Exponential Growth and Decay

Transcription

1 Exponential Growth and Decay In certain situations, the rate at which a thing grows or decreases is proportional to the amount present. When a substance undergoes radioactive decay, the release of decay particles precipitates additional decay as the particles collide with the atoms of the substance. The larger the mass, the larger the number of collisions per unit time. Consider a collection of organisms growing without environmental constraints. The larger the population, the larger the number of individuals involved in reproduction and hence, the higher the reproductive rate. Let be the amount of whatever is being measured the mass of the radioactive substance or the number of organisms, for example. Let t be the time elapsed since the measurements were started. Then = k. If k > 0, then increases with time (exponential growth); if k < 0, then decreases with time (exponential decay). I ll solve for in terms of t using separation of variables. First, formally move the s to one side and the t s to the other: = k. Integrate both sides and solve for : = k, ln = kt + C, e ln e kt+c, = e C e kt, = ±e C e kt. Let 0 = ±e C : = 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 20 initially and 40 after 2 days. How many are there after 4 days? If is the number of roaches after t days, then When t = 2, = 40: = 20e kt. 40 = 20e 2k, 7 = e 2k, ln7 = lne 2k = 2k, k = ln 7 2. Hence, When t = 4, = 20e (tln 7)/2. = 20e (4ln 7)/ roaches.

2 Example. A population of MU flu virus grows in such a way that it triples every 5 hours. If there were 00 initially, when will there be ? Let N be the number of the little rascals at time t. Then N = 00e kt. Since the amount triples in 5 hours, when t = 5, N = 300: Hence, Set N = : 300 = 00e 5k, 3 = e 5k, ln 3 = lne 5k = 5k, k = ln3 5. N = 00e (tln 3)/ = 00e (tln 3)/5, 0000 = e (tln 3)/5, ln0000 = lne (t ln 3)/5 = t ln 3 5, t = 5 ln0000 ln hours. Example. The half-life of a radioactive substance 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 2 same amount of time for 00 grams to decay to 50 grams as it takes for grams to decay to grams. The half-life of radium is 620 years. How long will it take 00 grams of radium to decay to gram? Let M be the amount of radium left after t years. Then When t = 620, M = 50: M = 00e kt. Hence, Set M = : 50 = 00e 620k, 2 = e620k, ln 2 = lne620k = 620k, k = M = 00e (tln 2)/620. ln = ln = 00e (tln 2)/620, 0.0 = e (tln 2)/620, ln 0.0 = lne (t ln 2)/620 = t ln 2 620, 620 ln0.0 t = years. ln 2 Example. A population of bacteria grows exponentially in such a way that there are 00 after 2 hours and 750 after 4 hours. How many were there initially? Let be the number of bacteria at time t. Then = 0 e kt. 2

3 There are 00 after 2 hours: There are 750 after 4 hours: 00 = 0 e 2k. 750 = 0 e 4k. I ll solve for k first. Divide the second equation by the first: lug this into the first equation: = 0e 4k 0 e 2k = e2k, ln7.5 = lne 2k = 2k, k = 0.5 ln = 0 e 2 0.5ln 7.5 = 0 e ln 7.5 = 7.5 0, 0 = There were 3 bacteria initially (if I round to the nearest bacterium). According to Newton s law of cooling, the rate at which a body heats up or cools down is proportional to the difference between its temperature and the temperature of its environment. If T is the temperature of the object and T e is the temperature of the environment, then dt = k(t T e). If k > 0, the object heats up (an oven). If k < 0, the object cools down (a refrigerator). I ll solve the equation using separation of variables. Formally move the T s to one side and the t s to the other, then integrate: Let C 0 = ±e C : dt T T e = k, dt = T T e k, 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. T T e = C 0 e kt, T = T e + C 0 e kt. Let T 0 be 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 20 bagel is placed in a 70 room to cool. After 0 minutes, the bagel s temperature is 90. When will its temperature be 80? In this case, T e = 70 and T 0 = 20, so When t = 0, T = 90: T = e kt. 90 = e 0k, 0.4 = e 0k, ln 0.4 = lne 0k = 0k, k = 3 ln

4 Hence, Set T = 80: T = e (tln 0.4)/0. 80 = e (tln 0.4)/0, 0.2 = e (t ln 0.4)/0, ln 0.2 = lne (tln 0.4)/0 = t ln0.4, 0 t = 0 ln0.2 ln min. Example. A pair of shoes is placed in a 300 oven to bake. The temperature is 20 after 0 minutes and 52.7 after 20 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 obtain When t = 0, T = 20: When t = 20, T = 52.7: T = (T 0 300)e kt. 20 = (T 0 300)e 0k, 80 = (T 0 300)e 0k = (T 0 300)e 20k, 47.3 = (T 0 300)e 20k. Divide 80 = (T 0 300)e 0k by 47.3 = (T 0 300)e 20k and solve for k: = e 0k, ln = lne 0k = 0k, k = 80 ln lug this back 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. The logistic growth model is described by the differential equation = k(b ). is the quantity undergoing growth for example, an animal population and t is time. k is the growth constant, and the constant b is called the carrying capacity; the reason for the name will become evident shortly. Notice that if is less than b and is small compared to b, then b b. The equation becomes kb, which is exponential growth. Notice that the derivative 4 is positive, so increases with time.

5 If is greater than b, then b is negative, so the derivative is negative. This means that decreases with time. It s possible to solve the logistic equation using separation of variables, though it will require a little trick (called a partial fraction expansion). Separate the variables: (b ) = k. Now (b ) = ( b b + ). (You can verify that this is correct by adding the fractions on the right over a common denominator.) Therefore, I have ( b b + ) = k, (ln ln b ) = kt + C, b ln b = bkt + bc, exp ln b = exp(bkt + bc), b = ebc e bkt. I can replace the s with a ± on the right, then set C 0 = ±e bc. This yields b = C 0e bkt. Now some routine algebra gives = bc 0e bkt + C 0 e bkt. Divide the top and bottom by C 0 e bkt, and set C = C 0 : = b C e bkt +. Suppose the initial population is 0. This means = 0 when t = 0: 0 = b C +, C = b 0. The equation is = ( b 0 5 b ). e bkt +

6 For example, consider the case where k = and b = 3. I ve graphed the equation for with 0 = 0.2,, 2, 3, 4, and 5: = 0.2 shows initial exponential growth. As the population increases, growth levels off, approaching = 3 asymptotically. 0 = and 0 = 2 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 = 5 yield populations that shrink, again approaching = 3 as t. You can see why b is called the carrying capacity. Think of it as the maximum number of individuals that the environment can support. If the initial population is smaller than b, 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 2005 by Bruce Ikenaga 6