STRAND: GRAPHS Unit 5 Growth and Decay

Transcription

1 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet STRAND: GRAPHS Unit 5 Growth and Deca TEXT Contents Section 5. Modelling Population 5. Models of Growth and Deca 5. Carbon Dating 5.4 Rate of Growth 5.5 Solving Eponential Equations 5.6 Problems of Logarithms

2 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet 5 Growth and Deca 5. Modelling Population Amoebae reproduce b dividing after a certain time. Radioactive substances have 'half lives' which are determined b the time it takes the radioactivit to halve. These are eamples of sstems which are modelled b 'eponential' functions. In 989 the world s human population was thought to be growing at about % per ear. That is, that at the end of each ear the population would be % more than it was at the start of the ear. In the first eample we form a model to describe this population growth, and then use it to predict the ear when the population would be twice the size it was in 989. The population at the end of 989 was approimatel 4.5 billion (4 5 ). If the population grew b % in 99, at the end of the ear it would be 45. billion. = 465. billion. The predicted population at the end of 99 can be found b the calculation 45. billion.. = billion Using this model, b the end of 99, the predicted population would be 45. (. ). Worked Eample For the model described above, calculate the population at the end of each ear, starting at 989, and continuing until the population has doubled from its value at 989. Plot the values on a graph. In general, the population is modelled b ( ) = 45. (. ) P f This gives the table with some ke data included, t Population (ears) (billions, to d.p.) t population has doubled b t = 4, i.e. ear In fact, the estimate of the world population for was 7.4 billion and there has been almost zero growth for some ears now!

3 5. CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet The graph below illustrates this growth. Population (billions) Time (ears) The model used in the eample above can be represented b the equation P= Pa () o when P is the population at the end of the ear number, P o is the initial population at ear = and a is a constant. So for the world population described above P= 45.. =,,,... Equations of the form () can also be used to describe populations that are declining. Some species are endangered because the have declining populations, for various reasons like hunting, habitat destruction, new predators or infertilit. Man marine species such as whales and some fish give cause for concern. Models are made to help predict future trends in fish stock levels, which take into account man features like fishing techniques and environmental conditions. In the following eercises, ou will produce a model of a fish population based on the assumption that it is declining b 5% each ear. You then use it to find the number of ears before the population becomes so low that it is in danger of being unable to sustain itself.

4 5. CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Eercises Assume that a particular fisher has a population of, fish and that current fishing methods cause this population to decline b 5% in a ear.. (a) Cop and complete the table below, and use it to help ou form a model of the population p in terms of the ears elapsed,, for =,,...,. Years elapsed Population.85 = (a) Using our data from Question, plot and draw the graph of the fish population for the first ears. (b) If the population falls below sa 5, the fish become quite widel separated. In these conditions it becomes difficult to find good catches, and the fish themselves breed at a much reduced rate. Therefore, using 5 as an 'action level' use our model to find the number of ears before which the population becomes dangerousl low. 5. Models of Growth and Deca The models above show how mathematics can be used to model growth and deca of populations. Another eample is bacteria, which divide in two ever minute. The growth in numbers is illustrated in the following table. Minutes since Number of bacteria introduction of bacteria ( = ) ( = ) = = 4 = = = = t t The table shows a 'pattern ' for the number of bacteria at an time after the bacteria was introduced. From this pattern, it is eas to see that after t minutes the number of bacteria will be t.

5 5. CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet t Therefore, n = is a model for the bacteria growth. The graph for this function is shown here, for values of t between and 4. n t Drawing a smooth curve through these discrete points gives a continuous model of growth, although in this eample, it onl makes sense to use whole numbers for t. However, man situations which have models like this have continuous not discrete domains, so the curve is tpical of this tpe of function. n n t = 4 5 t An function of the form a, where a is a positive constant is called an eponential function (as is the 'eponent' or power of a). Although the eample above (and the eamples developed in Activities and ) was onl defined for positive values of the eponent ( t ), eponential functions are defined for an real value. For instance, if f ( ) =, then f Worked Eample (a) ( ) = =, using the rules for indices. Use a graphic calculator or computer to help ou make sketches of these functions, (i) - (iv) below, using the same pair of aes. Use a range of the values between and +, and to + on the -ais. (i) = (ii) = (iii) = (iv) = ( 5. ) (b) (c) (d) Do the curves have a common point? What is the relationship between = and = ( 5. )? What happens as becomes large and positive or large and negative? 4

6 5. CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet (a) (i) = (iv) = ( 5. ) (iii) = (ii) = (b) Common point is (, ) (c) (d) = and = ( 5). are reflections in the -ais. As, and but = and (. 5). As, and but = and (. 5). Worked Eample (a) In a similar wa, illustrate the graphs of (i) = (ii) = (b) (c) Compare them with =,. Describe their behaviour for large, positive or negative. 5

7 5. CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet (a) (b) (c) = is the reflection of = in the -ais. = = = ( 5) = in the -ais. As, both and. so is illustrated in the previous graph, and is a reflection of As, both and Another eample of the use of the eponential function is in the modelling of Radon 9, which is an 'isotope' of the gaseous element Radon. It occurs naturall in some tpes of rock and its seepage from beneath buildings has been identified as a major concern in some parts of the countr. Radon 9 is radioactive, with a half life of about 4 seconds. This means that if there are atoms of Radon 9 in a sample of the gas, 4 seconds later there will be half this number left, 5. 4 more seconds later, and the number of Radon 9 atoms will halve again, to 5, and so on. This decaing sstem can be modelled with an eponential function, with a negative eponent. Time in seconds after Number of atoms sample is collected (seconds) left 4 ( 8 ( ( ) ) ) Since ever 4 seconds increases the power of the eponent b, ou can write the model equation as 4 t Nt () = ( ) Note that this can also be written as Nt () = ( ) = using the properties of indices. 4 4 t t 6

8 5. CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Worked Eample The government s statistical service made several predictions about the United Kingdom s population in 4. One of these was that the population would grow b % ever ear. If the population in 4 was 68 million, form a model for the UK s future population. Use it to draw a graph or spreadsheet of the projected population, and from this estimate the ear when the population will equal 8 million. ( ) = Pt t 68. when t is the number of ears after 4. The predicted values are given in the table below. t Pt ( ) (millions) So this predicts that b the ear, the UK's population will be above 8 million. The model used in the eample above is Pt ()= Pa t when P = 68 ( million) and a=.. So far ou have used eponential functions to model various populations and radioactive deca. This last application can be used to help date archaeological objects through Carbon dating. This section is completed with a summar of the general properties of the eponential functions ba = and = ba for a >. As is shown below, both curves pass through (, b) and the range of both functions is all real numbers greater than zero. a is the base and the eponent of the function. = ba, a > = ba, a > < b b 7

9 5. CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Eercises. Use the population model ( ) = Pt Pa t with P = 68 million and (a) a =. (b) a = 5. (c) a =. In each case, estimate when the population will reach 8 million.. In the question above, what factors in the population model have not been full taken into account? 5. Carbon Dating Carbon 4 ( 4 C) is an isotope of carbon with a half life of 57 ears. It eists in the carbon dioide in the atmosphere, and all living things absorb some Carbon 4 as the breathe. This remains in an animal or plant, and is constantl added to until the organism dies. After this time, the Carbon 4 decas, reducing to half the amount stored in the bod after 57 ears. The amount halves again after another 57 ears, and so on, with no new Carbon 4 absorbed. In 946 an American scientist, Williard Libb, developed a wa of 'dating' archaeological objects b measuring the Carbon 4 radiation present in them. This radioactivit is compared with that found in things living now. For instance, if bones of recentl dead animals produce becquerels per gram of bone carbon (a becquerel is the unit of radioactivit), and an old bone produces onl 5 becquerels, the radioactivit has halved since the animal which had the old bone died. As the half life of Carbon 4 is 57 ears, this would mean the animal died in 75 BC approimatel, as tested in 5. Eercises. Complete the table below. Age (in ears) Radiation (becs) 57 = 46 ( ) =

10 5. CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet. Use our model to produce a graph, showing radioactivit on the vertical ais and time, in ears, on the horizontal. Draw the graph for values of t up to 5 ears.. From our graph, estimate the ages of bones with these radioactivities: (a) (b) 8.5 becquerels per gram of carbon;. becquerels per gram of carbon. 5.4 Rate of Growth Suppose a colon of bacteria doubles in number ever minute as ever member of the colon divides in two. So if there are bacteria at the start of the colon, there will be 4 a minute later (an increase of in one minute), 8 two minutes later (an increase of 4 in one minute) and so on. As the number of bacteria increases, so the rate at which that number increases goes up. So the rate of increase of an eponential function is closel related to the value of the function at an point. This suggests that eponential functions and their derivatives are closel linked. The net eample will eplore these links. Worked Eample Plot and draw the curves below for values of between and + and on separate aes. (a) = (b) = Using a ruler to draw tangents to each of the curves, calculate the gradient of each curve at =, 5.,, 5. and. The figure opposite illustrates the method. Note that gradient = change in change in 4 = change change Gradient In this case all gradients are positive since a positive change in results in a positive change in. Plot the values for the gradient of each of the graphs and sketch in the gradient curve. How do the original graph and the gradient curve of each function seem to be related? The graphs are shown below with dashed curves to indicate the gradient curve. 9

11 5.4 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet gradient of gradient of gradient of = 9 = gradient = gradient of =.5.5 It is noted that the gradient curve of values for > whilst the gradient curve of slightl lower values. = is similar to = but with slightl larger = again is similar to = but with

12 5.4 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet If ou have access to a computer or calculator that is capable of showing the derivative, that is, the gradient, of a function, then ou can find an eponential function whose derivative eactl fits over its own graph b considering = a with a in the range 5. < a < 9.. The derivative of is alwas less than the value of the function itself, as is the derivative of 5., although it is a closer fit to the function than that of. =.5 = Derivative Derivative The derivative of has a greater value than the function. This suggests that there is an eponential function, with a base between.5 and, which has its derivative the same as itself. Derivative = e = (, ) Such a function would therefore be its own derivative. The base required for this to happen is denoted b the letter 'e'. Unfortunatel, its value cannot be given eactl - like π and it is irrational, and so it cannot be epressed eactl as a fraction or decimal. To five decimal places, it is.788. ( ) = e is often referred to as the eponential function. It is unique in The function f mathematics, in that it is its own derivative. This propert makes it etremel important in man branches of the subject. To summarise, we write this as d = e = e d when the derivative, d, is the gradient of the functions. d

13 5.4 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Worked Eample (a) Use a graphic calculator, computer or spreadsheet to make sketches of these graphs. (i) = e ( (ii) = e + ) ( (iii) = e ) (iv) = e + (v) = e (Note that calculators or computers ma use the epression = ep ( ) for = e.) (b) Compare each of the sketches with the graph = e, and state the relationship between each graph and that of = e. (c) Use the fact that the derivative of e is e, to work out the derivatives of each of the other functions. (a) (i) 4 = e (ii) 4 = ( ) e

14 5.4 (iii) CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet 4 = ( ) e (iv) 4 = e (v) = e (b) (ii) Translation of (iii) Translation of (iv) Translation of (v) Reflection in ais

15 5.4 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet (c) (ii) e + (iii) e + (iv) e (v) e The function f ( ) = e is a mapping from the set of real numbers, R, to the positive real ( ) = e has numbers. Its graph shows that it is a one to one function. This means that f an inverse function. The graph of this inverse function is a reflection in the line = of the graph of = e. The graph below shows e and its inverse function, which is usuall written as ln( ). This function is read as 'the natural (or Naperian) logarithm of ' or 'the logarithm to base e of '. (John Napier, 55-67, was a Scottish mathematician who originated the concept of logarithms.) = e = e = (, ) = ln The figure shows that ln ( ) is not defined for negative values of (or zero), as there is ( ). So ln( ), for instance, does not eist. The no graph to the left of the ais for ln range of ln( ), however, is the full set of real numbers. Worked Eample Find if e =. Give our answer to two d.p. Since e =, and = ln is the inverse function of e, = ln Using a calculator to find ln (), gives = 46. to d.p. To summarise, for a >, e = a = n a Note that the brackets round 'a' in ln a have been omitted and will be in future ecept where it might cause confusion. 4

16 5.4 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Worked Eample 4 Solve, to s.f. the equation e = 5, Since e = 5, then e = 5 Since e and ln are inverse functions, = ln = + ln = ( + ln ) =. 755 to s.f. Eercises. Solve e = 5 to d.p.. Solve e = to d.p.. Solve 4 e = to s.f. 4. Solve e = to d.p. e 5. Solve = 4 to s.f. 6. Solve e =.5 to d.p. e = 6 7. Solve 4 to d.p. 8. Solve 7e = to s.f. 9. Solve e e =to d.p.. Solve e = 4e to s.f. 5

17 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet 5.5 Solving Eponential Equations Earlier in this unit, ou have produced eponential functions as models, and then used graphs to estimate the solution to a problem. The logarithmic function allows ou to calculate rather than estimate these solutions as is shown in the eample below. Worked Eample A bacteria colon doubles in number ever minute, from a starting population of one. m The population model is P =, where P is the population and m the number of minutes since the colon was started. Find the time when the population first equals. The problem requires a solution to the equation P = or m = Taking log of each side of the equation ln m = n () Now, is a positive real number, so there is some number, call it n, such that e n = (see figure below). Then n = ln ( ). 69. = = e e n So m = ( e n ) m replacing b e n n m nm and since ( e ) = e, using the properties of indices, m in () above can be replaced b e nm, where n = ln(). This gives n e nm = n But ln is the inverse of e, so n e mn = n nm = nm. Hence ln ln Therefore m = = = minutes. n ln Hence m = 9 minutes, 58 seconds to the nearest second. 6

18 5.5 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet The eample illustrates how a general method for solving eponential equations works. This process can be made quicker b using the results developed below. Consider the function a, where a >. As a is a number greater than zero, there is a real number, n, such that e n = a (see figure below. This means that n = ln a, since ln is the inverse function for the eponential function e. = e= e a So a n = ( e ) replacing a b e n. n That is, a equation But n n = ln a so = e using laws of indices, and taking logarithms of both sides gives the n ln a = ln e = n n a = ln a This result is a great help in solving a wide variet of eponential equations. Worked Eample Solve = 5, giving our answer to d.p. Since = 5 ln( ) = ln 5 ( )ln= ln 5 ln ln= ln 5 ln ln 5 ln= ln ln 5= ln ( ln ln 5) = ln ln = = ln ln to d.p. 7

19 5.5 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Worked Eample A sample of wood has 4 C radioactivit of 6 becquerels per gram. New wood has 4 C radioactivit of 6.68 becquerels per gram of Carbon 4. The half life of 4 C is 57 ears;. Form a model based on the work in Section 5.4 for the 4 C radiation in wood, of the form R= ba t, where R is the radioactivit, b and a are constants, and t is the time in ears since the sample was formed. Use our equation to find to the nearest ear when R = 6 becquerels per gram of carbon. Assume a model of the form R = ba t. Since the half life of 4 C is 57 ears, we have at t =, R = ba = b at t = 57, R 57 ba = = R a 57 Hence a is given b 57 = a or 57 ln a = ln (. 5). For the wood test at time t after sample was formed Thus R = 668. and R( t) = 6 6 = 668. a t giving 6 = a t 668. tln a = ln So t = 6 ln 668. ln a = 6 ln ln ears ( ) So the sample was formed about 89 ears ago (to the nearest decade). 8

20 5.5 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Eercises. Solve = 5 to d.p.. Solve to d.p. =. Solve 4 = to s.f. 4. Solve = 5 to d.p. 5. Solve = 6 to s.f. 6. Solve = 4 to d.p. 7. Solve 5 to s.f. = + 8. Solve = 4 to d.p. 9. Solve 5 = e to d.p. + =. Solve 6 to d.p. 5.6 Properties of Logarithms As well as obeing the rule ln( a ) = ln a logarithms also obe, for an real numbers a, b, ln( ab) = ln a + ln b () and ( ) = () ln a ln a ln b b To prove the first result, (), note that a and b can be written in the form m a= e, b= e for some real numbers m and n. Then ln( ab) = ln( e e ) n = ln( e ) = m+ n Since ln is the inverse function of e, m n m+ n m e ln a= ln( e ) = m ln b= ln( e ) = n n 9

21 5.6 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet so that ln( ab) = ln a + ln b You will see how useful these results are in the following applications. Before the theor of gravitation was developed b Sir Isaac Newton, the best laws available to describe planetar motion were those formulated b Johann Kepler, a German astronomer. His laws were based on his own meticulous observations, and were used later as a 'benchmark test' for Newton s own theor. The net eample investigates Kepler's third law. Worked Eample This table shows how the average radius of a planet s orbit around the Sun, R, is related to the period of that orbit in ears, T. (The orbits are elliptical, not circular, so an average radius is used here). Onl the planets known to Kepler are included. Planet Radius, R (millions of km) Period, T (ears) Mercur Venus 8..6 Earth 49.6 Mars Jupiter Saturn Assuming that T and R are linked b a model of the form T = ar b, we can use the data to find estimates for the constants a and b, together with the properties of logarithms. Assuming a power law of the form T = ar b, taking logs of each side gives b ln T = ln( ar ) b = + ( ) ln a ln R ( using equation ()) = ln a+ bln R This equation resembles a straight line equation = m + c with replaced b ln T and b ln R. So a graph of ln T against ln R should give a straight line and the constants a and b can be estimated from the graph. The constant b will be the gradient of the line, and ln a will be the intercept on the vertical ais.

22 5.6 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Unfortunatel, for this data ln a is negative and difficult to estimate from our graph. Having estimated b (the gradient) ou can obtain an estimate of a b considering the point ln T =. ln T ln a gradient b Worked Eample For the data above, plot a graph of ln T against ln R, and use it to estimate the values of the constants a and b. ln R ln T ln R ln T (7.6,.8) (5., ) ln R Gradient = = =

23 5.6 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet To find an estimate of a, note that when ln T =, ln a + bln R = i.e. ln a 5. ln( 5. ) 4. Hence T = 4. R The note produced b a musical instrument is directl related to its frequenc (the number of times the air is caused to vibrate ever second). The higher the frequenc, the higher the note. In order to set the frets on a guitar in the correct place, the maker must know how the length of a string affects the frequenc of the note it produces. Worked Eample This relationship between length, l (cm), and frequenc, f (hz), can be found eperimentall. The table shows some data collected b eperiment for a particular tpe of string. Length l (cm) Frequenc f (hz) The relationship is assumed to be of the form f = al b where a and b are constant. Use logarithms to 'linearise' the relationship, as described previousl. Plot ln f on a vertical ais and ln l on the horizontal, and draw a line of best fit. Find the gradient and intercept with the vertical ais of this line, and so determine the values of a and b. If f then = al b for constants a and b, ln f = ln a + bln l To estimate values of a and b, we graph ln f against ln l. ln f ln l

24 5.6 ln f CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet ln l Gradient = b You can best find an estimate for a b substituting values, for eample, ln f = 6. when ln l = 9. This gives 6. = ln a i.e. ln a = 7. a 7. Hence ou can deduce that f l The frequencies produced are also affected b the tension in the string and so, even with frets correctl placed, the guitarist must still 'tune' the instrument b changing the tensions in the strings. 'Middle C' has a frequenc of 64 Hz. The final application in this section is based on the method used b forensic scientists to estimate the time of death of a bod. When a person dies, the bod's temperature begins to cool. The temperature of the bod at an time after death is governed b Newton's Law of Cooling, which applies to an cooling object: D= ae kt

25 5.6 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet D is the temperature difference between the cooling object and its surrounding, a and k are constants, and t is the time since the object started to cool. The values of a and k depend on the size, shape and composition of the object and the initial temperature difference. If D is plotted against the time, t, the graph will be similar to the curve shown opposite. D To find out the equation of the curve which applies to a dead bod, the values of a and k must be found. This will require two readings of its temperature. Worked Eample 4 The police arrive at the scene of a murder at 8 a.m. a t On arrival, the temperature of the bod and its surroundings are measured at 4 C and 7 C respectivel. This was taken to be the moment when the time, t, was equal to zero. At 9 a.m. when t =, the bod temperature was measured as C and the room temperature still as 7 C. Estimate the time of death. The two sets of data are D= 4 7 = 7 at t = D= 7 = 6 at t = Substituting in the governing equation gives D= ae kt k. 7 = ae = ae = a (since e = ); and Therefore and taking 'logs', k. k k 6 = ae = ae = 7e e k = 6 7 k = ln 6 = k =. 66 Hence D = t 7 66 e. () 4

26 5.6 CMM Subject Support Strand: GRAPHS Unit 5 Growth and Deca: Tet Now normal bod temperature is given b 6. 9 C, so the corresponding value of D is given b D = = 9. 9 Substituting this value of D into equation () and solving for t will give ou the estimated time of death; this gives = e 66 t. 66t 9. 9 e = 7 = t ln 7 t = 9. 9 ln =. 599 hours ( hours 6 minutes). So the estimated time of death is estimated at 5.4 am, or about 5. am. Eercises. A bod is found at. pm. The bod temperature at midnight is found to be C and at. am it is. 5 C. Assuming the surroundings are at a constant temperature of C, estimate the time of death.. A phsicist conducts an eperiment to discover the half life of an element. The radioactivit at one moment from a sample of the element is measured as becquerels. One hour later the radioactivit is just 8 becquerels. Assuming that the radioactivit is governed b a formula of the form R = a kt where R is the radioactivit in becquerels per gram, t the time in hours, and a and k are constants, find the values of a and k, and hence determine the half life in hours. 5