Introduction Growthequations Decay equations Forming differential equations Case studies Shifted equations Test INU0115/515 (MATHS 2)

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1 GROWTH AND DECAY CALCULUS 12 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Growth and Decay 1/ 24 Adrian Jannetta

2 Introduction Some of the simplest systems that can be modelled by differential equations are those involving growth and decay. In this presentation we ll revise some prior mathematics before studying differential equations in this specific context. Solving exponential equations. Exponential growth and decay Graphs of exponential functions Constant of proportionality Forming and solving differential equations Case studies: Population growth Radioactive decay Shifted growth and decay Growth and Decay 2/ 24 Adrian Jannetta

3 Recap: exponential equations An exponential equation has the form where a>0 and b are real numbers. The goal is to solve to find x. a x = b (1) There is also a relationship between indices and logarithms that is worth restating here: a b = c log a c= b (2) The usual approach to solving an exponential equation is to use logarithms. Growth and Decay 3/ 24 Adrian Jannetta

4 Solve 8 x = 60. Take logs of both sides: Move the power: Make x the subject: log8 x = log60 x log8=log60 x= log60 log8 We can evaluate with a calculator to get x= (to 3 D.P.). In that example, it didn t matter which base of logarithm was used. Also, we could have presented the answer as x= log 8 60 using the relationship given in equation 2. In other cases we ll find it useful to take logarithms in a particular base: usually base e. Growth and Decay 4/ 24 Adrian Jannetta

5 Solve e 2x = 3. This time we ll take logarithms in base e. lne 2x 2x ln e = ln3 = ln3 But lne=log e e=1 (by the definition given by equation 2) so that 2x=ln3 x= 1 2 ln3 We can evaluate, if needed, to get x= (to 3 D.P.). Growth and Decay 5/ 24 Adrian Jannetta

6 Rate of change Given an exponential equation in base e Differentiate: y= Ae kx dy dx = kaekx = ky This simple relationship between the function and its derivative shows us that, in general: y= Ae kx dy dx = ky We can verify the converse is true. Given dy = ky, separation of variables dx and integration will give the solution y= Ae kx. Growth and Decay 6/ 24 Adrian Jannetta

7 Growth equations Exponential growth equations take the form y y= a b x where a and b>0 are constants. For example y= 10 2 x is an equation in which the value of y grows to large values before x does. The graph of y= a b x is shown here (for x 0). a 0 Note the following: When x=0 then y= a. As x then y. x Growth and Decay 7/ 24 Adrian Jannetta

8 Exponential growth Solutions to differential equations of growth usually have the independant variable as time t and the base invariably turns out to be base e. N The solutions often have the form N= N 0 e kt where N is the dependant variable and k is a constant which controls how quickly the growth happens in time t. The graph is still exponential growth like the previous one! N 0 0 At t= 0 we have N= N 0. Therefore N 0 represents the initial value of N. t Growth and Decay 8/ 24 Adrian Jannetta

9 Decay equations Exponential growth equations take the form y= a b x where a and b>0 are constants. a y For example y= 10 2 x is an equation in which the value of y grows to large values before x does. The graph of y= a b x is shown here (for x 0). 0 Note the following: When x=0 then y= a. As x then y 0. x Growth and Decay 9/ 24 Adrian Jannetta

10 Exponential decay Solutions to differential equations of decay turn out to have the form N N= N 0 e kt N 0 where N is the dependant variable and k is a constant which controls how quickly the decay happens in time t. The graph is still exponential decay like the previous one. 0 At t= 0 we have N= N 0. Therefore N 0 represents the initial value of N. t Growth and Decay 10/ 24 Adrian Jannetta

11 Constant of proportionality If a is proportional to b the relationship is written a b This means that a increases as b increases. Furthermore, it means there is a constant k such that a=kb This relationship is linear; a graph of a against b would show a straight line of gradient k. If a is inversely proportional to b the relationship is written a 1 b This means that a increases as b decreases (or vice-versa). In this case it means there is a constant k such that a= k b The constant k is called a constant of proportionality. Growth and Decay 11/ 24 Adrian Jannetta

12 Forming a differential equation Given a description of a problem you should be able to form a differential equation to describe it. Write down differential equations describing the following situations. 1 At time t, the rate of increase of a population with P people is proportional to the number present at that time. 2 The densityρ of the atmosphere decreases with height h. The rate of change is proportional to height. 3 The rate of increase of ice thickness x on a lake at time t is inversely proportional to the thickness of ice already present. 4 The gradient of a curve is inversely proportional to the product of its coordinates. 5 Newton s law of heating or cooling states that an object at temperature T will change at a rate proportional to the difference between object temperature and the temperature of the surroundings T dp dt = kp dρ dh = kh 3 4 dx dt = k x dy dx = k xy 5 dt dt = k(t 0 T) Growth and Decay 12/ 24 Adrian Jannetta

13 Exponential growth A scientist initially counts 160 bacteria in a sample of water. Assuming the number of bacteria N increases at a rate proportional to the number present, write down a differential equation connecting N and the time t. If the rate of increase of the number is initially 80 per hour, how many are there after 3 hours? The differential equation for this dn dt N Rewriting with a constant of proportionality: dn dt = kn We are also given initial conditions N= 160 and dn = 80 when t= 0. dt Use them to find k: 80=160k k= 1 2 The differential equation becomes: dn dt = 1 2 N Solve by separating the variables: And integrate: 1 N dn = lnn= 1 2 t+ C 1 2 dt Growth and Decay 13/ 24 Adrian Jannetta

14 Exponentiate both sides: N= e 2 1 t+c = Ae 2 1 t The constant A can be found using N= 160 when t= =Ae =A The particular solution is therefore N= 160e 1 2 t. This solution can be used to find the number of bacteria at any given time t. When t= 3 hours: N= 160e So there are about 717 bacteria present after 3 hours. Growth and Decay 14/ 24 Adrian Jannetta

15 Population growth Case Study: Population growth In a simple model of population growth the number of members P increases at a rate proportional to the number present at time t. dp dt P dp dt = kp If there are initially P 0 members then: dp dt = kp, P(0)=P 0 Separate the variables: 1 P dp= k dt Integrate: ln P= kt+ C Substitute P= P 0 and t= 0: ln P 0 = C Put this back into the equation: ln P=kt+ ln P 0 Next, we ll combine the logarithm terms together: ln P ln P 0 P ln P 0 = kt = kt Remove the logarithm by exponentiating both sides: P P 0 = e kt The population function P(t) is therefore given by P= P 0 e kt To find the value of k we would need a further piece of information; another measurement of P at another time, or the value of P at some time. Growth and Decay 15/ 24 Adrian Jannetta

16 Radioactive decay Case Study: Radioactive decay Radioactive decay is a process in which an unstable atom loses energy by emitting particles. This is a depiction of alpha decay - one of several types of radioactivity. In doing so the original atom is transformed into an atom of another element. For example, a certain kind of Uranium is transformed into Lead by this process. It is impossible to predict when a particular atom will decay. However, if large numbers of atoms are present then the rate at which decay occurs is predictable. The physical law which governs radioactivity states that if there are N radioactive atoms present in a sample at time t then the rate of decay is directly proportional to N. Growth and Decay 16/ 24 Adrian Jannetta

17 Radioactive decay Formulating the decay equation Given that there are N radioactive atoms present in a sample at time t then the rate of decay is directly proportional to N. The rate of change is negative because N is decreasing as t increases. Therefore the differential equation governing radioactivity is: dn dt = kn (where k is a positive constant of proportionality whose value depends on the type of atom). Suppose we know that there are N 0 atoms present at t= 0. This is an initial condition (a type of boundary condition) which will provide information about the constant of integration. If we want to make predictions about the value of N at some future (or past) time t then we must solve the differential equation. Therefore the initial value problem to be solved is dn dt = kn, N(0)=N 0 Growth and Decay 17/ 24 Adrian Jannetta

18 Radioactive decay Solving the decay equation Solution is by separating the variables and integrating: 1 N dn = k dt lnn = kt+ C Using the rules of logs and indices we can express this more compactly: N= e kt+c = Ae kt Where we have changed the form of the constant of integration (A=e C ). Apply the initial condition N= N 0 at t= 0: N 0 = Ae 0 A=N 0 The solution to the differential equation is given by: N= N 0 e kt Growth and Decay 18/ 24 Adrian Jannetta

19 Radioactive decay Half-life of a radioactive substance It is useful to know the time taken for half of the atoms present to decay. We can find this very easily from the solution. N= N 0 e kt We are trying to find the time t for which N= 1 2 N 0: 1 2 N 0 = N 0 e kt 1 2 = e kt Take natural logs of both sides and simplify: ln 1 2 = kt t= ln 1 2 k This special value of t, called the half-life, is represented by the Greek letterτ( tau ) and it can be simplified further: τ=(ln 2)/k Uses of the Half-life Potassium-40 has a half-life of 1.25 billion years and it has been used to date the age of the Earth (about 4.5 billion years old). Iodine-128 has a half life of 25 minutes and is used to measure the performance of the thyroid gland in the human body. Growth and Decay 19/ 24 Adrian Jannetta

20 Radioactive decay N N N N N 0 τ 2τ 3τ t The number of atoms in a substance (and therefore the total mass) undergoing radioactive decay will fall to half of its initial value after a period ofτhas passed. After 2τ it will have halved again so that it is one-quarter of the initial value. After 3τ has passed the number of atoms will be just one-eighth of the initial value. Growth and Decay 20/ 24 Adrian Jannetta

21 Shifted growth and decay This type of situation occurs when the growth (or decay) function is modified to include some other growth or decay factor. dy dx = ky+c This equation describes many different situations. Some examples include: A growing population including a constant immigration or emigration. Newton s law of heating/cooling. Compound interest on a bank account with deposits or withdrawals. Mixing problems (rate of change depends on input and output rates). Growth and Decay 21/ 24 Adrian Jannetta

22 Assume a city population is growing naturally (exponentially) but 1,000 people per year are leaving the city. Given that the natural growth constant is k = 0.01 and that there are currently 200,000 people in the city, calculate the population in 10 years and 20 years from the present. The differential equation for this is: dp = 0.01P 1000 dt Separate the variables and integrate: P 1000 dt = Given t= 0 (the present) and P= dt 100ln 0.01P 1000 = t+ C Substitute the values to get: The particular solution is. C= 100ln 1000 Next we ll simplify and rearrange to find P. 100ln 0.01P 1000 =t+ 100ln 1000 Growth and Decay 22/ 24 Adrian Jannetta

23 Bring the log terms to the LHS 100ln 0.01P ln 1000 = t 100(ln 0.01P 1000 ln 100) = t Use log rules to simplify: 0.01P ln = t P ln = t 1000 Make P the subject: 0.01P = e t P 1000 = e 0.01t 1000 P = 100(1000e 0.01t ) Factorising gives: P= (e 0.01t + 1) A quick check: t= 0 gives P= (1+1)= as expected! After t= 10 years we will find P= After t= 20 years we calculate that P= Growth and Decay 23/ 24 Adrian Jannetta

24 Test yourself... Try to answer the following questions... 1 Given P= 2000e 2t where t is time; is this a growth or decay equation? 2 Solve dq dt = kq given that Q=Q 0 at t= 0. 3 In (2) given that Q=3 at t= 1 and Q=0.25 at t= 4, find the values of Q 0 and k. 4 The rate of decrease of gravitational force F with distance r from a mass is inversely proportional the cube of the distance. Write down a differential equation for this situation. 5 Solve dx = 0.1x+2 given that x(0)=50. What happens to x as t? dt Answers: 1 Growth; because of the positive exponent. 2 Q=Q 0 e kt 4 df dr = k r 3 3 k= 1 3 ln 12 ; Q 0= x= 70e t/ As t, x 20. Growth and Decay 24/ 24 Adrian Jannetta