Growth and Decay Models

Transcription

1 Growth and Decay Models --08 In certain situations, the rate at which a thing grows or decreases is proportional to the amount present. When a sustance undergoes radioactive decay, the release of decay particles precipitates additional decay as the particles collide with the atoms of the sustance. The larger the mass, the larger the numer of collisions per unit time. Consider a collection of organisms growing without environmental constraints. The larger the population, the larger the numer of individuals involved in reproduction and hence, the higher the reproductive rate. Let e the amount of whatever is eing measured the mass of the radioactive sustance or the numer of organisms, for example. Let t e the time elapsed since the measurements were started. Then d dt = k. If k > 0, then increases with time (exponential growth); if k < 0, then decreases with time (exponential decay). k is called the growth constant (or decay constant). I ll solve for in terms of t using separation of variales. First, formally move the s to one side and the t s to the other: d = kdt. Integrate oth sides and solve for : d = kdt Let 0 = ±e C. This gives ln = kt+c e ln = e kt+c = e C e kt = ±e C e kt = 0 e kt. Note that 0 is the initial amount: Setting t = 0, (0) = 0 e 0 = 0. Example. A population of roaches grows exponentially under Calvin s couch. There are 0 initially and 40 after days. How many are there after 4 days? If is the numer of roaches after t days, then = 0e kt. When t =, = 40: 40 = 0e k 7 = e k ln7 = lne k = k k = ln7

2 Hence, When t = 4, = 0e (tln7)/. = 0e (4ln7)/ roaches. Remark. We ll often have expressions of the form e (stuff), and this can e inconvenient to write if stuff is complicated. You can use the exponential function notation to avoid this prolem: exp(stuff) = e (stuff). For instance, in the last example, ( ) ln7 exp t = e (tln7)/. Example. A population of MU flu virus grows in such a way that it triples every hours. If there were 00 initially, when will there e ? Let N e the numer of the little rascals at time t. Then N = 00e kt. Since the amount triples in hours, when t =, N = 300: 300 = 00e k 3 = e k ln3 = lne k = k k = ln3 Hence, Set N = : N = 00e (tln3)/ = 00e (tln3)/ 0000 = e (tln3)/ ln0000 = lne (tln3)/ ln0000 = tln3 t = ln0000 ln hours The half-life of a radioactive sustance is the amount of time it takes for a given mass M to decay to M. Note that in radioactive decay, this time is independent of the amount M. That is, it takes the same amount of time for 00 grams to decay to 0 grams as it takes for grams to decay to grams. Example. The half-life of radium is 60 years. How long will it take 00 grams of radium to decay to gram?

3 Let M e the amount of radium left after t years. Then M = 00e kt. When t = 60, M = 0: 0 = 00e 60k = e60k ln = lne60k = 60k Hence, Set M = : k = ln 60 = ln 60 M = 00e (tln)/60. = 00e (tln)/ = e (tln)/60 ln0.0 = lne (tln)/60 = tln 60 t = 60ln years ln Example. A population of acteria grows exponentially in such a way that there are 00 after hours and 70 after 4 hours. How many were there initially? Let e the numer of acteria at time t. Then There are 00 after hours: There are 70 after 4 hours: = 0 e kt. 00 = 0 e k. 70 = 0 e 4k. I ll solve for k first. Divide the second equation y the first: lug this into the first equation: 70 = 0 e 4k 00 = 0 e k = 0e 4k 0 e k = ek ln7. = lne k = k k = 0.ln7. 00 = 0 e 0.ln7. = 0 e ln7. = 7. 0, 0 =

4 There were 3 acteria initially (if I round to the nearest acterium). Newton s Law of Cooling. AccordingtoNewton s law of cooling, therateatwhichaodyheatsuporcoolsdownisproportional to the difference etween its temperature and the temperature of its environment. If T is the temperature of the oject and T e is the temperature of the environment, then dt dt = k(t T e). If k > 0, the oject heats up (an oven). If k < 0, the oject cools down (a refrigerator). I ll solve the equation using separation of variales. Formally move the T s to one side and the t s to the other, then integrate: dt T T e = kdt, dt = T T e kdt, ln T T e = kt+c, e ln T Te = e kt+c, T T e = e C e kt, T T e = ±e C e kt. Let C 0 = ±e C : T T e = C 0 e kt, T = T e +C 0 e kt. Let T 0 e the initial temperature that is, the temperature when t = 0. Then T 0 = T e +C 0 e 0 = T e +C 0, C 0 = T 0 T e. Thus, T = T e +(T 0 T e )e kt. Example. A 0 agel is placed in a 70 room to cool. After 0 minutes, the agel s temperature is 90. When will its temperature e 80? In this case, T e = 70 and T 0 = 0, so When t = 0, T = 90: T = 70+0e kt. 90 = 70+0e 0k, 0.4 = e 0k, ln0.4 = lne 0k = 0k, k = ln Hence, T = 70+0e (tln0.4)/0. Set T = 80: 80 = 70+0e (tln0.4)/0, 0. = e (tln0.4)/0, ln0. = lne (tln0.4)/0 = tln0.4, 0 t = 0ln0. ln min. 4

5 Example. A pair of shoes is placed in a 300 oven to ake. The temperature is 0 after 0 minutes and.7 after 0 minutes. What was the initial temperature of the shoes? I set T e = 300 in T = T e +(T 0 T e )e kt to otain When t = 0, T = 0: When t = 0, T =.7: T = 300+(T 0 300)e kt. 0 = 300+(T 0 300)e 0k, 80 = (T 0 300)e 0k..7 = 300+(T 0 300)e 0k, 47.3 = (T 0 300)e 0k. Divide 80 = (T 0 300)e 0k y 47.3 = (T 0 300)e 0k and solve for k: = e 0k, ln = lne 0k = 0k, k = 80 ln lug this ack into 80 = (T 0 300)e 0k and solve for T 0 : 80 = (T 0 300)e [0 ( /0)ln(80/47.3)], 80 = (T 0 300)e ln(80/47.3), 80 = (T 0 300)e ln(47.3/80), 80 = (T 0 300) , = T 0 300, T 0 = In the real world, things do not grow exponentially without limit. It s natural to try to find models which are more realistic. Logistic Growth. The logistic growth model is descried y the differential equation d dt = k( ). is the quantity undergoing growth for example, an animal population and t is time. k is the growth constant, and the constant is called the carrying capacity; the reason for the name will ecome evident shortly. Notice that if is less than and is small compared to, then. The equation ecomes d dt d k, which is exponential growth. Notice that the derivative dt If is greater than, then is negative, so the derivative d is positive, so increases with time. is negative. This means that dt decreases with time. It s possile to solve the logistic equation using separation of variales, though it will require a little trick (called a partial fraction expansion). Separate the variales: Now d ( ) = kdt. ( ) = ( + ).

6 (You can verify that this is correct y adding the fractions on the right over a common denominator.) Therefore, I have ( + ) d = kdt (ln ln ) = kt+c ln = kt+c expln = exp(kt+c) = ec e kt I can replace the s with a ± on the right, then set C 0 = ±e C. This yields Now some routine algera gives = C 0e kt. = C 0e kt +C 0 e kt. Divide the top and ottom y C 0 e kt, and set C = C 0 : = C e kt +. Suppose the initial population is 0. This means = 0 when t = 0: Thus, the equation is 0 = C + = ( 0 so C = 0. ). e kt + For example, consider the case where k = and = 3. I ve graphed the equation for with 0 = 0.,,, 3, 4, and :

7 0 = 0. shows initial exponential growth. As the population increases, growth levels off, approaching = 3 asymptotically. 0 = and 0 = are already large enough that the population spends most of its time levelling off, rather than growing exponentially. An initial population 0 = 3 remains constant. 0 = 4 and 0 = yield populations that shrink, again approaching = 3 as t. You can see why is called the carrying capacity. Think of it as the maximum numer of individuals that the environment can support. If the initial population is smaller than, the population grows upward toward the carrying capacity. If the initial population is larger than, individuals die and the population decreases toward the carrying capacity. c 08 y Bruce Ikenaga 7