Part III: Advanced Math KEY
1.
Please note: Passing pretest III with an 85% or over
fulfills objectives for this section
2.
[pp. 73-89] Understand the concept of and be able
to work with logarithms
A logarithm is another way to discuss numbers that increase or
decrease exponentially. Like scientific notation, we can discuss
extremely large numbers.
Exponential form would be 102 = 100
As a log, this would be expressed as log10 100=2
‘the log of 100 to the base 10 = 2”
In the old days, scientists had to use slide rulers to arrive at a
logarithmic relationship; nowadays we simply use the calculator’s
LOG function. Input the number for which you need to find the log
and press LOG
Using a calculator’s log function, find the following logs of these
numbers.
The log of:
Is:
24
1.38
100
2
1.30
.13
The most common use of log relationships is the calculation of the
blood’s pH [level of acidity] based on the Henderson/Hasselbalch
equation.
Below is a simplification of this formula. X is actually calculated from
two other numbers that we will discuss later in this unit.
pH = 6.1 + the log of (x)
example x = 24
pH = 6.1 + the log of (24)
pH = 6.1 + 1.38
pH = 7.48
Because the numbers for which we need logs for the
Henderson/Hasselbalch equation are limited to a range of zero to 49, we
can use a modified log table for this action. Go to this link to find this
table.
Logs: http://wwwappskc.nhmccd.edu/programs/respcare/Log_H.doc
Using the table [or the calculator] find the pH, when x is known:
If x is:
+ 6.1 ,
the pH is:
15
1.18
7.28
16
1.2
7.30
20
1.3
7.40
21
1.32
7.42
22
1.34
7.44
24
1.38
7.48
FYI, to find the antilog of a number, you would input the number,
press the inverse key inv
3.
Log is
[pp 88-109] Be able to read, interpret and create
graphs:
To read scientific papers, the RCP will have to read & interpret
information in various types of graphs. A graph is a visual
interpretation of several numbers. The reader cannot easily make a
connection by just looking at a list of numbers.
Example Looking at the list of numbers: 23, 34, 43, 25, 46, 56, 87
yields us little information, but if these numbers were arranged on a
scale, we can make some assumptions about their relationships.
6
5
4
Series1
3
2
1
0
10
20
30
40
50
60
Once these same numbers are placed on a scale, one sees relationships
better. That item 4 and item 1 are both much smaller than the others and
that if item 4 changed places with 2, the columns would rise
symmetrically.
Scales of single lists of numbers give a little information; but if one is
comparing more than one data set the ability of tables and graphs to
impart information is enhanced.
For instance, you have three patients who get a drug that we suspect
affects their RAW [airway resistance] To plot this, you collect RAW at 1
hour before the medication, 1 hour afterwards and 4 hours later. Below are
the results of the experiment.
7
6
5
4
1 hr before
1 hr after
3
4 hr after
2
1
0
RAW A
RAW B
RAW C
1. At what time did most patients have the highest RAW?
At 1 hour after the medication
2. What could you say about the use of this medication if high
RAW were not desirable?
We may need to give a different medication to treat this
high RAW
3. What could you say about the duration of the effect of the drug
on these three patients?
Within 4 hours the affect is starting to drop back to pretreatment, but the RAW is still increased
Graphs have two dimensions: The vertical and the horizontal. We
call data that is plotted on the vertical, the Y axis and the horizontal as
the X axis. The Y axis is sometimes called the ordinate and the X axis
is called the abscissa.
A trick to remember y and x:
“ y do you stand up, when x lies across?’
During research, we will keep some parameters the same
[constants] but vary other parameters [variable.] When reporting
the results of our research, we will plot the variable [independent] on
the x axis and the results [dependent] along the y axis.
Coordinate graphs
A coordinate graph is one in which we incorporate the numbers line
of zero in the middle and positive numbers on the right side and
negative numbers on the left.
x -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
o
1 2 3 4 5 6 7 8 9 10 x
But we also have the same negative [down] and positive [up] number
lines on the vertical y axis. So that both number lines intersect at zero.
Go here for an example of this type of graph:
http://et.bgcbellevue.org/logo/img/cartesian-coordinate-system.png
The zero on both horizontal and vertical creates a cross which divides
the field of the graph into 4 parts called quadrants.
The 4 quadrants are in counter clockwise order:
Quadrant I is located between
12:00 and 3:00.
Quadrant II is between 12:00 and
9:00.
Quadrant III is between 9:00 and
6:00
Quadrant IV is the position between 6:00 and 3:00
You have collected data:
y
3
-4
-2
2
x
3
-2
3
-3
Please, plot these coordinates
II
I
3y, x3
2y,-3x
o
-2y,3x
-4y,-2x
III
IV
a. [pp. 90-91] Straight-line graphics
Relationships that are directly proportional are those in which as one
number changes the other number changes by the same proportions
In other words, in the relationship described by this formula:
VE = VT (f)
Calculate these:
VT
.500
.500
.500
f
5
10
20
VE
2.5
5
10
Plot the above table of VE and f.
15
14
13
12
11
10
9
8
7
6
5
4
3
2
f
5
10
As the f rises, what happens to the Ve?
As the f rises, the VE rises
15
20
25
b. [pp. 90-97] Linier and direct proportional & nondirect proportional; calculation of the sloop.
Any linier relationship can be expressed as a formula such as X = y (p) and
can be predicted. Anytime a scientist can find a linier relationship between two
variables, predictions become simple and extrapolation is possible.
Extrapolation is making predictions of unknown results based on unknown
data based on earlier relationships, but it is important to realize that
extrapolation can get you into trouble. Say you discover that increasing oxygen
from 205 to 30% and 40% raises the patient’s Sp02, you attempt to extrapolate
past this known data, by applying a linier regression to extrapolate what happens
to the Sp02 at Fi02 of 100%. This would get you into trouble, because if the
experiment had gone on longer by actually measuring the Fi02 and the Sp02, you
would have discovered that you cannot get past 100% [full is full]
Linier and direct proportional [pp.92]
Will always be straight line that will intersect the origin.
origin
Linier & non-direct proportional [pp. 93]
This relationship also produces a straight line but because the formula contains
other parts, it will not intersect the origin.
Calculation of the sloop. [pp. 92-93]
In the linier, none-directional graph, we can make predictions and extrapolations
based on calculation of the sloop. We use this calculation when we calibrate some
equipment by high and low parameters.
The formula for calculation of the sloop is below:
y = (m x) + b
y = the independent variable
x = dependent variable
m= sloop [Δ y / Δ x]
b = the y intersect
step 1 to get the Δ y / Δ x, you will have to divide the rise by the run.
The rise = y1- y2 while the run = x1-x2….so m = (y1- y2) / (x1-x2)
Go to page 93 and look at the graph to see how the rise and run were found.
Because this is a straight line we can assume that m is constant once we figure
out the Δ y / Δ x
Step 2 the b portion of the sloop equation is another constant, it is the point
where the x & y intersect when x is zero.
Now the formula looks like this:
y = [(Δ y / Δ x) x ] + b
y
x
15
100
25
250
35
400
From the above data, calculate y when we know that x = 5 and b =50
y = [(Δ y / Δ x) x ] + b
y = [(10 / 150) 5 ] + 50=
y = [(.0666) 5 ] + 50=
y = .33 + 50 =
y= 50.33
c. [pp. 98-103] parabolic and sigmoid curves.
Parabolic curve
Not all graphs are straight lines, some are based on intersection of a cone
and are curved. A famous respiratory therapy relationship is described by
a parabolic curve is the relationship between VD/VT and PaC02 and
VE.
Go to handout for this curve for reading and homework.
d. Sigmoid curve:
Another common curve used in respiratory care is the sigmoid curve.
This is a curve that is basically s shaped. An example of a sigmoid
curve is found on pp. 102-103
Notice that the line is flat at the bottom and at the top and that the
middle of the curve, we have the steepest portion of the sigmoid curve.
Go to handout for this curve for reading and homework.
e. [pp. 109-111] inflection points
In a sigmoid curve, at the point at the bottom where the flattened
portion turns into the steep portion, we have an inflection point.
This is a point at which significant changes occur; a turning point.
If the RCP wants to select a parameter change, he might want to get
more ‘bang for his buck’ by making the change at the infection point.
Go to handout for this curve for reading and homework.
4.
[pp. 124-132,144-154, ] Use algebra to calculate
formulae with variables and factors.
When using algebra, remember that we put complicated concepts into single letters
that stand for numbers s0 that we can [1] calculate the formula and [2] make general
statements about the relationships of these figures. In this section we will only
calculate these formulas. As time goes by the student will have to learn these formuli
and be able to calculate them to make determinations about patient status based on
data collected.
To solve for X [the unknown] we need to have numbers for the other parts of the
formula. For instance take the below formula. It is clear that one number is
subtracted from another.
We use symbols such as () and [] and {} to show us which parts of the formula needs
to be solved first. Some problems are so simple that they can be solved in one step
P(A-a)D02 which is also PA02-Pa02
Solve for P(A-a)D02 When: PA02 = 140 mmHg and Pa02 = 95 mmHg
PA02-Pa02
140 mmHg – 95 mmHg= 45 mmHg
The P(A-a)D02 = 45 mmHg
It would be impossible to calculate the P(A-a)D02 if you aren’t given both the
PA02and the Pa02.
Calculate these:
PA02 =
250
650
149
Pa02 =
102
450
65
P(A-a)D02
148 mmHg
200 mmHg
84 mmHg
Here is another formula that is simple enough to be completed in a single step.
a. Pi02 = (Pb ) Fi02
Solve for Pi02 , when Pb = 759 mmHg and Fi02 = .209
Pi02 = (Pb ) Fi02
Pi02 = (759 ) .209
Pi02 = 158.63
Calculate these:
Fi02 =
(Pb ) =
Pi02
700
.30
210 mmHg
350
.45
157.5 mmHg
730
1.00
730 mmHg
1460
.5
730 mmHg
b. Pa02/PA02 or a/A ratio
In this formula, it is clear that one figure is divided by the other figure
Solve for A/a ratio When Pa02 = 100 and PA02 = 200
Pa02/PA02
100/200
½ = .5
The a/A ratio is .5
Calculate these:
Pa02 :
PA02 :
A/a ratio is:
120
285
.42
150
155
.96
65
560
.116
c. C = ΔV/Δ P
FYI, the symbol Δ or ‘delta’ means ‘change.’ Solve for C when ΔV = 300 ml and Δ P
= 25 cmH20.
C = ΔV/Δ P
C = 300 ml/25 cmH20
C = 120ml/ cmH20
Calculate these :
ΔV
300
750
65
ΔP
30
50
25
C is:
10 ml/ cmH20
15 ml/ cmH20
2.6 ml/ cmH20
d. Raw = (P1 – P2)/ V
The () tells us to subtract the two numbers before the resulting number is divided
by V.
Solve for Raw when the P1 = 7 cmH20 and P2 = 5 cmH20 and V = 100 l/second
Raw = (P1 – P2)/ V
Raw = (7 – 5)/ 100
Raw = 2/ 100
Raw = .02 cmH20/l/second
Calculate these :
P1 :
P2:
V
RAW is
45
38
100
.07 cmH20/l/second
50
38
50
.24 cmH20/l/second
65
60
200
.025 cmH20/l/second
e. P = 2ST/r
This is a little bit more complex. It is obvious that P equal something divided by r,
but we must first turn 2STinto a single number. This formula takes an extra step.
Solve for P when ST = 5 and r = 15
P = 2ST/r
Step 1: P = 2(5)/15
Step 2: P = 10/15
P =.666
Solve these:
ST
r
P=
55
10
11
15
55
.545
45
50
1.8
f. PA02 = (Pi02 – PH2O)-[PaC02/.8]
The next formula is more complex. We have the symbols ( ) and [ ] to tell us to
reduce series of numbers into a single number before we can subtract one from the
other.
When Pi02 = 685 mmHg
PH2O = 47 mmHg [a constant]
PaC02 = 45 mmHg
Calculate these :
PaC02 :
38
50
65
PA02 = (Pi02 – PH2O)-[PaC02/.8]
PA02 = (685 – 47)-[45/.8]
PA02 = (638)-[56.25]
PA02 = 638 - 56.25
PA02 = 581.75 mmHg
Pi02:
150
178
350
PH2O
47
47
47
PA02 is
55.5
68.5
221.75
NOTE the PH20 in all cases turns out to be 47. When a number in a formula is
always the same number we call it a ‘constant’ sometimes symbolized by “k.”
g. VD/VT = (PaC02 – PeC02) / PaC02
The ( ) tells us to subtract these two before we divide the single number by
PaC02.
Solve for VD/VT when PaC02 = 45 and PeC02 = 43
VD/VT = (PaC02 – PeC02) / PaC02
VD/VT = (45 – 43) / 45
VD/VT = (45 – 43) / 45
VD/VT = 2/45
VD/VT =. 044
Calculate these:
PaC02
PeC02
VD/VT
45
36
.2
50
40
.25
35
25
.285
h. Henderson/Hasselbalch equation [entire equation]
pH = 6.1 + the log of [HC03/ (PaC02 x .03)]
Note that we need to reduce the numbers inside the () before we divide this by the
HC03 and then reduce this to a single number inside the [] before we can get the log of
the single number before we can add this single figure to the 6.1
Solve for the pH when the HC03 = 26 and the Pac02 = 45
pH = 6.1 + the log of [HC03/ (PaC02 x .03)]
pH = 6.1 + the log of [26/ (45 x .03)]
pH = 6.1 + the log of [26/ (1 .35)]
pH = 6.1 + the log of [26/ (1.35)]
pH = 6.1 + the log of [19.25]
pH = 6.1 + 1.28
pH = 7.38
Calculate these:
HC03
PaC02
pH
21
35
7.40
26
30
45
65
7.38
7.287
Solving problems with the use of several formulas in series.
In the real world, you will have a patient with several pages of data, frequently, you
will need to calculate one or even 3 or 4 formulas in order to calculate the final one.
You need to calculate the a/A ratio for your patient you have the following data on
them. NOTE: some of this data in this lab report is needed, some isn’t.
Fi02 35% [.35]
Pb 760 mmHg
PH20 47 mmHg
Pa02 is 55 mmHg
PaC02 45 mmHg
HC03- 24
To calculate the a/A ratio, you need the Pa02 which is 55 mmHg…. but the PA02 that
is not available on this lab report.
To get the PA02, we need to go back to another formula:
PA02 = (Pi02 – PH2O) - [PaC02/.8]
At this point, we need still another bit of data—the Pi20.
Pi02 = (Pb ) Fi02
So… to find the PA02, we perform several steps:
Step 1 Pi02 = (Pb ) Fi02 = 760 (.35) = 266 mmHg
Step 2
PA02 = (Pi02 – PH2O) - [PaC02/.8] = (266 -47) - [45 /.8]
PA02 = (219) - [56.25] = 162.75 mmHg
Step 3 a/A ratio = Pa02/PA02 = 55/162.75 = .337
Calculation of TLC
On page 131 of your textbook is a graphic representation of the
volumes of the lung. The TLC is the total lung capacity. It is made up
of all 4 volumes. Each of the 4 capacities contains 2 or more of the
lung volumes.
Answer the questions on pp. 131-132 (31-35) realizing that you will
have to perform several calculations before you can get your
answers.
5.
Be able to answer word problems based on the
math skills in Part III