SalasSV_08_09_ex - 504 CHAPTER 8 TECHNIQUES OF INTEGRATION...

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Unformatted text preview: 504 CHAPTER 8 TECHNIQUES OF INTEGRATION Suppose that the exposure to new people is proportional to the number of people who have not seen the product out of a total population of M viewers. Let P (t ) denote the number of viewers who have been exposed to the product at time t . The company has determined that no one was aware of the product at the start of the campaign [P (0) = 0] and that 30% of the viewers were aware of the product after 10 days. (a) Determine the differential equation that describes the number of viewers who are aware of the product at time t . (b) Determine the solution of the differential equation from part (a) that satisfies the initial condition P (0) = 0. (c) How long will it take for 90% of the population to be aware of the product? 44. A drug is fed intravenously into a patient’s bloodstream at the constant rate r . Simultaneously, the drug diffuses into the patient’s body at a rate proportional to the amount of drug present. (a) Determine the differential equation that describes the amount Q(t ) of the drug in the patient’s bloodstream at time t . (b) Determine the solution Q = Q(t ) of the differential equation found in part (a) that satisfies the initial condition Q(0) = 0. (c) Find lim Q(t ). t →∞ describes a population that undergoes periodic fluctuations. Assume that P (0) = 1000 and find P (t ). Use a graphing utility to draw the graph of P . (b) The differential equation dP = (2 cos 2π t )P + 2000 cos 2π t dt describes a population that undergoes periodic fluctuations as well as periodic migration. Continue to assume that P (0) = 1000 and find P (t ) in this case. Use a graphing utility to draw the graph of P and estimate the maximum value of P . c 46. The Gompertz equation dP = P (a − b ln P ), dt where a and b are positive constants, is another model of population growth. (a) Find the solution of this differential equation that satisfies the initial condition P (0) = P0 , where P0 > 0. HINT: Define a new dependent variable Q by Q = ln P . (b) Find lim P (t ). t →∞ c 45. (a) The differential equation dP = (2 cos 2π t )P dt (c) Determine the concavity of the graph of P . (d) Use a graphing utility to draw the graph of P in the case where a = 4, b = 2, and P0 = 1 e2 . Does the graph 2 confirm your result in part (c)? ∗ 8.9 INTEGRAL CURVES; SEPARABLE EQUATIONS Integral Curves If a function y = y(x) satisfies a first-order differential equation, then along the graph of the function, the numbers x, y, y are related as prescribed by the equation. In the case of a nonlinear differential equation, it is usually difficult (and sometimes impossible) to find functions which are explicitly defined and satisfy the differential equation. More often, we can find plane curves which, though not the graphs of functions, do have the property that along the curve the numbers x, y, y are related as prescribed by the differential equation. Such curves are called integral curves (solution curves) of the differential equation. To solve a nonlinear differential equation is to find the integral curves of the equation. Separable Equations A first-order differential equation is said to be separable if it can be written in the form (8.9.1) p(x) + q(y)y = 0. The functions p and q are assumed to be continuous where defined. ∗ 8.9 INTEGRAL CURVES; SEPARABLE EQUATIONS 505 Solution Method We expand equation (8.9.1) to read p(x) + q[ y(x)] y (x) = 0. Integrating this equation with respect to x, we find that p(x) dx + q[ y(x)] y (x) dx = C , where C is an arbitrary constant. From y = y(x), we have dy = y (x) dx. Therefore p(x) dx + q( y) dy = C . The variables have been separated. Now, if P is an antiderivative of p and Q is an antiderivative of q, then this last equation can be written (8.9.2) P (x ) + Q ( y ) = C . This equation represents a family of curves in the xy-plane. As we show below, all these curves are integral curves for Equation (8.9.1). To prove this statement,we must show that, if x and y are related by the equation P (x) + Q(y) = C , then x, y, y are related by the equation p(x) + q(y)y = 0. The proof is straightforward: if P (x ) + Q (y ) = C , then, by implicit differentiation with respect to x, we have p(x) + q(y) y = p(x) + q[ y(x)] y (x) = d P (x ) + Q [ y (x ) ] dx d d [P (x) + Q(y)] = [C ] = 0. = dx dx Example 1 The differential equation x + yy = 0 is separable. We can find the integral curves by writing x dx + Carrying out the integration, we have 12 x 2 y dy = C . + 1 y2 = C , 2 which, since C is arbitrary, we can write as x2 + y2 = C . For this equation to give us a curve, we must take C ≥ 0. For C > 0, the integral √ curves are circles of radius C centered at the origin. For C = 0, the integral curve degenerates to a single point, the origin. 506 CHAPTER 8 TECHNIQUES OF INTEGRATION Example 2 Show that the differential equation y= xy − y y+1 is separable and find an integral curve which satisfies the initial condition: y(2) = 1. SOLUTION To show that the differential equation is separable, we try to write it in the form (8.9.1): (y + 1)y = y(x − 1) y(1 − x) + (y + 1)y = 0. The next step is to divide this equation by y. As you can verify, the horizontal line y = 0 is an integral curve. However, we can ignore it because it doesn’t satisfy the initial condition y(2) = 1. Therefore, with y = 0, we can write 1−x+ y+1 y = 0, y 1 y = 0. y 1−x+ 1+ The equation is separable. Writing (1 − x) dx + y 1+ 1 dy = C , y we find that the integral curves take the form 2 x − 1 x2 + y + ln |y| = C . 2 1 (2, 1) The condition y(2) = 1 forces 3 x 1 2 2 − 1 (2)2 + 1 + ln (1) = C , 2 which implies C = 1. Figure 8.9.1 The integral curve that satisfies the initial condition is the curve x − 1 x2 + y + ln y = 1 2 Example 3 (Figure 8.9.1) Show that the differential equation y = xey−x is separable and find the integral curves. SOLUTION Since ey−x = ey e−x , the equation can be written xe−x ey − y = 0, and thus xe−x − e−y y = 0. ∗ 8.9 INTEGRAL CURVES; SEPARABLE EQUATIONS 507 The equation is separable. Writing xe−x dx − we have −xe−x − e−x + e−y = C and therefore e−y = xe−x + e−x + C . y 2 1 C = – 0.3 C = 0.3 1 2 3x C = 0.5 e−y dy = C , These are the integral curves. In this case, we can solve for y and express the curves as functions of x: y = − ln (xe −x +e −x + C ). The graphs of several of the integral curves are shown in Figure 8.9.2. Figure 8.9.2 Applications In the mid-nineteenth century, the Belgian biologist P. F. Verhulst used the differential equation (8.9.3) dy = k y (M − y), dt where k and M are positive constants, to study the population growth of various countries. This equation is now known as the logistic equation, and its solutions are called logistic functions. Life scientists have used this equation to model the spread of a disease through a population, and social scientists have used it to study the flow of information. In the case of a disease, if M denotes the total number of people in the population and y(t ) is the number of infected people at time t , then the differential equation states that the rate of growth of the disease is proportional to the product of the number of people who are infected and the number who are not. The differential equation is separable since it can be written in the form k− Integrating this equation, we have k dt − kt − and therefore 1 1 ln |y| − ln |M − y| = kt + C . M M It is a good exercise in manipulating logarithms, exponentials, and arbitrary constants to show that this equation can be solved for y, and thus the solution can be expressed as a function of t . The result can be written as y= CM . C + e−Mkt (not the same C as above) 1 y = 0. y (M − y ) 1 dy = C , y (M − y ) (partial fraction decomposition) 1/M 1/ M + dy = C , y M −y 508 CHAPTER 8 TECHNIQUES OF INTEGRATION If y = y(t ) satisfies the initial condition y(0) = R, R < M , then R= and we have (8.9.4) CM , C+1 which implies C= R , M −R y (t ) = MR R + (M − R)e−Mkt This particular solution is shown graphically in Figure 8.9.3. Note that y is increasing for all t ≥ 0. In the Exercises you are asked to show that the graph is concave up on [0, t1 ] and concave down on [t1 , ∞). In the case of a disease, this means that the disease is spreading at an increasing rate up to time t = t1 ; and after t1 the disease is still spreading, but at a decreasing rate. Note, also, that y(t ) → M as t → ∞. y y=M M R t1 t Figure 8.9.3 Example 4 A rumor spreads through a population of 5000 people at a rate proportional to the product of the number of people who have heard it and the number who have not. Suppose that 100 people initiate the rumor and that a total of 500 people have heard the rumor after two days. How long will it take for half the people to hear the rumor? SOLUTION Let y (t ) denote the number of people who know the rumor at time t . Then y satisfies the logistic equation with M = 5000 and the initial condition R = y(0) = 100. Thus, by (8.9.4), y (t ) = 5000 100(5000) = . −5000 kt 100 + 4900e 1 + 49e−5000 kt The constant of proportionality k can be determined from the condition that y(2) = 500. We have 500 = 5000 . 1 + 49e−10,000 k 1 + 49e−10,000 k = 10, 9 e−10,000 k = , 49 −10, 000 k = ln (9/49) and thus k ∼ 0. 00017. = Therefore, y (t ) = 5000 . 1 + 49e−0.85 t ∗ 8.9 INTEGRAL CURVES; SEPARABLE EQUATIONS 509 To determine how long it will take for half the population to hear the rumor, we solve the equation 2500 = for t . We get 1 + 49e−0.85t = 2, e−0.85t = 1 , 49 t= ln(1/49) ∼ = 4. 58 −0. 85 5000 1 + 49e−0.85t It will take slightly more than 4 1 days for half the population to hear the rumor. 2 EXERCISES *8.9 Find the integral curves 1. y = y sin (2x + 3). 2. y = (x2 + 1)(y2 + y). 3. y = (xy)3 . 5. y = − sin 1/x . x2 y cos y x+y where f (x, y) is an expression in x and y, generates a family of curves: all the integral curves of that particular equation. The orthogonal trajectories of this family are the integral curves of the differential equation 4. y = 3x2 (1 + y2 ). 6. y = y +1 . y + yx 2 y =− 1 . f (x , y ) 7. y = x e . 8. y = xy2 − x − y2 + 1. (y + 1)2 9. (y ln x) y = . x 10. ey sin 2x dx + cos x (e2y − y) dy = 0. 11. (y ln x) y = y2 + 1 . x 12. y = 1 + 2y 2 . y sin x The figure shows the family of a parabolas y = Cx2 . The orthogonal trajectories of this family of parabolas are ellipses. y Solve the initial value problem. 13. y = x 14. y = 1 − y2 , 1 − x2 y(0) = 0. x ex−y , 1 + ex y(1) = 0. We can establish this by starting with the equation y = Cx2 and differentiating. Differentiation gives y(2) = 0. y = 2Cx. Since y = Cx , we have C = y/x2 , and thus 2 x2 y − y 15. y = , y(3) = 1. y+1 16. x2 y = y − xy, y( − 1) = −1. y(0) = π/4. 17. (xy2 + y2 + x + 1) dx + (y − 1) dy = 0, 18. cos y dx + (1 + e−x ) sin y dy = 0, 19. y = 6e2x−y , y(0) = 0. y(1) = 1. 20. xy − y = 2x2 y, Orthogonal trajectories If two curves intersect at right angles, one with slope m1 and the other with slope m2 , then m1 m2 = −1. A curve that intersects every member of a family of curves at right angles is called an orthogonal trajectory for that family of curves. A differential equation of the form y = f (x, y), y 2y x= . x2 x The orthogonal trajectories are the integral curves of the differential equation x y =− . 2y y =2 As you can check, the solutions of this equation are all of the form x2 + 2y2 = K . 510 CHAPTER 8 TECHNIQUES OF INTEGRATION In Exercises 21–26, find the orthogonal trajectories for the following families of curves. In each case draw several of the curves and several of the orthogonal trajectories. 21. 2x + 3y = C . 23. xy = C . 25. y = Cex . 22. y = Cx. 24. y = Cx3 . 26. x = Cy4 . 32. A descending parachutist is acted on by two forces: a constant downward force mg and the upward force of air resistance, which (within close approximation) is of the form −β v2 where β is a positive constant. (In this problem we are taking the downward direction as positive.) (a) Express t in terms of the velocity v, the initial velocity √ v0 , and the constant vc = mg /β . (b) Express v as a function of t . (c) Express the acceleration a as a function of t . Verify that the acceleration never changes sign and in time tends to zero. (d) Show that in time v tends to vc . (This number vc is called the terminal velocity.) c In Exercises 27 and 28, show that the given family is selforthogonal. Use a graphing utility to graph at least four members of the family. x2 y2 28. 2 + 2 = 1. 27. y2 = 4C (x + C ). C C −4 29. Suppose that a chemical A combines with a chemical B to form a compound C . In addition, suppose that the rate at which C is produced at time t varies directly with the amounts of A and B present at time t . With this model, if A0 grams of A are mixed with B0 grams of B, then dC = k (A0 − C )(B0 − C ). dt (a) Find the amount of compound C present at time t if A0 = B0 . (b) Find the amount of compound C present at time t if A0 = B0 . c 33. A flu virus is spreading rapidly through a small town with a population of 25,000. The disease is spreading at a rate proportional to the product of the number of people who have it and the number who don’t. Suppose that 100 people had the disease initially, and 400 had it after 10 days. (a) How many people will have the disease at an arbitrary time t ? How many will have it after 20 days? (b) How long it will take for half the population to have the flu? (c) Use a graphing utility to sketch the graph of your solution in part (a). 34. Consider the logistic equation (8.9.3). Show that dy/dt is increasing if y < M /2 and is decreasing if y > M /2. What can you conclude about dy/dt at the instant y = M /2? Explain. c 30. A mathematical model for the growth of a certain strain of bacteria is dP = 0. 0020 P (800 − P ), dt where P = P (t ) denotes the number of bacteria present at time t . There are 100 bacteria present initially. (a) Find the population P = P (t ) at any time t . (b) Use a graphing utility to sketch the graph of P and dP /dt . (c) Approximate the time at which the bacterial culture experiences its most rapid growth rate. That is, locate the maximum value of dP /dt . Use three decimal place accuracy. What point on the graph of P corresponds to the maximum growth rate? 31. When an object of mass m is moving through air or a viscous medium, it is acted on by a frictional force that acts in a direction opposite to its motion. This frictional force depends on the velocity of the object and (within close approximation) is given by F (v ) = − α v − β v 2 , where α and β are positive constants. (a) From Newton’s second law, F = ma, we have dv = − α v − β v2 . dt Solve this differential equation to find v = v(t ). (b) Find v if the object has initial velocity v(0) = v0 . (c) What is lim v(t )? m t →∞ c 35. A rescue package whose mass is 100 kilograms is dropped from an airplane at a height of 4000 meters. As it falls, the air resistance is equal to twice its velocity. After 10 seconds, the package’s parachute opens and the air resistance is now four times the square of its velocity. (a) What is the package’s velocity at the instant the parachute opens? (b) Determine an expression for the velocity of the package at time t after the parachute opens. (c) What is the terminal velocity of the package? HINT: There are two differential equations that govern the package’s velocity and position: one for the free-fall period and one for the period after the parachute opens. 36. It is known that m parts of chemical A combine with n parts of chemical B to produce a compound C . Suppose that the rate at which C is produced varies directly with the product of the amounts of A and B present at that instant. Find the amount of C produced in t minutes from an initial mixing of A0 pounds of A with B0 pounds of B, given that: (a) n = m, A0 = B0 , and A0 pounds of C are produced in the first minute. (b) n = m, A0 = 1 B0 , and A0 pounds of C are produced in 2 the first minute. (c) n = m, A0 = B0 , and A0 pounds of C are produced in the first minute. ∗ 8.9 INTEGRAL CURVES; SEPARABLE EQUATIONS 511 HINT: Denote by A(t ), B(t ), and C (t ) the amounts of A, B, and C present at time t . Observe that C (t ) = kA(t )B(t ). Then note that A0 − A(t ) = m C (t ) m+n and B0 − B(t ) = n C (t ) m+n and thus C (t ) = k A0 − m C (t ) m+n B0 − n C (t ) . m+n CHAPTER HIGHLIGHTS 8.1 Integral Tables and Review Important integral formulas (p. 446) A table of integral appears on the inside covers. 8.2 Integration by Parts A ( x − α )k and Bx + C (x2 + β x + γ )k u dv = uv − v du. (p. 451) called the partial fraction decomposition (p. 474). To integrate an improper rational functions, express it as a polynomial plus a proper rational function by dividing the denominator into the numerator (p. 474). *8.6 Some Rationalizing Substitutions Success with the technique depends on choosing u and dv so that vdu is easier to integrate than udv. The integral of ln x and the integrals of the inverse trigonometric functions are calculated using integration by parts (p. 453). 8.3 Powers and Products of Trigonometric Functions For integrals involving n f (x) for some function f , let un = f (x), nun−1 du = f (x) dx. For rational expressions in sine and cosine, let u = tan (x/2); then sin x = 8.7 Integrals of the form sin x cos x dx can be calculated by using the basic identity sin2 x cos2 x = 1 and the double-angle formulas for sine and cosine Reduction formulas: 1 n−1 sinn x dx = − sinn−1 x cos x + n n cosn x dx = 1 n−1 cosn−1 x sin x + n n sinn−2 x dx, cosn−2 x dx. m n 2u 1 − u2 2 , cos x = , dx = du (p. 484). 2 1+u 1 + u2 1 + u2 Numerical Integration Left-endpoint, right-endpoint, and midpoint estimates; trapezoidal rule; Simpson’s rule (p. 487) The theoretical error in the trapezoidal rule varies as 1/n2 (p. 491). The theoretical error in Simpson’s rule varies as 1/n4 (p. 492). *8.8 Differential Equations; First-Order Linear Equations The main tools for calculating such integrals as tanm x secn x dx are the identities 1 + tan2 x = sec2 x and integration by parts. 8.4 Integrals Involving a2 − x2 , a2 + x2 , x2 − a2 differential equation (p. 496) order (p. 496) solution (p. 496) general solution (p. 497) particular solution (p. 497) initial-value problem integrating factor (p. 497) (p. 499) Newton’s law of cooling (p. 499) A first-order differential equation is linear if it can be written in the form y + p(x)y = q(x), where p and q are continuous functions on some interval I . The general solution is y = e−H (x) where H (x) = p(x)dx. eH (x) q(x) dx + C , Such integrals may be calculated by a trigonometric substitution: for for for a2 − x 2 a2 x2 + − x2 a2 set set set a sin u = x, a tan u = x, a sec u = x. *8.9 Integral Curves; Separable Equations It may be necessary to complete the square under the radical before making a trigonometric substitution (p. 470). 8.5 Partial Fractions Integral curves (p. 504) A first-order differential equation is separable if it can be written in the form p(x) + q(y)y = 0. The integral curves are of the form p(x)dx + The logistic equation (p. 507). q(y)dy = x. A proper rational function may be integrated by writing it as a sum of fractions of the form ...
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