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# Mathematics Review

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### Mathematics Review

1. 1. Mathematics Review CHAPTER 2 Author: Michael Makoid, Phillip Vuchetich and John Cobby Reviewer: Phillip Vuchetich BASIC MATHEMATICAL SKILLS OBJECTIVES 1. Given a data set containing a pair of variables, the student will properly construct (III) various graphs of the data. 2. Given various graphical representations of data, the student will calculate (III) the slope and intercept by hand as well as using linear regression. 3. The student shall be able to interpret (V) the meaning of the slope and intercept for the various types of data sets. 4. The student shall demonstrate (III) the proper procedures of mathematical and algebraic manipulations. 5. The student shall demonstrate (III) the proper calculus procedures of integration and differentiation. 6. The student shall demonstrate (III) the proper use of computers in graphical simu- lations and problem solving. 7. Given information regarding the drug and the pharmacokinetic assumptions for the model, the student will construct (III) models and develop (V) equations of the ADME processes using LaPlace Transforms. 8. The student will interpret (IV) a given model mathematically. 9. The student will predict (IV) changes in the final result based on changes in vari- ables throughout the model. 10. The student will correlate (V) the graphs of the data with the equations and mod- els so generated. 2-1 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
2. 2. Mathematics Review 2.1 Concepts of Mathematics Pharmacokinetcs is a challenging field involving the application of mathematical concepts to real situations involving the absorbtion, distribution, metabolism and excretion of drugs in the body. In order to be successful with pharmacokinetics, a certain amount of mathematical knowledge is essential. This chapter is meant to review the concepts in mathematics essential for under- This is just a review. Look it over. You should standing kinetics. These concepts are generally taught in other mathematical be able to do all of these courses from algebra through calculus. For this reason, this chapter is presented as manipulations. a review rather than new material. For a more thorough discussion of any particu- lar concept, refer to a college algebra or calculus text. Included in this section are discussions of algebraic concepts, integration/differen- tiation, graphical analysis, linear regression, non-linear regression and the LaPlace transform. Pk Solutions is the computer program used in this course. A critical concept introduced in this chapter is the LaPlace transform. The LaPlace Something new - LaPlace transforms. transform is used to quickly solve (integrate) ordinary, linear differential equa- Useful tool. tions. The Scientist by Micromath Scientific Software, Inc.1 is available for work- ing with the LaPlace transform for problems throughout the book. 1. MicroMath Scientific Software, Inc., P.O. Box 21550, Salt Lake City, UT 84121-0550, 2-2 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
3. 3. Mathematics Review 2.2 Mathematical Preparation 2.2.1 ZERO AND INFINITY Any number multiplied by zero equals zero. Any number multiplied by infinity ( ∞ ) equals infinity. Any number divided by zero is mathematically undefined. Any number divided by infinity is mathematically undefined. 2.2.2 EXPRESSING LARGE AND SMALL NUMBERS Large or small numbers can be expressed in a more compact way using indices. 5 316000 becomes 3.16 × 10 Examples: How Does Scientific Notation Work? –3 0.00708 becomes 7.08 ×10 In general a number takes the form: n A × 10 Where A is a value between 1 and 10, and n is a positive or negative integer The value of the integer n is the number of places that the decimal point must be moved to place it immediately to the right of the first non-zero digit. If the decimal point has to be moved to its left then n is a positive integer; if to its right, n is a negative integer. Because this notation (sometimes referred to as “Scientific Notation”) uses indi- ces, mathematical operations performed on numbers expressed in this way are sub- ject to all the rules of indices; for these rules see Section 2.2.4. A shorthand notation (AEn) may be used, especially in scientific papers. This may n be interpreted as A × 10 , as in the following example: 4 2.28E4 = 2.28 ×10 = 22800 2-3 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
4. 4. Mathematics Review 2.2.3 SIGNIFICANT FIGURES A significant figure is any digit used to represent a magnitude or quantity in the place in which it stands. The digit may be zero (0) or any digit between 1 and 9. For example: TABLE 2-1 Significant Figures Number of Significant Significant Value Figures Figures (a) 572 2,5,7 3 (b) 37.10 0,1,3,7 4 4 (c) 0,1,6,5 4 10.65 x 10 (d) 0.693 3,6,9 3 (e) 0.0025 2,5 2 Examples (c) to (e) illustrate the exceptions to the above general rule. The value 10 How do I determine the number of significant raised to any power, as in example (c), does not contain any significant figures; figures? hence in the example the four significant figures arise only from the 10.65. If one or more zeros immediately follow a decimal point, as in example (e), these zeros simply serve to locate the decimal point and are therefore not significant figures. The use of a single zero preceding the decimal point, as in examples (d) and (e), is a commendable practice which also serves to locate the decimal point; this zero is therefore not a significant figure. Significant figures are used to indicate the precision of a value. For instance, a What do significant fig- ures mean? value recorded to three significant figures (e.g., 0.0602) implies that one can reli- ably predict the value to 1 part in 999. This means that values of 0.0601, 0.0602, and 0.0603 are measurably different. If these three values cannot be distinguished, they should all be recorded to only two significant figures (0.060), a precision of 1 part in 99. After performing calculations, always “round off” your result to the number of sig- nificant figures that fairly represent its precision. Stating the result to more signifi- cant figures than you can justify is misleading, at the very least! 2-4 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
5. 5. Mathematics Review 2.2.4 RULES OF INDICES n An index is the power to which a number is raised. Example: A where A is a What is an index? number, which may be positive or negative, and n is the index, which may be pos- itive or negative. Sometimes n is referred to as the exponent, giving rise to the general term, “Rules of Exponents”. There are three general rules which apply when indices are used. (a) Multiplication An n+m =  --  × B n+m n m n m A ×A = A A ×B -  B (b) Division n n n n–m ------ =  A × B A- A- ------ = An – m -- -  B m m A B (c) Raising to a Power nm nm (A ) =A There are three noteworthy relationships involving indices: (i) Negative Index 1- –n –n = ----- As n tends to infinity ( n → ∞ ) then A → 0 . A n A (ii) Fractional Index 1 -- - n n A= A (iii) Zero Index 0 A =1 2-5 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
6. 6. Mathematics Review 2.2.5 LOGARITHMS Some bodily processes, such as the glomerular filtration of drugs by the kidney, What is a logarithm? are logarithmic in nature. Logarithms are simply a way of succinctly expressing a number in scientific notation. In general terms, if a number (A) is given by n then log ( A ) = n A = 10 where ‘log’ signifies a logarithm to the base 10, and n is the value of the logarithm of (A). 5 Example: 713000 becomes 7.13 × 10 , ( 5 + 0.85 ) 0.85 0.85 5 5.85 × 10 = 10 and 7.13 = 10 , thus 713000 becomes 10 = 10 and log ( 713000 ) = 5.85 Logarithms to the base 10 are known as Common Logarithms. The transformation of a number (A) to its logarithm (n) is usually made from tables, or on a scientific calculator; the reverse transformation of a logarithm to a number is made using anti-logarithmic tables, or on a calculator. The number before the decimal point is called the characteristic and tells the place- What is the characteris- tic? the mantissa? ment of the decimal point (to the right if positive and to the left if negative). The number after the decimal is the mantissa and is the logarithm of the string of num- bers discounting the decimal place. 2.2.6 NATURAL LOGARITHMS Instead of using 10 as a basis for logarithms, a natural base (e) is used. This natural What is a natural loga- rithm? base is a fundamental property of any process, such as the glomerular filtration of a drug, which proceeds at a rate controlled by the quantity of material yet to undergo the process, such as drug in the blood. To eight significant figures, the value of the transcendental function, e, is ∞ 1 Where x is an inte- ∑ ---- e = 2.7182818 ... Strictly speaking, e = 1 + x! x=1 ger ranging from 1 to infinity ( ∞ ) , 2-6 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
7. 7. Mathematics Review ∞ ∑ denotes the summation from x = 1 to x = ∞ , and x=1 ! is the factorial (e.g., 6! = 6x5x4x3x2x1= 720) n In general terms, if a number (A) is given by A = e , then by definition, ln ( A ) = n Where, ‘ln’ signifies the natural logarithm to the base e , and n is the value of the natural logarithm of A . Natural logarithms are sometimes known as Hyperbolic or Naperian Logarithms; again tables are available and scientific calculators can do this automatically. The anti-logarithm of a natural logarithm may be found from exponential tables, which n give the value of e for various values of n. Common and natural logarithms are related as follows: How are natural loga- rithms ln x and common ln ( A ) = 2.303 × log ( A ) , and logarithms log x related? log ( A ) = 0.4343 × ln ( A ) Because logarithms are, in reality, indices of either 10 or e , their use and manipu- lation follow the rules of indices (See Section 2.2.4). (a) Multiplication: n m n m n+m To multiply N × M , where N = e and M = e ; NM = e × e =e . ln ( NM ) = n + m ; but By definition, n = ln ( N ) and m = ln ( M ) , hence ln ( NM ) = ln ( N ) + ln ( M ) Thus, to multiply two numbers (N and M) we take the natural logarithms of each, add them together, and then take the anti-logarithm (the exponent, in this case) of the sum. 2-7 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
8. 8. Mathematics Review (b) Division ln  ---- = ln ( N ) – ln ( M ) N -  M (c) Number Raised to a Power m ln ( N ) = m × ln ( N ) There are three noteworthy relationships involving logarithms: (i) Number Raised to a Negative Power ) = – m × ln ( N ) = m × ln  --- 1 –m ln ( N -  N –m As m tends to infinity ( m → ∞ ) , then ln ( N ) → –∞ (ii) Number Raised to a Fractional Power 1  ---  - 1 m ln ( m N ) = ln  N  = --- × ln ( N ) -  m (iii) Logarithm of Unity ln ( 1 ) = log ( 1 ) = 0 2-8 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
9. 9. Mathematics Review 2.2.7 NEGATIVE LOGARITHMS The number 0.00713 may be expressed as: –3 7.13 ×10 , or 0.85 –3 × 10 , or 10 – 2.15 10 . Hence, log ( 0.00713 ) = – 2.15 , which is the result generated by most calculators. However, another representation of a negative logarithm (generally used by refer- encing a log table): log (0.00713) = 3.85 The 3 prior to the decimal point is known as the characteristic of the logarithm; it can be negative (as in this case) or positive, but is never found in logarithmic tables. The .85 following the decimal point is known as the mantissa of the loga- rithm; it is always positive, and is found in logarithmic tables. In fact 3 is a symbolic way of writing minus 3 (-3) for the characteristic. In every case the algebraic sum of the characteristic and the mantissa gives the correct value for the logarithm. Example: log (0.00713) = 3.85 Add -3 and 0.85 Result is -2.15, which is the value of log ( 0.00713 ) The reason for this symbolism is that only positive mantissa can be read from anti- logarithmic tables, and hence a positive mantissa must be the end result of any log- arithmic manipulations. Note that while there are negative logarithms (when N < 1), they do not indicate that number itself is negative; the sign of a number (e.g., - N) is determined only by inspection following the taking of anti-logarithms. 2-9 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
10. 10. Mathematics Review 2.2.8 USING LOGARITHMIC AND ANTI-LOGARITHMIC TABLES Though the preferred method to using logarithms is with a calculator or computer, the understanding of how the number is being manipulated may be important in understanding the use of logarithms. (See the end of this chapter for Logarithm tables). (a) Find the log of (62.54): 1 62.54 = 6.254 ×10 ; convert 62.54 to scientific notation ---> 1. Look up the mantissa for 6254 in a table of logarithms: it is 7962. 2. 1 0.7962 1 1.7962 Hence, 6.254 ×10 × 10 = 10 log ( 62.54 ) = 1.7962 = 10 and 3. (b) Find the log of (0.00329) –3 0.00329 = 3.29 ×10 1. The mantissa for 329 is 5172 2. Hence, log(0.00329) = 3.5172. 3. Note that in both examples the value of the characteristic is the integer power to which 10 is raised when the number is written in scientific notation. (c) Multiply 62.54 by 0.00329 How do I multiply using logarithms? log (62.54) = 1.7972 log (0.00329) = 3.5172 log (62.54 + log (0.00329) = 1.7962+3.5172 = 1.7962-3+ 0.5172=-0.6866 0.6866=1.3134 (d) We wish to find anti-log (1.3134) Look up the anti-log for the 0.3134 (man- tissa) in a table: it is 2058. –1 Antilog (1.3134) = 2.058 ×10 Hence, antilog (1.3134) = 0.2058 log (62.54) - log (0.00329) = 1.796 - 3.5172=1.796 +3 - 0.5172=4.2788 How do I divide using logarithms? antilog 4.2788 = 19002 2-10 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
12. 12. Mathematics Review TABLE 2-3 Scale of Metric system and SI Name Symbol Multiplication Factor Name Symbol Multiplication Factor exa- E deci- d –1 18 10 10 peta- P centi- c –2 15 10 10 tera- T milli- m –3 12 10 10 µ giga- G micro- –6 9 10 10 mega- M nano- n –9 6 10 10 kilo- k pico- p –12 3 10 10 hecto- h femto- f –15 2 10 10 deca- da atto- a –18 1 10 10 TABLE 2-4 Dimensional Unit Dimension Symbol Unit Symbol Volume V liter l Concentration C grams/liter g/l 2-12 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
13. 13. Mathematics Review 2.2.10 DIMENSIONAL ANALYSIS It is a general rule that the net dimensions (and units) on the two sides of any equa- How are units useful? tion should be equal. If this is not so, the equation is necessarily meaningless. Consider the following equation which defines the average concentration of a drug FD- in blood after many repeated doses, ( C b )∞ = ---------- VKτ Where: F is the fraction of the administered dose ultimately absorbed (Dimensions: none), • • D is the mass of the repeated dose (Dimension: M), • V is the apparent volume of distribution of the drug (Dimension: V = L ) –1 • K is the apparent first-order rate constant for drug elimination (Dimension: T ), • and τ is the dosing interval (Dimension: T ) Writing the dimensions relating to the properties of the right-hand side of the equa- tion gives: ----------------------- = M M- ---- - –1 V V⋅T ⋅T M Thus ( C b )∞ has the dimensions of ---- , which are correctly those of concentration. - V Sometimes dimensional analysis can assist an investigator in proposing equations which relate several properties one with the other. If the units cancel, and you end up with the correct unit of measure, you probably did it right. If you obtain units that do not make sense, it’s guaranteed sure that you did it wrong. 2-13 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
14. 14. Mathematics Review 2.3 Calculus Calculus concerns either the rate of change of one property with another (differen- What is Calculus? tial calculus), such as the rate of change of drug concentrations in the blood with time since administration, or the summation of infinitesimally small changes (inte- gral calculus), such as the summation of changing drug concentrations to yield an assessment of bioavailability. In this discussion a few general concepts will be pro- vided, and it is suggested an understanding of graphical methods should precede this discussion. 2.3.1 DIFFERENTIAL CALCULUS 2.3.2 NON-LINEAR GRAPHS 3 Consider the following relationship: y = x TABLE 2-5 x, y sample data x 0 1 2 3 4 y 0 1 8 27 64 As can be seen from the graph (Figure 2-1), a non-linear plot is produced, as expected. y=x3 FIGURE 2-1. 70 60 50 40 30 20 10 0 1 2 3 4 (Question: How could the above data be modified to give a linear graph?) 2-14 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
15. 15. Mathematics Review 2.3.3 SLOPE OF NON-LINEAR GRAPH As with a linear graph, ∆y y2 – y1 --------------- = ----- - - ∆x x2 – x1 Where ∆y is the incremental change in y and ∆x is the incremental change in x But, as can be seen (Figure 1), the slope is not constant over the range of the graph; it increases as x increases. The slope is a measure of the change in y for a given change in x. It may then be stated that: “the rate of change of y with respect to x varies with the value of x.” 2.3.4 VALUE OF THE SLOPE 3 We need to find the value of the slope of the line y = x when x = 2 (See Figure 1). Hence, we may choose incremental changes in x which are located around x ≈ 2. FIGURE 2-2. ∆y / ∆x when x ≈ 2 ∆y ----- - ∆x ∆y ∆x x1 x2 y1 y2 0 4 4 0 64 64 16.000 1 3 2 1 27 26 13.000 1.5 2.5 1.0 3.375 15.625 12.250 12.250 1.8 2.2 0.4 5.832 10.648 4.816 12.040 1.9 2.1 0.2 6.859 9.261 2.042 12.010 1.95 2.05 0.1 7.415 8.615 1.200 12.003  ∆y  As may be seen, the value of the slope tends towards a value of 12.000 as the ----- -  ∆x  magnitude of the incremental change in x becomes smaller around the chosen value of 2.0. Were the chosen incremental changes in x infinitesimally small, the true value of the slope (i.e., 12.000) would have appeared in the final column of the above table. 2-15 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
16. 16. Mathematics Review Calculus deals with infinitesimally small changes. When the value of ∆x is infini- tesimally small it is written dx and is known as the derivative of x. Hence, dy ----- = f ( x ) - dx Where dy/dx is the derivative of y with respect to x and f ( x ) indicates some func- tion of x. 2.3.5 DIFFERENTIATION FROM FIRST PRINCIPLES Differentiation is the process whereby the derivative of y with respect to x is found. Thus the value of dy/dx, in this case, is calculated. (a) Considering again the original expression: 3 y=x (b) Let the value of y increase to y + dy because x increases to x + dx . Hence, 3 y + dy = ( x + dx ) (EQ 2-13) Multiplying out: 3 2 2 3 y + dy = x + 3x ( dx ) + 3x ( dx ) + ( dx ) (EQ 2-14) (c) The change in y is obtained by subtraction of the original expression from the last expression. (i.e., Eq. 2 - Eq. 1) 2 2 3 dy = 3x ( dx ) + 3x ( dx ) + ( dx ) (EQ 2-15) Dividing throughout to obtain the derivative, dy ----- = 3x 2 + 3x ( dx ) + ( dx )2 - dx When dx is infinitesimally small, its magnitude tends to zero ( dx → 0 ) . The limit- ing value of this tendency must be dx = 0 . At this limit, 2-16 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
17. 17. Mathematics Review dy ----- = 3x 2 - (EQ 2-16) dx 2 Hence the derivative of y with respect to x at any value of x is given by 3x . (d) In section 2.3.4 we saw how the true value of the slope (i.e., dy/dx) would be 12.0 when x = 2 . This is confirmed by substituting in Equation 1-16. dy ----- = 3x 2 = 3 ( 2 2 ) = 12 - dx 2.3.6 RULE OF DIFFERENTIATION Although the rate of change of one value with respect to another may be calculated as above, there is a general rule for obtaining a derivative. Let x be the independent variable value, y be the dependent variable value, A be a constant, and n be an exponential power. The general rule is: n If y = Ax then dy ----- = nAx n – 1 - dx The Rules of Indices may need to be used to obtain expressions in the form n y = Ax 5 (e.g., if y = x) 2.3.7 THREE OTHER DERIVATIVES 0 (a) If y = Ax , then y = A (i.e., y is constant) 2-17 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
18. 18. Mathematics Review dy Hence, ----- = 0 - dx Thus the derivative of a constant is always zero. (b) Accept that if y = ln ( x ) dy 1 then ----- = -- . - - dx x This derivative is important when considering apparent first-order processes, of which many bodily processes (e.g., excretion of drugs) are examples. Ax (c) Accept that if y = Be where B and A are constants, and e is the natural dy Ax base then ----- = ABe - dx This derivative will be useful in pharmacokinetics for finding the maximum and minimum concentrations of drug in the blood following oral dosing. 2.3.8 A SEEMING ANOMALY Consider the following two expressions: n (a) If y = Ax , then dy ----- = nAx n – 1 - dx n (b) If y = Ax + A , dy n–1 n–1 then ----- = nAx - + 0 = nAx dx Both of the original expressions, although different, have the same derivative. This fact is recognized later when dealing with integral calculus. 2-18 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
19. 19. Mathematics Review 2.3.9 INTEGRAL CALCULUS Generally integral calculus is the reverse of differential calculus. As such it is used to sum all the infinitesimally small units (dy) into the whole value (y). Thus, ∫ dy ∫ is the symbol for integration. = y , where 2.3.10 RULE OF INTEGRATION The derivative expression may be written: dy ----- = Ax n , or - dx n dy = Ax ⋅ dx To integrate, ∫ dy ∫ Ax dx = A ∫ x dx n n y= = A general rule states: n+1 Ax - A ∫ x dx = --------------- + A n n+1 Where A is the constant of integration However, there is one exception - the rule is not applicable if n = – 1 dy 2 Example: If ----- = 3x (See section 2.3.5), - dx 2+1 3x - then y = -------------- + A , and 2+1 3 y = x +A 2-19 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
20. 20. Mathematics Review 2.3.11 THE CONSTANT OF INTEGRATION There has to be a constant in the final integrated expression because of the seeming 3 3 anomaly referred to in section 2.3.8. As mentioned, both y = x and y = x + A dy 2 will give, on differentiation, ----- = 3x . - dx So whether or not a constant is present and, if so, what is its value, can only be decided by other knowledge of the expression. Normally this other knowledge takes the form of knowing the value of y when x = 0 . In the case of our graphical example we know that when x = 0 , then y = 0 . The integrated expression for this particular case is: 3 y = x + A , therefore 3 0 = 0 + A , thus A = 0 In some examples, such as first-order reaction rate kinetics, the value of A is not zero. 2.3.12 THE EXCEPTION TO THE RULE It occurs when n = – 1 1 y = A ∫ x dx = A ∫ -- dx –1 - x Upon integration, y = A ⋅ ln ( x ) + A This is the reverse of the derivative stated in section 2.3.10 (b). 2.3.13 A USEFUL INTEGRAL Accept that if, dy ----- = Be Ax - dx 2-20 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
21. 21. Mathematics Review then, Ax Be y = ----------- + A A This integral will be useful for equations which define the bioavailability of a drug product. 2.3.14 EXAMPLE CALCULATIONS (a) Consider, 2 c = 3t ( t – 2 ) + 5 Where c is the drug concentration in a dissolution fluid at time t . Then, multiplying out, 3 2 c = 3t – 6t + 5 The rate of dissolution at time t is dc ----- = 9t 2 – 12t - dt So at any time, the rate may be calculated. dc 2 (b) Consider, ----- = 3t ( t – 4 ) = 9t – 12t - dt Then rearranging, 2 dc = 9t ⋅ dt – 12 ⋅ dt The integral of c is: ∫ dc 3 2 c= = 3t + A – 6t + B 2-21 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
22. 22. Mathematics Review where B is a second constant. Adding the two constants together, 3 2 c = 3t = 6t + D where D = A + B We know, from previous work, that when t = 0 , then c = 5 Substituting 5 = D , the final expression becomes: 2 2 c = 3t + 6t + 5 Which is the initial expression in example (a) above. (c) Following administration of a drug as an intravenous injection, – dC p ------------ = KC p - dt Where C p is the plasma concentration of a drug at time t K is the apparent first-order rate constant of elimination. Rearranging, 1- – K ⋅ dt = ----- ⋅ dC p Cp 1 – Kt = – K ∫ dt = ∫ ------ ⋅ dCp Cp This integral is the exception to the rule (see section 2.3.12). – Kt = ln ( C p ) + A We know that when t = 0 , C p = ( C p ) 0 . 2-22 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
23. 23. Mathematics Review Substituting, 0 = ln ( C p )0 + A Or, A = – ln ( C p )0 Hence – Kt = ln ( C p ) – ln ( C p ) 0 or, ln ( C p ) = ln ( C p ) 0 – Kt or, – Kt Cp = ( C p )0 ⋅ e 2-23 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/