Chapter Eleven. Chapter Eleven

Transcription

1 Chapter Eleven Chapter Eleven

2 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons ConcepTests and Answers and Comments for Section. For Problems, which of the following functions satisf the given differential equation? (a) e (b) e (c) e (d) e (e) e. = (a) and (c) It ma be useful to ask a similar question with functions such as e, e, e, and e with the differential equations =, =, =, = +, = +, and = + e.. = (d) and (e) See the Comment for Problem.. + = (a), (b), and (c) See the Comment for Problem.. + = e (b) See the Comment for Problem. For Problems 5, which of the following functions satisf the given differential equation? (a) sin (b) cos (c) e sin (d) e cos (e) e 5. = (e) It ma be useful to ask a similar question with functions such as sin(), cos(), e sin(), e cos(), with the differential equations =, =, + = cos(), + = 5.. = (a) and (b) See the Comment for Problem = (c) and (d) See the Comment for Problem 5.. = e cos (c) See the Comment for Problem 5.

3 CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons 7 9. Which of (a) (d) is a graph of a function that is equal to its own derivative, i.e. f () = f() for all? (a) (b) (c) (d) (c). Because f () = f(), the graph of f will be increasing when f is positive and decreasing when f is negative. This problem and the following one are repeats of earlier problems. The are meant as warm-up questions for Concept Questions - of the net section.

4 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons 0. Which of (a) (d) is a graph of a function that is equal to the negative of its own derivative, i.e. f() = f ()? (a) (b) 5 5 (c) 5 (d) (b). From f() = f () we see that the graph will have negative slopes where it is positive and positive slopes where it is negative. This eliminates (c) and (d). Also, larger values of f() require steeper slopes. Students ma find other reasons for contradictions.

5 ConcepTests and Answers and Comments for Section. CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons 9. Which of the following graphs could be that of a solution to the differential equation =? (a) (c) (b) (d) (a). Notice that solutions have positive slopes if < 0 and > 0, and negative slopes if > 0 and > 0. Also the slope of the solution curve is zero if = 0 or = 0. You could have students eliminate other choices b finding regions of the graph that contradict the fact that its slope is given b.

6 0 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons. Which of the following graphs could be that of a solution to the differential equation =? (a) (I) (b) (II) (c) (III) (d) (IV) (e) (I) and (IV) (f) (I), (II), and (IV) (g) (I) and (III) (h) (I), (II), and (III) (i) (I), (II), (III), and (IV) (I) (II) (III) (IV) (g). From = we see that solutions have positive slopes for < 0 and > 0 and negative slopes for > 0 and > 0. This eliminates (II). The remaining three graphs have zero slope at = 0. In addition, the slope at (, 7) in (IV) should be, rather than almost zero as it is in (IV). This eliminates (IV). Students ma find other reasons for eliminating choices.

7 CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons. Which of the following graphs could be that of a solution to the differential equation = /? (a) (b) (c) (d) (a). From = /, we see that solutions have positive slopes when and have the same sign. This eliminates (b). The slope when = = must be which eliminates (c) and (d). Students ma find other reasons for eliminating choices.

8 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons. Eplain wh the slope field in Figure. could onl be that of eactl one of the following. (a) = (b) = / (c) = / (d) = Figure. (a). The slope field shows positive slopes in quadrants I and III, which eliminates (d). The slope field suggests horizontal tangents when = 0, which eliminates (b). The slope field shows increasing slopes when increases for a fied, which eliminates (c). The slope field is consistent with (a). You could repeat this question showing the slope field of =. 5. Eplain wh the slope field in Figure. could onl be that of eactl one of the following. (a) = (b) = / (c) = / (d) = Figure. (b). The slope field shows positive slopes in quadrants I and III, which eliminates (d). The slope field suggests horizontal tangents when = 0, which eliminates (c). The slope field suggests vertical tangents when = 0, which eliminates (a). The slope field is consistent with (b). You could repeat this question showing the slope field of = /.

9 ConcepTests and Answers and Comments for Section. CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons. For which of the following differential equations will Euler s method for approimating a solution which starts at (0, ) give an underestimate? (a) = (b) = (c) = (d) = (a) and (b). For (a), differentiating gives = =, and for (b), we have = =. Thus solutions of these two equations that start at (0, ) are concave up, so tangent lines are below the curve, giving underestimates. You could also ask which differential equations lead to overestimates. ConcepTests and Answers and Comments for Section.. Which of the following are separable differential equations? (a) = ( + ) (b) = + (c) = e + (d) = sin( + ) (e) = ln + ln (f) = cos( ) (g) = arcsin (a), (c), (d), (f), and (g). The equations in (a), (d), (f), and (g) have the form = g()f() or = f()g(), and so the are separable. (c) can be written as = e e, which is also separable. (b) can be written as = ( + ) and (e) as = ln(), neither of which are separable. Follow-up Question. Determine the solution of the separable equations. Answer. (f) does not have a simple antiderivative, and (g) will require integration b parts.

10 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons ConcepTests and Answers and Comments for Section.5. Which of the following slope fields appear to have at least one equilibrium solution? (a) (b) (c) (d) (a) and (b). On a graph, an equilibrium solution is a horizontal line. Thus on a slope field we should look for short horizontal lines the same distance from the -ais. This question ma be repeated with other choices.

11 CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons 5. A am is put in an oven where its temperature, H, is modeled b dh dt = k(h H 0), where t is time and k and H 0 are positive constants. Figure. shows four solutions of this differential equation. Which curve(s) correspond(s) to the (a) Warmest oven? (b) Lowest initial temperature? (c) Same initial temperature? (d) Largest value of k? (I) 5 (II) 5 0 t 0 t (III) 5 (IV) 5 0 t 0 t Figure. (a) The warmest oven results in the largest final temperature, therefore (I). (b) The lowest initial temperature is shown b (IV). (c) (II) and (III) have the same initial temperature. (d) Because dh dt = k(h H 0), and at t = 0 the values of H H 0 are equal to for the curves in (I), (II) and (III), the curve with the largest initial slope will correspond to the largest value of k. The curve (III) has the largest initial slope of these three curves. The value of H H 0 at t = 0 for the curve (IV) is smaller than and the initial slope is slightl less than that in curve (III), so the value of k for curve (IV) must be smaller than that of curve (III). You could point out that we can onl make these comparisons because the scales on both aes are identical.

12 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons ConcepTests and Answers and Comments for Section.. A chemical is produced at a rate which is directl proportional to both (a ) and (t + b), and inversel proportional to ( + c), where is the amount of chemical at time t, and a, b, and c are positive constants. (a) Write the differential equation that describes this process. (b) Give all equilibrium solutions. d k(a )(t + b) (a) =, where k is a positive constant. dt + c (b) There is onl one equilibrium solution, = a. You could ask students to eplain wh k > 0. ConcepTests and Answers and Comments for Section.7. Consider solutions to a differential equation modeling eponential growth and solutions to a differential equation modeling logistic growth. (a) Give at least two similarities between these solutions. (b) Give at least two differences between these solutions. (a) Solutions modeling eponential growth are increasing and concave up, which is also true for small values of the dependent variable for logistic growth. Both are alwas positive. (b) Eponential solutions are unbounded, logistic solutions are bounded. Eponential solutions are concave up everwhere, logistic solutions are concave down as the approach the carring capacit from below. Eponential solutions have no inflection points, logistic solutions have an inflection point if their initial value is positive and less than half the carring capacit. To illustrate, ou could graph solutions to two specific equations with the same initial conditions.. The logistic model for US oil production is dp ( dt = kp P ) L where P is measured in billions of barrels of oil and t represents the ear. Which is the meaning of P(t)? (a) The quantit of oil produced in the US in the ear t. (b) The total quantit of oil produced in the US in all ears up to and including the ear t. (c) The quantit of oil still in the ground in the US in the ear t. (d) The fraction of oil originall in the ground in the US that is produced in the ear t. (b) This is a good opportunit to review the meanings of both P and dp/dt and to make certain that students are clear that peak oil production is about a maimum for dp/dt, not for P.

13 CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons 7. The logistic model for US oil production is dp ( dt = kp P ) L where P is measured in billions of barrels of oil and t represents the ear. Which are the units of L and k? (a) L: billions of barrels; k: billions of barrels per ear. (b) L: billions of barrels per ear; k: billions of barrels. (c) L: billions of barrels; k: billions of barrels. (d) L: billions of barrels; k: ear (e) L: billions of barrels; k: no units attached. (d) Since L is an upper bound for the quantit of oil that will ever be produced, its units are billions of barrels. The units of kp( P/L) are the same as the units of the derivative, dp/dt, namel (billions of barrels)/(ear). Since P has units of billions of barrels and ( P/L) has no units attached, the constant k must have units (ear). Ask the students for the meaning of k. In the earl ears, when P/L is near zero, we have dp/dt kp, so P grows approimatel eponentiall with growth rate k. At that time, ever ear P increases approimatel b a factor of e k. Of course later on the growth rate (dp/dt)/p decreases dramaticall, approaching zero as oil reserves are ehausted.. The Gompertz differential equation, dp/dt = kp ln(p/l), where k and L are positive constants, is used to model tumors in animals. List at least two characteristics of solutions of this differential equation that are common to solutions of the logistic differential equation, dp/dt = kp( P/L). Both have an equilibrium solution at P = L, and are increasing for 0 < P < L. For the Gompertz equation, d ( ) P dp dt = k dp ln(p/l) + = k(ln(p/l) + ) dp dt dt dt, so solutions to both equations are concave down for values of P just below the equilibrium solution at L. To illustrate, ou could graph solutions to two specific equations with the same initial conditions.

14 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons ConcepTests and Answers and Comments for Section.. Figure. shows a phase trajector for a SI-phase plane modeling an epidemic. From this figure, estimate (a) The size of this population. (b) The maimum number infected, and the number who have never been infected at this time. (c) The number that never were infected during this epidemic. 00 I S Figure. (a) 000 (b) 0 and 00. Some have alread recovered b this time, so these do not add to 000. (c) 00 In estimating numbers from a graph, a range of answers is possible for choice part (b).

15 ConcepTests and Answers and Comments for Section.9 CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons 9. Graph nullclines and determine all equilibrium points for the following differential equations. (a) d/dt = ( )( ), d/dt = ( )( ) (b) d/dt = ( )( ), d/dt = ( ) (a) The nullclines where trajectories are horizontal are =, and =, those where the trajectories are vertical are = and =. Thus the equilibrium points are (, ) and (,). This is shown in Figure.5. Figure.5 Figure. (b) The nullclines where trajectories are horizontal are = 0 and =, those where the trajectories are vertical are = and =. Thus the equilibrium points are (, 0), (, 0), (, ), and (,). This is shown in Figure.. Using other products of functions of and allows ou to ask the same question for other differential equations. ConcepTests and Answers and Comments for Section.0. The motion of a mass on a spring is governed b the initial value problem + p = 0, (0) = A, (0) = B, where p, A, and B are positive constants. (a) What is the effect of increasing p? (b) What is the effect of increasing B? (c) What is the effect of increasing A? (d) What is the effect of making B a negative number? (e) What is the effect of making A a negative number? (a) Increasing p decreases the period of the oscillations. (b) Increasing B for B > 0 increases the initial velocit, and therefore the amplitude of the oscillations if A = 0. (c) Increasing A for A > 0 increases the initial displacement, and therefore the amplitude of the oscillations, if B = 0. (d) If B is a negative number and A positive, the initial motion will be toward the equilibrium position. If B and A are both negative, the initial motion will be awa from the equilibrium point. (e) If A is a negative number, the initial displacement is on the opposite side of the equilibrium position than when it is positive. You could repeat this questions having A < 0, B > 0; then A > 0, B < 0; and finall, A < 0, B < 0.

16 0 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons. Which of the following graphs is that of a function that satisfies the relationship f () = f()? (a) (b) (c) (d) (a). Because f () = f(), solution curves will be concave down when the solution is positive, and concave up where the solution is negative. You could also ask what f () = f() implies about the location of the inflection points.

17 CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons. Which of the following graphs is that of a function that satisfies the relationship f () = f()? (a) (I), (II), and (III) (b) (I) and (II) (c) (II) and (III) (d) (I) and (III) (e) (I) (f) (II) (g) (III) (h) None of these (I) (II) (III) (g). Because f () = f(), solution curves will be concave down when the solution is positive and concave up when the solution is negative. You might point out that while functions that satisf f () = f() are periodic, the converse is not true.

18 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons ConcepTests and Answers and Comments for Section.. Match the following differential equations with the solutions shown in Figure.7. (a) + = 0 (b) = 0 (c) = 0 (d) + 5 = 0 (I) (II) t t (III) (IV) t t Figure.7 (a) corresponds to (IV) as it describes undamped motion, (b) corresponds to (II) as there are no oscillations, (c) corresponds to (III) as it describes underdamped motion, (d) corresponds to (I) as the motion is oscillator, but with increasing amplitude. You could have students provide the analtical solution to these differential equations.

19 CHAPTER ELEVEN Hughes Hallett et al c 009, John Wile & Sons. The motion of a mass on a spring is governed b + b + 9 = 0 where b is a positive constant. (a) For what values of b is the motion underdamped? (b) For what values of b is the motion overdamped? (c) For what values of b is the motion criticall damped? The characteristic equation is r + br + 9 = 0, with the solutions r = b ± b 9 = b ± b 9. Thus motion is underdamped if b 9 < 0, i.e. 0 < b <. The motion is overdamped if b 9 > 0, i.e. b >, and criticall damped if b =. Follow-up Question. What would happen if b was a negative number?

20 CHAPTER ELEVEN Hughes Hallett et al c 005, John Wile & Sons