugs is equivalent to reducing N to (1 f)n as far as the fraction of the population that can participate in the disease cycle. What fraction of the population should be so treated to eliminate this disease. (Hint: consider the basic reproductive ratio for the disease, r 0 = Nβ/r. Avoid confusing r 0 with the recovery rate parameter, r.) (a) The terms proportional to SI are the rates of new infections produced through contact between susceptible and infected people. The terms ri are the rate of formation of new susceptibles when the infected people recover. (b) Adding the equations leads to This means that N is constant. dn = di + ds = 0. (c) The mean duration of the disease is 5 days, and thus r 1/5 = 0.2. (d) Since N = 100, β = 0.01, r = 0.2, the value of r 0 is r 0 = Nβ/r = 100(0.01)/0.2 = 5. We need to reduce r 0 to just below 1 to make eliminate the disease. Let r 0 = (1 f)r 0 = (1 f)5 = 1. Solving for the fraction f leads to (1 f) = 1/5 = 0.2 so f = 0.8. Thus 80% of the population would have to be treated to eradicate this disease. (e) 6.9 Disease dynamics with gradual loss of immunity Consider the disease dynamics shown in Figure 6.6, where R(t) are recovered, immune individuals, and γ > 0 is some constant per capita rate at which these individuals lose immunity and become susceptible to the disease again. γ S β I r R Figure 6.6: Figure for problem 6.9. (a) Write down a set of differential equations that describe the system shown in Figure 6.6. v November 26,

11 Science 1 Problems Chapter 6 (b) Show that the total population, N = S(t) + I(t) + R(t) is constant. (c) Eliminate R from the differential equations for S and I using part (b). (d) Find the steady state solutions for S, I. (e) Show that the disease will be endemic in the population provided some inequality is satisfied, and argue that this inequality is equivalent to r 0 > 1, where r 0 is the basic reproductive parameter for the disease. (a) ds di di = βsi + γr, = βsi ri, = ri γr. (b) Adding the equations leads to dn/ = 0 since all terms cancel on the left hand side. Thus, N is constant. (c) R(t) = N S(t) I(t) so ds = βsi + γ(n S I), di = βsi ri. (d) The steady states are 0 = βsi + γ(n S I), 0 = βsi ri. From the second equation, I(βS r) = 0, so either I = 0 (no disease) in which case R = 0, S = N, or else (βs r) = 0, so S = r/β. Plugging this into the first equation leads to 0 = ri + γ(n r β I). This can be solved for I to get I = γ(n r/β). γ + r The value of R is then obtained by plugging S = r/β and this expression for I into R = N S I. This state (provided it is positive) is the endemic disease steady state. (e) The steady state found in (d) exists only provided (N r/β) > 0. This means that we require the inequality r 0 βn/r > 1 for the disease to be endemic. v November 26,

12 Science 1 Problems (set by L. Keshet) Chapter Predators and Prey The model shown below has been used to described the dynamics of two species, a population of predators interacting with a population of prey: dx = rx pxy, dy = qxy my, where r, p, q, m > 0 are constants, t is time in days, and x, y are in number of animals in a certain African National Preserve. (a) Which variable represents the population of predators, and which are their prey? Explain every term in the equations, and make a list of assumptions that have been used to derive this model. (b) Determine the units for each of the constants in this model. (c) Find the steady state values of each variable. (d) The ranger in the African National Preserve is forced to allow some hunting of the prey population (gazelles) by Safari Tours. The Safari Tours contract specifies that gazelles will be kept at a constant population level, G by a combination of hunting and restocking (i.e. releasing gazelles bred in captivity). What will happen to the lion population under this scenario? (a) x=prey, y=predator, since contact causes x to decrease and y to increase. The model assumes (1) that the prey and predators are all identical as far as growth/mortality, (2) that prey have a constant per capita birth rate r in the absence of predators, and negligible mortality, (3) that without prey, the predators would die with mortality rate m, and (4) that contact between predators and prey leads to a decrease in the prey population and an increase (e.g. through providing nourishment for mating and births) in the predator population. (b) r, m have units of 1/day. p, q have units of per individual per day. (c) 0 = rx pxy = x(r py), 0 = qxy my = y(qx m). The solutions are x = 0, y = 0 or x = m/q, y = r/p. (d) Gazelles are prey, so the hunting by Safari Tours results in x = G=constant. Thus the gazelles are no longer determined by their own growth and predation-mortality, so we need consider only the lion population. The population of lions then satisfies dy = qgy my = (qg m)y. Thus, if G > m/q, the lion population will grow, whereas if G < m/q, the lion population will decline. If G = m/q, the lion population will stay constant. v November 26,