Mid Biomathematics PDF

# Biomathematics Biomathematics is the application of mathematical, statistical, and computational methods in biology. Traditionally this was mainly in developing mathematical models of biological systems, from the molecular and cellular levels through to whole organism models, and population and e...

# Biomathematics Biomathematics is the application of mathematical, statistical, and computational methods in biology. Traditionally this was mainly in developing mathematical models of biological systems, from the molecular and cellular levels through to whole organism models, and population and ecological studies. A mathematical model is a description of the behavior of a system using mathematical language. ## Objectives of this course It is an introduction to the application of mathematical models in the study of biological systems. The main objectives of this course is to introduce 1. Some basic techniques for developing mathematical models of biological processes. 2. Some basic mathematical knowledge and skills required in the analysis of the models. Many examples of the models introduced in this course are taken from population biology. Why should you also study mathematics and statistics if you are interested in biology or medicine? Mathematics as a language is often far more suitable for precise description of the characteristics of various biological processes. # Indices and Logarithms ## Indices Indices or powers show how many times a number is multiplied by itself. | Index (or Power) | Base | Basic Numeral (or Number) | |---|---|---| | 2³ = 8 | 2 | 8 | ### Properties of Indices 1. a * a = a² 2. a^m/a^n= a^(m-n) 3. (a^m)^n = a^(m*n) 4. (ab)^m= a^m*b^m 5. (a/b)^m = a^m/b^m 6. a^0 = 1 7. a^-m = 1/a^m 8. a^(1/n) = n√a ## Exponential Functions Are functions where *f(x) = a^x + B* where a is any real constant and B is any expression. For example, *f(x) = e^-x - 1* is an exponential function. The exponential function *exp(t)* is given by *exp(t) = e^t*, where *e = 2.71828...* ## Logarithmic Functions Are the inverse of exponential functions. For example, the inverse of *y = a^x* is *y=log_a(x)*, which is the same as *x = a^y*. (Logarithms written without a base are understood to be base 10.) Ex. Convert to logarithmic form: 8 = 2^x Solution: Remember that the logarithm is the exponent. x = log_2(8) # Logarithms and Solving Equations The logarithm of x to the base a is written "log_a(x)" and is defined as follows. *log_a(x) = b if and only if a^b = x* where a > 0 (a ≠ 1), and x > 0. Note that the base must be positive and different from 1, and the expression that you are taking the logarithm of must also be positive. Some logarithms can be evaluated easily. For example, log_4(64) = 3 since 4³ = 64. ## Special Bases If the log base is 10, then the log is called the common logarithm and we write "log" for log_10. If the log base is the number "e", then the logarithm is called the natural logarithm and we write "ln" for log_e . The "ln" key on a scientific calculator gives values for the natural logarithm. | log form | exponential form | |---|---| | y = ln(x) | e^y = x | | y = log(x) | a^y = x | ## Properties of Logarithms Here, a > 0 and a ≠ 1. 1. log_a(1) = 0 2. log_a(a) = 1 3. log_a(a^x) = x 4. a^(log_a(x)) = x, if x > 0 5. log_a(x) + log_a(y) = log_a(xy) if x, y > 0 6. log_a(x)-log_a(y) = log_a(x/y) if x, y > 0 7. log_a(x^r) = r log_a(x) if x > 0 **Example 1:** Find parameters a and b so that f(0) = 0 and f(1) = 2, where f is a logarithmic function given by f(x) = a*Log_2(x + b) **Solution to Example 1:** Use the fact that f(0) = 0 to obtain 0 = a * Log_2( 0 + b ) Divide both sides by a to obtain Log_2(b) = 0 Solve for b b = 2^0 b = 1 Function f can be written as a*Log_2(x + 1) Use the fact that f(1) = 2 to obtain 2 = a * Log_2(1 + 1) Simplify, Log_2(2) = 1 a = 2 Function f is given by f(x) = 2*Log_2(x + 1) # Problem: Solve for x: 3^x = 8 **Solution:** Take the logarithm of both sides. log(3^x) = log(8) Use theorem 7 to simplify the equation. x* log(3) = log(8) Solve for x by dividing each side by log(3). x = log(8)/log(3) A decimal approximation may be found if desired - x=1.8929. # Problem: Solve log_a(5x + 7) = 2 for x. **Solution:** Write an equivalent exponential expression. 5x + 7 = a^2 5x + 7 = 9 Solve for x. 5x = 2 x = 2/5 # Problem: Solve log_a(x) = -3. **Solution:** Convert the logarithm to exponential form a^-3 = x x = 1/8 # Example: Apply the laws of logarithms to log(abc/d^3) **Answer:** According to the 5,6 laws, log(abc/d^3) = log(abc) - log(d^3) = log(a) + log(b) + log(c) - log(d^3) = log(a) + log(b) + 2log(c) - 3log(d) # Logarithms and Modeling Many phenomena in nature seem to follow the law that an amount A varies with time according to the formula A = A_0 * e^(kt) Where A_0 is the original amount (the amount at time t=0) and k is a non-zero constant. If k is positive, then the amount A gets larger or grows with time, and the amount is said to be experiencing exponential growth. If k is negative, then the amount A gets smaller or diminishes with time, and the amount is said to be experiencing exponential decay. When solving exponential growth/decay problems, two situations often occur: - Time is given; you are to find the amount at that given time. This usually just involves evaluating the amount function. - The amount is given; you are to determine at what time this amount occurs. This usually involves solving an exponential equation, which means logarithms will be needed. ## Modeling Exponential Decay - Using Logarithms A common example of exponential decay is radioactive decay. Radioactive materials, and some other substances, decompose according to a formula for exponential decay. That is, the amount of radioactive material A present at time t is given by the formula A = A_0 * e^(kt) where k < 0. A radioactive substance is often described in terms of its half-life, which is the time required for half the material to decompose. # Problem After 500 years, a sample of radium-226 has decayed to 80.4% of its original mass. Find the half-life of radium-226. **Solution** Let A = the mass of radium present at time t (t=0 corresponds to 500 years ago). We want to know for what time t is A = (1/2)A_0. However, we do not even know what k is yet. Once we know what k is, we can set A in the formula for exponential decay to be equal to (1/2)A_0, and then solve for t. First we must determine k. We are given that after 500 years, the amount present is 80.4% of its original mass. That is, when t=500, A= 0.804 A_0. Substituting these values into the formula for exponential decay, we obtain: 0.804 A_0 = A_0 * e^(k * 500) Dividing through by A_0 gives us 0.804 = e^(500k) Which is an exponential equation. To solve this equation, we take natural logs (ie. ln) of both sides. (Common logs could be used as well.) ln(0.804) = ln(e^(500k)) We know that ln(e^(500k)) = 500k by the cancellation properties of ln and e. So the equation becomes ln(0.804) = 500k and k = (ln(0.804))/500. This is the exact solution; evaluate the natural log with a calculator to get the decimal approximation k = -0.000436.