Applications of First Order Equations

Transcription

1 Applications of First Orer Equations Viscous Friction Consier a small mass that has been roppe into a thin vertical tube of viscous flui lie oil. The mass falls, ue to the force of gravity, but falls more slowly than it woul in a flui lie water because the oil is thicer than water, (or, as we say, the oil is more viscous than the water). One of Newton s laws asserts that the time rate of change of the momentum of the mass is equal to the sum of the external forces acting on the mass, t mvt F where m is the mass of the object an vt is the mass velocity as a function of time. The forces acting on the mass are the force of gravity, F g an the friction or viscous force of the oil, F f. Then F g mg an F f vt, where we have assume that the friction force is proportional to the velocity (i.e., the faster the object moves, the more the oil retars it). The constant of proportionality,, is assume to be constant an then the negative sign appears in the efinition of F f so that the force acts in the irection opposite to the irection of the velocity. Then we have mvt mg vt, v v, 1 where v enotes the initial velocity of the mass. We can rewrite the equation as vt g m vt. 2 Before solving this equation we will first carry out a qualitative analysis on the equation. We note that if g m vt, i. e., vt mg then v t so vt is ecreasing if g m vt, i. e., vt mg then v t so vt is increasing if vt mg then v t so vt is constant Then a solution curve that begins at a point above the value mg will ecrease towar this value while a solution curve starting out below this value will increase towars mg. In aition, we can ifferentiate the equation (2) to obtain v"t m v t m g m vt, 1

2 from which we conclue that if vt mg then v t an if vt mg then v t. Then the solution curves above the value mg all have positive curvature an ecrease towar the limiting value v mg. Those curves that begin below the limiting value have negative curvature an increase towar v. This constant coul be interprete as the terminal velocity of the projectile falling in the flui.. The curve lying above the constant v mg is consistent with an object that is fire into the oil with a large initial velocity in which case the projectile graually slows own as it continues to fall only uner the effect of gravity an the viscous rag of the oil. The curve lying below the constant represents the case where the object is introuce into the tube with a small or zero initial velocity. In this case, the velocity increases towar the terminal velocity mg ue to gravity an friction. In orer to now solve for v(t), we note that equation (2) is equivalent to v g m v t, from which it follows that or Then an m lng m v t C lng m v C 1. g m v C 2 e, vt mg C 3 e. The initial conition is satisfie if mg C 3 v so, finally, vt mg mg v e v v v e, where we introuce the notation, v mg for the terminal velocity. This is an explicit formula for the solution curves escribe above. Note that vt x t where xt enotes the position as a function of time. Then this position function can be obtaine by integrating the function vt with respect to t. Exercise- Fin xt if the initial position is x.. Discuss how you might solve for if the parameters m, g are assume nown. Can you thin of an experiment for fining v? If so, then can be foun from the nowlege of v. 2

3 A Dissolving Pill A spherical pill is roppe into a flui where it begins to issolve at a rate that is proportional to the surface area expose to the flui. We can express this last statement in mathematical terms by writing Volume K Surface Area 3 t where K enotes a positive constant of proportionality an the minus sign inicates that the volume ecreases as long as the surface area is positive. For a sphere, we have V 4 3 Rt3 an SA 4Rt 2 where Rt enotes the raius of the sphere (which varies with t). Then an or t Volume t 4Rt 2 R t K 4Rt 2, R t K. 4 3 Rt3 4Rt 2 R t Then Rt Kt C an Rt R Kt, where R enotes the initial raius of the pill. If the initial volume of the pill is V, then R 3V 4 1/3 an Rt 3V 4 1/3 Kt from which it is easy to fin the time to completely issolve if K is nown. How might you go about fining K from an experiment? Exercise- Fin out the issolving time for a cubical pill, assuming that the pill maintains its cubical shape as it issolves. Which issolves in less time, the cubical pill or the spherical pill? Water Heating Strategy Suppose a cylinrical water heater contains a heating element that electrically heats the water in the tan. We have alreay seen this moel, which we can write in the form t Tt T f Tt, where T f enotes the temperature to which the water will be eventually heate if the heating element is left on constantly, an is a positive constant that reflects the proportionality 3

4 between the time rate of change in the water temperature an the ifference between T f an Tt. Note that as long as Tt is less than T f, T t is positive so the temperature of the water increases. We must also account for the fact that there is heat lost to the surrounings which we will assume is at a uniform temperature we will enote by S. This loss will prouce a rate of change of temperature proportional to the ifference between the tan temperature an the temperature of the surrounings. That is, Tt KS Tt t Here K enotes a positive constant so this equation asserts that as long as Tt is greater than S, the temperature of the water will ecrease. Now the water in the tan is being simultaneously heate by the heating element an coole by loss of heat to the surrounings so our equation that reflects this must have the form, t Tt KS Tt T f Tt KS T f KTt. Initially we have, T T in, where T in is the temperature at which the water enters the tan. It will be more convenient if we write this equation in the form t Tt KM Tt where M KS T f K 4 We now that we can solve this equation by writing T M T Kt which leas to lnm Tt Kt C an Tt M C 1 e Kt. It follows from the initial conition that Tt M T in Me Kt. 5 This solution preicts that the temperature will increase from T in towar the limiting value M at an exponential rate that is etermine by the constants an K. Note that M is a temperature between T f an S an that M is close to T f if K is very small, while M is closer to S if is small compare to K. A small value of K correspons to a tan that is well insulate to there is little heat loss to the surrounings. Now suppose that when the water temperature reaches some value T max M, we turn off the water heater. Note that we can use equation 5 to etermine the time t a at which the temperature reaches T max ; i.e. we solve Tt a M T in Me Kta T max. 4

5 for t a. This gives e Kta T max M T in M an t a 1 K ln T max M T in M 1 K ln M T in M T max. Since the heater is turne off, the equation 4 is replace by the equation t Tt KS Tt with Tt a T max 6 This escribes how the water will graually cool from a temperature of T max, as a result of heat loss to the exterior. The solution of 6 is foun in the usual way to be, Tt S T max Se Ktta 7 We can now suppose that when the temperature of the water reaches some selecte temperature T min with S T min T max, we turn the heater bac on again. The time t b when the temperature reaches T min is obtaine by solving Tt b S T max Se Kt bt a T min to get e Kt bt a T min S T max S or t b t a 1 K ln T min S T max S We now solve 4 on the interval t t b with the initial conition, Tt b T min. The water temperature will increase towar M an we can then turn the heater off again when the temperature reaches T max at some time t c t b. In this way, the temperature is maintaine between limits T min an T max without the heater having to be on constantly. Absorption of Meications When you tae a pill to obtain meication, the pill first goes into your stomach an the meication passes into your GI tract. From there the meication is absorbe into your bloostream an circulate through your boy before being eliminate from the bloo by the ineys an other organs. If we let xt enote the amount of meication in your GI tract at time t, then we can moel the movement of the meication out of the GI tract with the equation x t 1 xt, x A. 8 This is the assertion that after taing the pill, an amount A of meication is in the GI tract an it ecreases at a rate proportional to the amount currently present in the GI tract. If the amount of meication in the bloostream at time t is enote by yt, then y t 1 xt 2 yt, y, 9 expresses the fact that meication is coming into the bloostream at exactly the rate it is leaving the GI tract an it is leaving the bloostream at some rate expresse by the proportionality constant 2. Also, we are assuming that there is no meication in the bloostream initially. Now this is two equations for the two unnown functions xt an yt, 5

6 but the first equation can be solve inepenently an the solution substitute into the equation (9). The solution of (8) is easily foun to be xt A e 1 t, an then y t 2 yt 1 xt 1 A e 1 t. We will fin a particular solution for the y-equation by the metho of unetermine coefficients. We guess that y p t a e 1 t an then y p t 2 y p t 1 a e 1 t 2 a e 1 t 1 A e 1 t. This leas to Then a 1 A 2 1 an y p t yt C e 2t 1 A 2 1 e 1 t 1 A 2 1 e 1 t. an, using the initial conition to evaluate C, we get yt 1 A 2 1 e 1 t e 2t Plotting x(t) an y(t) versus t for some representative values of the constants gives the following figure y x xt an yt vs t If we wish to consier a moel that represents taing a continuously acting pill, (a pill that releases meication continuously so as to maintain a constant level of meication in the GI tract for a sustaine perio of time), we might moify the previous moel 8,9 to rea x t X 1 xt, x. y t 1 xt 2 yt, y. 6

7 In this case, we fin xt X 1 1 e 1t Plotting these solutions gives, yt e 2t 2 e 1t y x xt an yt vs t Clearly this prouces a longer constant level of meication in the bloostream. Of course, eventually, the level of meication in the GI tract will go to zero an the level in the bloostream will then also ecrease to zero. A moel which coul escribe the perioic taing of a sequence of time release pills woul loo lie x t X t 1 xt, x. y t 1 xt 2 yt, y, where X t enotes a piecewise constant function that alternates between a positive value an zero. We will iscuss an effective way to solve equations involving such terms when we iscuss the Laplace transform. 7