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# Mathematics Review

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Mathematics Review

Mathematics Review

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## Mathematics ReviewDocument Transcript

• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• 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/
• Mathematics Review Plot of Perfect vs. Real data FIGURE 2-3. 6 4 2 0 -2 -4 -6 -3 -2 -1 0 1 2 x Simply looking at the columns x and y (real) it might be difficult to see the rela- tionship between the two variables. But looking at the graph, the relationship becomes apparent. Thus, the graph is a great aid to clear thinking. For every graph relating variables, there is an equation and, conversely for every equation there is a graph. The plotting of graphs is comparatively simple. The reverse process of find- ing an equation to fit a graph drawn from experimental data is more difficult, except in the case of straight lines. 2.4.1 GRAPHICAL CONVENTIONS Certain conventions have been adopted to make the process of rendering a data set How are graphs made? to a graphical representation extremely simple. The ‘y’ variable, known as the dependent variable, is depicted on the vertical axis (ordinate); and the ‘x’ variable, known as the independent variable, is depicted on the horizontal axis (abscissa). It is said that ‘y’ varies with respect to ‘x’ and not ‘x’ varies with ‘y’. A decision as to which of the two related variables is dependent can only be made be considering the nature of the experiment. To illustrate, the plasma concentration of a drug given by IV bolus depends on time. Time does not depend on the plasma concentration. Consequently, plasma concentration would be depicted on the ‘y’ axis and time on the ‘x’ axis. 2-25 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review and consequently a plot of ln y′ (the dependent variable) versus x′ (the indepen- dent variable) will yield a straight line with a slope of m and an intercept of ln b′ . Expressions of any other form are non linear. For example: An expression relating the plasma concentration of a drug ( C p ) over time ( t ) . – Kt C p = C p0 e this relationship put in linear perspective yields: ln C p = ln C p0 – Kt , which is in the form y = b + mx The graphs that yield a straight line are the ones with the ordinate being ln C p0 , and the abscissa being t . Any other combination of functions of C p and t will be non-linear, e.g., • C p versus t • C p versus ln t • ln C p versus ln t The appropriate use of a natural logarithm in this case serves to produce linearity. However, the use of logarithms does not automatically straighten a curved line in all examples. Some relationships between two variables can never be resolved into ( n – m) (n – m + 1) x + … + ( k n )x a single straight line, e.g., y = k 0 + k 1 x + k2 x where n ≥ 2 ;n = m + 1 or K a FD –K t Kt C p = ------------------------ ⋅ ( e – e a ) - V ( Ka – K ) (It is possible to resolve this equation into the summation of two linear graphs which will be shown subsequently.) 2-27 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.4.3 THE SLOPE OF A LINEAR GRAPH (M) From the equation a prediction may be made as to whether the slope is positive or What is the slope of a straight line? negative. In the previous example, the slope is negative, i.e: m = – K TABLE 2-9 Sample data of caffeine elimination µg C p  ------- -  mL ln C p t (min) 12 3.75 1.322 40 2.80 1.030 65 2.12 0.751 90 1.55 0.438 125 1.23 0.207 173 0.72 -0.329 The differences in both the y-values and the x-values may be measured graphically to obtain the value of the slope, m. Then knowing the value of m, the value of K may be found. ( C p ) of caffeine over time Plasma Concentration FIGURE 2-4. 101 g/mL) u Caffeine Concentration ( 100 10-1 0 50 100 150 200 Time (min) 2-28 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.4.4 LINEAR REGRESSION: OBTAINING THE SLOPE OF THE LINE The equation for a straight line is: y = m⋅x+b • y is the dependent variable • x is the independent variable • m is the slope of the line • b is the intercept of the line The equation for the slope of the line using linear regression is: ( Σ( x ) ⋅ Σ ( y ) ) – ( n ⋅ Σ ( x ⋅ y )- ) m = -------------------------------------------------------------------- 2 2 [ Σ ( x ) ] – ( n ⋅ Σ( x ) ) And the intercept is b = y – ( m ⋅ x ) Linear Regression for data in table 2-9 TABLE 2-10 2 X⋅Y X Y X 12 1.322 144 15.864 40 1.030 1600 41.2 65 0.751 4225 48.815 90 0.438 8100 39.42 125 0.207 15625 25.875 173 -0.329 29929 -56.917 ΣX = 505 ΣY = 3.239 ΣXY = 114.257 2 ΣX = 59623 2 ( ΣX ) = 255025 Σx Σy x = ----- = 4.167 - y = ----- = 0.5398 - n n Using the data from table 2-10 in the equation for the slope of the line ( 505 ⋅ 3.239 ) – ( 6 ⋅ 114.257 ) · m = --------------------------------------------------------------------- = – 0.01014 255025 – ( 6 ⋅ 59623 ) 0.5398 – ( – 0.01014 ⋅ 4.167 ) = 1.4229 . and the intercept would be b = Note that this is ln C . In oder to find the C p0 , the anti-ln of must be taken. i.e. b b 1.4229 Cpo = e = e = 4.15 2-29 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review It is important to realize that you may not simply take any two data pairs in the data set to get the slope. In the above data, if we simply took two successive data pairs from the six data pairs in the set, this would result in five different slopes ( ∆x ⁄ ∆y ) ranging from -0.0066 to -0.0125 as shown in table 2-11. Clearly, this is unacceptable. Even to guess, you must plot the data, eyeball the best fit line by placing your clear straight edge through the points so that it is as close to the data as possible and look to make sure that there are an equal number of points above the line as below. Then take the data pairs from the line, not the data set. TABLE 2-11 Sample slope data from figure 2-4 ∆y ----- - ∆x ∆y ∆x Time (x) ln Conc. (y) 12 1.322 -28 0.292 -0.0104 40 1.030 -25 0.28 -0.0112 65 0.751 -25 0.312 -0.0125 90 0.438 -35 0.231 -0.0066 125 0.207 -48 0.536 -0.0112 173 -0.329 2.4.5 PARALLEL LINES Two straight lines are parallel if they have the same slope. Calculating for the intercept of a linear graph (b): (a) Not knowing the value of m; The graph may be extrapolated, or calcu- lations performed, at the situation where t = 0 . In this case b = ln C p0 . (b) Knowing the value of m; • There are two ways: for any point on the graph: y 1 = mx 1 + b and b = y 1 – mx 1 Hence, b may be calculated from a knowledge of y 1 and x 1 . • Secondly, the graph may be extrapolated or calculations performed, at the situation where t = 0 . In this case, b = ln C p 2-30 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.4.6 GRAPHICAL EXTRAPOLATIONS It is dangerous to extrapolate on non-linear graphs, and it is unwise to extrapolate How far can I predict? too far on linear graphs. Most often extrapolation is used to find the value of y at a selected value of x. If the size of the graph does not permit physical extrapolation to the desired value, the required result may be obtained by calculation. The values of m and b must be found as shown above. Then: y' = mx' + b , where x' is the selected value of x, and y' is the new calculated value for y. 2.4.7 SIGNIFICANCE OF THE STRAIGHT LINE The more closely the experimental points fit the best line, and the higher the num- ber of points, the more significant is the relationship between y and x. As you may expect, statistical parameters may be calculated to indicate the significance. By using all the experimental data points, calculations may be made to find the What good is a straight line? optimum values of the slope m, and the intercept, b. From these values the corre- lation coefficient (r).and the t-value may be obtained to indicate the significance. Exact details of the theory are available in any statistical book, and the calculations may most easily be performed by a computer using The Scientist or PKAnalyst in this course. The advantage of computer calculation is that it gives the one and only best fit to the points, and eliminates subjective fitting of a line to the data. 2.4.8 GRAPHICAL HONESTY Any graph drawn from 2 points is scientifically invalid. Preferably, straight-line How many points are needed? graphs should have at least 3 - 5 points, and non-linear graphs a few points more. As a graph is a visual representation which enables the experimenter to average Can I discard points that don’t fit? out the small deviations in results from the “perfect” result, no one result can be unjustifiably ignored when the best fitting line is drawn. Still, an “errant” point may be justifiably ignored if there were unusual experimental circumstances which may have caused the deviation. Thus it is not justifiable to omit a point solely because it “does not fit”. 2.4.9 AXES WITH UNEQUAL SCALES In mathematical studies, the scales of the x and y are almost always equal but very often in plotting chemical relations the two factors are so very different in magni- 2-31 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review tude that this can not be done. Consequently, it must be borne in mind that the rela- tionship between the variables is given by the scales assigned to the abscissa and ordinate rather than the number of squares counted out from the origin. FIGURE 2-5. y = 0.1 x 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 0 2 4 6 8 10 10 8 6 4 2 0 0 2 4 6 8 10 For example (shown in Figure 2-5), these two parabolic curves represent the same equation the only difference is the scales are different along the y axis. Frequently it is not convenient to have the origin of the graph coincide with the lower left hand corner of the coordinate paper. Full utilization of the paper with suitable intervals is the one criteria for deciding how to plot a curve from the experimental data. For example, the curve below (Figure 2-6) is poorly planned, where the following (Figure 2-7) is a better way of representing the gas law PV = nRT 2-32 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review Poorly presented graph FIGURE 2-6. 50 40 30 20 10 0 0 4 8 12 16 20 Well arranged graph FIGURE 2-7. 25 20 15 10 5 0 1 2 3 4 5 6 7 8 9 10 P ( ) 2.4.10 GRAPHS OF LOGARITHMIC FUNCTIONS 2 Previously variables were raised to constant powers; as y = x . In this section x constants are raised to variable powers; as y = 2 . Equations of this kind in which the exponent is a variable are called (naturally) exponential equations. The most x important exponential equation is where e is plotted against x . 2-33 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.4.11 SEMILOGARITHMIC COORDINATES Exponential or logarithmic equations are very common in physical chemical phe- nomenon. One of the best ways of determining whether or not a given set of phe- nomenon can be expressed by a logarithmic or exponential equation is to plot the logarithm of one property against another property. Frequently a straight line is obtained and its equation can be readily found. For example: In the following table the plasma concentration ( C p ) of the immunosuppressant cyclosporine was measured after a single dose (4mg/kg) as a function of time. TABLE 2-12 Plasma concentration of cyclosporine ng ----- - Concentration ml Time (hours) 0.25 1900 .75 1500 1.5 1300 4 900 6 600 8 390 D’mello et al., Res. Comm. Chem. Path. Pharm. 1989: 64 (3):441-446 These can be illustrated in three different ways (Figures 2-8, 2-9, 2-10), • Concentration vs. time directly • Log concentration vs. time directly • Log concentration vs. time with concentration plotted directly on to log scale of ordinate. Concentration (ng/ml) vs. time (hr) FIGURE 2-8. 2000 1800 1600 1400 1200 1000 800 600 400 200 0 1 2 3 4 5 6 7 8 2-34 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review Log Concentration vs. time FIGURE 2-9. 3.200 Log Concentration - Time Curve 3.100 3.000 2.900 2.800 2.700 2.600 2.500 2.400 2.300 0 1 2 3 4 5 6 7 8 Log concentration (on log scale) vs. time FIGURE 2-10. 10000 1000 100 0 1 2 3 4 5 6 7 8 Graphing is much easier because the graph paper itself takes the place of a loga- rithmic table, as shown in Figure 1-10. Only the mantissa is designated by the graph paper. Scaling of the ordinate for the characteristic is necessary. The general equation y = Be ax can be expressed as a straight line by basic laws of indices. ax ln y = ln B + ln ( e ) → ln y = ln B + ax or ln y = ax + ln B One axis is printed with logarithmic spacing, and the other with arithmetic spac- ing. It is used when a graph must be plotted as in the example (Figure 1-4) y = log [ C p ] and x = t . 2-35 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review In this example, the vertical logarithmic axis is labelled “Plasma concentration of cyclosporine” and the values plotted are the ordinary values of [ C p ] . Thus, there is no need to use logarithmic tables, because the logarithmic spacing is responsible for obtaining a straight line. Two problems may occur when graphing on a logarithmic mantissa: a) there are not enough cycles to incorporate all the data b) obtaining the value of the slope is difficult. In this instance the slope is given by: ln [ C p ] 2 – ln [ C p ] 1 y2 – y1 m = --------------- = -------------------------------------------- - x2 – x1 t2 – t1 Hence, before calculating the value of m, the two selected values of [ Cp ] 1 and [ Cp ] 2 must be converted, using a calculator, to ln [ Cp ] 1 and ln [ C p ] 2 in order to satisfy the equation. The same problem may arise in obtaining the intercept value, b. The two problems may be avoided by plotting the same data on ordinary paper, in which case the vertical axis is labelled “log plasma concentration”. However, in this instance the ordinary values of [ C p ] must be converted to ln [ Cp ] prior to plot- ting. It is the ln [ Cp ] values which are then plotted. The calculation of the slope is direct in this case, as the values of y 1 and y 2 may be read from the graph. Hence, one must consider the relative merits of semilogarith- mic and ordinary paper before deciding which to use when a log plot is called for. In the case of semilog graphs the slope may be found in a slightly different manner, i.e., taking any convenient point on the line ( y 1 ) we usually take the as the second point, ( y 2) one half of ( y 1 ) . Thus, y1 ln  -------------------  ln  --------- 1- -  ( 1 ⁄ 2 )y 1  1 ⁄ 2 ln y 1 – ln ( ( 1 ⁄ 2 )y 1 ) ln 2 - 0.693 m = ---------------------------------------------- = ------------------------------ = -------------------- = ------------- = ------------ - - - - t 1 – t2 t 1 – t2 t 1 – t2 t1 – t2 –t1 ⁄ 2 t1 < t2 , (in which case, t 2 – t 1 is called the half-life t½ ). Since then t1 – t2 = –t1 ⁄ 2 0.693 0.693 because and k = ------------ . m = ------------ = – k - - –t1 ⁄ 2 t1 ⁄ 2 2-36 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.4.12 LOG - LOG COORDINATES a Functions of the type y = Bx give straight lines when plotted with logarithms along both axis. i.e., equation in logarithmic form is: log y = log B + a log x or log y = a log x + log b which is in the form y = mx + b . This is directly applicable to parabolic and hyperbolic equations previously discussed (see Figure 1-5). 2.4.13 PITFALLS OF GRAPHING: POOR TECHNIQUE The utility of these procedures requires proper graphing techniques. The picture that we draw can cause formation of conceptualizations and correlations of the data that are inconsistent with the real world based simply on a bad picture. Conse- quently the picture must be properly executed. The most common error is improper axes labelling. On a single axis of rectilinear coordinate paper (standard graph paper), a similar distance between two points corresponds to a similar difference between 2 numbers. Thus, Graphing using standard number spacing FIGURE 2-11. 40 30 20 10 0 0 5 10 15 20 25 30 2-37 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review Nonstandard (incorrect) graph FIGURE 2-12. 40 30 20 10 1 0 2 5 20 0 10 30 Obviously, the distance (Time) on the graph 12 between 0 and 2 hours should not be the same as the distance between 10 and 20 hours. It is, and therefore Figure 2- 12 is wrong. Similarly, the use of similar paper may result in some confusion. With logarithms the mantissa for any string of numbers, differing only by decimal point placement, is the same. What differentiates one number from another, in this case, is the char- acteristic. Thus, TABLE 2-13 Logarithmic graphing Number Mantissa Characteristic Log 234 .3692 2 2.3692 23.4 .3692 1 1.3692 2.34 .3692 0 0.3692 0.234 .3692 -1 1.3692 The paper automatically determines the relationship between strings of numbers (mantissa) by the logarithmic differences between the numbers on the axis within a cycle. The student must determine the order of magnitude (characteristic) to be rel- egated to each cycle. 2-38 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review Logarithmic mantissa FIGURE 2-13. Logarithmic Plot 103 234 102 23.4 101 Y axis (units) 2.34 100 0.234 10-1 1.0 1.5 2.0 2.5 3.0 3.5 4.0 X axis (units) Thus, we see, in Figure 2-13, the cycle on the semilog paper to relate to orders of magnitude (e.g., 1, 10, 100, 1000, etc.) and consequently the characteristic of the exponent. The third common problem is labelling the log axis as log “y”. This is improper. It is obvious from the spacing on the paper that this function is logarithmic, and thus the axis is simply labelled “y”. There are almost as many different errors as there are students and it is impossible to list them all. These few examples should alert you to possible problems. 2.4.14 GRAPHICAL ANALYSIS We will look at several different types of plots of data: Straight line going down on semi-log paper FIGURE 2-14. 0 1 2 3 4 5 2-39 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review Find the slope by taking any two values on the Y axis such that the smaller value is one half of the larger. The time that it takes to go from the larger to the smaller is the half-life. Dividing 0.693 by the half-life yields the rate constant. Extrapolating the line back to t = 0 yields the intercept. Curved line which plateaus on semi-log paper. FIGURE 2-15. 0 2 4 6 8 10 12 Curved line which goes up and then straight down on semi-log paper. FIGURE 2-16. 0 5 10 15 20 25 Find the terminal slope by taking any two values on the Y axis such that the smaller value is one half of the larger. The time that it takes to go from the larger to the smaller is the half-life. Dividing 0.693 by the half-life yields the rate constant. Plot type one is reasonably easily evaluated. There are 2 important things that can be obtained: Slope and Intercept. However, the slope and intercept have different meanings dependent on the data set type plotted. The slope is the summation of all the ways that the drug is eliminated, -K. 2-40 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review TABLE 2-14 Plot type 1 examples Data Type Y axis X axis Slope Intercept IV Bolus Parent Drug Conc. parent compound Time -K dose C p0 = ---------- - Vd Kr ⋅ X 0 IV Bolus Parent Time -K dXu --------- urine rate of excretion - (mid) dt parent compound IV Bolus Parent Time -K Xu ∞ – Xu Cumulative urine kr -------------- K ⋅ X0 data Plot type two is not usually evaluated in its present form as only the plateau value can be obtained easily. But again it has different meanings dependent on the data plotted. TABLE 2-15 Plot Type 2 examples Data Type Y axis X axis Plateau Value IV Bolus Parent Time Xu Cumulative urine kr Xu ∞ = -------------- K ⋅ X0 data parent compound IV Infusion Drug concentration Time Q- Q ( C p )ss = ----------- = --- - Parent parent compound K⋅V cl Usually urine data of this type (parent compound - IV bolus) is replotted and eval- uated as plot 1 (above). Infusion data can be replotted using the same techniques, but usually is not. Plot type 3 must be stripped of the second rate constant from the early time points, thus: There are 3 things that can be obtained from the plot: the terminal slope (the smaller rate constant), the slope of the stripped line (the larger rate constant) and the intercept. The rate constants obtained from a caternary chain (drug moving from one box to another in sequence in compartmental modeling) are the summa- tion of all the ways that the drug is eliminated from the previous compartment and all the ways the drug is eliminated from the compartment under consideration. See LaPlace Transforms for further discussion. Again, dependent on the data set type being plotted they will have different values. 2-41 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review TABLE 2-16 Plot Type 3 examples Data Type Y axis X axis S1 S2 Intercept km ⋅ X 0 IV Bolus Metabolite conc. Time -K small -Klarge ------------------------------------------------------- Parent ( K l arg e – K smal l ) ⋅ V dm k mu ⋅ k m ⋅ X 0 IV Bolus Time -K small -Klarge dXmu -------------- excretion - ----------------------------------- - Parent (mid) dt K l arg e – K smal l rate of metabolite into urine k a ⋅ fX 0 Oral Drug conc. Time -K small -Klarge --------------------------------------------------- - ( K l arg e – K smal l ) ⋅ V d 2-42 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.5 Pharmacokinetic MoKdeling It has been observed that, after the administration of a drug, the concentration of the drug in the body appear to be able to be described by exponential equations. Thus, it appears that, even though the processes by which the drug is absorbed. dis- tributed, metabolized and excreted (ADME) may be very complex, the kinetics (math) which mimics these processes is made up of relatively simple first order processes and is called first order pharmacokinetics. A second observation is that the resulting concentration is proportional to dose. When this is true, the kinetics is called linear. When this math is applied to the safe and effective therapeutic man- agement of an individual patient, it is called clinical pharmacokinetics. Thus, in clinical pharmacokinetics, we monitor plasma concentrations of drugs and suggest dosage regimens which will keep the concentration of drug within the desired ther- apeutic range. Pharmacodynamics refers to the relationship between the drug con- centration at the receptor and the intensity of pharmacological (or toxicological) response. It is important to realize that we want to control the pharmacological response. We do that indirectly by controlling the plasma concentration. In order for this to work, we assume kinetic homogeneity, which is that there is a predict- able relationship between drug concentration in the plasma (which we can mea- sure) and drug concentration at the receptor site (which we can not measure). This assumption is the basis for all clinical therapeutics. Models are simply mathematical constructs (pictures) which seem to explain the relationship of concentration with time (equations) when drugs are given to a per- son (or an animal). These models are useful to predict the time course of drugs in the body and to allow us to maintain drug concentration in the therapeutic range (optimize therapy). The simplest model is the one used to explain the observations. We model to summarize data, to predict what would happen to the patient given a dosage regimen, to conceptualize what might be happening in disease states and to compare products. In every case, the observations come first and the explanation next. Given that a data set fits a model, the model can be used to answer several different types of questions about the drug and how the patient handles the drug (its disposition), for example: if the drug were to be given by an oral dose, how much is absorbed and how fast? Are there things which might affect the absorp- tion, such as food or excipients in the dosage form itself. What would happen if the drug were to be given on a multiple dose regimen? What if we increased the dose? etc. 2-43 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review You should be able to: • be facile in the use of the equations. You should be able to graphically manipulate data sets and extract pharmacokinetic parameters, applying the appropriate equations or variations of them. • define all new words used in this section. e g.: Succinctly define, stating rigorously the meaning of any symbols used and the dimensions of measurement. • compare and contrast new concepts used in this section. e. g.: rate and rate constant, zero and first order kinetics, bolus and infusion methods, excretion and elimination, the assumptions made in pharmacokinetic models with physiological reality. Why can these assumptions be made? • pictorially represent any two variables (graph) one vs. the other. e.g. for each of the following pairs of variables (ordinate against abscissa), draw a graph illustrating the qualitative profile of their relationship. Where appropriate, indicate the nature of important slopes, intercepts, and values. Unless you specifically indicate on your plot that semi-log paper is being considered (write “S-L”), it will be assumed that rectilinear paper is being considered. Graphs are for a drug given by IV Bolus where applicable. 2.5.1 MAKING A MODEL The differential equations used result from the ka model which is our conceptualization of what is X happening to the drug in the body. The box (compartment) is the area of interest. We want to find out how the mass of drug, X , changes with time in that compartment, the rate, and how the rates change with time, the differential equations. The picture that we build is made up of building blocks, consisting of the arrow How do we make a dif- ferential equation? and what the arrow touches. The arrow demonstrates how quickly the mass of drug, X , declines. The arrow times the box that the arrow touches = the rate. Rates can go in, i.e. arrows pointing to a box mean drug is going in (+ rate). Rates can go out, i.e. arrows pointing away means drug is going out (- rate). Rate = rate constant (arrow) times mass of drug (box). So the arrow and box really is a pictorial repre- sentation of a rate where the rate is the rate constant on the top of the arrow times what the tail of the arrow touches. k⋅X. Again, the rate constant, k , tells the magnitude of the rate, Consider the following simple chain: 2-44 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review k12 k23 X2 X1 X3 The building blocks are k 12 ⋅ X1 and k 23 ⋅ X2 . Every arrow that touches the compart- ment of interest becomes part of the differential equation. If the arrow goes to the box, it’s positive; if it goes away from the box, it’s negative. To find dX1 ⁄ dt (the rate of change of X 1 with time), we simply add up all of the rates which affect X1 (all of the arrows that touch X1 ) dX 1 -------- = – k 12 ⋅ X 1 - dt and thus: dX 2 -------- = k 12 ⋅ X 1 – k 23 ⋅ X2 - dt dX 3 -------- = k 23 ⋅ X 2 - dt (Note: the first subscript of the rate constant and the subscript of the box from which it originates are the same.) You should be able to develop the series of interdependent differential equations which would result from any model. The integration of those equations by use of the Laplace Table is done by transforming each piece of the equation into the Laplace domain (looking it up on the table and substituting). The algebra per- formed solves for the time dependent variable: put everything except the variable (including the operator, s) on the right side and put the variable on the left. Find the resulting relationship on the left side of the table. The corresponding equation on the right side of the table in the integrated form. You should be able to integrate any differential equation developed from any model (within reason) that we can conceptualize. (Note: Each subsequent variable is dependent on the ones that precedes it. In fact, the solutions to the preceding variables are substituted into the differential to remove all but one of the time dependent functions - the one that we are currently attempting to solve.) 2-45 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.5.2 ONE COMPARTMENT OPEN MODEL A simplified picture (mathematical construct) of the way the body handles drug is one where the body can be conceived to be a rapidly stirred beaker of water (a sin- gle compartment). We put the drug in and the rate at which the drug goes away is proportional to how much is present (first order). Thus the assumptions are: • Body homogeneous (one compartment) • Distribution instantaneous • Concentration proportional to dose (linear) • Rate of elimination proportional to how much is there. (First order) It is important to note that we know some of these assumptions are not true. It is of little consequence, as the data acts as if these were true for many drugs. The visual image which is useful is one of a single box and a single arrow going out of the box depicting one compartment with linear kinetics. The dose is placed in the box and is eliminated by first order processes. In many cases, more complicated mod- els (more boxes) are necessary to mathematically mimic the observed plasma ver- sus time profile when one or more of these assumptions are not accurate. For example, the two compartment (or multi-compartment) model results when the body is assumed to not be homogeneous and distribution is not instantaneous. 2-46 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review integrated, the final equation will describe the mass of drug actually in the body at any time. The procedure used is to replace the Independent variable (time) by a function containing the LaPlace Operator, whose symbol is “s”. In doing so we have replaced the time domain by a complex domain. This is analogous to replacing a number by its logarithm. Once in the complex domain, the transformed function may be manipulated by regular algebraic methods. Then the final expression in the complex domain is replaced by its equivalent in the time domain, yielding the inte- grated equation. This ultimate process is analogous to taking an antilogarithm. 2.6.1 TABLE OF LAPLACE TRANSFORMS The replacement of expressions in one domain by their equivalents in another is A table of useful LaPlace transforms is given in accomplished by reference to tables. One column shows time domain expressions, Section 2.7. Page 2-56. stated as f ( t ) , and second column shows the corresponding complex domain expressions, stated as the LaPlace Transform. Note that “ f ( t ) ” simply means – at “some function of time”. For example, when f ( t ) is Be , then the LaPlace Transform is B ⁄ ( s + a ) , where “B” is a constant and “a” is a rate constant , then f ( t ) is At . 2 A⁄s For example, when the LaPlace Transform is 2.6.2 SYMBOLISM For simplicity in writing transformed rate expressions (and to distinguish them from untransformed (time domain) expressions), the following symbolism will be employed: “a bar will be placed over the dependent variable which is being transformed”. Example: If X is the mass of unchanged drug in the body at any time, then X is the LaPlace Transform of this mass. 2-48 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review Using the pharmacokinetic symbolism from chapter one, the compartments are named and placed: metabolites below (or above the plane of the parent com- pound): compounds going into the urine, to the right; and compounds going into the feces, to the left of the compounds in the body. The rate constants connecting the compartments also follow the symbolism from chapter one. In the above flow chart, K, the summation of all the ways that X is removed from the body, is ku + kf + km while K1, the summation of all the ways that Xm is removed from the body, is kmu +kmf. Only those compartments are used which correspond to the drug’s pharmacoki- netic description, thus when a drug is given by IV bolus and is 100% metabolized with the metabolite being 100% excreted into the urine the model would look like this: Dose X km kmu Xm Xmu Thus in this flow chart, K, the summation of all the ways that X is removed from the body, is km while K1, the summation of all the ways that Xm is removed from the body, is kmu. 2-50 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review Drugs sometimes are metabolized to two (or more) different metabolites. In the first case, the drug is metabolized by two separate pathways resulting in this flow chart: Xmf1 kmf1 Xm1 kmu1 Xmu1 Dose km1 ku kf X Xu Xf km2 kmf2 kmu2 Xmu2 Xmf2 Xm2 In this flow chart, K, the summation of all the ways that X is removed from the body, is ku + kf + km1 + km2 while K1, the summation of all the ways that Xm1 is removed from the body, is kmu1 +kmf1 and K3, the summation of all the ways that Xm2 is removed from the body, is kmu2 + kmf2. While in a second case, the drug is metabolized and the metabolite is further metabolized resulting in this flow chart: Dose kf ku Xu X Xf km1 kmf1 kmu1 Xmf1 Xm1 Xmu1 km2 kmf2 kmu2 Xmu2 Xmf2 Xm2 2-51 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review In this flow chart, K, the summation of all the ways that X is removed from the body, is ku + kf + km1 while K1, the summation of all the ways that Xm1 is removed from the body, is kmu1 +kmf1+ km2 and K3, the summation of all the ways that Xm2 is removed from the body is kmf2 + kmu2. Both of these flow charts result in very different end equations, so it is imperative that the flow charts accurately reflect the fate of the drug. 2.6.4 STEPS FOR INTEGRATION USING THE LAPLACE TRANSFORM • Draw the model, connect the boxes with the arrows depicting where the drug goes. • The building blocks of the differential rate equations are the arrows and what the tail touches. • Write the differential rate equation for the box in question. The box is on the left side of the equal sign and the building blocks are on the other. If the arrow goes away from the box, the building block is negative, if it is going towards the box, the building block is positive. • Take the LaPlace Transform of each side of the differential rate equation, using the table where necessary. • Algebraically manipulate the transformed equation until an equation having only one trans- formed dependent variable on the left-hand side is obtained. • Convert the transformed expression back to the time domain, using the table where necessary to yield the Integrated equation. 2-52 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.6.5 EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM Following an intravenous injection of a drug (bolus dose), its excretion may be represented by the following pharmacokinetic scheme: (Scheme I) ku X Xu Where X is the mass of unchanged drug in the body at any time. X u is the cumulative mass of unchanged drug in the urine up to any time, and k u is the apparent first-order rate constant for excretion of unchanged drug. Consider how the body excretes a drug a. The building block is the arrow and what it touches. This first box (compart- ment) of interest is [ X ] . The arrow ( k u) is going out, therefore, the rate is going out and is negative, thus dX ------ = –k u X - (EQ 2-17) dt The negative sign indicates loss from the body. Taking the LaPlace Transform of each side of equation 2-17: sX – X0 = – k u X (EQ 2-18) Note that because the independent variable (time) did not appear on the right-hand side of equation 2-17, neither did the LaPlace Operator, s, appear there in equation 2-18. All that was necessary was to transform the dependent variable ( X ) into X . Hence, the table was only required for transforming the left-hand side of equation 2-18. Manipulating the transformed equation: (X) 1. Get only one variable which changes with time 2. Get on the left and everything else on the right. X 2-53 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review sX + k u X = X0 X ( s + k u ) = X0 X0 X = ------------- (EQ 2-19) s + ku A- X = --------------- LetX0 = A Letk u = a (EQ 2-20) (s + a) Note that X is the only transformed dependent variable and is on the left-hand side of equation 2-19. Converting back to the time domain: –ku t X = X0e (EQ 2-21) A- Note that the right-hand side of equation 2-21 was analogous to --------------- in the (s + a) table, because X0 is a constant (the initial dose administered). The left-hand side of equation 2-21 could be converted back without the table. The final expression is the familiar first-order integrated expression. 2-54 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.6.6 SECOND EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM Look at Scheme I again. Consider how the drug goes from the body into the urine. The next box of interest is Xu . The arrow is coming in, therefore the rate is com- ing in and is positive. thus, dXu -------- = k u X - (EQ 2-22) dt (b) Taking the LaPlace Transform of each side of equation 2-22: sX u – ( X u ) 0 = k u X (EQ 2-23) But, at zero time, the cumulative mass of unchanged drug in the urine was zero: that is ( Xu ) 0 = 0 . sX u = k u X (EQ 2-24) (c) Manipulating the transformed equation: ku X Xu = -------- - (EQ 2-25) s Note that there are two transformed dependent variables. One of them ( X ) can be replaced by reference to equation 2-19. ku X0 Xu = -------------------- - (EQ 2-26) s(s + ku) A- Let ( k u X0 ) = A Let ( k u ) = a X = ------------------ (EQ 2-27) s(s + a) (d) Converting back to the time domain: –k t kuX0 ⋅ ( 1 – e u ) X u = --------------------------------------- - ku 2-55 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review Where k u X 0 and k u are analogous to “A” and “a” respectively in the table. Sim- plifying, – kut Xu = X0 ( 1 – e ) (EQ 2-28) 2.6.7 THIRD EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM During the intravenous infusion of a drug, its excretion may be represented by the following pharmacokinetic scheme: (Scheme II) ku Infusion Q X Xu Where Q is the zero-order infusion rate constant (the drug is entering the body at a constant rate and the rate of change of the mass of drug in the body is governed by the drug entering the body by infusion and the drug leaving the body by excretion). The drug entering the body does so at a constant (zero-order) rate. dX ------ = Q – k u X - (EQ 2-29) dt (b) Taking the LaPlace Transform of each side of equation 2-29: Q sX – X 0 = --- – k u X - s Note that because Q is a rate, and is therefore a function of the independent vari- able (time), its transformation yields the LaPlace Operator. In this case, Q was analogous to “A” in the table. But, at zero time, the mass of unchanged drug in the body was zero: that is, X0 = 0 Q sX = --- – k u X - (EQ 2-30) s (c) Manipulating the transformed equation: Q- X = -------------------- (EQ 2-31) s ( s + ku ) 2-56 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review A- Let ( Q = A ) Let ( k u = a ) X = ------------------ (EQ 2-32) s( s + a) (d) Converting back to the time domain: –k t Q(1 – e u ) ---------------------------- - X= (EQ 2-33) ku 2.6.8 CONCLUSIONS The final integrations (Eqs. 24, 24, and 28) are not the ultimate goal of pharmaco- kinetics. From them come the concepts of: (a) elimination half-life 1. (b) apparent volume of drug distribution 1. (c) plateau drug concentrations 2. These, and other concepts arising from still other equations, are clinically useful. Once the method of LaPlace Transforms is mastered, it becomes easy to derive equations given only the required pharmacokinetic scheme. Under such circum- stances, it no longer becomes necessary to remember a multitude of equations, many of which, though very similar, differ markedly in perhaps one minute detail. As with any new technique, practice is required for its mastery. In this case, mas- tery will banish the “calculus blues.” It is also possible to see certain patterns which begin to emerge from the derivation of the equations. For example, for a drug given by IV bolus the equation is monoexponential, with the exponent being K, summation of all the ways that the drug is removed form the body. A graph of the data (Cp v T on semi-log paper) results in a straight line the slope of which is K, always. If the drug is entirely metabolized K = km. If the drug is entirely excreted unchanged into the urine, K = ku. If the drug is metabolized and excreted unchanged into the urine, K = km +ku. thus K can have different meanings for different drugs, depending on how the body removes the drug. Following the drug given by IV bolus a second example of a pattern would be that of the data of the metabolite of the drug. From the LaPlace, the equation for the plasma concentration of the metabolite of the drug has in it K and K1, the summation of all the ways that the metabolite is removed from the body, always. K1 would have different meanings depending on how the metabolite is removed from the body. 2-57 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review After several years teaching, I was fortunate to have a resident rotate through our pharmacokinetic site. She had come with a strong Pharmacokinetcs background and during our initial meeting, she had told me that she had a copy of John Wag- ner’s new textbook on pharmacokinetcs. She was excited that, finally, there was a compilation of all the equations used in pharmacokinetics in one place. “There are over 500 equations in the new book and I know every one,” she said. “I’m not sure which one goes with which situation, though.” OOPS! Throughout this text and on each exam, each equation is derived from first princi- ples using scientific method, modeling and LaPlace Transforms in the hopes that memorization will be minimized and thought (and consequently proper interpreta- tion) would be maximized. 2-58 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.6.9 TABLE OF LAPLACE TRANSFORMS A, B are constants a, b, c are rate constants ( a ≠ b ≠ c ) s is the LaPlace Operator is a variable, dependent on time ( t ) x m is a power constant TABLE 2-17 Table of LaPlace Transforms Time Function, F ( t ) LaPlace Transform, f ( s ) A A -- - s At A --- - 2 s A ( m! - ) m At -------------- m+1 s – at A- Ae ---------- s+a m – at A At e -------------------------- - m+1 (s + a) – at A A(1 – e ) ------------------ - -------------------------- s (s + a ) a – at A- ---- – A ( 1 – e ) At -------------------------- -------------------- - 2 s (s + a) 2 a a – at As – B- B( 1 – e ) ------------------ – at – -------------------------- s (s + a ) Ae a – at As – B -  A + B  1 – e - – Bt -------------------- -- ----------------- - ---- -  a  a  a 2 s (s + a) 2-59 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review TABLE 2-17 Table of LaPlace Transforms Time Function, F ( t ) LaPlace Transform, f ( s ) – bt – at A A( e – e ) -------------------------------- - -------------------------------- - (s + a)(s + b) (a – b) – bt – at A ----------------  1 e - –  ----------------- A – 1–e - ----------------------------------- - ----------------- s( s + a )( s + b ) (a – b)  b   a  A  1 – e –bt  – at ------------------------------------- - ----- – ----------------  ----------------- –  1 – e - At A - - ----------------- 2 s ( s + a) ( s + b) ab ( a – b )  b2   a2  1 As – B - ---------------- [ ( B + Aa )e – at – ( B + Ab )e –bt ] -------------------------------- ( a – b) (s + a)(s + b) As – B -   – bt – at ---------------- A ( e– bt – e –at ) – B  ( 1 – e ) – ( 1 – e )  ----------------------------------- 1 s( s + a )( s + b ) ---------------------- ---------------------- (a – b) b a   As – B  B  ----------------------   – bt – at ----------------  A + --  ( 1 – e ) –  A + B ( 1 – e ) – ----- ------------------------------------- - 1 Bt - -- ---------------------- - - 2 s ( s + a) ( s + b) (a – b)  b b a a ab  – ct – bt – at A e e e ------------------------------------------------- - A -------------------------------- + -------------------------------- + -------------------------------- - - - (s + a)(s + b)(s + c) (a – c)(b – c) (a – b )(c – b) (b – a )(c – a) – ct – bt – at A (1 – e ) - (1 – e ) - (1 – e ) - ---------------------------------------------------- A ----------------------------------- + ------------------------------------ + ------------------------------------ s( s + a )( s + b )( s + c ) c( a – c )( b – c ) b( a – b)( c – b ) a( b – a )( c – a) A – ct – bt – at (1 – e ) (1 – e ) - (1 – e ) - ------------------------------------------------------ - At ----- – A -------------------------------------- + -------------------------------------- + -------------------------------------- - 2 s ( s + a) ( s + b ) ( s + c ) 2 2 2 bc c ( a – c) ( b – c ) b ( a – b) ( c – b) a ( b – a )( c – a) dX sX – X 0 ------ - dt (m + 1 ) A At ----- ------------------- m (m + 1) s 2-60 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.6.10 LAPLACE TRANSFORM PROBLEMS By means of the LaPlace Transform, find the equation for: The amount of drug in the body when the drug is given by IV Bolus (assume no metabolism). 1. The amount of drug in the urine when the drug is given by IV Bolus (assume no metabolism). 2. The amount of metabolite in the body when the drug is given by IV Bolus (assume no parent 3. drug excretion) The amount of metabolite of the drug in the urine when the drug is given by IV Bolus (assume 4. no parent drug excretion) The amount of metabolite of the drug in the urine when the drug is given by IV Bolus (assume 5. both parent drug and metabolite excretion) The amount of drug in the body when the drug is given by IV infusion (assume no metabolism). 6. The amount of drug in the urine when the drug is given by IV infusion (assume no metabolism). 7. The amount of metabolite in the body when the drug is given by IV infusion (assume no parent 8. drug excretion). The amount of metabolite in the urine when the drug is given by IV infusion (assume no parent 9. drug excretion). The Rate of excretion of the metabolite into the urine for a drug given by IV bolus when 10. km+ku=kmu. The amount of the principle metabolite (Xm1) when the drug is eliminated by several pathways 11. (Xu, Xm1,Xm2,Xm3,etc) X- The concentration of drug, ----- , in the body when the drug is given orally by a delivery system 12. Vd which is zero order. What is the concentration at equilibrium ( T ∞ ). The amount of metabolite of a drug in the body when the drug is given by IV Bolus and con- 13. comitant IV infusion. Disopyramide (D) is a cardiac antiarrythmic drug indicated for the suppression and prevention 14. of ectopic premature ventricular arrythmias and ventricular tachycardia. It appears that disopy- ramide is metabolized by a single pathway to mono-dealkylated disopyramide (MND). In a recent study, the pharmacokinetics of disopyramide were attempted to be elucidated by means of a radioactive tracer. Since both D and MND would be labeled by the tracer, any equations showing the time course of the label would show both the D and MND. By means of the laPlace transform, find the equation for the rate of appearance of tracer into the urine if the drug were given by IV Bolus. 2-61 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2.6.11 LAPLACE TRANSFORM SOLUTIONS 1. The amount of drug in the body when the drug is given by IV bolus (assume no metabolism). Dose X ku Xu X = Xo At time zero, all of the IV bolus is in the com- partment. Here K = ku dX ------ = – k u X - dt sX – X0 = –kuX sX + kuX = X o X ( s + k u )= Xo Xo X = ------------- s + ku A- Let ( X0 ) = A , k u = a ,X = --------------- (s + a) – kut X = Xo e 2-62 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 2. the amount of a drug in the urine when the drug is – kut Xu= Xo ( 1 – e ) given by IV bolus (assume the drug is NOT metabolized Dose ku X Xu NOTE: You must start where the drug is at time = 0. Again K = ku dX ------ = – k u X - dt sX – X0 = –kuX sX + kuX = X o X ( s + k u )= Xo Xo X = ------------- You only need to go this far because you s + ku need to know X to put it in the equation for Xu. Thus: dXu -------- = k u X - dt sX u – Xuo = k u X (Xuo = 0. No drug in urine at time = 0) sX u = k u X Substituting from above for X. X o ku sX u = ----------------- - (s + ku) Xo ku X u = -------------------- - s(s + ku) A- Let ( X0 k u ) = A , k u = a , X u = ------------------ s( s + a ) – kut ku X o ( 1 – e ) X u = ------------------------------------ - ku 2-63 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 3. the amount of metabolite of a drug in the body km X 0 sX m = -------------- – k mu Xm Substituting for X from above when the drug is given by IV bolus (assume no parent s + km drug excretion). km ⋅ Xo X m = ----------------------------------------- - ( s + k mu ) ( s + k m ) Dose Let ( k m X 0 ) = A , k m = a = K , k mu = b = K1 A X m = -------------------------------- - (s + a)(s + b ) X ( km ⋅ X o ) –k t –k t X m = ------------------------ ⋅ ( e mu – e m ) or - ( k m – k mu ) km ( km ⋅ Xo ) – K 1t –K t X m = --------------------- ⋅ ( e – e ) in general terms. Xm kmu Xmu - ( K – K1 ) Remember: NOTE: We could also: You must start where the drug is at time = 0. Let ( k m X 0 ) = A , k m = b , k mu = a Here K = km and K1 = kmu and then dx ( km ⋅ X o ) ----- = – k m X - –k t –k t dt X m = ------------------------ ⋅ ( e m – e mu ) or - ( k mu – k m ) sX – Xo = – k m X ( km ⋅ Xo ) –Kt – K1t X m = --------------------- ⋅ ( e – e ) - sX + k m X = Xo ( K1 – K ) X ( s + km ) = Xo Both of those equations are identical. Xo X = -------------- s + km A- Let ( X0 ) = A , k m = a , X = --------------- (s + a) – kmt X = Xo e dXm --------- = k m X – k mu X m - dt sX m – Xm0 = k m X – kmu X m 2-64 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 4. the amount of metabolite of a drug in the urine X om = 0 when the drug is given by IV bolus (assume the parent compound is not excreted). sX mu = k mu X m Here, again, K = km and K1 = kmu ( k mu ) ( k m X o ) sX mu = ----------------------------------------- - ( s + k mu ) ( s + k m ) Dose ( k mu ) ( k m ) ( X o ) X mu = -------------------------------------------- - s ( s + kmu ) ( s + k m ) X k mu k m Xo  1 – e –kmt 1 – e –kmut X mu = --------------------  -------------------- – ----------------------- - - k mu – k m  k m k mu  km Xm kmu Xmu dXm ---------- = k m X – k mu X m - dt sX m – Xom = k m X – kmu X m sX m + k mu Xm = k m X X o substitute previously solved X = -------------- s + km km Xo X m ( s + k mu ) = -------------- s + km km X o X m = ----------------------------------------- - ( s + k mu ) ( s + k m ) – kmt – kmut km Xo ( e ) –e X m = -------------------------------------------------- - ( k mu – k m ) dXmu ------------ = k mu Xm - dt sX mu – Xom = k muX m Remember at time zero, 2-65 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 5. the amount of metabolite of a drug excreted in dX mu ------------ = k mu Xm - the urine when both the parent and metabolite are dt excreted. Here K = km + ku and K1 = kmu sX mu – Xom = k muX m = K1Xm sX mu = K1X m Dose ( k mu ) ( kmXo ) sX mu = -------------------------------------- ( s + K1 ) ( s + K ) ku Xu ( k mu ) ( k m ) ( Xo ) X X mu = ---------------------------------------- - s ( s + K1 ) ( s + K ) km k mu k m X o 1 – e – Kt 1 – e – K1t X m u = ---------------------  ------------------ – --------------------- - - K1 – K  K K1  Xm kmu Xmu dX ------ = – k u X – k m X - dt sX – Xo = – k u X – k m X sX + k u X + k m X = X o X ( s + ku + km ) = Xo ( ku + km ) = K Xo X = ----------- - s+K – Kt X = Xo e dXm ----------- = k m X – k mu Xm = k m X – K1Xm - dt sXm – X om = k m X – K1Xm sX m + K1Xm = k m X km Xo X m = -------------------------------------- ( s + K1 ) ( s + K ) 2-66 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 6. The amount of drug in the body from a drug given by IV infusion (assume no metabolism). Q X ku Xu At time zero, all the drug is still in the IV bag, therefore there is no drug in the body. X0 = 0 Here K = ku dX ------ = Q – k u X - dt Q sX – X0 = --- – K u X - s Q sX + k u X = --- - s Q X = -------------------- - s ( s + ku ) Q –k t X = ---- ( 1 – e u ) or X = Q ( 1 – e –Kt ) - --- - ku K 2-67 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 7. the amount of drug in the urine when the drug is given by infusion (assume the drug is NOT metabolized). Q X ku Xu Here K = ku dX ------ = Q – k u X - dt Q sX – Xo = --- – k u X - s Q sX + k u X = --- - s Q X ⋅ ( s + k u ) = --- - s Q- X = ------------------------ s ⋅ (s + ku ) dXu -------- = k u X - dt sX u – Xo = k u X sX u = k u X ku Q sX u = ------------------------ - s ⋅ (s + ku) ku Q X u = --------------------------- - 2 s ⋅ ( s + ku )  1 – e –kut k u Qt X u = ----------- – Qk u  ------------------- -  k2  ku u – kut – Kt (1 – e ) (1 – e ) X u = Qt – Q ------------------------ or Xu = Qt – Q ---------------------- - - ku K 2-68 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 8. the amount of metabolite of a drug in the body – bt – at X m = ----------------  ----------------- –  ----------------- 1–e - 1–e - A from a drug given by IV infusion (assume no parent drug (a – b)  b   a  excretion) Here K = km and K1 = kmu    1 – e – k mu t –km t km Q ------------------------  1 – e - –  --------------------- or - ------------------- Xm = ( k mu – k m )  k m   k mu  Q X – Kt – K1t km Q ---------------------  ------------------ –  --------------------  1–e 1–e - - Xm = ( K1 – K )  K   K1  km Xm kmu Xmu . dX ------ = Q – k m X - dt Q sX – Xo = --- – k m X - s Q sX + k m X = --- - s Q X ( s + k m ) = --- - s Q X = --------------------- - s ( s + km ) – kmt (1 – e ) X = Q ------------------------- - km dXm --------- = k m X – k mu X m - dt sX m – Xo = k m X – k mu Xm sX m = k m X – k mu Xm km Q X m = -------------------------------------------- - s ( s + k mu ) ( s + k m ) 2-69 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 9. the amount of metabolite of a drug in the urine dX mu ------------ = k mu Xm - from a drug given by IV infusion (assume that the parent dt compound is not excreted). Here K = km and K1 = kmu sX mu – Xomu = k mu X m substitute X m k mu k m Q X mu = ---------------------------------------------- - Q 2 s ( s + k m ) ( s + k mu ) X k mu k m Qt k mu k m Q  ( 1 – e –kmut ) ( 1 – e – k mt ) X mu = --------------------- –  --------------------   ---------------------------- – -------------------------- or km - - - k m ⋅ k mu  k m – k m u   2 2 k k mu m k mu k m Q  ( 1 – e –Kt ) ( 1 – e –K1t ) Xm kmu Xmu k mu k m Qt X mu = --------------------- –  -------------------   ----------------------- – -------------------------  - - - -  K1 – K   K1 ⋅ K  2 2 K K1 dX ------ = Q – k m X - dt Q sX – Xo = --- – k m X - s Q sX + k m X = --- - s Q X ( s + k m ) = --- - s Q X = --------------------- - s ( s + km ) dXm --------- = k m X – k mu X m - dt sX m – Xo = k m X – k mu Xm sX m + k mu X m = k m X km Q X m ( s + k mu ) = --------------------- - s ( s + km ) km Q X m = -------------------------------------------- - s ( s + k m ) ( s + k mu ) km Q  – kmut  – kmt --------------------  1 – e ----------------------- – 1 – e - -------------------- Xm = k m – k mu  k mu   k m  2-70 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 10. the rate of excretion of the metabolite into the dX m --------- = ( k m X – k muX m ) - urine for a drug given by IV bolus when dt k m + k u = k mu sX m – Xom = ( k m X – K1Xm ) In this case, K = ku +km and K1 = kmu and thus K = K1. This is not normal but could happen. The problem arises sX m + K1X m = k m X when we get to the LaPlace that assumes the rate con- stants are different (i.e. a ≠ b ) because for this special km X o X m ( s + K1 ) = ----------- - case a = b . s+K km Xo X m = -------------------------------------- (remember- K1 = K) ( s + K ) ( s + K1 ) Dose km Xo X m = ----------------------------------------- - ( s + K1 ) ( s + K1 ) ku Xu km Xo X X m = ---------------------- - (kmX0 = A) 2 ( s + K1 ) km – K1t X m = k m Xo te Xm kmu Xmu dX mu ------------ = k mu Xm - dt dX mu ------------ = k mu k m Xo te –K1t dX - . ------ = – k u X – k m X - dt dt sX – Xo = – k u X – k m X sX + k u X + k m X = X o X ( s + ku + km ) = Xo Xo X = ------------------------------ - ( s + k u + km ) K = ( ku + km ) k mu = K1 Xo X = ---------------- - (s + K) – Kt X = Xo e 2-71 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review the principal metabolite ( X m1 ) when the drug is 11. k m1 Xo X m1 ( s + K1 ) = ---------------- - cleared by several pathways ( X u, X m 1, X m 2 ) (s + K) In this case K = km1 + km2 + ku, K1 = kmu1 and K2 = k m1 X o X m1 = -------------------------------------- kmu2 ( s + K1 ) ( s + K ) k m1 Xo – K1t –Kt X m1 = ---------------- ( e –e ) - K – K1 Xm1 kmu1 Xmu1 Dose km1 ku X Xu km2 kmu2 Xmu2 Xm2 dX ------ = – k u X – k m1 X – k m2 X - dt sX – Xo = – k u X – k m1 X – k m2 X sX + k u X + k m1 X + km2 X = Xo X ( s + k u + k m1 + k m2 ) = Xo Let K = ku + km1 + km2 Xo X = ---------------- - (s + K) K = ( k u + k m1 + k m2 ) and K1 = k mu1 dXm1 ------------ = k m1 X – K1X m1 - dt sX m1 – Xm1o = k m1 X – K1X m1 X m1o = 0 k m1 Xo sX m1 + K1X m1 = ---------------- - (s + K) 2-72 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review the concentration of drug X ⁄ Vd in the body 12. k a Q  ( 1 – e –Kt ) ( 1 – e –kat )  X = -------------------  ----------------------- – ------------------------  - - - when the drug is given orally by a delivery system which ( ka – K )  K ka  is zero order. What is the concentration in the body at equilibrium ( t ∞ )  ( 1 – e –Kt ) ( 1 – e –kat )  ka Q Here K = ku C = --------------------------  ----------------------- – ------------------------  - - - ( k a – K )Vd  K ka  Q Xa ka ku Xu   kQ X - 1 1- If t= ∞ , then e –kt = 0 , thus C = --------------------------  --- – ----  a - ( k a – K )Vd  K ka  Q simplified yields: C = ---------- - KVd dXa -------- = Q – k a X a - dt Q sX a – Xao = --- – k a Xa - s X a0 = 0 sX a + k a Xa = ( Q ⁄ s ) X a ( s + ka ) = ( Q ⁄ s ) Q- X a = -------------------- s(s + ka) – kat Q( 1 – e ) X a = ----------------------------- ka dX ------ = k a X a – KX - dt sX – Xo = k a Xa – KX Xo= 0 ka Q sX + KX = -------------------- - s ( s + ka ) ka Q X ( s + K ) = -------------------- - s ( s + ka ) ka Q X = -------------------------------------- - s ( s + ka ) ( s + K ) 2-73 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review the metabolite of a drug in the body Xm given 13 k m Q  ( 1 – e –K1t ) ( 1 – e –Kt )  X m = ---------------------  ------------------------- – ----------------------  by IV infusion and concomitant IV bolus dose. - - - ( K – K1 )  K1 K  Infusion: IV Bolus: Here K = km and K1 = kmu dX ------ = – KX - dt Dose sX – Xo = – KX sX + KX = X o Q X X ( s + K )= Xo km Xo X = ----------- - s+K Xm kmu Xmu dX m --------- = k m X – K1X m - dt dX sX m – Xo = k m X – K1X m ------ = Q – KX - dt sX m + K1X m = k m X Q sX – Xo = --- – KX - s km X o X m ( s + K1 ) = ----------- - Q s+K sX + KX = --- - s km Xo X m = -------------------------------------- Q ( s + K1 ) ( s + K ) X ( s + K ) = --- - s km Xo – Kt – K 1t X m = --------------------- ( e – e ) - Q- ( K1 – K ) X = ------------------- s( s + K ) Thus, dXm --------- = k m X – K1X m - km X o X m =  --------------------- ( e – e - –Kt –K 1t ) + … dt  ( K1 – K )  sX m – Xo = k m X – K1Xm  k m Q  ( 1 – e –Kt ) ( 1 – e –K1t )   ----------------  ---------------------- – -------------------------  - - - sX m + K1X m = k m X  K1 – K   K K1 km Q X m ( s + K1 ) = ------------------- - s( s + K ) km Q X m = ---------------------------------------- - s ( s + K1 ) ( s + K ) 2-74 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/
• Mathematics Review 14. By means of the LaPlace transform, find the km X o X m ( s + K1 ) = ----------- - equation for the rate of appearance of the tracer in the s+K urine if the drug were given by IV bolus. km Xo Here K = ku + km and K1 = kmu X m = -------------------------------------- ( s + K ) ( s + K1 ) Dose km Xo - –K1t – e –Kt } X m = --------------------- { e ( K – K1 ) k mu k m Xo dX mu ku Xu ------------ = k mu Xm = --------------------- { e –K1t – e – Kt } - - X ( K – K1 ) dt k mu k m X o dX u dX mu -------- + ------------ =  k u ( X o e ) +  --------------------- { e – e } – Kt –K1t – Kt km - - -   ( K – K1 )  dt dt Xm kmu Xmu dX ------ = – k u X – k m X - dt sX – Xo = – k u X – k m X sX + k u X + k m X = X o X ( s + ku + km ) = Xo ( ku + km ) = K Xo X = ---------------- - (s + K) – Kt X = Xo e dXu -------- = k m X = k u ( Xo e –Kt ) - dt dXm --------- = k m X – K1X m - dt sX m – Xo = k m X – k1Xm km Xo sX m + K1X m = ---------------- - (s + K) 2-75 Basic Pharmacokinetics REV. 00.1.6 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/