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18. A pond forms as water collects in a conical depression of ra a and depth h. Suppose that water flows in at a constant rate k and is lost through evaporation at a rate proportional to the surface area. a. Show that the volume V(1) of water in the pond at time t satisfies the differential equation dV =k-απ(3α/πh)2/3 2/3, dt where a is the coefficient of evaporation. b. Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically stable? c. Find a condition that must be satisfied if the pond is not to overflow.

18. A pond forms as water collects in a conical depression of ra a and depth h. Suppose that water flows in at a constant rate k and is lost through evaporation at a rate proportional to the surface area. a. Show that the volume V(1) of water in the pond at time t satisfies the differential equation dV =k-απ(3α/πh)2/3 2/3, dt where a is the coefficient of evaporation. b. Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically stable? c. Find a condition that must be satisfied if the pond is not to overflow.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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68 CHAPTER 2 First-Order Differential Equations 18. A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k and is lost through evaporation at a rate proportional to the surface area. a. Show that the volume V(t) of water in the pond at time t satisfies the differential equation =k-aπ(3a/πh) 2/3v2/3, dV dt where a is the coefficient of evaporation. b. Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically stable? c. Find a condition that must be satisfied if the pond is not to overflow. Harvesting a Renewable Resource. Suppose that the population y of a certain species of fish (for example, tuna or halibut) in a given area of the ocean is described by the logistic equation dy dt y r(1-²) y. r(1– K dy = r Although it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level and possibly even driven to extinction. Problems 19 and 20 explore some of the questions involved in formulating a rational strategy for managing the fishery. 16 19. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population y: the more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by Ey, where E is a positive constant, with units of 1/time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is replaced by (₁ Epi con Ber mat diff sim Sim and 21. tho wh sus the sic dy tha the x =
Transcribed Image Text:68 CHAPTER 2 First-Order Differential Equations 18. A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k and is lost through evaporation at a rate proportional to the surface area. a. Show that the volume V(t) of water in the pond at time t satisfies the differential equation =k-aπ(3a/πh) 2/3v2/3, dV dt where a is the coefficient of evaporation. b. Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically stable? c. Find a condition that must be satisfied if the pond is not to overflow. Harvesting a Renewable Resource. Suppose that the population y of a certain species of fish (for example, tuna or halibut) in a given area of the ocean is described by the logistic equation dy dt y r(1-²) y. r(1– K dy = r Although it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level and possibly even driven to extinction. Problems 19 and 20 explore some of the questions involved in formulating a rational strategy for managing the fishery. 16 19. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population y: the more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by Ey, where E is a positive constant, with units of 1/time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is replaced by (₁ Epi con Ber mat diff sim Sim and 21. tho wh sus the sic dy tha the x =
Problems Problems 1 through 4 involve equations of the form dy/dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions in the ty-plane. G 1. dy/dt = ay+by2, a> 0, b>0, -∞0 < y0 <∞0 G 2. dy/dt = y(y-1)(y-2), 20 G 3. dy/dt = e) - 1, -∞< % < ∞ -∞<3<∞ 4. dy/dt = e) -1, 5. Semistable Equilibrium Solutions. Sometimes a constant. equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it (see Figure 2.5.9). In this case the equilibrium solution is said to be semistable. a. Consider the equation dy/dt = k(1- y)², (19) where k is a positive constant. Show that y = 1 is the only critical point, with the corresponding equilibrium solution (t) = 1. Gb. Sketch f(y) versus y. Show that y is increasing as a function of t for y< 1 and also for y> 1. The phase line has upward-pointing arrows both below and above y = 1. Thus solutions below the equilibrium solution approach it, and those above it grow farther away. Therefore, o(t) = 1 is semistable. c. Solve equation (19) subject to the initial condition y(0) = yo and confirm the conclusions reached in part b. y k o(t) = k y viszoq #alo odw to nothogong these o(t) = k k t (a) (b) SS FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤0; (b) dy/dt > 0. Problems 6 through 9 involve equations of the form dy/dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 5). Draw the phase line, and sketch several graphs of solutions in the ty-plane. G 6. dy/dt = y²(y²-1), G 7. dy/dt = y(1-y²), -∞0<yo <∞ -∞o <yo<∞ -∞<yo <∞ -∞o <yo <∞ G 8. dy/dt = y²(4- y²), 9. dy/dt = y²(1-y)², nge For and Lad boid. A cso 10. Complete the derivation of the explicit formula for the solution (11) of the logistic model by solving equation (10) for y. 11. In Example 1, complete the manipulations needed to arrive at equation (13). That is, solve the solution (11) for t. 12. Complete the derivation of the location of the vertical asymptote in the solution (15) when yo > T. That is, derive formula (16) by finding the value of when the denominator of the right-hand side of equation (15) is zero. 13. Complete the derivation of formula (18) for the locations of the inflection points of the solution of the logistic growth model with a threshold (17). Hint: Follow the steps outlined on p. 66. 14. Consider the equation dy/dt = f(y) and suppose that y, is a critical point that is, f(y₁) = 0. Show that the constant equilibrium solution (1) = y₁ is asymptotically stable if f'(y₁) < 0 and unstable if f'(y₁) > 0. 15. Suppose that a certain population obeys the logistic equation dy/dt = ry(1-(y/K)). a. If yo K/3, find the time has doubled. Find the value of year. b. If yo/K = a, find the time T at which y(T)/K = 3, where 0 a, 3 < 1. Observe that T→ ∞o as a → 0 or as 3 1. Find the value of T for r = 0.025 per year, a = 0.1, and 30.9. G 16. Another equation that has been used to model population growth is the Gompertz¹5 equation dy dt at which the initial population corresponding to r = 0.025 per =ry ln Rigoloid where r and K are positive constants. a. Sketch the graph of f(y) versus y, find the critical points, and determine whether each is asymptotically stable or unstable. b. For 0 ≤ y ≤ K, determine where the graph of y versus t is concave up and where it is concave down. c. For each y in 0 ≤ y ≤ K, show that dy/dt as given by the Gompertz equation is never less than dy/dt as given by the logistic equation. ble 17. a. Solve the Gompertz equation pitchfork bifurcatio doradi bas y In (). ald T dy =ry ln dt st subject to the initial condition y(0) = yo. maldo Hint: You may wish to let u = ln(y/K). ad lo obroba b. For the data given in Example 1 in the text (r= 0.71 per year, K = 80.5 x 106 kg, yo/K = 0.25), use the Gompertz model to find the predicted value of y(2). c. For the same data as in part b, use the Gompertz model to find the time at which y(7) = 0.75K. 15 Benjamin Gompertz (1779-1865) was an English actuary. He developed his model for population growth, published in 1825, in the course of constructing mortality tables for his insurance company.
Transcribed Image Text:Problems Problems 1 through 4 involve equations of the form dy/dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions in the ty-plane. G 1. dy/dt = ay+by2, a> 0, b>0, -∞0 < y0 <∞0 G 2. dy/dt = y(y-1)(y-2), 20 G 3. dy/dt = e) - 1, -∞< % < ∞ -∞<3<∞ 4. dy/dt = e) -1, 5. Semistable Equilibrium Solutions. Sometimes a constant. equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it (see Figure 2.5.9). In this case the equilibrium solution is said to be semistable. a. Consider the equation dy/dt = k(1- y)², (19) where k is a positive constant. Show that y = 1 is the only critical point, with the corresponding equilibrium solution (t) = 1. Gb. Sketch f(y) versus y. Show that y is increasing as a function of t for y< 1 and also for y> 1. The phase line has upward-pointing arrows both below and above y = 1. Thus solutions below the equilibrium solution approach it, and those above it grow farther away. Therefore, o(t) = 1 is semistable. c. Solve equation (19) subject to the initial condition y(0) = yo and confirm the conclusions reached in part b. y k o(t) = k y viszoq #alo odw to nothogong these o(t) = k k t (a) (b) SS FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤0; (b) dy/dt > 0. Problems 6 through 9 involve equations of the form dy/dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 5). Draw the phase line, and sketch several graphs of solutions in the ty-plane. G 6. dy/dt = y²(y²-1), G 7. dy/dt = y(1-y²), -∞0<yo <∞ -∞o <yo<∞ -∞<yo <∞ -∞o <yo <∞ G 8. dy/dt = y²(4- y²), 9. dy/dt = y²(1-y)², nge For and Lad boid. A cso 10. Complete the derivation of the explicit formula for the solution (11) of the logistic model by solving equation (10) for y. 11. In Example 1, complete the manipulations needed to arrive at equation (13). That is, solve the solution (11) for t. 12. Complete the derivation of the location of the vertical asymptote in the solution (15) when yo > T. That is, derive formula (16) by finding the value of when the denominator of the right-hand side of equation (15) is zero. 13. Complete the derivation of formula (18) for the locations of the inflection points of the solution of the logistic growth model with a threshold (17). Hint: Follow the steps outlined on p. 66. 14. Consider the equation dy/dt = f(y) and suppose that y, is a critical point that is, f(y₁) = 0. Show that the constant equilibrium solution (1) = y₁ is asymptotically stable if f'(y₁) < 0 and unstable if f'(y₁) > 0. 15. Suppose that a certain population obeys the logistic equation dy/dt = ry(1-(y/K)). a. If yo K/3, find the time has doubled. Find the value of year. b. If yo/K = a, find the time T at which y(T)/K = 3, where 0 a, 3 < 1. Observe that T→ ∞o as a → 0 or as 3 1. Find the value of T for r = 0.025 per year, a = 0.1, and 30.9. G 16. Another equation that has been used to model population growth is the Gompertz¹5 equation dy dt at which the initial population corresponding to r = 0.025 per =ry ln Rigoloid where r and K are positive constants. a. Sketch the graph of f(y) versus y, find the critical points, and determine whether each is asymptotically stable or unstable. b. For 0 ≤ y ≤ K, determine where the graph of y versus t is concave up and where it is concave down. c. For each y in 0 ≤ y ≤ K, show that dy/dt as given by the Gompertz equation is never less than dy/dt as given by the logistic equation. ble 17. a. Solve the Gompertz equation pitchfork bifurcatio doradi bas y In (). ald T dy =ry ln dt st subject to the initial condition y(0) = yo. maldo Hint: You may wish to let u = ln(y/K). ad lo obroba b. For the data given in Example 1 in the text (r= 0.71 per year, K = 80.5 x 106 kg, yo/K = 0.25), use the Gompertz model to find the predicted value of y(2). c. For the same data as in part b, use the Gompertz model to find the time at which y(7) = 0.75K. 15 Benjamin Gompertz (1779-1865) was an English actuary. He developed his model for population growth, published in 1825, in the course of constructing mortality tables for his insurance company.
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