Basic Pharmacokinetics-ch2: Mathematics Review

CHAPTER 2

Mathematics Review

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 simulations 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 variables throughout the model.

10.

The student will correlate (V) the graphs of the data with the equations and models so generated.

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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 is just a review. Look it over. You should be able to do all of these manipulations.

This chapter is meant to review the concepts in mathematics essential for understanding kinetics. These concepts are generally taught in other mathematical courses from algebra through calculus. For this reason, this chapter is presented as a review rather than new material. For a more thorough discussion of any particular concept, refer to a college algebra or calculus text. Included in this section are discussions of algebraic concepts, integration/differentiation, graphical analysis, linear regression, non-linear regression and the LaPlace transform. Pk Solutions is the computer program used in this course.

Something new LaPlace transforms. Useful tool.

A critical concept introduced in this chapter is the LaPlace transform. The LaPlace transform is used to quickly solve (integrate) ordinary, linear differential equations. The Scientist by Micromath Scientific Software, Inc.1 is available for working with the LaPlace transform for problems throughout the book.

1. MicroMath Scientific Software, Inc., P.O. Box 21550, Salt Lake City, UT 84121-0550,

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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.

How Does Scientific Notation Work?

316000 becomes 3.16 × 10

Examples:

0.00708 becomes 7.08 ×10

5

–3

In general a number takes the form: A × 10

n

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 indices, mathematical operations performed on numbers expressed in this way are subject 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

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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

Value

Figures

Significant Figures

Number of Significant Figures

(a)

572

2,5,7

3

(b)

37.10

0,1,3,7

4

(c)

10.65 x 104

0,1,6,5

4

(d)

0.693

3,6,9

3

(e)

0.0025

2,5

2

How do I determine the number of significant figures?

Examples (c) to (e) illustrate the exceptions to the above general rule. The value 10 raised to any power, as in example (c), does not contain any significant 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.

What do significant figures mean?

Significant figures are used to indicate the precision of a value. For instance, a value recorded to three significant figures (e.g., 0.0602) implies that one can reliably 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 significant figures that fairly represent its precision. Stating the result to more significant figures than you can justify is misleading, at the very least!

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2.2.4

RULES OF INDICES

What is an index?

n

An index is the power to which a number is raised. Example: A where A is a number, which may be positive or negative, and n is the index, which may be positive 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 n

m

A ×A = A

n+m

n

A ×B

m

n+m A n =  --- × B B

(b) Division n

An–m -----= A m A

n

n n–m AA -------- × B = m  B B

(c) Raising to a Power n m

(A )

= A

nm

There are three noteworthy relationships involving indices: (i) Negative Index A

–n

1 –n = -----n- As n tends to infinity ( n → ∞ ) then A → 0 . A

(ii) Fractional Index 1 --n

A =

n

A

(iii) Zero Index 0

A = 1

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2.2.5

LOGARITHMS

What is a logarithm?

Some bodily processes, such as the glomerular filtration of drugs by the kidney, 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 A = 10

n

then log ( A ) = n

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 , and 7.13 = 10

0.85

, thus 713000 becomes 10

0.85

5

× 10 = 10

( 5 + 0.85 )

= 10

5.85

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. What is the characteristic? the mantissa?

2.2.6

The number before the decimal point is called the characteristic and tells the placement 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 numbers discounting the decimal place.

NATURAL LOGARITHMS

What is a natural logarithm?

Instead of using 10 as a basis for logarithms, a natural base (e) is used. This natural 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 ∞

e = 2.7182818 ... Strictly speaking, e = 1 +

1 Where x is an inte-

∑ ---x!x=1

ger ranging from 1 to infinity ( ∞ ) ,

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∞

∑

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. How are natural logarithms ln x and common logarithms log x related?

Common and natural logarithms are related as follows: ln ( A ) = 2.303 × log ( A ) , and log ( A ) = 0.4343 × ln ( A ) Because logarithms are, in reality, indices of either 10 or e , their use and manipulation follow the rules of indices (See Section 2.2.4). (a) Multiplication: n

m

n

To multiply N × M , where N = e and M = e ; NM = e × e By definition,

m

= e

n+m

.

ln ( NM ) = n + m ; but

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.

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(b) Division N ln  ----- = ln ( N ) – ln ( M )  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 ln ( N

–m

1 ) = – m × ln ( N ) = m × ln  ---- N

As m tends to infinity ( m → ∞ ) , then ln ( N

–m

) → –∞

(ii) Number Raised to a Fractional Power 1

ln ( m

 --m 1 N ) = ln  N  = ---- × ln ( N ) m  

(iii) Logarithm of Unity ln ( 1 ) = log ( 1 ) = 0

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2.2.7

NEGATIVE LOGARITHMS The number 0.00713 may be expressed as: –3

7.13 ×10 , or 10 10

0.85

–3

× 10 , or

– 2.15

.

Hence, log ( 0.00713 ) = – 2.15 , which is the result generated by most calculators. However, another representation of a negative logarithm (generally used by referencing 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 logarithm; 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 antilogarithmic tables, and hence a positive mantissa must be the end result of any logarithmic 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.

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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 ;

1.

convert 62.54 to scientific notation --->

2.

Look up the mantissa for 6254 in a table of logarithms: it is 7962.

3.

1

Hence, 6.254 ×10

= 10

0.7962

1

× 10 = 10

1.7962

and

log ( 62.54 ) = 1.7962

(b) Find the log of (0.00329) –3

1.

0.00329 = 3.29 ×10

2.

The mantissa for 329 is 5172

3.

Hence, log(0.00329) = 3.5172.

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. How do I multiply using logarithms?

(c) Multiply 62.54 by 0.00329 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 (mantissa) in a table: it is 2058. Antilog (1.3134) = 2.058 ×10

–1

Hence, antilog (1.3134) = 0.2058 How do I divide using logarithms?

log (62.54) - log (0.00329) = 1.796 - 3.5172=1.796 +3 - 0.5172=4.2788 antilog 4.2788 = 19002

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2.2.9

DIMENSIONS

What is a unit?

There are three fundamental dimensions which are used in various combinations to express the properties of matter. Each of these dimensions has been assigned a definite basic unit, which acts as a reference standard. TABLE 2-2 Dimensions

Dimension

Dimensional Symbol

Unit

Unit Symbol

Length

L

meter

m

Mass

M

gram

g

Time

T

second

sec

How are units made bigger and smaller?

In the metric system, which emerged from the French Revolution around 1799, there are various prefixes which precede the basic units and any derived units. The prefixes indicate the factor by which the unit is multiplied. When the index of the factor is positive the prefixes are Greek and have hard, consonant sounds. In contrast, when the index is negative, the prefixes are Latin and have soft, liquid sounds. (see Table 2-3).

How big is big?

Examples: An average adult male patient is assumed to have a mass of 70 kilograms (70 kg). An average adult male patient is assumed to have a height of 180 centimeters (180 cm). A newly minted nickel has a mass of 5.000 g. Doses of drugs are in the mg (10-3 g) range (occasionally g) never Kg (103 g) or larger. Students have told me that the dose that they have calculated for their patient is 108 g (converting to common system - ~ 100 tons). I doubt it. Get familiar with this system. Note that the plural of Kg or cm is Kg or cm; do not add an “s”. In pharmacy there are two derived units which are commonly used, even though they are related to basic units. The Liter (L) is the volume measurement and is a cube 10 cm on a side (1L = (10cm)3 = 1000 cm3 ) while the concentration measurement and has the units of Mass per Volume.

Why should I use units?

Whenever the magnitude of a measured property is stated, it is imperative to state the units of the measurement. Numbers are useless by themselves. Example: The procainamide concentration range is 4-8 µ g/ml; stating the range without units may lead to a potentially lethal error in which procainamide is administered in a sufficient dose to attain a range of 4-8 mg/ml, which is 1000 times too large and would give rise to cardiac arrest.

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TABLE 2-3 Scale

Name

Symbol

exa-

E

peta-

P

tera-

T

giga-

G

mega-

M

kilo-

k

hecto-

h

deca-

da

of Metric system and SI Multiplication Factor 10 10 10

Name

Symbol

18

deci-

d

15

centi-

c

12

milli-

m

9

micro-

µ

6

nano-

n

3

pico-

p

2

femto-

f

1

atto-

a

10 10 10 10 10

Multiplication Factor 10 10 10 10 10 10 10 10

–1 –2 –3 –6 –9

–12 –15 –18

TABLE 2-4

Dimension

Dimensional Symbol

Unit

Unit Symbol

Volume

V

liter

l

Concentration

C

grams/liter

g/l

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2.2.10

DIMENSIONAL ANALYSIS

How are units useful?

It is a general rule that the net dimensions (and units) on the two sides of any equation 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 ) • K is the apparent first-order rate constant for drug elimination (Dimension: T

–1

),

• and τ is the dosing interval (Dimension: T )

Writing the dimensions relating to the properties of the right-hand side of the equation 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.

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2.3 Calculus What is Calculus?

Calculus concerns either the rate of change of one property with another (differential calculus), such as the rate of change of drug concentrations in the blood with time since administration, or the summation of infinitesimally small changes (integral calculus), such as the summation of changing drug concentrations to yield an assessment of bioavailability. In this discussion a few general concepts will be provided, and it is suggested an understanding of graphical methods should precede this discussion.

2.3.1

DIFFERENTIAL CALCULUS

2.3.2

NON-LINEAR GRAPHS 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

3

As can be seen from the graph (Figure 2-1), a non-linear plot is produced, as expected. FIGURE 2-1.

y=x3

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?)

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2.3.3

SLOPE OF NON-LINEAR GRAPH As with a linear graph, y2 – y1 ∆y ---------------- = -----x2 – x1 ∆x 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

x1

x2

∆x

y1

y2

∆y

∆y -----∆x

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

As may be seen, the value of the slope

∆ -----y-  ∆x 

tends towards a value of 12.000 as the

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.

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Calculus deals with infinitesimally small changes. When the value of ∆x is infinitesimally 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 function 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: y = x

3

(b) Let the value of y increase to y + dy because x increases to x + dx . Hence, y + dy = ( x + dx )

3

(EQ 2-13)

Multiplying out: 3

2

2

y + dy = x + 3x ( dx ) + 3x ( dx ) + ( dx )

3

(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

dy = 3x ( dx ) + 3x ( dx ) + ( dx )

3

(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 limiting value of this tendency must be dx = 0 . At this limit,

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dy ------ = 3x 2 dx

(EQ 2-16)

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: If y = Ax

n

then dy ------ = nAx n – 1 dx The Rules of Indices may need to be used to obtain expressions in the form y = Ax

n

(e.g., if y =

2.3.7

5

x)

THREE OTHER DERIVATIVES 0

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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. (c) Accept that if y = Be Ax dy base then ------ = ABe dx

Ax

where B and A are constants, and e is the natural

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 n–1 ------ = nAx 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.

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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 2.3.10

= y , where

∫ is the symbol for integration.

RULE OF INTEGRATION The derivative expression may be written: dy ------ = Ax n , or dx n

dy = Ax ⋅ dx To integrate, y =

∫ dy

=

∫ Ax

n

dx = A ∫ x dx n

A general rule states: n+1

n Ax A ∫ x dx = ---------------- + A 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

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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 –1 y = A ∫ x dx = A ∫ --- dx 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

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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: c =

∫ dc

3

2

= 3t + A – 6t + B

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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 – Kt = – K ∫ dt =

1

- ⋅ dC p ∫ ----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 .

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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, Cp = ( C p )0 ⋅ e

– Kt

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2.4 Graphs Why do we graph?

We would like to organize the chaotic world around us so that we can predict (see into the future) and retrodict (see into the past) what will happen or has happened. Our recorded observations are collectively known as data. We make a theory about what we think is happening and that theory is expressed in an equation. That determines our paradigm of how we see the world. This paradigm is expressed as a graph. The language of science is mathematics and graphs are its pictures.

TABLE 2-6

English

What is a graph?

Science

Observations

Data

Theory

Equations

Paradigms (pictures)

Graphs

A graph is simply a visual representation showing how one variable changes with alteration of another variable. The simplest way to represent this relationship between variables is to draw a picture. This pictorializing also is the simplest way for the human mind to correlate, remember, interpolate and extrapolate perfect data. An additional advantage is it enables the experimenter to average out small deviations in experimental results (non-perfect, real data) from perfect data. For example: TABLE 2-7 Perfect

vs. Real data

Perfect

Real

-3

-5

-4.6

-2

-3

-3.4

-1

-1

-0.6

0

+1

+0.8

+1

+3

+3.4

+2

+5

+4.4

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FIGURE 2-3.

Plot of Perfect vs. Real data

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 relationship 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 finding 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

How are graphs made?

Certain conventions have been adopted to make the process of rendering a data set 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.

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Any point in the defined space of the graph has a unique set of coordinates: 1) the ‘x’ value which is the distance along the ‘x’ axis out from the ‘y’ axis and always comes first; and 2) the ‘y’ value which is the distance, along the ‘y’ axis up or down from the ‘x’ axis, and always comes second. Several points are shown in Figure 2. For example, (0,1) is on the line and (1,0) is not. The intersection of x and y axis is the origin with the coordinates of (0,0). In two dimensional spaces, the graph is divided into 4 quadrants from (0,0), numbered with Roman numerals from I through IV. It should be readily apparent that the coordinates for all points within a particular quadrant are of the same sign type i.e., TABLE 2-8 Quadrants

on a cartesian graph

Quadrant II (-x, +y)

Quadrant I (+x, +y)

Quadrant III (-x, -y)

Quadrant IV (+x, -y)

A line (or curve) on a graph is made up of an infinite number of points, each of which has coordinates that satisfy a given equation. For example, each point on the line in Figure 2 is such at its coordinates fit the equation y = 2x + 1 . That is for any value of x (the independent variable), multiplying the x value by 2 and adding 1 results in the y value (the dependent variable).

2.4.2

STRAIGHT LINE GRAPHS

What is a straight line?

A graph is a straight line (linear) only if the equation from which it is derived has the form y = mx + b Where: • y = dependent variable • x = independent variable -----y• m = slope of the straight line = ∆ ∆x

• b = the y intercept (when x = 0) • or if the equation can be “linearized”, e.g.,

y′ = b′e

mx′

is not linear. However. ln y′ = ln b′ + mx′

is of the same general form as: y = b + mx

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and consequently a plot of ln y′ (the dependent variable) versus x′ (the independent 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 ) . C p = C p0 e

– Kt

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 a single straight line, e.g., y = k 0 + k 1 x

( n – m)

+ k2 x

(n – m + 1)

+ … + ( k n )x

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.)

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2.4.3

THE SLOPE OF A LINEAR GRAPH (M)

What is the slope of a straight line?

From the equation a prediction may be made as to whether the slope is positive or negative. In the previous example, the slope is negative, i.e: m = – K TABLE 2-9 Sample

data of caffeine elimination

t (min)

µg C p  --------  mL

ln C p

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. FIGURE 2-4.

Plasma Concentration

( C p ) of caffeine over time

Caffeine Concentration (

g/mL) u

101

100

10-1 0

50

100

150

200

Time (min)

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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 )

TABLE 2-10

Linear Regression for data in table 2-9 2

X⋅Y

X

Y

X

12

1.322

144

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

ΣX = 59623

2

15.864

ΣXY = 114.257

2

( ΣX ) = 255025 Σy y = ------ = 0.5398 n

Σx x = ------ = 4.167 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 )

and the intercept would be b = In oder to find the ln C . b

Cpo = e = e

1.4229

0.5398 – ( – 0.01014 ⋅ 4.167 ) = 1.4229 . C p0 ,

the anti-ln of

b

Note that this is must be taken. i.e.

= 4.15

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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

Time (x)

2.4.5

slope data from figure 2-4

ln Conc. (y)

∆x

∆y

∆y -----∆x

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

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 calculations 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

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2.4.6

GRAPHICAL EXTRAPOLATIONS

How far can I predict?

It is dangerous to extrapolate on non-linear graphs, and it is unwise to extrapolate 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 number 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.

What good is a straight line?

By using all the experimental data points, calculations may be made to find the optimum values of the slope m, and the intercept, b. From these values the correlation 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

How many points are needed?

Any graph drawn from 2 points is scientifically invalid. Preferably, straight-line graphs should have at least 3 - 5 points, and non-linear graphs a few points more.

Can I discard points that don’t fit?

As a graph is a visual representation which enables the experimenter to average 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-

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tude that this can not be done. Consequently, it must be borne in mind that the relationship 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

0

2

4

6

8

10

10 8 6 4 2 0

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

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FIGURE 2-6.

Poorly presented graph

50 40 30 20 10 0 0

FIGURE 2-7.

4

8

12

16

20

Well arranged graph

25 20 15 10 5 0 1

2

3

4

5

P

2.4.10

6

7

(

8

9 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 .

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2.4.11

SEMILOGARITHMIC COORDINATES Exponential or logarithmic equations are very common in physical chemical phenomenon. One of the best ways of determining whether or not a given set of phenomenon 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

Concentration Time (hours) 0.25

1900

.75

1500

1.5

1300

4

900

6

600

8

390

ng -----ml

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. FIGURE 2-8.

Concentration (ng/ml) vs. time (hr)

2000 1800 1600 1400 1200 1000 800 600 400 200 0

1

2

3

4

5

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6

7

8

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FIGURE 2-9.

Log Concentration vs. time

3.200 3.100 3.000 2.900 2.800 2.700 2.600 2.500 2.400 2.300

Log Concentration - Time Curve

0

FIGURE 2-10.

1

2

3

4

5

6

7

8

5

6

7

8

Log concentration (on log scale) vs. time

10000

1000

100 0

1

2

3

4

Graphing is much easier because the graph paper itself takes the place of a logarithmic 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 spacing. It is used when a graph must be plotted as in the example (Figure 1-4) y = log [ C p ] and x = t .

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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 plotting. 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 semilogarithmic 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  1 ln  ------------------ln  ----------   ln y 1 – ln ( ( 1 ⁄ 2 )y 1 ) ( 1 ⁄ 2 )y 1 1⁄2 ln 2 0.693 - = ------------------------------ = --------------------- = ------------m = ---------------------------------------------= ------------t 1 – t2 t 1 – t2 t 1 – t2 t1 – t2 –t1 ⁄ 2 (in which case, t 2 – t 1 is called the half-life t½ ). Since because

0.693 m = ------------- = – k –t1 ⁄ 2

and

t1 < t2 ,

then

t1 – t2 = –t1 ⁄ 2

0.693 k = ------------- . t1 ⁄ 2

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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. Consequently 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, FIGURE 2-11.

Graphing using standard number spacing

40 30 20 10 0 0

5

10

15 20

25

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FIGURE 2-12.

Nonstandard (incorrect) graph

40 30 20 10 1 0 2

0

5

10

20

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 212 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 characteristic. 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 relegated to each cycle.

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FIGURE 2-13.

Logarithmic mantissa Logarithmic Plot 103 234

102

Y axis (units)

23.4

101 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: FIGURE 2-14.

0

Straight line going down on semi-log paper

1

2

3

4

5

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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.

FIGURE 2-15.

0

2

FIGURE 2-16.

0

Curved line which plateaus on semi-log paper.

4

6

8

10

12

Curved line which goes up and then straight down on semi-log paper.

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.

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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

IV Bolus Parent

dXu ---------- urine rate of excretion dt parent compound

Time (mid)

-K

Kr ⋅ X 0

IV Bolus Parent

Xu ∞ – Xu Cumulative urine

Time

-K

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

Xu Cumulative urine data parent compound

Time

kr Xu ∞ = -------------K ⋅ X0

IV Infusion Parent

Drug concentration parent compound

Time

Q Q ( C p )ss = ------------ = ---K⋅V cl

Usually urine data of this type (parent compound - IV bolus) is replotted and evaluated 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 summation 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.

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TABLE 2-16 Plot

Data Type

Type 3 examples Y axis

X axis

S1

S2

Intercept

IV Bolus Parent

Metabolite conc.

Time

-K small

-Klarge

km ⋅ X 0 ------------------------------------------------------( K l arg e – K smal l ) ⋅ V dm

IV Bolus Parent

dXmu --------------- excretion dt rate of metabolite into urine

Time (mid)

-K small

-Klarge

k mu ⋅ k m ⋅ X 0 ----------------------------------K l arg e – K smal l

Oral

Drug conc.

Time

-K small

-Klarge

k a ⋅ fX 0 --------------------------------------------------( K l arg e – K smal l ) ⋅ V d

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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. distributed, 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 management 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 therapeutic range. Pharmacodynamics refers to the relationship between the drug concentration 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 predictable relationship between drug concentration in the plasma (which we can measure) 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 person (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 absorption, 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.

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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

X

ka

The differential equations used result from the model which is our conceptualization of what is 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. How do we make a differential equation?

The picture that we build is made up of building blocks, consisting of the arrow 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 representation of a rate where the rate is the rate constant on the top of the arrow times what the tail of the arrow touches. Again, the rate constant, k , tells the magnitude of the rate,

k⋅X.

Consider the following simple chain:

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X1

k12

X2

k23

X3

The building blocks are k 12 ⋅ X1 and k 23 ⋅ X2 . Every arrow that touches the compartment 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 performed 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.)

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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 single 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 models (more boxes) are necessary to mathematically mimic the observed plasma versus 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.

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2.6 The LaPlace Transform Why do we need to know the LaPlace transform?

One of the important facets of biopharmaceutics is a familiarity with the principles of pharmacokinetics. This latter discipline describes the study of the dynamic processes by which the body “handles” or “disposes of” an administered drug. These processes (absorption, distribution, metabolism, and excretion (ADME) are dynamic in that they represent the time-dependent changes occurring to the drug. Thus, in pharmacokinetics the time course of these changes, which overall describe the fate of the administered drug, is described mathematically. If the mathematical principles are understood, it is then possible to use pharmacokinetics in clinical practice, such as the design of rational dosage regimens (T.S. Foster and D.U.A. Bourne, Amer. J. Hosp. Pharm., 34, 70-75 (1977). Understanding (Bloom’s level 4) is not simply memorizing (Bloom’s level 1) nor calculating using a memorized equation (Bloom’s level 3). The authors believe that the proper conceptualizing of the process and the subsequent derivation of the appropriate equations will lead to an understanding of the mathematical principles, and thus, a better, more optimal dosing regimen. Since a mathematical description of the time-dependent ADME processes is required, it becomes necessary to deal with their corresponding rate equations. Inevitably this will involve calculus (mainly integral calculus). However, the LaPlace Transform provides a method whereby calculus can be performed with minimal trauma. If a conscientious effort to learn the method is made and applied, a potentially serious obstacle (the fear of calculus) to the understanding and appreciation of biopharmaceutics will be removed. Indeed, many students will find they no longer fear integration and are thus free to comprehend the principles underlying pharmacokinetics, which, after all, is the primary aim. So, the LaPlace Transform is a tool which is of great assistance in pharmacokinetics; its utility and importance should not be lightly disregarded.

The LaPlace Transform: What Is It?

There is, of course, a theoretical background to the LaPlace Transform. However, it can be used without recourse to a complete theoretical discussion, though appropriate pharmaceutical use of the method is found in the following references: M. Mayersohn and M. Gibaldi, Amer. J. Pharm. Ed., 34, 608-614 (1970). M. Gibaldi and D. Perrier, “Pharmacokinetics”, Marcel Dekker, pp. 267-272 (1975). Basically, the LaPlace Transform is used to solve (integrate) ordinary, linear differential equations. In pharmacokinetics such equations are zero and first-order rate equations in which the independent variable is time. For instance, if a differential equation describing the rate of change of the mass of drug in the body with time is

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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 integrated equation. This ultimate process is analogous to taking an antilogarithm.

2.6.1

TABLE OF LAPLACE TRANSFORMS

A table of useful LaPlace transforms is given in Section 2.7. Page 2-56.

The replacement of expressions in one domain by their equivalents in another is accomplished by reference to tables. One column shows time domain expressions, 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 For example, when the LaPlace Transform is

2.6.2

A⁄s

2

, then f ( t ) is At .

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.

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2.6.3

CONVENTIONS USED IN DRAWING PHARMACOKINETIC SCHEMA. When drugs enter the body, they will encounter many different fates. It is important to set up the possible fates of the drug by creating a well thought out flow chart or scheme in order to follow all the events that are occurring in the body as described by the pharmacokinetic description of the drug. For example, a drug may be excreted unchanged or may undergo hepatic metabolism to yield active or inactive metabolites. All of these components are part of pharmacokinetics, which by definition, includes ADME (the Absorption, Distribution, Metabolism and Excretion of drugs), and must be considered. This flow chart becomes the backbone or the framework upon which to build the equations which describe the pharmacokinetics of the drug. The differential equations result as a direct consequence of the flow chart. Using Laplace transforms, the integration of these differential equations are simplified and provide the pharmacokineticist to (easily?) keep track of all of the variables in the equation. If the drug scheme or flow chart is set up incorrectly, this would have a definite negative impact or the expected equations (as well as the answers and your grade). Below are two examples of how to construct a flow chart. Note that not all drugs follow the same flow chart and it is quite possible that you will need only to use a portion of these examples when construction your own. In general, schema are relatively consistent in the placement of the compartments in relationship to one another. You might consider, for example a drug, given by IV bolus, which is metabolized and both the metabolite and the parent compound are excreted unchanged as shown below: Feces

Body

Urine

Dose kf

Parent Compound Xf

ku X

Xu

km kmf Metabolite

Xmf

kmu Xm

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Xmu

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Using the pharmacokinetic symbolism from chapter one, the compartments are named and placed: metabolites below (or above the plane of the parent compound): 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 pharmacokinetic 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.

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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

Dose

Xf

kf

kmu1

Xmu1

km1 ku

X

Xu

km2 Xmf2

kmf2

Xm2

kmu2

Xmu2

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 Xf

kf

ku

X

Xu

km1 Xmf1

kmf1

Xm1

kmu1

Xmu1

km2 Xmf2

kmf2

Xm2

kmu2

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Xmu2

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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 transformed 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.

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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)

X

ku

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 (compartment) 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 dt

(EQ 2-17)

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: 1. Get only one variable which changes with time 2. Get

X

(X)

on the left and everything else on the right.

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sX + k u X = X0 X ( s + k u ) = X0 X0 X = ------------s + ku LetX0 = A

Letk u = a

(EQ 2-19)

A X = ---------------(s + a)

(EQ 2-20)

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:

X = X0e

–ku t

(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.

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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 coming in and is positive. thus, dXu --------- = k u X dt

(EQ 2-22)

(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 = -------s

(EQ 2-25)

Note that there are two transformed dependent variables. One of them ( X ) can be replaced by reference to equation 2-19. ku X0 Xu = -------------------s(s + ku) Let ( k u X0 ) = A

Let ( k u ) = a

(EQ 2-26)

A X = ------------------s(s + a)

(EQ 2-27)

(d) Converting back to the time domain: –k t

kuX0 ⋅ ( 1 – e u ) X u = ---------------------------------------ku

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Where k u X 0 and k u are analogous to “A” and “a” respectively in the table. Simplifying, Xu = X0 ( 1 – e

2.6.7

– kut

)

(EQ 2-28)

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)

Infusion

Q

ku

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 dt

(EQ 2-29)

(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 variable (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 s

(EQ 2-30)

(c) Manipulating the transformed equation: Q X = --------------------s ( s + ku )

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(EQ 2-31)

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Let ( Q = A )

A X = ------------------s( s + a)

Let ( k u = a )

(EQ 2-32)

(d) Converting back to the time domain: –k t

Q(1 – e u ) X = ----------------------------ku

2.6.8

(EQ 2-33)

CONCLUSIONS The final integrations (Eqs. 24, 24, and 28) are not the ultimate goal of pharmacokinetics. From them come the concepts of: 1.

(a) elimination half-life

1.

(b) apparent volume of drug distribution

2.

(c) plateau drug concentrations

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 circumstances, 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, mastery 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.

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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 Wagner’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 principles using scientific method, modeling and LaPlace Transforms in the hopes that memorization will be minimized and thought (and consequently proper interpretation) would be maximized.

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2.6.9

TABLE OF LAPLACE TRANSFORMS

TABLE 2-17 Table

A, B

are constants

a, b, c

are rate constants ( a ≠ b ≠ c )

s

is the LaPlace Operator

x

is a variable, dependent on time ( t )

m

is a power constant

of LaPlace Transforms Time Function, F ( t )

LaPlace Transform, f ( s )

A

A --s

At

A--2 s

m

A ( m! -) -------------m+1 s

– at

A---------s+a

m – at

A -------------------------m+1 (s + a)

At Ae

At e

– at

A ------------------s (s + a )

A(1 – e ) -------------------------a – at

At (1 – e ) ----- – A -------------------------2 a a

A -------------------2 s (s + a)

– at

As – B-----------------s (s + a )

Ae

– at

B( 1 – e ) – -------------------------a – at

1–e A + B ---  ------------------ – Bt ----   a a  a

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As – B -------------------2 s (s + a)

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TABLE 2-17 Table

of LaPlace Transforms Time Function, F ( t ) – bt

LaPlace Transform, f ( s )

– at

A -------------------------------(s + a)(s + b)

A ( e – e )-------------------------------(a – b) – bt

– at

– e -  ----------------1 – e - A  1-------------------------------–   b a  (a – b)

A ----------------------------------s( s + a )( s + b )

 1 – e –bt  1 – e –at At- ---------------A ----–  ------------------ –  ------------------ ab ( a – b )  b2   a2 

A ------------------------------------2 s ( s + a) ( s + b)

1 – at – bt ---------------[ ( B + Aa )e – ( B + Ab )e ] ( a – b)

As – B -------------------------------(s + a)(s + b)

 ( 1 – e –bt ) ( 1 – e – at )  1 – bt – at ---------------A ( e – e ) – B  ---------------------- – ----------------------  (a – b) b a  

As – B ----------------------------------s( s + a )( s + b )

  ( 1 – e –bt )  ( 1 – e – at ) 1 B B Bt --------------------------------------A + – A + – ----- ---------------------( a – b )  b  b a a ab  

As – B -------------------------------------2 s ( s + a) ( s + b)

– ct

– bt

– at

e e e A --------------------------------- + --------------------------------- + --------------------------------(a – c)(b – c) (a – b )(c – b) (b – a )(c – a) – ct

– bt

– at

(1 – e ) (1 – e ) (1 – e ) A ------------------------------------ + ------------------------------------- + ------------------------------------c( a – c )( b – c ) b( a – b)( c – b ) a( b – a )( c – a)

A ------------------------------------------------(s + a)(s + b)(s + c) A ---------------------------------------------------s( s + a )( s + b )( s + c )

At(1 – e ) ( 1 – e ) - -------------------------------------(1 – e ) ----– A -------------------------------------+ -------------------------------------+ 2 2 2 bc c ( a – c) ( b – c ) b ( a – b) ( c – b) a ( b – a )( c – a)

A -----------------------------------------------------2 s ( s + a) ( s + b ) ( s + c )

dX ------dt

sX – X 0

– ct

– bt

– at

(m + 1 )

A ----m s

At ------------------(m + 1)

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2.6.10

LAPLACE TRANSFORM PROBLEMS By means of the LaPlace Transform, find the equation for: 1.

The amount of drug in the body when the drug is given by IV Bolus (assume no metabolism).

2.

The amount of drug in the urine when the drug is given by IV Bolus (assume no metabolism).

3.

The amount of metabolite in the body when the drug is given by IV Bolus (assume no parent drug excretion)

4.

The amount of metabolite of the drug in the urine when the drug is given by IV Bolus (assume no parent drug excretion)

5.

The amount of metabolite of the drug in the urine when the drug is given by IV Bolus (assume both parent drug and metabolite excretion)

6.

The amount of drug in the body when the drug is given by IV infusion (assume no metabolism).

7.

The amount of drug in the urine when the drug is given by IV infusion (assume no metabolism).

8.

The amount of metabolite in the body when the drug is given by IV infusion (assume no parent drug excretion).

9.

The amount of metabolite in the urine when the drug is given by IV infusion (assume no parent drug excretion).

10.

The Rate of excretion of the metabolite into the urine for a drug given by IV bolus when km+ku=kmu.

11.

The amount of the principle metabolite (Xm1) when the drug is eliminated by several pathways (Xu, Xm1,Xm2,Xm3,etc)

12.

X The concentration of drug, ------ , in the body when the drug is given orally by a delivery system Vd

which is zero order. What is the concentration at equilibrium ( T ∞ ). 13.

The amount of metabolite of a drug in the body when the drug is given by IV Bolus and concomitant IV infusion.

14.

Disopyramide (D) is a cardiac antiarrythmic drug indicated for the suppression and prevention of ectopic premature ventricular arrythmias and ventricular tachycardia. It appears that disopyramide 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.

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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 compartment. 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) X = Xo e

– kut

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2. the amount of a drug in the urine when the drug is given by IV bolus (assume the drug is NOT metabolized

Xu= Xo ( 1 – e

– kut

)

Dose X

ku

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:

dX --------u- = 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

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3. the amount of metabolite of a drug in the body when the drug is given by IV bolus (assume no parent drug excretion).

km X 0 sX m = -------------– k mu Xm Substituting for X from above s + km 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 - ⋅ ( e mu – e m ) or X m = -----------------------( k m – k mu )

km Xm

kmu

Xmu

( km ⋅ Xo ) - ⋅ ( e – K 1t – e –K t ) in general terms. X m = --------------------( 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

dx ------ = – k m X dt sX – Xo = – k m X

and then

( km ⋅ X o ) –k t –k t X m = ------------------------- ⋅ ( e m – e mu ) or ( k mu – k m )

sX + k m X = Xo

( km ⋅ Xo ) - ⋅ ( e – K t – e –K1t ) X m = --------------------( 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) X = Xo e

– kmt

dX ---------m- = k m X – k mu X m dt sX m – Xm0 = k m X – kmu X m

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4. the amount of metabolite of a drug in the urine when the drug is given by IV bolus (assume the parent compound is not excreted). Here, again, K = km and K1 = kmu

Dose

X om = 0 sX mu = k mu X m ( k mu ) ( k m X o ) sX mu = ----------------------------------------( s + k mu ) ( s + k m ) ( 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 s + km

o substitute previously solved X = --------------

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,

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5. the amount of metabolite of a drug excreted in the urine when both the parent and metabolite are excreted.

dX mu ------------- = k mu Xm dt

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

X

Xu

km Xm

( k mu ) ( k m ) ( Xo ) X mu = ---------------------------------------s ( s + K1 ) ( s + K ) k mu k m X o 1 – e – Kt 1 – e – K1t X m u = ----------------------  ------------------- – --------------------- K1  K1 – K  K

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 X = Xo e

– Kt

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 )

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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 ) –k t Q X = ----- ( 1 – e u ) or X = Q ---- ( 1 – e –Kt ) ku K

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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 ) dX --------u- = 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 ) k u Qt  1 – e –kut - X u = ----------– Qk u  ------------------ku  k2  u

– kut

– Kt

(1 – e ) (1 – e ) X u = Qt – Q ------------------------- or Xu = Qt – Q ----------------------ku K

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8. the amount of metabolite of a drug in the body from a drug given by IV infusion (assume no parent drug excretion) Here K = km and K1 = kmu

– bt

– at

1–e 1–e A X m = ----------------  ------------------ –  ------------------ b a (a – b) km Q  1 – e – k m t  1 – e – k mu t X m = -------------------------  -------------------- –  --------------------- or ( k mu – k m )  k m   k mu 

Q

X

k m Q  1 – e –Kt  1 – e –K1t - ------------------ – --------------------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 dX ---------m- = 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 )

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9. the amount of metabolite of a drug in the urine from a drug given by IV infusion (assume that the parent compound is not excreted). Here K = km and K1 = kmu

dX mu ------------- = k mu Xm dt sX mu – Xomu = k mu X m substitute X m

Q

k mu k m Q X mu = ---------------------------------------------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 = ---------------------- –  ---------------------  ----------------------------- – -------------------------- 2 2 k m ⋅ k mu  k m – k m u   k k

km

mu

Xm

kmu

Xmu

m

k mu k m Qt k mu k m Q  ( 1 – e –Kt ) ( 1 – e –K1t ) X mu = ---------------------- –  --------------------  ------------------------ – -------------------------- 2 2  K1 ⋅ K K1 – K    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 ) dX ---------m- = 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 ) k m Q  1 – e –kmut  1 – e –kmt X m = ------------------- ----------------------- – --------------------- k m – k mu  k mu   k m 

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Mathematics Review

10. the rate of excretion of the metabolite into the urine for a drug given by IV bolus when

k m + k u = k mu In this case, K = ku +km and K1 = kmu and thus K = K1. This is not normal but could happen. The problem arises when we get to the LaPlace that assumes the rate constants are different (i.e. a ≠ b ) because for this special case a = b .

sX m – Xom = ( k m X – K1Xm ) sX m + K1X m = k m X km X o X m ( s + K1 ) = -----------s+K km Xo X m = -------------------------------------( s + K ) ( s + K1 )

Dose Xu

km Xo X m = ----------------------2( s + K1 )

km

X m = k m Xo te Xm

(remember- K1 = K)

km Xo X m = -----------------------------------------( s + K1 ) ( s + K1 ) ku

X

dX m ---------- = ( k m X – k muX m ) dt

kmu

Xmu

dX

. ------- = – k u X – k m X dt

(kmX0 = A)

– K1t

dX mu ------------ = k mu Xm dt dX mu ------------ = k mu k m Xo te –K1t 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) X = Xo e

– Kt

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the principal metabolite ( X m1 ) when the drug is

11.

cleared by several pathways ( X u, X m 1, X m 2 ) In this case K = km1 + km2 + ku, K1 = kmu1 and K2 = kmu2

Xm1 Dose

kmu1

Xmu1

k m1 Xo X m1 ( s + K1 ) = ---------------(s + K) k m1 X o X m1 = -------------------------------------( s + K1 ) ( s + K ) k m1 Xo – K1t –Kt -(e X m1 = ---------------–e ) K – K1

km1 ku

X

Xu

km2 Xm2

kmu2

Xmu2

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)

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12. the concentration of drug X ⁄ Vd in the body when the drug is given orally by a delivery system which is zero order. What is the concentration in the body at equilibrium ( t ∞ )

k a Q  ( 1 – e –Kt ) ( 1 – e –kat )  - ------------------------ – -------------------------  X = ------------------( k a – K )  K ka 

Here K = ku

ka Q  1 – e –Kt ) ( 1 – e –kat )  -  (----------------------- – -------------------------  C = -------------------------( k a – K )Vd  K ka 

Q Xa

ka

X

ku

Xu

k Q  ( k a – K )Vd  K

 ka 

1- ---1 a -  --If t= ∞ , then e –kt = 0 , thus C = -------------------------– - Q simplified yields: C = ---------KVd

dX --------a- = 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 )

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13 the metabolite of a drug in the body Xm given by IV infusion and concomitant IV bolus dose. Infusion:

k m Q  ( 1 – e –K1t ) ( 1 – e –Kt )  - -------------------------- – -----------------------  X m = --------------------K1 K ( K – K1 )  

Here K = km and K1 = kmu

IV Bolus:

dX ------- = – KX dt

Dose

sX – Xo = – KX

Q

sX + KX = X o

X

X ( s + K )= Xo km Xm

Xo X = ----------s+K kmu

Xmu

dX m ---------- = k m X – K1X m dt dX ------- = Q – KX dt

sX m – Xo = k m X – K1X m

Q sX – Xo = ---- – KX s

sX m + K1X m = k m X

Q sX + KX = ---s Q X ( s + K ) = ---s Q X = -------------------s( s + K ) dX ---------m- = k m X – K1X m dt sX m – Xo = k m X – K1Xm sX m + K1X m = k m X

km X o X m ( s + K1 ) = -----------s+K km Xo X m = -------------------------------------( s + K1 ) ( s + K ) km Xo – Kt – K 1t -(e – e X m = --------------------) ( K1 – K ) Thus,

km X o – Kt – K 1t  ) + … X m =  ---------------------- ( e – e ( K1 – K )  k m Q  ( 1 – e –Kt ) ( 1 – e –K1t )   -----------------  ----------------------- – --------------------------  K K1  K1 – K  

km Q X m ( s + K1 ) = ------------------s( s + K ) km Q X m = ---------------------------------------s ( s + K1 ) ( s + K ) Basic Pharmacokinetics

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14. By means of the LaPlace transform, find the equation for the rate of appearance of the tracer in the urine if the drug were given by IV bolus.

km X o X m ( s + K1 ) = ----------s+K

Here K = ku + km and K1 = kmu

km Xo X m = -------------------------------------( s + K ) ( s + K1 )

Dose

km Xo - { e –K1t – e –Kt } X m = --------------------( K – K1 ) ku

X

Xu

k mu k m X o dX dX mu  – Kt –K1t – Kt --------u- + ------------ = k u ( X o e ) +  --------------------- { e – e }    ( K – K1 ) dt dt

km Xm

k mu k m Xo –K1t – Kt dX mu ------------- = k mu Xm = ---------------------{e –e } dt ( K – K1 )

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) X = Xo e

– Kt

dX --------u- = k m X = k u ( Xo e –Kt ) dt dX ---------m- = k m X – K1X m dt sX m – Xo = k m X – k1Xm km Xo sX m + K1X m = ---------------(s + K)

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