Math 155: Calculus for the Biological Sciences

Transcription

1 Math 155: Calculus for the Biological Sciences Eric Hanson August 19, 2010 Abstract Lecture notes on Calculus for the Biological Sciences based on Modeling the Dynamics of Life the Second Edition by Frederick R. Adler Contents 1 Introduction to Discrete-Time Dynamical Systems Biology and Dynamics Models of Malaria Dynamical Systems Variables, Parameters, and Functions in Biology Variables and Parameters Functions Combining Functions: Sum, Product, and Composition Inverse Functions The Units and Dimensions of Measurements and Functions Converting Units Dimensions vs. Units Linear Functions and Their Graphs Proportional Relations Slope Equations of Lines Discrete-Time Dynamical Systems (DTDS) DTDS and Updating Functions A Word On Updating Functions DTDS and Units/Dimensioins Equations for Solutions to DTDS DTDS Solutions and Exponential Functions (Adler 1.7) Generalizing the Bacterial Population Growth DTDS

2 Bibliography 22 2

3 Day 1 Preliminaries: Go over the course information sheet. Note the exam dates and times (Thursday evenings) and state that no makeup exams are given, with a few exceptions if an alternate exam time request form is submitted at least a week in advance. Tutoring and office hours are held in the Great Hall of the TILT building. Students are encouraged to meet there to work on practice problems. In this chapter, we review linear, trigonometric power, logarithmic, and exponential functions, so that we can study functions and modeling. Functions and modeling are the main tools needed to use mathematics to study biology. Modeling is the art of taking a description of a biological phenomenon and converting it into mathematical form. Our goal is to quantify the dynamics (change) of living things with functions. Note: Please read Sections 1.1 to 1.4 in Adler. Each section should be entirely review of topics from the prerequisites for this course. We will cover them quickly, so have them read by tomorrow and be ready to ask any questions that you may have. 1.1 Biology and Dynamics 1 Introduction to Discrete-Time Dynamical Systems Note: Please read Section 1.1 there may be a question on the quiz related to the material. Living systems change; they are dynamic. What is life, anyway? One thing that it involves is change. Taking a dynamical approach to studying biology is necessarily mathematical, because describing dynamics requires quantifying measurements. Answering questions such as: What is changing? How fast is it changing? What is it changing into? To answer these questions we will develop the tools of mathematics and use the language of mathematics. Step Quantify the measurements Describe a dynamical rule Develop a model Find a solution Write a Simulation Definition The numerical values that describe the system A description of how the basic measurements change A mathematical translation of the observations Use of mathematical methods to predict behavior Use a computer to predict behavior 3

4 1.1.1 Models of Malaria Early in the 20th century, it was discovered that malaria is transmitted by mosquitoes. (It is only female mosquitoes, by the way, that transmit the disease.) Question 1. Can the spread of disease be controlled by killing mosquitoes, even if not every mosquito is killed? Some argued that it couldn t. (You may have them discuss if it could or not and maybe see why it s not so obvious.) Sir Ronald Ross believed that malaria could be controlled in this way. The following poem Sir Ronald Ross wrote as he was researching malaria: In this, O Nature, yield to me. I pace and pace, and think and think, and take with fever d hands, and note down all I see. That some distant light may haply break. The painful faces ask, can we not cure? We answer, No, not yet; we seek the laws. O reveal through all this thy obscure The unseen, small, million-murdering cause. Sir Ross built his mathematical model based on the following assumptions: 1. An infected person becomes infected when bitten by an infected mosquito and 2. an uninfected mosquito becomes infected after biting an infected person Diagram 1. The dynamics of malaria Based on this model Sir Ronald Ross, thought about how population sizes change with time, and showed that the disease could be eradicated even without killing every mosquito. Evidence of this is seen in the United States, where malaria has been virtually eliminated even thought the mosquitoes capable of transmitting the disease persist in many areas. 4

5 1.1.2 Dynamical Systems Dynamical systems encode how things change. 1. Discrete-time dynamical systems arise if measurements are made at equally spaced intervals (such as every second, every minute, every year...). Examples: The population (of, for example, bacteria or fish or tigers,...) in a given year (or second,...) as a function of the population in the previous year (or second,...) The number of mutant alleles present in one generation as a function of the number of mutant alleles present in the previous generation 2. Continuous-time dynamical systems arise if measurements are made continuously Example: If you walk from one end of the board to the other, your position p(t) depends on time t, where t is now a continuous variable, any positive real number. The position function can be determined from i) your initial position p(0) and ii) your speed, your rate of change. Your speed may be positive or negative, constant or variable. Knowing your initial position and speed, you can sketch your position as a function of continuous time. 5

6 1.2 Variables, Parameters, and Functions in Biology Variables and Parameters Definition 1. A variable is a symbol that represents a measurement (numerical quantity) that can change during the course of an experiment Example 1. Bacterial population growth Time(t) Bacteria(b) Definition 2. A parameter is a symbol that represents a measurement (numerical quantity) that does not change during an experiment. Example 2. Variables( b) and parameters( T ) t b when T = 5 b when T = Recall the Cartesian Coordinate plane and its parts: origin, axes, and labels Functions Functions describe relations between objects. Definition 3. A function is a mathematical object that takes an input, performs an operation on it, and returns a unique new object as an output. The input is called the argument or independent variable and the out put is called the value or dependent variable. The sets of objects that a function relates are defined as follows: Definition 4. The set of all possible inputs a function accepts is called the domain. The set of all possible things a function can return as as an output is called the codomain, and the set of all things that the function does return as outputs is called the range. Note: The subtle differences between the definitions. 6

7 Diagram 2. Venn diagram for functions Example 3. A simple function for population growth over 3 weeks. t Population (P ) Formula for the function: P (t) = 10 x 2. The inputs are in weeks, in this case only positive values of time make sense, so the domain is t 0. The range is P 10, but the codomain is all real numbers (R). Graph of the function (relation): Recall The vertical line test is a graphical method for recognizing relations that are not functions. If some vertical line crosses the graph two or more time, the relation fails the vertical line test and is not a function. Example 4. A relation that fails the vertical line test: 7

8 1.2.3 Combining Functions: Sum, Product, and Composition Definition 5. The sum of f + g of the functions f, g is the function defined by (f + g)(x) = f(x) + g(x) Example 5. Let f(x) = 2x and g(x) = 3x 2 + x then Evaluating for x = 2 we have (f + g)(x) = f(x) + g(x) = 2x + 3x 2 + x = 3x 2 + 3x (f + g)(2) = 3(2) 2 + 3(2) = 3(4) + 6 = 18. Note that f(2) = 2(2) = 4 and g(2) = 3(2) 2 + (2) = = 14. Consider that gives us the same answer. f(2) + g(2) = = 18 Remark 1. Graphically the sum of two function is a vertical sum across its inputs. Definition 6. The product f g of the functions f, g is the function defined by (f g)(x) = f(x) g(x) Example 6. Let f(x) = 2x and g(x) = 3x 2 + x then (f g)(x) = f(x) g(x) = 2x(3x 2 + x) = 6x 3 + 2x 2 Evaluation for x = 2 we have (f g)(2) = 6(2) 3 + 2(2) 2 = = 56 8

9 Note that f(2) = 2(2) = 4 and g(2) = 3(2) 2 + (2) = = 14. Consider that gives us the same answer. f(2) g(2) = 4 14 = 56 Remark 2. Graphically the product of two function is a vertical product across its inputs. Definition 7. The composition f g of function f, g is the function defined by (f g)(x) = f(g(x)) Example 7. Let f(x) = 2x and g(x) = 3x 2 + x then Evaluation for x = 2 we have (f g)(x) = f(g(x)) = f(3x 2 + x) = 2(3x 2 + x) = 6x 2 + 2x (f g)(x) = 6(2) 2 + 2(2) = 6(4) + 4 = 28 Note that g(2) = 3(2) 2 + (2) = = 14 and then we have which is the same answer as above. f(g(2)) = f(14) = 2(14) = 28 Remark 3. Composition, f g, is the same as evaluating g at x and then evaluating f at the output. In this situation f is called the outer function and g is called the inner function. 9

10 Diagram 3. Venn diagram for composition of functions Note that the domain and range of the functions in function composition must match up and that in general f g is different form g f. Example 8. Composition of functions in Biology Let P (T ) 1 10 T be the population of squirrels per square mile in a forest in terms of the number of trees per square mile, T, and assume the population of trees per square mile is related to the age of the forest by T (t) = 10, 000 2t 2 + t where t is years beginning in According to the model, what is the population of squirrels this year. Solution: P (t) = P (T (t)) = 1 10 (10, 000 2t2 + t) = 1, t t 2010 corresponds to t = 20, so the number of squirrels is Inverse Functions P (20) = 1, (20) (20) = 1, (400) + 2 = 1, = 982 Inverse functions reverse the process that functions apply to inputs. That is given an output to a particular function, the inverse of the function determines the input that produced the output. Diagram 4. Venn diagram for inverse functions 10

11 Definition 8. The function f 1 is the inverse of f if f(f 1 (x)) = x and f 1 (f(x)) = x Algorithm 1. Finding the Inverse of a Function 1. Write the equation y = f(x) 2. Solve for x in terms on y 3. The inverse function is the operation done to y Remark 4. In applied mathematics we do not switch the x and y. Even though this looks weird, in applied mathematics variables have meaning, that is they represent specific objects. Thus switching names can cause confusion. Example 9. Let y = 5 6 x then the inverse is x = 6 5 y Some functions do not have an inverse, consider the following: Example 10. What is the inverse parabola, see graph below: Any function that has the same output for more than one input does not have an inverse. The horizontal line test is a graphical method for determining if a function has an inverse. If the graph of a function intersects any horizontal line in two or more points then the function does not have an inverse. However if a function does not have an inverse, it might have an inverse on part of it s domain. 11

12 Example 11. Suppose a population of bacteria is modeled by b(t) = t using Algorithm 1 we find, b = t b 55 = t (b 55) = t 2 10(b 55) = t t = ± 10(b 55) If we restrict the domain to positive values of t, then b(t) passes the horizontal line test and t = b 1 (t) = 10(b 55) Remark 5. The function f(x) = x 5 +x+1 passes the horizontal line test but we cannot solve for the inverse algebraically. In fact French mathematician Evariste Galois proved that there is no formula for the solution of a general polynomial with degree greater than 4. (Galois was only 20 years old). However, in mathematical modeling just knowing the existence of an inverse is important, and the the computation of an inverse can then be done numerically. 1.3 The Units and Dimensions of Measurements and Functions Question 2. Why are units and dimensions important? Example 12. Consider = 2. Is this wrong? Not if we have the proper labels for the units, 2Na + + 2Cl = 2NaCl. This is now a standard chemistry formula Converting Units Example 13. Converting miles to centimeters. We use the basic identities: these imply 5280ft = 1mile 12in = 1ft 2.54cm = 1in 12

13 so we have 5280 ft mile = 1 12 in ft = cm in = 1 1mile = 1mile = 1mile 5280 ft mile 12in ft 2.54cm in = 160, 934.4cm Example 14. Convert 1 C (cup) to ml given that 1 pint = liters and that 1 pint = 2 C. solution: 1 C = ml Note: There is an algorithm for the conversion procedure on page 26 of Adler Dimensions vs. Units Definition 9. Dimensions describe underlying quantities that can be measured in different units. Units are a particular standard of measure that describe dimensions. Quantity Dimensions Sample Units length length meter, micron, inch duration time second, minute, day mass mass gram, kilogram area length 2 square meter, acre volume length 3 liter, cubic meter, gallon speed length/time meters/second, mph acceleration length/time 2 meters/second 2 force mass length/time 2 dynes, pounds density mass/length 3 grams/liter Fundamental relations are formulas for translation between dimensions, some are stated below: 13

14 Volume of a Sphere Surface Area of a Sphere V = 4π 3 r3 S = 4πr 2 Area of a Circle A = πr 2 Perimeter of a circle P = 2πr Volume, Area, and Thickness V = AT Total Number and Mass m = µb Mass, Density, and Volume M = ρv Note: Please review scaling and shifting of functions, pages There may be a quiz question on this topic. 1.4 Linear Functions and Their Graphs Proportional Relations Definition 10. A proportional relation is a relation where the output is proportional to the output, that is the ratio of the output to the input is a constant. The general formula for a proportional relations is: f(x) = ax where a is some constant value, called the constant of proportionality. Clearly, for x 0. a = ax x = output input Example 15. b out = 3b in clearly a = 3. Suppose b in = 10 then b out = 30, now consider b out b in = = 3 = a It is easy to see that the constant of proportionality is the slope of the graph of a proportional relation. Now we recall how to find the slope of general linear relation Slope Definition 11. Most generally slope is output input. The slope of a line is the change between two data points. Suppose we call the points (x 1, y 1 ) and (x 2, y 2 ) then the slope, m, is given by:. m = output input = y 2 y 1 x 2 x 1 14

15 Example 16. Find the slope between (1, 14) and (20, 25). Then m = y 2 y 1 x 2 x 1 = = 24 6 = 4 Note: Notice the relationship between slope of linear relations and slope of proportional relations (i.e. constant of proportionality). Example 17. f(x) = x + 4 is not a proportional relation. Can you see why? Equations of Lines Definition 12. A line passing through the point (x 0, y 0 ) with slope m has formula: 1. Point-Slope Form of a Line 2. Slope-Intercept Form for a Line where b is the y-intercept. y = m(x x 0 ) + y 0 y = mx + b Example 18. Finding line equation examples: 1. m = 2 and (0, 4), then we have line y = 2x + 4. Why? 2. (0, 4) and (4, 2), then we have m = = 2 4 = 1 2 y = 1 2 x (2, 4) and (4, 2), then we have m = 4 2 y = 1(x 2) = 2 2 Example 19. Does the following data line on a line. time Mites on a lizard y and the line equation is = 1 and the line equation is

16 In order for the data to lie on a line the slope must be the same between all data points (each row is a data point). Clearly the data does on line on a line since the slope between t = 0 and t = 1 (m = 5) is different from the slope between t = 4 and t = 5 (m = 10). Now suppose we have the following data, time Mites on a lizard, y Clearly this data lies on a line with equation y = 5t We can then interpolate, that is estimate for untested inputs. Suppose we want to estimate for t = 1.5 and t = 5, then we find there are an estimated 17.5 and 35 mites respectively. Recall that when solving for the intersection of two lines, parallel lines do not intersect and identical lines have infinitely many solutions. Read pages for examples of word problems solved using linear equations. 1.5 Discrete-Time Dynamical Systems (DTDS) In a Discrete-Time Dynamical Systems (DTDS) a function is applied to an input to obtain an output. In biology we often use DTDS s to examine populations, so with a population as an input we can obtain an output that is the new population say a year later. DTDS provide a way to predict what will happen to a population in the long run. However, if the DTDS has an output that is the population value one year later, that same system cannot tell us what the value is half a year later. This is the concept of discrete vs. continuous time. Question 3. What is the difference between discrete and continuous time? Later we will learn a way to solve DTDS so that we can obtain solutions that are not restricted to the discrete time step. Diagram 5. Venn diagram for a DTDS. 16

17 1.5.1 DTDS and Updating Functions We can think of a DTDS as describing the relation between a quantity measured at the beginning and the end of an experiment or a time interval. We now more formally define a DTDS: Let m be a measurement, then we denote m t as the measurement at the beginning of a time interval. We think of t as the current time and then t + 1 is the time one step into the future. Then we can describe the relation between the initial measurement m t and the next measurement, m t+1, by a DTDS: Definition 13. A DTDS takes a measurement m t and applies an updating function f to determine the value of the measurement one time step later, m t+1. In mathematical notation we have m t+1 = f(m t ) Example 20. Suppose a population of bacteria triples every month and that the population is initially 1000 bacteria. Clearly we know how to calculate the population one month later, 1000(3) = In the notation of discrete time dynamical systems we have, b 0 = 1000 as an initial condition and the population the next year is given by the relation b t+1 = 3b t. We can now iterate the DTDS to determine the population values for the next 5 months. Month Population 1 b 1 = 3b 0 = 3(1000) = b 2 = 3b 1 = 3(3000) = b 3 = 3b 2 = 3(9000) = b 4 = 3b 3 = 3(27000) = b 5 = 3b 4 = 3(81000) = Example 21. Suppose it is spring time for a male (bull) moose, so his antlers are beginning to grow. Moose antlers grow to full size in 3 to 5 months, making them the fastest growing animal organ. Suppose that Marvin the moose s antlers grow 6 inches a month. How big will they be in 5 months? (information obtained from Initial condition: a 0 = 0 DTDS: a t+1 = a t

18 Month Antler Length 1 a 1 = a = = 6 2 a 2 = a = = 12 3 a 3 = a = = 18 4 a 4 = a = = 24 5 a 5 = a = = 30 Thus Marvin s antlers grow to 30 inches. Example 22. Suppose we know that each day a patient uses up half of the medication in her bloodstream. However he is given a new dose sufficient to raise the concentration in the bloodstream by 1.0 milligrams per liter. Then we have the DTDS M t+1 = 0.5M t + 1. If the initial condition for the system is M 0 = 0 can you graph the outputs of the system. (Note the updating function is linear) A Word On Updating Functions Updating functions are FUNCTIONS, this means that any of the operations that we can perform on function we can also perform on updating functions. We can sum (subtract), multiply (divide**?), compose updating functions, and find inverses (if they exist). Example 23. Consider the composition of the bacteria growth updating function, P (b t ) = 3b t. (P P )(b t ) = P (P (b t )) = P (3b t ) = 3(3b t ) = 9b t Now consider the initial condition b 0 = 1000 under our new updating function. Then our next value is 9000 which is b 2. If we continue with more iterations it become clear that the composition of the updating function corresponds to a two-step updating function, that is we have b t+2 = 9b t. Question 4. Can you find the inverse of the updating function in Example 22? DTDS and Units/Dimensioins Please read pages on units and dimensions. 18

19 1.5.4 Equations for Solutions to DTDS Definition 14. The sequence of values m t for t = 1, 2, 3,... is called the solution to the DTDS m t+1 = f(m t ) starting with initial condition m 0 The graph of a solution is the discrete set of points (0, m 0 ), (1, m 1 ), (2, m 2 ),... (t, m t ) It is possible to find a formula for the solution to simple DTDS s, but not in many complicated cases. However, if we can find this formula it is very useful, since with it we can easily calculate values for large values of t. Also we can interpolate for all values of t not just for values in the discrete set of solutions (i.e. t = 1.5). Example 24. A Solution to the Bacteria DTDS Suppose again that we have DTDS b t+1 = 3b t with initial condition b 0 = Then we have the following solutions: b 1 = 3b 0 = 3(1000) b 2 = 3b 1 = 3(3b 0 ) = = b 3 = 3b 2 = 3(3b 1 ) = 3(3(3b 0 )) = b t = 3 t 1000 b t = 3 t b 0 is called the closed form solution to the DTDS. Using this form we can calculate the solution for any value of t. Example 25. Now consider b t+1 = 3b t with a different initial condition b 0 = 50. The we have the following solution: b 1 = 3b 0 = 3(50) b 2 = 3b 1 = 3(3b 0 ) = = b 3 = 3b 2 = 3(3b 1 ) = 3(3(3b 0 )) = b t = 3 t 50 Clearly this is a different solution than what we found above, so it is apparent the the solution to a DTDS is dependent on both the initial condition and the updating function. Note: That while the closed form solution provides a continuous set the solution set to the DTDS is still discrete and a graph of the solution should be a set of points. Unless we wish to connect the points for emphasis of a pattern. 19

20 Example 26. Can we find a closed form solution to the DTDS for the length of Marvin the Moose s antlers? Recall that Marvin s antlers growth follows the DTDS, a t+1 = a t + 6 with initial condition, a 0 = 0 and that a moose s antlers grow for at most 5 months. So we only have the values of a t for t = 0, 1, 2, 3, 4, 5, thus the domain of the closed form solution is the set, dom(f) = {t 0 t 5}. Then the solution to the DTDS is: a 1 = a = = 6 a 2 = a = (0 + 6) + 6 = a 3 = a = ((0 + 6) + 6) + 6 = a t = 0 + 6t What would the solution be if a 0 = 1? Example 27. Let s find a solution to the the Medication DTDS. Recall m t+1 = 0.5m t +1 and let the initial condition for the system be M 0 = 5.0 The we have solutions: m 1 = 0.5(5) + 1 = 3.5 m 2 = 0.5(3.25) + 1 = 2.75 m 3 = 0.5(2.75) + 1 = m 4 = 0.5(2.375) + 1 = The solution values appear to be approaching 2. In fact each step the solutions move halfway closer to 2. Consider the following: m 0 2 = 5 2 = 3 m 1 2 = = 1.25 = m 2 2 = = 0.75 = m 3 2 = = = m 4 2 = = =

21 These observations can lead us to a solution: m 0 = m 1 = m 2 = m 3 = m 4 = m t = t 3 Question 5. Consider finding a closed form for the DTDS in the Example 27, using initial condition m 0 = 1 (the solution is on page 63). Remark 6. In general finding closed form solutions to a DTDS is not easy. Consider x t+1 = 2x t +30 with initial condition x 0 = 10 and you will find that a pattern is not very obvious in this case. In fact given a random DTDS it is likely a closed form solution is very hard to find. 1.6 DTDS Solutions and Exponential Functions (Adler 1.7) Recall the bacterial population growth DTDS, b t+1 = 3b t with initial condition b 0 = 1 has closed form solution b t = 3 t This is an example of an exponential function. Suppose we want to solve for b t = 300 then we must be able to solve 300 = 3 t Question 6. How do we do this? Answer: We use the natural logarithm Generalizing the Bacterial Population Growth DTDS Thus far we have assumed that all bacteria in our model live through each cycle. Clearly this is a big assumption, so lets now suppose that only a fraction of the new bacteria survive. We will call the fraction σ. Let ω be the number of offspring per bacteria. Then in the DTDS b t+1 = 3b t, ω = 3. Let r be the per capita production, that is the number of surviving offspring per 21

22 Bibliography References [1] F.R. Adler, Modeling the Dynamics of Life: Calculus and Probability for Life Sciences Second Edition, Thomson Brooks/Cole, Belmont CA, (2005) 22