BASIC PHARMACOKINETICS-Ch2: Mathematics Review

CHAPTER 2 Mathematics Review

Author: Michael Makoid, Phillip Vuchetich and John CobbyReviewer: Phillip Vuchetich

BASIC MATHEMATICAL SKILLS OBJECTIVES

1. Given a data set containing a pair of variables, the student will properly construct (III) various graphs of the data.

2. Given various graphical representations of data, the student will calculate (III) the slope and intercept by hand as well as using linear regression.

3. The student shall be able to interpret (V) the meaning of the slope and intercept for the various types of data sets.

4. The student shall demonstrate (III) the proper procedures of mathematical and algebraic manipulations.

5. The student shall demonstrate (III) the proper calculus procedures of integration and differentiation.

6. The student shall demonstrate (III) the proper use of computers in graphical simu-lations and problem solving.

7. Given information regarding the drug and the pharmacokinetic assumptions for the model, the student will construct (III) models and develop (V) equations of the ADME processes using LaPlace Transforms.

8. The student will interpret (IV) a given model mathematically.

9. The student will predict (IV) changes in the final result based on changes in vari-ables throughout the model.

10. The student will correlate (V) the graphs of the data with the equations and mod-els so generated.

Basic Pharmacokinetics REV. 00.1.6 2-1Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf/

Mathematics Review

2.1 Concepts of Mathematics

Pharmacokinetcs is a challenging field involving the application of mathematicalconcepts to real situations involving the absorbtion, distribution, metabolism andexcretion of drugs in the body. In order to be successful with pharmacokinetics, acertain 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 under-standing kinetics. These concepts are generally taught in other mathematicalcourses from algebra through calculus. For this reason, this chapter is presented asa review rather than new material. For a more thorough discussion of any particu-lar concept, refer to a college algebra or calculus text.

Included in this section are discussions of algebraic concepts, integration/differen-tiation, graphical analysis, linear regression, non-linear regression and the LaPlacetransform. 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 LaPlacetransform is used to quickly solve (integrate) ordinary, linear differential equa-

tions. The Scientist by Micromath Scientific Software, Inc.1 is available for work-ing with the LaPlace transform for problems throughout the book.

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


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indi-re sub-

ay

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?

Examples: 316000 becomes

0.00708 becomes

In general a number takes the form:

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 bemoved to place it immediately to the right of the first non-zero digit. If the decimalpoint has to be moved to its left then n is a positive integer; if to its right, n is anegative integer.

Because this notation (sometimes referred to as “Scientific Notation”) uses ces, mathematical operations performed on numbers expressed in this way aject 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 m

be interpreted as , as in the following example:

2.28E4 =

∞( )

3.16 105×

7.083–×10

A 10n×

A 10n×

2.284×10 22800=


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ig-nifi-

2.2.3 SIGNIFICANT FIGURES

A significant figure is any digit used to represent a magnitude or quantity in theplace in which it stands. The digit may be zero (0) or any digit between 1 and 9.For example:

How do I determine the number of significant figures?

Examples (c) to (e) illustrate the exceptions to the above general rule. The value 10raised 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 oneor more zeros immediately follow a decimal point, as in example (e), these zerossimply 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), isa commendable practice which also serves to locate the decimal point; this zero istherefore not a significant figure.

What do significant fig-ures mean?

Significant figures are used to indicate the precision of a value. For instance, avalue recorded to three significant figures (e.g., 0.0602) implies that one can reli-ably predict the value to 1 part in 999. This means that values of 0.0601, 0.0602,and 0.0603 are measurably different. If these three values cannot be distinguished,they should all be recorded to only two significant figures (0.060), a precision of 1part in 99.

After performing calculations, always “round off” your result to the number of snificant figures that fairly represent its precision. Stating the result to more sigcant figures than you can justify is misleading, at the very least!

TABLE 2-1 Significant Figures

ValueSignificant

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


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apply

2.2.4 RULES OF INDICES

What is an index? An index is the power to which a number is raised. Example: where A is anumber, which may be positive or negative, and n is the index, which may be pos-itive or negative. Sometimes n is referred to as the exponent, giving rise to thegeneral term, “Rules of Exponents”. There are three general rules which when indices are used.

(a) Multiplication

(b) Division

(c) Raising to a Power

There are three noteworthy relationships involving indices:

(i) Negative Index

As n tends to infinity then .

(ii) Fractional Index

(iii) Zero Index

An

An

Am

A=n m+

× An

Bm× A

B---

nB×

n m+=

An

Am

------- An m–

=A

n

Bm

------- AB---

nB×

n m–=

An( )

mA

nm=

An– 1

An

------= n ∞→( ) An–

0→

A

1n---

An=

A0

1=


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ithm

ationtific

using

place-Thenum-

naturaln of a

dergoof the

-

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 anumber in scientific notation.

In general terms, if a number (A) is given by

then

where ‘log’ signifies a logarithm to the base 10, and is the value of the logarof (A).

Example: 713000 becomes ,

and , thus 713000 becomes

and

Logarithms to the base 10 are known as Common Logarithms. The transformof a number (A) to its logarithm (n) is usually made from tables, or on a sciencalculator; the reverse transformation of a logarithm to a number is made anti-logarithmic tables, or on a calculator.

What is the characteris-tic? the mantissa?

The number before the decimal point is called the characteristic and tells the ment of the decimal point (to the right if positive and to the left if negative). number after the decimal is the mantissa and is the logarithm of the string of bers discounting the decimal place.

2.2.6 NATURAL LOGARITHMS

What is a natural loga-rithm?

Instead of using 10 as a basis for logarithms, a natural base (e) is used. This base is a fundamental property of any process, such as the glomerular filtratiodrug, which proceeds at a rate controlled by the quantity of material yet to unthe process, such as drug in the blood. To eight significant figures, the value transcendental function, e, is

... Strictly speaking, Where is an inte

ger ranging from 1 to infinity ,

A 10n

= A( )log n=

n

7.13 105×

7.13 100.85

= 100.85

105× 10

5 0.85+( )10

5.85= =

713000( )log 5.85=

e 2.7182818= e 1 1x!----

x 1=

∞

∑+= x

∞( )


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f the

hms;. Thehich

nipu-

ach,se) of

denotes the summation from , and

! is the factorial (e.g., 6! = 6x5x4x3x2x1= 720)

In general terms, if a number (A) is given by , then by definition,

Where, ‘ln’ signifies the natural logarithm to the base , and is the value o

natural logarithm of .

Natural logarithms are sometimes known as Hyperbolic or Naperian Logaritagain tables are available and scientific calculators can do this automaticallyanti-logarithm of a natural logarithm may be found from exponential tables, w

give the value of for various values of n.

How are natural loga-rithms and common logarithms related?

Common and natural logarithms are related as follows:

, and

Because logarithms are, in reality, indices of either 10 or , their use and malation follow the rules of indices (See Section 2.2.4).

(a) Multiplication:

To multiply , where and ; .

By definition, ; but

and , hence

Thus, to multiply two numbers (N and M) we take the natural logarithms of eadd them together, and then take the anti-logarithm (the exponent, in this cathe sum.

x 1=

∞

∑ x 1 to x ∞= =

A en

=

A( )ln n=

e n

A

en

xln

xlog A( )ln 2.303 A( )log×=

A( )log 0.4343 A( )ln×=

e

N M× N en

= M em

= NM en

em× e

n m+= =

NM( )ln n m+=

n N( )ln= m M( )ln= NM( )ln N( )ln M( )ln+=


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(b) Division

(c) Number Raised to a Power

There are three noteworthy relationships involving logarithms:

(i) Number Raised to a Negative Power

As tends to infinity , then

(ii) Number Raised to a Fractional Power

(iii) Logarithm of Unity

NM-----

ln N( )ln M( )ln–=

Nm( )ln m N( )ln×=

Nm–( )ln m N( )ln×– m

1N----

ln×= =

m m ∞→( ) Nm–( )ln ∞–→

Nm( )ln N

1m----

ln1m---- N( )ln×= =

1( )ln 1( )log 0= =


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2.2.7 NEGATIVE LOGARITHMS

The number 0.00713 may be expressed as:

, or

, or

.

Hence, , which is the result generated by most calculators.

However, another representation of a negative logarithm (generally used by refer-encing a log table):

log (0.00713) = 3.85

The 3 prior to the decimal point is known as the characteristic of the logarithm; itcan be negative (as in this case) or positive, but is never found in logarithmictables. The .85 following the decimal point is known as the mantissa of the loga-rithm; it is always positive, and is found in logarithmic tables.

In fact 3 is a symbolic way of writing minus 3 (-3) for the characteristic. In everycase the algebraic sum of the characteristic and the mantissa gives the correctvalue for the logarithm.

Example: log (0.00713) = 3.85

Add -3 and 0.85

Result is -2.15, which is the value of

The reason for this symbolism is that only positive mantissa can be read from anti-logarithmic tables, and hence a positive mantissa must be the end result of any log-arithmic manipulations. Note that while there are negative logarithms (when N <1), they do not indicate that number itself is negative; the sign of a number (e.g., -N) is determined only by inspection following the taking of anti-logarithms.

7.133–×10

100.85

103–×

102.15–

0.00713( )log 2.15–=

0.00713( )log


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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 inunderstanding the use of logarithms. (See the end of this chapter for Logarithmtables).

(a) Find the log of (62.54):

1. convert 62.54 to scientific notation ---> ;

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

3. Hence, and

(b) Find the log of (0.00329)

1.

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 towhich 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 (man-tissa) in a table: it is 2058.

Antilog (1.3134) =

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

62.54 6.2541×10=

6.2541×10 10

0.796210

1× 101.7962

= = 62.54( )log 1.7962=

0.00329 3.293–×10=

2.0581–×10


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acyre

be 10

nt

ange is000

2.2.9 DIMENSIONS

What is a unit? There are three fundamental dimensions which are used in various combinations toexpress the properties of matter. Each of these dimensions has been assigned a def-inite basic unit, which acts as a reference standard.

How are units made big-ger 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. Theprefixes indicate the factor by which the unit is multiplied. When the index of thefactor is positive the prefixes are Greek and have hard, consonant sounds. In con-trast, when the index is negative, the prefixes are Latin and have soft, liquidsounds. (see Table 2-3).

How big is big? Examples: An average adult male patient is assumed to have a mass of 70 kilo-grams (70 kg). An average adult male patient is assumed to have a height of 180centimeters (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. Stu-

dents 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 sys-tem. Note that the plural of Kg or cm is Kg or cm; do not add an “s”. In pharmthere are two derived units which are commonly used, even though they arelated to basic units. The Liter (L) is the volume measurement and is a cu

cm on a side (1L = (10cm)3 = 1000 cm3 ) while the concentration measuremeand 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 statethe units of the measurement. Numbers are useless by themselves.

Example: The procainamide concentration range is 4-8 g/ml; stating the rwithout units may lead to a potentially lethal error in which procainamideadministered in a sufficient dose to attain a range of 4-8 mg/ml, which is 1times too large and would give rise to cardiac arrest.

TABLE 2-2 Dimensions

DimensionDimensional

Symbol Unit Unit Symbol

Length L meter m

Mass M gram g

Time T second sec

µ


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TABLE 2-3 Scale of Metric system and SI

Name Symbol Multiplication Factor Name Symbol Multiplication Factor

exa- E deci- d

peta- P centi- c

tera- T milli- m

giga- G micro-

mega- M nano- n

kilo- k pico- p

hecto- h femto- f

deca- da atto- a

TABLE 2-4

DimensionDimensional

Symbol UnitUnit

Symbol

Volume V liter l

Concentration C grams/liter g/l

1018

101–

1015

102–

1012

103–

109 µ 10

6–

106

109–

103

1012–

102

1015–

101

1018–


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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 equa-tion should be equal. If this is not so, the equation is necessarily meaningless.

Consider the following equation which defines the average concentration of a drug

in blood after many repeated doses,

Where:

• is the fraction of the administered dose ultimately absorbed (Dimensions: none),

• is the mass of the repeated dose (Dimension: M),

• is the apparent volume of distribution of the drug (Dimension: )

• is the apparent first-order rate constant for drug elimination (Dimension: ),

• and is the dosing interval (Dimension: )

Writing the dimensions relating to the properties of the right-hand side of the equa-tion gives:

Thus has the dimensions of , which are correctly those of concentration.

Sometimes dimensional analysis can assist an investigator in proposing equationswhich relate several properties one with the other. If the units cancel, and you endup with the correct unit of measure, you probably did it right. If you obtain unitsthat do not make sense, it’s guaranteed sure that you did it wrong.

Cb( )∞FDVKτ-----------=

F

D

V V L=

K T1–

τ T

M

V T1–

T⋅ ⋅------------------------ M

V-----=

Cb( )∞MV-----


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2.3 Calculus

What is Calculus? Calculus concerns either the rate of change of one property with another (differen-tial calculus), such as the rate of change of drug concentrations in the blood withtime since administration, or the summation of infinitesimally small changes (inte-gral calculus), such as the summation of changing drug concentrations to yield anassessment of bioavailability. In this discussion a few general concepts will be pro-vided, and it is suggested an understanding of graphical methods should precedethis discussion.

2.3.1 DIFFERENTIAL CALCULUS

2.3.2 NON-LINEAR GRAPHS

Consider the following relationship:

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

FIGURE 2-1. y=x3

(Question: How could the above data be modified to give a linear graph?)

TABLE 2-5 x, y sample data

x 0 1 2 3 4

y 0 1 8 27 64

y x3

=

0

10

20

30

40

50

60

70

1 2 3 4


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gure

as the

en

n of

2.3.3 SLOPE OF NON-LINEAR GRAPH

As with a linear graph,

Where is the incremental change in y and 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 givenchange 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

We need to find the value of the slope of the line when (See Fi1). Hence, we may choose incremental changes in x which are located around

.

As may be seen, the value of the slope tends towards a value of 12.000

magnitude of the incremental change in x becomes smaller around the chosvalue of 2.0. Were the chosen incremental changes in x infinitesimally small, thetrue value of the slope (i.e., 12.000) would have appeared in the final columthe above table.

y2 y1–

x2 x1–---------------- y∆

x∆------=

y∆ x∆

FIGURE 2-2. / when

0 4 4 0 64 64 16.000

1 3 2 1 27 26 13.000

1.5 2.5 1.0 3.375 15.625 12.250 12.250

1.8 2.2 0.4 5.832 10.648 4.816 12.040

1.9 2.1 0.2 6.859 9.261 2.042 12.010

1.95 2.05 0.1 7.415 8.615 1.200 12.003

y x3

= x 2=

x 2≈

y∆ x∆ x 2≈

x1 x2 x∆ y1 y2 y∆y∆x∆

------

y∆x∆

------


Mathematics Review

Calculus deals with infinitesimally small changes. When the value of is infini-tesimally small it is written dx and is known as the derivative of x. Hence,

Where dy/dx is the derivative of y with respect to x and indicates some func-tion of x.

2.3.5 DIFFERENTIATION FROM FIRST PRINCIPLES

Differentiation is the process whereby the derivative of y with respect to x isfound. Thus the value of dy/dx, in this case, is calculated.(a) Considering again the original expression:

(b) Let the value of y increase to because x increases to .

Hence,

(EQ 2-13)

Multiplying out:

(EQ 2-14)

(c) The change in y is obtained by subtraction of the original expression from thelast expression. (i.e., Eq. 2 - Eq. 1)

(EQ 2-15)

Dividing throughout to obtain the derivative,

When dx is infinitesimally small, its magnitude tends to zero . The limit-

ing value of this tendency must be . At this limit,

x∆

dydx------ f x( )=

f x( )

y x3

=

y dy+ x dx+

y dy+ x dx+( )3=

y dy+ x3

3x2

dx( ) 3x dx( )2dx( )3

+ + +=

dy 3x2

dx( ) 3x dx( )2dx( )3

+ +=

dydx------ 3x

23x dx( ) dx( )2

+ +=

dx 0→( )dx 0=


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(EQ 2-16)

Hence the derivative of y with respect to x at any value of x is given by .

(d) In section 2.3.4 we saw how the true value of the slope (i.e., dy/dx) would be12.0 when . This is confirmed by substituting in Equation 1-16.

2.3.6 RULE OF DIFFERENTIATION

Although the rate of change of one value with respect to another may be calculatedas 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 aconstant, and n be an exponential power.

The general rule is:

If

then

The Rules of Indices may need to be used to obtain expressions in the form

(e.g., if )

2.3.7 THREE OTHER DERIVATIVES

(a) If ,

then (i.e., is constant)

dydx------ 3x

2=

3x2

x 2=

dydx------ 3x

23 2

2( ) 12= = =

y Axn

=

dydx------ nAx

n 1–=

y Axn

=

y x5=

y Ax0

=

y A= y


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Hence,

Thus the derivative of a constant is always zero.

(b) Accept that if

then .

This derivative is important when considering apparent first-order processes, ofwhich many bodily processes (e.g., excretion of drugs) are examples.

(c) Accept that if where B and A are constants, and e is the natural

base then

This derivative will be useful in pharmacokinetics for finding the maximum andminimum concentrations of drug in the blood following oral dosing.

2.3.8 A SEEMING ANOMALY

Consider the following two expressions:

(a) If , then

(b) If ,

then

Both of the original expressions, although different, have the same derivative. Thisfact is recognized later when dealing with integral calculus.

dydx------ 0=

y x( )ln=

dydx------ 1

x---=

y BeAx

=

dydx------ ABe

Ax=

y Axn

=

dydx------ nAx

n 1–=

y Axn

A+=

dydx------ nAx

n 1–0+ nAx

n 1–= =


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2.3.9 INTEGRAL CALCULUS

Generally integral calculus is the reverse of differential calculus. As such it is usedto sum all the infinitesimally small units (dy) into the whole value (y).

Thus,

, where the symbol for integration.

2.3.10 RULE OF INTEGRATION

The derivative expression may be written:

, or

To integrate,

A general rule states:

Where A is the constant of integration However, there is one exception - the ruleis not applicable if

Example: If (See section 2.3.5),

then , and

yd∫ y= is∫

dydx------ Ax

n=

dy Axn

dx⋅=

y yd∫ Axn

xd∫ A xn

xd∫= = =

A xn

xd∫ Axn 1+

n 1+---------------- A+=

n 1–=

dydx------ 3x

2=

y3x

2 1+

2 1+--------------- A+=

y x3

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

anomaly referred to in section 2.3.8. As mentioned, both and

will give, on differentiation, .

So whether or not a constant is present and, if so, what is its value, can only bedecided by other knowledge of the expression. Normally this other knowledgetakes the form of knowing the value of y when .

In the case of our graphical example we know that when , then . Theintegrated expression for this particular case is:

, therefore

, thus

In some examples, such as first-order reaction rate kinetics, the value of A is notzero.

2.3.12 THE EXCEPTION TO THE RULE

It occurs when

Upon integration,

This is the reverse of the derivative stated in section 2.3.10 (b).

2.3.13 A USEFUL INTEGRAL

Accept that if,

y x3

= y x3

A+=

dydx------ 3x

2=

x 0=

x 0= y 0=

y x3

A+=

0 03

A+= A 0=

n 1–=

y A x1–

xd∫ A1x--- xd∫= =

y A x( )ln A+⋅=

dydx------ Be

Ax=


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then,

This integral will be useful for equations which define the bioavailability of a drugproduct.

2.3.14 EXAMPLE CALCULATIONS

(a) Consider,

Where is the drug concentration in a dissolution fluid at time .

Then, multiplying out,

The rate of dissolution at time t is

So at any time, the rate may be calculated.

(b) Consider,

Then rearranging,

The integral of is:

yBe

Ax

A----------- A+=

c 3t2

t 2–( ) 5+=

c t

c 3t3

6t2

– 5+=

dcdt------ 9t

212t–=

dcdt------ 3t t 4–( ) 9t

212t–= =

dc 9t2

dt 12 dt⋅–⋅=

c

c cd∫ 3t3

A 6t2

– B+ += =


Mathematics Review

where is a second constant.

Adding the two constants together,

where

We know, from previous work, that when , then

Substituting , the final expression becomes:

Which is the initial expression in example (a) above.

(c) Following administration of a drug as an intravenous injection,

Where is the plasma concentration of a drug at time

is the apparent first-order rate constant of elimination.

Rearranging,

This integral is the exception to the rule (see section 2.3.12).

We know that when , .

B

c 3t3

6t2

D+= =

D A B+=

t 0= c 5=

5 D=

c 3t2

6t2

5+ +=

dCp–

dt------------- KCp=

Cp t

K

K– dt⋅ 1Cp

------ dCp⋅=

Kt– K td∫– 1Cp

------∫ Cpd⋅= =

Kt– Cp( )ln A+=

t 0= Cp Cp( )0=


Mathematics Review

Substituting,

Or,

Hence

or,

or,

0 Cp( )0ln A+=

A Cp( )0ln–=

Kt– Cp( )ln Cp( )0ln–=

Cp( )ln Cp( )0ln Kt–=

Cp Cp( )0 eKt–⋅=


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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.

What is a graph? A graph is simply a visual representation showing how one variable changes withalteration of another variable. The simplest way to represent this relationshipbetween variables is to draw a picture. This pictorializing also is the simplest wayfor the human mind to correlate, remember, interpolate and extrapolate perfectdata. An additional advantage is it enables the experimenter to average out smalldeviations in experimental results (non-perfect, real data) from perfect data. Forexample:

TABLE 2-6

English Science

Observations Data

Theory Equations

Paradigms (pictures) Graphs

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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l axisd on not

maderationlasmae ‘y’

FIGURE 2-3. Plot of Perfect vs. Real data

Simply looking at the columns x and y (real) it might be difficult to see the rela-tionship between the two variables. But looking at the graph, the relationshipbecomes apparent. Thus, the graph is a great aid to clear thinking. For every graphrelating variables, there is an equation and, conversely for every equation there is agraph. The plotting of graphs is comparatively simple. The reverse process of find-ing an equation to fit a graph drawn from experimental data is more difficult,except in the case of straight lines.

2.4.1 GRAPHICAL CONVENTIONS

How are graphs made? Certain conventions have been adopted to make the process of rendering a data setto a graphical representation extremely simple.

The ‘y’ variable, known as the dependent variable, is depicted on the vertica(ordinate); and the ‘x’ variable, known as the independent variable, is depictethe horizontal axis (abscissa). It is said that ‘y’ varies with respect to ‘x’ and‘x’ varies with ‘y’.

A decision as to which of the two related variables is dependent can only be be considering the nature of the experiment. To illustrate, the plasma concentof a drug given by IV bolus depends on time. Time does not depend on the pconcentration. Consequently, plasma concentration would be depicted on thaxis and time on the ‘x’ axis.

x

-3 -2 -1 0 1 2

-6

-4

-2

0

2

4

6


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aysp orwn in

twobered thee i.e.,

h ofon the fording

has

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 alwcomes first; and 2) the ‘y’ value which is the distance, along the ‘y’ axis udown from the ‘x’ axis, and always comes second. Several points are shoFigure 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). Indimensional spaces, the graph is divided into 4 quadrants from (0,0), numwith Roman numerals from I through IV. It should be readily apparent thatcoordinates for all points within a particular quadrant are of the same sign typ

A line (or curve) on a graph is made up of an infinite number of points, eacwhich has coordinates that satisfy a given equation. For example, each point line in Figure 2 is such at its coordinates fit the equation . That isany value of x (the independent variable), multiplying the x value by 2 and ad1 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 the form

Where:

• y = dependent variable

• x = independent variable

• m = slope of the straight line =

• b = the y intercept (when x = 0)

• or if the equation can be “linearized”, e.g.,

is not linear. However.

is of the same general form as:

TABLE 2-8 Quadrants on a cartesian graph

Quadrant II (-x, +y) Quadrant I (+x, +y)

Quadrant III (-x, -y) Quadrant IV (+x, -y)

y 2x 1+=

y mx b+=

y∆x∆

------

y′ b′emx′= y′ln b′ln mx′+=

y b mx+=


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and consequently a plot of (the dependent variable) versus (the indepen-

dent variable) will yield a straight line with a slope of and an intercept of .

Expressions of any other form are non linear. For example:

An expression relating the plasma concentration of a drug over time .

this relationship put in linear perspective yields:

, which is in the form

The graphs that yield a straight line are the ones with the ordinate being ,

and the abscissa being .

Any other combination of functions of and will be non-linear, e.g.,

• versus

• versus

• versus

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 inall examples. Some relationships between two variables can never be resolved into

a single straight line, e.g.,

where or

(It is possible to resolve this equation into the summation of two linear graphswhich will be shown subsequently.)

y ′ln x′m b′ln

Cp( ) t( )

Cp Cp0eKt–

=

Cpln Cp0ln Kt–=

y b mx+=

Cp0ln

t

Cp t

Cp t

Cp tln

Cpln tln

y k0 k1xn m–( )

k2xn m– 1+( ) … kn( )x

x+ + + +=

n 2 n;≥ m 1+=

Cp

KaFD

V Ka K–( )------------------------- e

Kte

Kat––( )⋅=


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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 ornegative. In the previous example, the slope is negative, i.e:

The differences in both the y-values and the x-values may be measured graphicallyto obtain the value of the slope, m. Then knowing the value of m, the value of may be found.

FIGURE 2-4. Plasma Concentration of caffeine over time

TABLE 2-9 Sample data of caffeine elimination

(min)

12 3.75 1.322

40 2.80 1.030

65 2.12 0.751

90 1.55 0.438

125 1.23 0.207

173 0.72 -0.329

m K–=

tCp

µgmL--------

Cpln

K

Cp( )

0 50 100 150 200

Time (min)

10-1

100

101

Caffein

e C

oncentr

ation (

ug/m

L)


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2.4.4 LINEAR REGRESSION: OBTAINING THE SLOPE OF THE LINE

The equation for a straight line is:

• 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:

And the intercept is

Using the data from table 2-10 in the equation for the slope of the line

and the intercept would be . Note that this is. In oder to find the , the anti-ln of must be taken. i.e.

TABLE 2-10 Linear Regression for data in table 2-9

12 1.322 144 15.864

40 1.030 1600 41.2

65 0.751 4225 48.815

90 0.438 8100 39.42

125 0.207 15625 25.875

173 -0.329 29929 -56.917

y m x b+⋅=

mΣ x( ) Σ y( )⋅( ) n Σ x y⋅( )⋅( )–

Σ x( )[ ]2n Σ x

2( )⋅( )–---------------------------------------------------------------------=

b y m x⋅( )–=

X Y X2

X Y⋅

ΣX 505= ΣY 3.239= ΣX2

59623= ΣXY 114.257=

ΣX( )2255025=

xΣxn

------ 4.167= = yΣyn

------ 0.5398= =

m505 3.239⋅( ) 6 114.257⋅( )–

255025 6 59623⋅( )–--------------------------------------------------------------------- 0.01014

·–= =

b 0.5398 0.01014– 4.167⋅( )– 1.4229= =

Cln Cp0 b

Cpo eb

e1.4229

4.15= = =


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It is important to realize that you may not simply take any two data pairs in thedata set to get the slope. In the above data, if we simply took two successive datapairs from the six data pairs in the set, this would result in five different slopes

ranging from -0.0066 to -0.0125 as shown in table 2-11. Clearly, this isunacceptable. Even to guess, you must plot the data, eyeball the best fit line byplacing your clear straight edge through the points so that it is as close to the dataas possible and look to make sure that there are an equal number of points abovethe line as below. Then take the data pairs from the line, not the data set.

2.4.5 PARALLEL LINES

Two straight lines are parallel if they have the same slope.

Calculating for the intercept of a linear graph (b):

(a) Not knowing the value of m; The graph may be extrapolated, or calcu-lations performed, at the situation where . In this case .

(b) Knowing the value of m;

• There are two ways: for any point on the graph: and

Hence, b may be calculated from a knowledge of and .

• Secondly, the graph may be extrapolated or calculations performed, at the situation where

. In this case,

TABLE 2-11 Sample slope data from figure 2-4

Time (x) ln Conc. (y)

12 1.322 -28 0.292 -0.0104

40 1.030 -25 0.28 -0.0112

65 0.751 -25 0.312 -0.0125

90 0.438 -35 0.231 -0.0066

125 0.207 -48 0.536 -0.0112

173 -0.329

∆x ∆y⁄( )

x∆ y∆y∆x∆

------

t 0= b Cp0ln=

y1 mx1 b+= b y1 mx1–=

y1 x1

t 0= b Cpln=


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n beintncesoint

t verygni-

2.4.6 GRAPHICAL EXTRAPOLATIONS

How far can I predict? It is dangerous to extrapolate on non-linear graphs, and it is unwise to extrapolatetoo far on linear graphs. Most often extrapolation is used to find the value of y at aselected 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 befound as shown above. Then: , where is the selected value of x,

and is the new calculated value for y.

2.4.7 SIGNIFICANCE OF THE STRAIGHT LINE

The more closely the experimental points fit the best line, and the higher the num-ber of points, the more significant is the relationship between y and x. As you mayexpect, 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 theoptimum values of the slope m, and the intercept, b. From these values the corre-lation coefficient (r).and the t-value may be obtained to indicate the significance.Exact details of the theory are available in any statistical book, and the calculationsmay most easily be performed by a computer using The Scientist or PKAnalyst inthis course.

The advantage of computer calculation is that it gives the one and only best fit tothe 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-linegraphs 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 averageout the small deviations in results from the “perfect” result, no one result caunjustifiably ignored when the best fitting line is drawn. Still, an “errant” pomay be justifiably ignored if there were unusual experimental circumstawhich may have caused the deviation. Thus it is not justifiable to omit a psolely 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 buoften in plotting chemical relations the two factors are so very different in ma

y' mx' b+= x'

y'


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tude that this can not be done. Consequently, it must be borne in mind that the rela-tionship between the variables is given by the scales assigned to the abscissa andordinate rather than the number of squares counted out from the origin.

FIGURE 2-5.

For example (shown in Figure 2-5), these two parabolic curves represent the sameequation 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 thelower left hand corner of the coordinate paper. Full utilization of the paper withsuitable intervals is the one criteria for deciding how to plot a curve from theexperimental 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

0 2 4 6 8 10

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

y 0.1 x=

0 2 4 6 8 10

0

2

4

6

8

10

PV nRT=


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FIGURE 2-6. Poorly presented graph

FIGURE 2-7. Well arranged graph

2.4.10 GRAPHS OF LOGARITHMIC FUNCTIONS

Previously variables were raised to constant powers; as . In this section

constants are raised to variable powers; as . Equations of this kind in whichthe exponent is a variable are called (naturally) exponential equations. The most

important exponential equation is where is plotted against .

0 4 8 12 16 20

0

10

20

30

40

50

1 2 3 4 5 6 7 8 9 10

P ( )

0

5

10

15

20

25

y x2

=

y 2x

=

ex

x


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2.4.11 SEMILOGARITHMIC COORDINATES

Exponential or logarithmic equations are very common in physical chemical phe-nomenon. One of the best ways of determining whether or not a given set of phe-nomenon can be expressed by a logarithmic or exponential equation is to plot thelogarithm of one property against another property. Frequently a straight line isobtained and its equation can be readily found. For example:

In the following table the plasma concentration of the immunosuppressant

cyclosporine was measured after a single dose (4mg/kg) as a function of time.

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)

TABLE 2-12 Plasma concentration of cyclosporine

Time (hours)Concentration

0.25 1900

.75 1500

1.5 1300

4 900

6 600

8 390

D’mello et al., Res. Comm. Chem. Path. Pharm. 1989: 64 (3):441-446

Cp( )

ngml------

200

400600

800

10001200

1400

16001800

2000

0 1 2 3 4 5 6 7 8


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FIGURE 2-9. Log Concentration vs. time

FIGURE 2-10. Log concentration (on log scale) vs. time

Graphing is much easier because the graph paper itself takes the place of a loga-rithmic table, as shown in Figure 1-10.

Only the mantissa is designated by the graph paper. Scaling of the ordinate for the

characteristic is necessary. The general equation can be expressed as astraight line by basic laws of indices.

or

One axis is printed with logarithmic spacing, and the other with arithmetic spac-ing. It is used when a graph must be plotted as in the example (Figure 1-4)

and .

Log Concentration - Time Curve

2.300

2.4002.500

2.600

2.7002.800

2.900

3.0003.100

3.200

0 1 2 3 4 5 6 7 8

100

1000

10000

0 1 2 3 4 5 6 7 8

y Beax

=

yln Bln eax( )ln+= yln Bln ax+=→ yln ax Bln+=

y Cp[ ]log= x t=


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n ofre isnsible

n by:

the

er, inr, inplot-

ay bearith- for.

nner,cond

In this example, the vertical logarithmic axis is labelled “Plasma concentratiocyclosporine” and the values plotted are the ordinary values of . Thus, theno need to use logarithmic tables, because the logarithmic spacing is respofor 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 give

Hence, before calculating the value of m, the two selected values of and

must be converted, using a calculator, to and in order to satisfy

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 papwhich case the vertical axis is labelled “log plasma concentration”. Howevethis instance the ordinary values of must be converted to prior to ting. It is the values which are then plotted.

The calculation of the slope is direct in this case, as the values of and mread from the graph. Hence, one must consider the relative merits of semilogmic and ordinary paper before deciding which to use when a log plot is called

In the case of semilog graphs the slope may be found in a slightly different mai.e., taking any convenient point on the line we usually take the as the se

point, one half of . Thus,

(in which case, is called the half-life ). Since , then

because and .

Cp[ ]

my2 y1–

x2 x1–----------------

Cp[ ]2

ln Cp[ ]1

ln–

t2 t1–--------------------------------------------= =

Cp[ ]1 Cp[ ]2

Cp[ ]1

ln Cp[ ]2

ln

Cp[ ] Cp[ ]ln

Cp[ ]ln

y1 y2

y1( )

y2( ) y1( )

myln 1 1 2⁄( )y1( )ln–

t1 t2–-----------------------------------------------

y1

1 2⁄( )y1--------------------

ln

t1 t2–-------------------------------

11 2⁄----------

ln

t1 t2–--------------------- 2ln

t1 t2–-------------- 0.693

t1 2⁄–-------------= = = = =

t2 t1– t½ t1 t2< t1 t2– t1 2⁄–=

m0.693

t1 2⁄–------------- k–= = k

0.693t1 2⁄

-------------=


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2.4.12 LOG - LOG COORDINATES

Functions of the type give straight lines when plotted with logarithmsalong both axis.

i.e., equation in logarithmic form is:

or which is in the form

. This is directly applicable to parabolic and hyperbolic equationspreviously discussed (see Figure 1-5).

2.4.13 PITFALLS OF GRAPHING: POOR TECHNIQUE

The utility of these procedures requires proper graphing techniques. The picturethat we draw can cause formation of conceptualizations and correlations of thedata that are inconsistent with the real world based simply on a bad picture. Conse-quently the picture must be properly executed.

The most common error is improper axes labelling. On a single axis of rectilinearcoordinate paper (standard graph paper), a similar distance between two pointscorresponds to a similar difference between 2 numbers. Thus,

FIGURE 2-11. Graphing using standard number spacing

y Bxa

=

ylog Blog a xlog+= ylog a xlog blog+=

y mx b+=

0 5 10 15 20 25 30

0

10

20

30

40


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FIGURE 2-12. Nonstandard (incorrect) graph

Obviously, the distance (Time) on the graph 12 between 0 and 2 hours should notbe the same as the distance between 10 and 20 hours. It is, and therefore Figure 2-12 is wrong.

Similarly, the use of similar paper may result in some confusion. With logarithmsthe mantissa for any string of numbers, differing only by decimal point placement,is the same. What differentiates one number from another, in this case, is the char-acteristic. Thus,

The paper automatically determines the relationship between strings of numbers(mantissa) by the logarithmic differences between the numbers on the axis within acycle. The student must determine the order of magnitude (characteristic) to be rel-egated to each cycle.

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

00

2 5 10 20 30

1

10

20

30

40


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r. It thus

ossible

FIGURE 2-13. Logarithmic mantissa

Thus, we see, in Figure 2-13, the cycle on the semilog paper to relate to orders ofmagnitude (e.g., 1, 10, 100, 1000, etc.) and consequently the characteristic of theexponent.

The third common problem is labelling the log axis as log “y”. This is impropeis obvious from the spacing on the paper that this function is logarithmic, andthe axis is simply labelled “y”.

There are almost as many different errors as there are students and it is impto 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. Straight line going down on semi-log paper

234

23.4

2.34

0.234

1.0 1.5 2.0 2.5 3.0 3.5 4.0

X axis (units)

10-1

100

101

102

103

Y a

xis

(u

nits)

Logarithmic Plot

0 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 isone half of the larger. The time that it takes to go from the larger to the smaller isthe half-life. Dividing 0.693 by the half-life yields the rate constant.

Extrapolating the line back to yields the intercept.

FIGURE 2-15. Curved line which plateaus on semi-log paper.

FIGURE 2-16. Curved line which goes up and then straight down on semi-log paper.

Find the terminal slope by taking any two values on the Y axis such that thesmaller value is one half of the larger. The time that it takes to go from the larger tothe 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 canbe obtained: Slope and Intercept. However, the slope and intercept have differentmeanings dependent on the data set type plotted. The slope is the summation of allthe ways that the drug is eliminated, -K.

t 0=

0 2 4 6 8 10 12

0 5 10 15 20 25


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Plot type two is not usually evaluated in its present form as only the plateau valuecan be obtained easily. But again it has different meanings dependent on the dataplotted.

Usually urine data of this type (parent compound - IV bolus) is replotted and eval-uated as plot 1 (above). Infusion data can be replotted using the same techniques,but usually is not.

Plot type 3 must be stripped of the second rate constant from the early time points,thus:

There are 3 things that can be obtained from the plot: the terminal slope (thesmaller rate constant), the slope of the stripped line (the larger rate constant) andthe intercept. The rate constants obtained from a caternary chain (drug movingfrom one box to another in sequence in compartmental modeling) are the summa-tion of all the ways that the drug is eliminated from the previous compartment andall the ways the drug is eliminated from the compartment under consideration. SeeLaPlace Transforms for further discussion.

Again, dependent on the data set type being plotted they will have different values.

TABLE 2-14 Plot type 1 examples

Data Type Y axis X axis Slope Intercept

IV Bolus Parent Drug Conc. parent compound Time -K

IV Bolus Parent urine rate of excretion

parent compound

Time (mid)

-K

IV Bolus Parent Cumulative urine

data

Time -K

TABLE 2-15 Plot Type 2 examples

Data Type Y axis X axis Plateau Value

IV Bolus Parent Cumulative urine data parent compound

Time

IV Infusion Parent

Drug concentration parent compound

Time

Cp0doseVd

-----------=

dXudt

---------- Kr X0⋅

Xu∞ Xu– krK X0⋅--------------

XuXu∞

krK X0⋅--------------=

Cp( )ssQ

K V⋅------------ Q

cl----= =


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TABLE 2-16 Plot Type 3 examples

Data Type Y axis X axis S1 S2 Intercept

IV Bolus Parent

Metabolite conc. Time -Ksmall -Klarge

IV Bolus Parent excretion

rate of metabolite into urine

Time (mid)

-Ksmall -Klarge

Oral Drug conc. Time -Ksmall -Klarge

km X0⋅Kl earg Ksmall–( ) Vdm⋅

-------------------------------------------------------

dXmudt

--------------- kmu km X0⋅ ⋅Kl earg Ksmall–------------------------------------

ka fX0⋅Kl earg Ksmall–( ) Vd⋅

----------------------------------------------------


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2.5 Pharmacokinetic MoKdeling

It has been observed that, after the administration of a drug, the concentration ofthe drug in the body appear to be able to be described by exponential equations.Thus, it appears that, even though the processes by which the drug is absorbed. dis-tributed, metabolized and excreted (ADME) may be very complex, the kinetics(math) which mimics these processes is made up of relatively simple first orderprocesses and is called first order pharmacokinetics. A second observation is thatthe resulting concentration is proportional to dose. When this is true, the kinetics iscalled linear. When this math is applied to the safe and effective therapeutic man-agement of an individual patient, it is called clinical pharmacokinetics. Thus, inclinical pharmacokinetics, we monitor plasma concentrations of drugs and suggestdosage regimens which will keep the concentration of drug within the desired ther-apeutic range. Pharmacodynamics refers to the relationship between the drug con-centration at the receptor and the intensity of pharmacological (or toxicological)response. It is important to realize that we want to control the pharmacologicalresponse. We do that indirectly by controlling the plasma concentration. In orderfor this to work, we assume kinetic homogeneity, which is that there is a predict-able relationship between drug concentration in the plasma (which we can mea-sure) and drug concentration at the receptor site (which we can not measure). Thisassumption is the basis for all clinical therapeutics.

Models are simply mathematical constructs (pictures) which seem to explain therelationship of concentration with time (equations) when drugs are given to a per-son (or an animal). These models are useful to predict the time course of drugs inthe 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 adosage regimen, to conceptualize what might be happening in disease states and tocompare products. In every case, the observations come first and the explanationnext. Given that a data set fits a model, the model can be used to answer severaldifferent 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, howmuch is absorbed and how fast? Are there things which might affect the absorp-tion, such as food or excipients in the dosage form itself. What would happen if thedrug were to be given on a multiple dose regimen? What if we increased the dose?etc.


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a

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 drug given by IV Bolus where applicable.

2.5.1 MAKING A MODEL

The differential equations used result from themodel which is our conceptualization of what ishappening to the drug in the body.

The box (compartment) is the area of interest. Wewant to find out how the mass of drug, , changes

with time in that compartment, the rate, and how the rates change with time, thedifferential equations.

How do we make a dif-ferential equation?

The picture that we build is made up of building blocks, consisting of the arrowand what the arrow touches. The arrow demonstrates how quickly the mass ofdrug, , declines. The arrow times the box that the arrow touches = the rate. Ratescan go in, i.e. arrows pointing to a box mean drug is going in (+ rate). Rates can goout, i.e. arrows pointing away means drug is going out (- rate). Rate = rate constant(arrow) times mass of drug (box). So the arrow and box really is a pictorial repre-sentation of a rate where the rate is the rate constant on the top of the arrow timeswhat the tail of the arrow touches.

Again, the rate constant, , tells the magnitude of the rate, .

Consider the following simple chain:

Xka

X

X

k k X⋅


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the

from

tionsse ofo the per-riabled then on

any

In fact,

ently

The building blocks are and . Every arrow that touches the compart-

ment of interest becomes part of the differential equation. If the arrow goes to thebox, it’s positive; if it goes away from the box, it’s negative.

To find (the rate of change of with time), we simply add up all of rates which affect (all of the arrows that touch )

and thus:

(Note: the first subscript of the rate constant and the subscript of the box which it originates are the same.)

You should be able to develop the series of interdependent differential equawhich would result from any model. The integration of those equations by uthe Laplace Table is done by transforming each piece of the equation intLaplace domain (looking it up on the table and substituting). The algebraformed solves for the time dependent variable: put everything except the va(including the operator, s) on the right side and put the variable on the left. Finresulting relationship on the left side of the table. The corresponding equatiothe right side of the table in the integrated form.

You should be able to integrate any differential equation developed frommodel (within reason) that we can conceptualize.

(Note: Each subsequent variable is dependent on the ones that precedes it. 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 currattempting to solve.)

X1 X3X2k12 k23

k12 X1⋅ k23 X2⋅

dX1 dt⁄ X1

X1 X1

dX1

dt--------- k12 X1⋅–=

dX2

dt--------- k12 X1 k23 X2⋅–⋅=

dX3

dt--------- k23 X2⋅=


Mathematics Review

2.5.2 ONE COMPARTMENT OPEN MODEL

A simplified picture (mathematical construct) of the way the body handles drug isone where the body can be conceived to be a rapidly stirred beaker of water (a sin-gle compartment). We put the drug in and the rate at which the drug goes away isproportional 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 oflittle consequence, as the data acts as if these were true for many drugs. The visualimage which is useful is one of a single box and a single arrow going out of thebox depicting one compartment with linear kinetics. The dose is placed in the boxand is eliminated by first order processes. In many cases, more complicated mod-els (more boxes) are necessary to mathematically mimic the observed plasma ver-sus time profile when one or more of these assumptions are not accurate. Forexample, the two compartment (or multi-compartment) model results when thebody is assumed to not be homogeneous and distribution is not instantaneous.


Mathematics Review

heseare

drug.verallf theineticser anddingngroperpriateus, a

es istions.he withplied,ppre-

theyderly-rans-and

ever,appro-

72

r ratentialme is

2.6 The LaPlace Transform

Why do we need to know the LaPlace trans-form?

One of the important facets of biopharmaceutics is a familiarity with the principlesof pharmacokinetics. This latter discipline describes the study of the dynamic pro-cesses by which the body “handles” or “disposes of” an administered drug. Tprocesses (absorption, distribution, metabolism, and excretion (ADME) dynamic in that they represent the time-dependent changes occurring to theThus, in pharmacokinetics the time course of these changes, which odescribe the fate of the administered drug, is described mathematically. Imathematical principles are understood, it is then possible to use pharmacokin clinical practice, such as the design of rational dosage regimens (T.S. FostD.U.A. Bourne, Amer. J. Hosp. Pharm., 34, 70-75 (1977). Understan(Bloom’s level 4) is not simply memorizing (Bloom’s level 1) nor calculatiusing a memorized equation (Bloom’s level 3). The authors believe that the pconceptualizing of the process and the subsequent derivation of the approequations will lead to an understanding of the mathematical principles, and thbetter, more optimal dosing regimen.

Since a mathematical description of the time-dependent ADME processrequired, it becomes necessary to deal with their corresponding rate equaInevitably this will involve calculus (mainly integral calculus). However, tLaPlace Transform provides a method whereby calculus can be performedminimal trauma. If a conscientious effort to learn the method is made and apa potentially serious obstacle (the fear of calculus) to the understanding and aciation of biopharmaceutics will be removed. Indeed, many students will find no longer fear integration and are thus free to comprehend the principles uning pharmacokinetics, which, after all, is the primary aim. So, the LaPlace Tform is a tool which is of great assistance in pharmacokinetics; its utility importance should not be lightly disregarded.

The LaPlace Trans-form: What Is It?

There is, of course, a theoretical background to the LaPlace Transform. Howit can be used without recourse to a complete theoretical discussion, though priate 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. 267-2(1975).

Basically, the LaPlace Transform is used to solve (integrate) ordinary, linear differ-ential equations. In pharmacokinetics such equations are zero and first-ordeequations in which the independent variable is time. For instance, if a differeequation describing the rate of change of the mass of drug in the body with ti


Mathematics Review

aveing a

ction in theinte-.

her issions,

omain

eans

ce

eml be

.

integrated, the final equation will describe the mass of drug actually in the body atany time.

The procedure used is to replace the Independent variable (time) by a functioncontaining the LaPlace Operator, whose symbol is “s”. In doing so we hreplaced the time domain by a complex domain. This is analogous to replacnumber by its logarithm. Once in the complex domain, the transformed funmay be manipulated by regular algebraic methods. Then the final expressioncomplex domain is replaced by its equivalent in the time domain, yielding the grated 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 anotaccomplished by reference to tables. One column shows time domain expresstated as , and second column shows the corresponding complex d

expressions, stated as the LaPlace Transform. Note that “ ” simply m

“some function of time”. For example, when is , then the LaPlaTransform is , where “B” is a constant and “a” is a rate constant

For example, when the LaPlace Transform is , then is .

2.6.2 SYMBOLISM

For simplicity in writing transformed rate expressions (and to distinguish thfrom untransformed (time domain) expressions), the following symbolism wilemployed:

“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 LaPlaceTransform of this mass.

f t( )f t( )

f t( ) Beat–

B s a+( )⁄

A s2⁄ f t( ) At


Mathematics Review

2.6.3 CONVENTIONS USED IN DRAWING PHARMACOKINETIC SCHEMA.

When drugs enter the body, they will encounter many different fates. It is impor-tant to set up the possible fates of the drug by creating a well thought out flowchart or scheme in order to follow all the events that are occurring in the body asdescribed by the pharmacokinetic description of the drug. For example, a drugmay be excreted unchanged or may undergo hepatic metabolism to yield active orinactive metabolites. All of these components are part of pharmacokinetics, whichby definition, includes ADME (the Absorption, Distribution, Metabolism andExcretion of drugs), and must be considered. This flow chart becomes the back-bone or the framework upon which to build the equations which describe the phar-macokinetics of the drug. The differential equations result as a direct consequenceof the flow chart. Using Laplace transforms, the integration of these differentialequations are simplified and provide the pharmacokineticist to (easily?) keep trackof all of the variables in the equation. If the drug scheme or flow chart is set upincorrectly, 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 con-struct a flow chart. Note that not all drugs follow the same flow chart and it isquite possible that you will need only to use a portion of these examples when con-struction your own.

In general, schema are relatively consistent in the placement of the compartmentsin relationship to one another. You might consider, for example a drug, given byIV bolus, which is metabolized and both the metabolite and the parent compoundare excreted unchanged as shown below:

X XuXf

Xm XmuXmf

Parent Compound

Metabolite

Feces Body Urine

kukf

km

kmf kmu

Dose


Mathematics Review

coki-lized like

romrom

Using the pharmacokinetic symbolism from chapter one, the compartments arenamed and placed: metabolites below (or above the plane of the parent com-pound): compounds going into the urine, to the right; and compounds going intothe feces, to the left of the compounds in the body. The rate constants connectingthe compartments also follow the symbolism from chapter one. In the above flowchart, 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 pharmanetic description, thus when a drug is given by IV bolus and is 100% metabowith the metabolite being 100% excreted into the urine the model would lookthis:

Thus in this flow chart, K, the summation of all the ways that X is removed fthe body, is km while K1, the summation of all the ways that Xm is removed fthe body, is kmu.

X

Xm Xmu

km

kmu

Dose


Mathematics Review

Drugs sometimes are metabolized to two (or more) different metabolites. In thefirst case, the drug is metabolized by two separate pathways resulting in this flowchart:

In this flow chart, K, the summation of all the ways that X is removed from thebody, is ku + kf + km1 + km2 while K1, the summation of all the ways that Xm1 isremoved from the body, is kmu1 +kmf1 and K3, the summation of all the waysthat Xm2 is removed from the body, is kmu2 + kmf2.

While in a second case, the drug is metabolized and the metabolite is furthermetabolized resulting in this flow chart:

X

Xm1

Xm2

km1

km2

Xmu1

Xu

Xmu2

ku

kmu1

kmu2kmf2

kf

kmf1Xmf1

Xf

Xmf2

Dose

X

Xm1

Xm2

km1

km2

Xmu1

Xu

Xmu2

ku

kmu1

kmu2kmf2

kf

kmf1Xmf1

Xf

Xmf2

Dose


Mathematics Review

In this flow chart, K, the summation of all the ways that X is removed from thebody, is ku + kf + km1 while K1, the summation of all the ways that Xm1 isremoved from the body, is kmu1 +kmf1+ km2 and K3, the summation of all theways that Xm2 is removed from the body is kmf2 + kmu2.

Both of these flow charts result in very different end equations, so it is imperativethat the flow charts accurately reflect the fate of the drug.

2.6.4 STEPS FOR INTEGRATION USING THE LAPLACE TRANSFORM

• Draw the model, connect the boxes with the arrows depicting where the drug goes.

• The building blocks of the differential rate equations are the arrows and what the tail touches.

• Write the differential rate equation for the box in question. The box is on the left side of the equal sign and the building blocks are on the other. If the arrow goes away from the box, the building block is negative, if it is going towards the box, the building block is positive.

• Take the LaPlace Transform of each side of the differential rate equation, using the table where necessary.

• Algebraically manipulate the transformed equation until an equation having only one trans-formed dependent variable on the left-hand side is obtained.

• Convert the transformed expression back to the time domain, using the table where necessary to yield the Integrated equation.


Mathematics Review

2.6.5 EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM

Following an intravenous injection of a drug (bolus dose), its excretion may berepresented by the following pharmacokinetic scheme:

(Scheme I)

Where is the mass of unchanged drug in the body at any time.

is the cumulative mass of unchanged drug in the urine up to any time, and is

the apparent first-order rate constant for excretion of unchanged drug.

Consider how the body excretes a drug

a. The building block is the arrow and what it touches. This first box (compart-ment) of interest is . The arrow is going out, therefore, the rate is going outand is negative, thus

(EQ 2-17)

The negative sign indicates loss from the body.

Taking the LaPlace Transform of each side of equation 2-17:

(EQ 2-18)

Note that because the independent variable (time) did not appear on the right-handside 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 into .Hence, the table was only required for transforming the left-hand side of equation2-18.

Manipulating the transformed equation:

1. Get only one variable which changes with time

2. Get on the left and everything else on the right.

X Xuku

X

Xu ku

X[ ] ku( )

dXdt------- kuX–=

sX X0– kuX–=

X( ) X

X( )

X


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(EQ 2-19)

(EQ 2-20)

Note that is the only transformed dependent variable and is on the left-hand sideof equation 2-19.

Converting back to the time domain:

(EQ 2-21)

Note that the right-hand side of equation 2-21 was analogous to in the

table, because 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.

sX kuX+ X0=

X s ku+( ) X0=

XX0

s ku+-------------=

LetX0 A= Letku a= X A

s a+( )----------------=

X

X X0ekut–

=

As a+( )

----------------

X0


Mathematics Review

2.6.6 SECOND EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM

Look at Scheme I again. Consider how the drug goes from the body into the urine.The next box of interest is . The arrow is coming in, therefore the rate is com-

ing in and is positive. thus,

(EQ 2-22)

(b) Taking the LaPlace Transform of each side of equation 2-22:

(EQ 2-23)

But, at zero time, the cumulative mass of unchanged drug in the urine was zero:that is .

(EQ 2-24)

(c) Manipulating the transformed equation:

(EQ 2-25)

Note that there are two transformed dependent variables. One of them can bereplaced by reference to equation 2-19.

(EQ 2-26)

(EQ 2-27)

(d) Converting back to the time domain:

Xu

dXu

dt--------- kuX=

sXu Xu( )0– kuX=

Xu( )0 0=

sXu kuX=

XukuX

s---------=

X( )

Xu

kuX0

s s ku+( )---------------------=

Let kuX0( ) A= Let ku( ) a= X A

s s a+( )-------------------=

Xu

kuX0 1 ek– ut–( )⋅

ku

----------------------------------------=


Mathematics Review

im-

y the

y at aned bytion).

t vari- wasin the

Where and are analogous to “A” and “a” respectively in the table. S

plifying,

(EQ 2-28)

2.6.7 THIRD EXAMPLE INTEGRATION USING THE LAPLACE TRANSFORM

During the intravenous infusion of a drug, its excretion may be represented bfollowing pharmacokinetic scheme:

(Scheme II)

Where Q is the zero-order infusion rate constant (the drug is entering the bodconstant rate and the rate of change of the mass of drug in the body is goverthe drug entering the body by infusion and the drug leaving the body by excreThe drug entering the body does so at a constant (zero-order) rate.

(EQ 2-29)

(b) Taking the LaPlace Transform of each side of equation 2-29:

Note that because Q is a rate, and is therefore a function of the independenable (time), its transformation yields the LaPlace Operator. In this case, Qanalogous to “A” in the table. But, at zero time, the mass of unchanged drug body was zero: that is,

(EQ 2-30)

(c) Manipulating the transformed equation:

(EQ 2-31)

kuX0 ku

Xu X0 1 ekut–

–( )=

X XukuQInfusion

dXdt------- Q kuX–=

sX X0–Qs---- kuX–=

X0 0=

sXQs---- kuX–=

XQ

s s ku+( )---------------------=


Mathematics Review

vationn ist theper)

irely, K = +ku. the

mple the

drugoved the

(EQ 2-32)

(d) Converting back to the time domain:

(EQ 2-33)

2.6.8 CONCLUSIONS

The final integrations (Eqs. 24, 24, and 28) are not the ultimate goal of pharmaco-kinetics. From them come the concepts of:

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 deriveequations given only the required pharmacokinetic scheme. Under such circum-stances, it no longer becomes necessary to remember a multitude of equations,many of which, though very similar, differ markedly in perhaps one minute detail.As with any new technique, practice is required for its mastery. In this case, mas-tery will banish the “calculus blues.”

It is also possible to see certain patterns which begin to emerge from the deriof the equations. For example, for a drug given by IV bolus the equatiomonoexponential, with the exponent being K, summation of all the ways thadrug is removed form the body. A graph of the data (Cp v T on semi-log paresults in a straight line the slope of which is K, always. If the drug is entmetabolized K = km. If the drug is entirely excreted unchanged into the urineku. If the drug is metabolized and excreted unchanged into the urine, K = kmthus K can have different meanings for different drugs, depending on howbody removes the drug. Following the drug given by IV bolus a second exaof a pattern would be that of the data of the metabolite of the drug. FromLaPlace, the equation for the plasma concentration of the metabolite of thehas in it K and K1, the summation of all the ways that the metabolite is remfrom the body, always. K1 would have different meanings depending on howmetabolite is removed from the body.

Let Q A=( ) Let ku a=( ) X A

s s a+( )-------------------=

XQ 1 e

k– ut–( )ku

-----------------------------=


Mathematics Review

was a

said.

rinci-s thatreta-

After several years teaching, I was fortunate to have a resident rotate through ourpharmacokinetic site. She had come with a strong Pharmacokinetcs backgroundand during our initial meeting, she had told me that she had a copy of John Wag-ner’s new textbook on pharmacokinetcs. She was excited that, finally, there 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“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 pples using scientific method, modeling and LaPlace Transforms in the hopememorization will be minimized and thought (and consequently proper interption) would be maximized.


Mathematics Review

2.6.9 TABLE OF LAPLACE TRANSFORMS

are constants

are rate constants

is the LaPlace Operator

is a variable, dependent on time

is a power constant

A B,

a b c, , a b c≠ ≠( )

s

x t( )

m

TABLE 2-17 Table of LaPlace Transforms

Time Function, LaPlace Transform, F t( ) f s( )

A As---

At A

s2

----

Atm A m!( )

sm 1+

---------------

Aeat– A

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

Atm

eat– A

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

A 1 eat–

–( )a

--------------------------A

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

Ata----- A 1 e

at––( )a

2--------------------------–

A

s2

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

Aeat– B 1 e

at––( )a

--------------------------–As B–

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

ABa---+

1 eat–

–a

------------------ Bt

a-----–

As B–

s2

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


Mathematics Review

TABLE 2-17 Table of LaPlace Transforms

Time Function, LaPlace Transform, F t( ) f s( )

A ebt–

eat–

–( )a b–( )

---------------------------------A

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

Aa b–( )

---------------- 1 ebt–

–b

------------------ 1 e

at––a

------------------ –

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

Atab------ A

a b–( )---------------- 1 e

bt––

b2

------------------ 1 e

at––

a2

------------------

––A

s2

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

1a b–( )

---------------- B Aa+( )eat–

B Ab+( )ebt–

–[ ] As B–s a+( ) s b+( )

---------------------------------

1a b–( )

---------------- A ebt–

eat–

–( ) B1 e

bt––( )b

---------------------- 1 eat–

–( )a

----------------------–

–As B–

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

1a b–( )

---------------- ABb---+

1 e

bt––( )

b---------------------- A

Ba---+

1 e

at––( )a

----------------------–Btab------–

As B–

s2

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

Ae

ct–

a c–( ) b c–( )--------------------------------- e

bt–

a b–( ) c b–( )--------------------------------- e

at–

b a–( ) c a–( )---------------------------------+ +

As a+( ) s b+( ) s c+( )

--------------------------------------------------

A1 e

ct––( )

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

bt––( )

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

at––( )

a b a–( ) c a–( )-------------------------------------+ +

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

Atbc------ A

1 ect–

–( )c

2a c–( ) b c–( )

-------------------------------------- 1 ebt–

–( )b

2a b–( ) c b–( )

--------------------------------------- 1 eat–

–( )a

2b a–( ) c a–( )

---------------------------------------+ +–A

s2

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

dXdt------- sX X0–

Atm 1+( )

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

A

sm

-----


Mathematics Review

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

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

13. The amount of metabolite of a drug in the body when the drug is given by IV Bolus and con-comitant 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 disopy-ramide is metabolized by a single pathway to mono-dealkylated disopyramide (MND). In a recent study, the pharmacokinetics of disopyramide were attempted to be elucidated by means of a radioactive tracer. Since both D and MND would be labeled by the tracer, any equations showing the time course of the label would show both the D and MND. By means of the laPlace transform, find the equation for the rate of appearance of tracer into the urine if the drug were given by IV Bolus.

XVd

------

T∞


Mathematics Review

2.6.11 LAPLACE TRANSFORM SOLUTIONS

1. The amount of drug in the body when the drug is givenby IV bolus (assume no metabolism).

At time zero, all of the IV bolus is in the com-

partment.

Here K = ku

X Xuku

Dose

X Xo=

dXdt------- kuX–=

sX X0 kuX–=–

sX kuX Xo=+

X s ku+( ) Xo=

XXo

s ku+-------------=

Let X0( ) A ku a XA

s a+( )----------------=,=,=

X Xoekut–

=

Basic Pharmacokinetics RECopyright © 1996-2000 Michael C. Makoid All Rights Reserved h

V. 00.1.6 2-62ttp://pharmacy.creighton.edu/pha443/pdf/

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

NOTE: You must start where the drug is at time = 0.Again K = ku

You only need to go this far because you

need to know X to put it in the equation for Xu. Thus:

(Xuo = 0. No drug in urine at time = 0)

Substituting from above for X.

X Xuku

Dose

dXdt------- kuX–=

sX X0 kuX–=–

sX kuX Xo=+

X s ku+( ) Xo=

XXo

s ku+-------------=

dXu

dt--------- kuX=

sXu Xuo kuX=–

sXu kuX=

sXu

Xoku

s ku+( )------------------=

Xu

Xoku

s s ku+( )---------------------=

Let X0ku( ) A ku a= X uA

s s a+( )-------------------=,,=

Xu

kuXo 1 ekut–

–( )ku

-------------------------------------=


Xu Xo 1 ekut–

–( )=


Mathematics Review

3. the amount of metabolite of a drug in the bodywhen the drug is given by IV bolus (assume no parentdrug excretion).

Remember:

You must start where the drug is at time = 0.

Here K = km and K1 = kmu

X

Xm Xmu

km

kmu

Dose

dxdt------ kmX–=

sX Xo kmX–=–

sX kmX Xo=+

X s km+( ) Xo=

XXo

s km+--------------=

Let X0( ) A km a X A

s a+( )----------------=,=,=

X Xoekmt–

=

dXm

dt---------- kmX= kmuXm–

sXm Xm0 kmX kmuXm–=–


Substituting for X from above

or

in general terms.

NOTE: We could also:

and then

or

Both of those equations are identical.

sXm

kmX0

s km+-------------- kmuXm–=

Xmkm Xo⋅

s kmu+( ) s km+( )------------------------------------------=

Let kmX0( ) A km a K kmu b K1==,==,=

XmA

s a+( ) s b+( )---------------------------------=

Xm

km X⋅o

( )km kmu–( )

------------------------- ek– mut

ek– mt

–( )⋅=

Xm

km X⋅ o( )K K1–( )

---------------------- eK– 1t

eK– t

–( )⋅=

Let kmX0( ) A km b kmu a=,=,=

Xm

km X⋅ o( )kmu km–( )

------------------------- ek– mt

ek– mut

–( )⋅=

Xm

km X⋅o

( )K1 K–( )

---------------------- eK– t

eK1t–

–( )⋅=


Mathematics Review

4. the amount of metabolite of a drug in the urinewhen the drug is given by IV bolus (assume the parentcompound is not excreted).

Here, again, K = km and K1 = kmu

substitute previously solved

Remember at time zero,

X

Xm Xmu

km

kmu

Dose

dXmdt

----------- kmX kmuXm–=

sXm Xom kmX kmuXm–=–

sXm kmuXm kmX=+

XXo

s km+--------------=

Xm s kmu+( )kmXo

s km+--------------=

XmkmXo

s kmu+( ) s km+( )------------------------------------------=

XmkmXo e

kmt–e

kmut––( )

kmu km–( )---------------------------------------------------=

dXmu

dt------------- kmuXm=

sXmu Xom kmuXm=–


Xom 0=

sXmu kmuXm=

sXmukmu( ) kmXo( )

s kmu+( ) s km+( )------------------------------------------=

Xmukmu( ) km( ) Xo( )

s s kmu+( ) s km+( )---------------------------------------------=

Xmu

kmukmXo

kmu km–--------------------- 1 e

kmt––

km

--------------------- 1 ekmut–

–kmu

-----------------------–

=


Mathematics Review

5. the amount of metabolite of a drug excreted inthe urine when both the parent and metabolite areexcreted.

Here K = km + ku and K1 = kmu

X

Xm Xmu

km

kmu

Dose

Xuku

dXdt------- kuX– kmX–=

sX Xo kuX– kmX–=–

sX kuX kmX Xo=+ +

X s ku km+ +( ) Xo=

ku km+( ) K=

XXo

s K+------------=

X XoeKt–

=

dXmdt

------------ kmX kmuXm kmX K1Xm–=–=

sXm Xom kmX K1Xm–=–

sXm K1Xm kmX=+

XmkmXo

s K1+( ) s K+( )--------------------------------------=


dXmu

dt------------- kmuXm=

sXmu Xom kmuXm K1Xm==–

sXmu K1Xm=

sXmu

kmu( ) kmXo( )s K1+( ) s K+( )

--------------------------------------=

Xmu

kmu( ) km( ) Xo( )s s K1+( ) s K+( )-----------------------------------------=

Xmu

kmukmXo

K1 K–---------------------- 1 e

Kt––

K------------------- 1 e

K1 t––K1

---------------------– =


Mathematics Review

6. The amount of drug in the body from a drug given byIV infusion (assume no metabolism).

At time zero, all the drug is still in the IV bag, thereforethere is no drug in the body. X0 = 0

Here K = ku

or

X XukuQ


sX X0–Qs---- KuX–=

sX kuX+Qs----=

XQ

s s ku+( )---------------------=

XQku

----- 1 ekut–

–( )= XQK---- 1 e

Kt––( )=



Mathematics Review

7. the amount of drug in the urine when the drug isgiven by infusion (assume the drug is NOT metabolized).

Here K = ku

or

X XukuQ


sX XoQs---- kuX–=–

sX kuXQs----=+

X s ku+( ) Qs----=⋅

XQ

s s ku+( )⋅-------------------------=

dXu

dt--------- kuX=

sXu Xo kuX=–

sXu kuX=

sXukuQ

s s ku+( )⋅-------------------------=

XukuQ

s2

s ku+( )⋅----------------------------=

Xu

kuQt

ku

----------- Qku1 e

kut––

ku2

--------------------

–=

Xu Qt Q1 e

kut––( )

ku

-------------------------–= Xu Qt Q1 e

Kt––( )K

-----------------------–=



Mathematics Review

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

.

X

Xm Xmu

km

kmu

Q

dXdt------- Q kmX–=

sX XoQs---- kmX–=–

sX kmXQs----=+

X s km+( ) Qs----=

XQ

s s km+( )----------------------=

X Q1 e

kmt––( )

km

--------------------------=

dXm

dt---------- kmX kmuXm–=

sXm Xo kmX kmuXm–=–

sXm kmX kmuXm–=

Xm

kmQ

s s kmu+( ) s km+( )---------------------------------------------=


or

XmA

a b–( )----------------=

1 ebt–

–b

------------------ 1 e

at––

a------------------

–

Xm

kmQ

kmu km–( )-------------------------=

1 ekmt–

–km

-------------------- 1 e

kmut––kmu

---------------------

–

Xm

kmQ

K1 K–( )----------------------=

1 eKt–

–K

------------------ 1 e

K1t––K1

--------------------- –


Mathematics Review

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

X

Xm Xmu

km

kmu

Q

dXdt------- Q kmX–=

sX XoQs---- kmX–=–

sX kmXQs----=+

X s km+( ) Qs----=

XQ

s s km+( )----------------------=

dXm

dt---------- kmX kmuXm–=

sXm Xo kmX kmuXm–=–

sXm kmuXm kmX=+

Xm s kmu+( )kmQ

s s km+( )----------------------=

Xm

kmQ

s s km+( ) s kmu+( )---------------------------------------------=

Xm

kmQ

km kmu–-------------------- 1 e

kmut––

kmu

----------------------- 1 e

kmt––

km

---------------------

–=


substitute

or

dXmu

dt------------- kmuXm=

sXmu Xomu kmuXm=– Xm

XmukmukmQ

s2

s km+( ) s kmu+( )-----------------------------------------------=

Xmu

kmukmQt

km kmu⋅----------------------

kmukmQ

km kmu–---------------------

–1 e

kmut––( )

k2

mu

----------------------------- 1 ekmt–

–( )

k2

m

--------------------------–

=

Xmu

kmukmQt

K1 K⋅----------------------

kmukmQ

K1 K–--------------------

–1 e

Kt––( )

K2

------------------------ 1 eK1 t–

–( )

K12

--------------------------–

=


Mathematics Review

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

In this case, K = ku +km and K1 = kmu and thus K = K1.This is not normal but could happen. The problem ariseswhen we get to the LaPlace that assumes the rate con-stants are different (i.e. ) because for this specialcase .

.

km ku+ kmu=

a b≠a b=

X

Xm Xmu

km

kmu

Dose

Xuku


sX Xo ku– X kmX–=–

sX kuX kmX Xo=+ +

X s ku km+ +( ) Xo=

XXo

s ku km+ +( )-------------------------------=

K ku km+( )=

kmu K1=

XXo

s K+( )-----------------=

X XoeKt–

=


(remember- K1 = K)

(kmX0 = A)

dXm

dt---------- kmX kmuXm–( )=

sXm Xom kmX K1Xm–( )=–

sXm K1Xm kmX=+

Xm s K1+( )kmXo

s K+------------=

Xm

kmXo

s K+( ) s K1+( )--------------------------------------=

Xm

kmXo

s K1+( ) s K1+( )------------------------------------------=

Xm

kmXo

s K1+( )2-----------------------=

Xm kmXoteK1t–

=

dXmu

dt------------- kmuXm=

dXmu

dt------------- kmukmXote

K1t–=


Mathematics Review

11. the principal metabolite when the drug is

cleared by several pathways

In this case K = km1 + km2 + ku, K1 = kmu1 and K2 =kmu2

Let K = ku + km1 + km2

and

Xm1( )

Xu Xm1 Xm2, ,( )

X

Xm1

Xm2

km1

km2

Xmu1

Xu

Xmu2

ku

kmu1

kmu2

Dose

dXdt------- kuX– km1X– km2X–=

sX Xo kuX– km1X– km2– X=–

sX kuX km1X km2X Xo=+ + +

X s ku km1 km2+ + +( ) Xo=

XXo

s K+( )-----------------=

K ku km1 km2+ +( )= K1 kmu1=

dXm1

dt------------- km1X K1Xm1–=

sXm1 Xm1o km1X K1Xm1–=–

Xm1o 0=

sXm1 K1Xm1km1Xo

s K+( )-----------------=+


Xm1 s K1+( )km1Xo

s K+( )-----------------=

Xm1km1Xo

s K1+( ) s K+( )--------------------------------------=

Xm1

km1Xo

K K1–----------------- e

K1t–e

Kt––( )=


Mathematics Review

12. the concentration of drug in the bodywhen the drug is given orally by a delivery system whichis zero order. What is the concentration in the body atequilibrium

Here K = ku

X Vd⁄

t∞( )

X XukuQ Xa ka

dXa

dt--------- Q kaXa–=

sXa XaoQs---- kaXa–=–

Xa0 0=

sXa kaXa Q s⁄( )=+

Xa s ka+( ) Q s⁄( )=

XaQ

s s ka+( )---------------------=

XaQ 1 e

kat––( )ka

-----------------------------=

dXdt------- kaXa KX–=

sX Xo kaXa KX–=–

Xo 0=

sX KXkaQ

s s ka+( )---------------------=+

X s K+( )kaQ

s s ka+( )---------------------=

XkaQ

s s ka+( ) s K+( )---------------------------------------=


If t= , then , thus

simplified yields:

XkaQ

ka K–( )-------------------- 1 e

Kt– )–(K

------------------------ 1 ekat–

–( )ka

-------------------------–

=

CkaQ

ka K–( )Vd--------------------------- 1 e

Kt– )–(K

------------------------ 1 ekat–

–( )ka

-------------------------–

=

∞ ekt–

0= CkaQ

ka K–( )Vd--------------------------- 1

K---- 1

ka

-----–

=

CQ

KVd-----------=


Mathematics Review

13 the metabolite of a drug in the body givenby IV infusion and concomitant IV bolus dose.

Infusion:

Here K = km and K1 = kmu

Xm

X

Xm Xmu

km

kmu

Q

Dose

dXdt------- Q KX–=

sX XoQs---- KX–=–

sX KXQs----=+

X s K+( ) Qs----=

XQ

s s K+( )--------------------=

dXm

dt---------- kmX K1Xm–=

sXm Xo kmX K1Xm–=–

sXm K1Xm kmX=+

Xm s K1+( )kmQ

s s K+( )--------------------=

XmkmQ

s s K1+( ) s K+( )-----------------------------------------=


IV Bolus:

Thus,

Xm

kmQ

K K1–( )---------------------- 1 e

K1t––( )K1

-------------------------- 1 eKt–

–( )K

-----------------------–

=

dXdt------- KX–=

sX Xo KX–=–

sX KX Xo=+

X s K+( ) Xo=

XXo

s K+------------=

dXm

dt---------- kmX K1Xm–=

sXm Xo kmX K1Xm–=–

sXm K1Xm kmX=+

Xm s K1+( )kmXo

s K+------------=

Xm

kmXo

s K1+( ) s K+( )--------------------------------------=

Xm

kmXo

K1 K–( )---------------------- e

Kt–e

K– 1t–( )=

Xm

kmXo

K1 K–( )---------------------- e

Kt–e

K– 1t–( )

…+=

kmQ

K1 K–----------------- 1 e

Kt––( )K

----------------------- 1 eK1t–

–( )K1

--------------------------–


Mathematics Review

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

Here K = ku + km and K1 = kmu

X

Xm Xmu

km

kmu

Dose

Xuku


sX Xo kuX– kmX–=–

sX kuX kmX Xo=+ +

X s ku km+ +( ) Xo=

ku km+( ) K=

XXo

s K+( )-----------------=

X XoeKt–

=

dXu

dt--------- kmX ku Xoe

Kt–( )==

dXm

dt---------- kmX K1Xm–=

sXm Xo kmX k1Xm–=–

sXm K1Xm

kmXo

s K+( )-----------------=+


Xm s K1+( )kmXo

s K+------------=

XmkmXo

s K+( ) s K1+( )--------------------------------------=

Xm

kmXo

K K1–( )---------------------- e

K1t–e

Kt––{ }=

dXmu

dt------------- kmuXm

kmukm

Xo

K K1–( )---------------------- e

K1t–e

Kt––{ }==

dXu

dt---------

dXmu

dt------------- ku Xoe

Kt–( )kmuk

mXo

K K1–( )----------------------

eK1 t–

eKt–

–{ }+ =+


BASIC PHARMACOKINETICS-Ch2: Mathematics Review

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