leastsomeof the problems. Unit 72 also contains a model exam and according.to* oierall,economic

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-TITLE
UMAP Modules-Units 71, 72, 73, 74, 75, 81-83, 234.
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DESCRIPTORS
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4
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Applications; Mathematical Erfrichment; Mathematical
Moddls;'Mathematics Instruction; MedicarEducition;
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IDENTIFIERS
Radioactivity
ABSTRACT
The first four units cover aspects of medical
applications of calcylus: 71-Measuring Cardiac.. Output; 72-Prescribing
Safe and'Effective Dosage; 73-Epidemics; and 74-Tracer Methods in.
.
Permiabi]ity. All units include
a set'of exercises and answers to at
leastsomeof the problems. Unit 72 also contains
a model exam and
anwes to this exam. The fifthlisit in this set covers, applications
to,,economics: 7§-,,Feldman's Model. This mathematicalmodel describes
the beha*iv overtime of
a two-Sector economy in which sectoral
investmentvallocations are controlled by
a cential-au0Ority
according.to* oierall,economic.plan. The unit includes
exercises
-
aniwers. The next three moduleslfoOus
on Graphical and Numerical.
Solution-of 'bifferential Equatior0: 81r-Problems Leading
to
Differential Equations; 82-Solving-Diff'erential Equations
Graphically; and 83-Solving Differentialjquitions Wimefically.
The
three-unit group contains a total of five quizzes and
one exam, and
answers are provided for ,all in 4pepdices. The last unit
covers
.
7epplications or'calculus to'chemistry: 234-RadiOactive
Chains-Parents
and Daughters, (MP)
'1..
4.
I
-
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Raproduptions
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*
'supplied by. EDRS are,thevbest that can be made,:.
*
from the'.original document.
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.
c!)
S
umap
MEASURING CARDIAC OUTPUT
UNIT 71
so
S
by,
MODULES AND MON OGRAPHS IN UNDERGRADUATE
Brindell Hotelick and Sinan Koont
MATHEMATICS AND ITS APPLICATIONS PROJECT
Department of Mathematics
.University of Maryland BaltiMore County
Baltimore, Maryland
21228
MEASURING CARDIAC OUTPUT
the Brindell Horelidc and Sinan Koont
'TABLE OF CONTENTS
THE TECHNIQUE OF DYE DILUTION
I.
1
.
.
e
,
sin
THE FORMULA FOR OARIAS OUTPUT
2.
3
.
i
2.1
Preliminary Illustration
3
.
2.2
Rectafl'gular Approximation
4
2.3
The Definite Integral
,
,
7
....
3.
COMPUTATION OF CARDIAC OUTPUT
8
,
.
3.1
Antidi fferenti a ti on
8
3.2
Numerical Method%
8
J2
sin
x.)
e.°
.
6.
4.
EXERCISES
11
C
o
5..
et,
ANSWERS TO EXERCISES
13,
MEDICAL APPLI ATIONS OF CALCULUS
N.
14.S. DEPARTMENT OF EDUCATION
Units 71-74
'1
NATIONAL INSTITUTE OF EDUCATION
.
--, -47 7 ^:^7 -7
.
EDUCATIONAL RESOURCES INFORMATION
,
'CENTER IERIC)
"PERMISSION TO REPRODUCTHIS
This
has been reproduced as
MATERIAL IN MICROFICHE
received f
the person or organization
NLY
originating it.
HAS BEEN GRANTED BY
0 Minor changes have been made to improve
reproduction quality.
t
edc/urnap /55cflapel stinewton,
mass:02160 -
t 4
,
Points of view or opinions stated in this docu-
1
ment do not necessarily represent official NW
position or policy,
16-mg EDUCATIONAL RESOURCES
o,
INFORMATION CENTER (ERIC)."
3
6,
(
lnteruodular Description Sheet:. UMAP
Unit 71
Title:
MEASURING CARDIAC OUTPUT
MODULES
ND MONOGRAPHS IN UNDERGRADUATE
Atithor:
MATHEMATICS AND ITS APPLICATIONS.
Brindell Horelick and
Sinan Koont
PROJECT (UMAP)
Department of Mathematics
(
,
University of Maryland Baltimore County
The goal of UMAP is to develop, through
a cconiuriffy of users
Baltimore, Maryland
21228
,
and developers, a system of instructional
modules in undergraduate
C
mathematics and its applications which
may be.uied to supplement
Review Stage/Date:
IV
revision 5/15/79
existing courses and from which complete
courses may-eventually
be built.
Classificalion:
MED APPL CALC/CARDIACcOUTPUT (U,71)'
The Project is guided by a.National
Suggested Support
Steering Committee, of
Materials:
mathematicians, scientists and educators.
UMAP is funded by a
grant from the National Acience Foundation
to Education Development
References:
Center, tinc.,, a publicly supported,
nonprofit corporation engaged
Guyton, A,C. (1956), Textbook of Medical
Physiology, W.B. Saunders,
in educational research in the U.S
and abroad.
Philadelphia.
Hackett, E. (1973), Blood, Saturday
&
Review Press, New York.
PROJECT STAFF
Simon, W. (1972), Mathematical Techniques
for. Physiology and
Ross 1. Finney
Medicine, Academic PressNew 'York.
.
Director
'
-
Schwartz, A. (1974); Calculus and
Solomon Garfunkel
Analytic Geometry, Holt,
Associate Director/Consortium
RinehaPt, and Kinston,'New
York.,
coordinator
,
.
Felit)a DeMay
Associate Director for Admjnistration
`4»
'Prerequisile Skills:
Barbara Keiczewski
n
Coordinator for Materials
Prodmctionm,
L.,
Be able to identify lim
1.
Dianne Lally
f(x.)Ax as a definite:integral.
Prdject Secretary
=
iE
1
1
Paula.M.'Santillo
Administrative Assistant
4
2., Know how to define the area under
a curve.
Be able to
Carol Forray
Pnpduction Assistant
evaluate definite integrals by antidifferentiition.
Zachayy Zevi tags
St
f Assistant
Be familiar with the Oaptzoidal
3.
and .parabolic (Simpson's)
.
rule.
'
.
NATIONAL STEERING COMMITTE
.
.
-).
.
Out1,ut
W.T. Martin
MO.T.(Chairman)
1.
Be able to describe how dye
Steven J. Brams
New York University
dilution technique is used to
determine cardiac output.
J.layron Clarkson
Texas Southern University
2.
Be able to explain the setting
Ernest -J! Henley
up of the Riemann
University of Houton
sum herein
of
and the derivation
Equations (2) and (3)
--"-William-HOgaii-
Harvard University,
Donald A. Larson
at Buffalo
William F. Lucas
.Cornell University
%
Other_Related Units:
R. Duncan Luce;
.
Harvard University.
_prescribing Safe and-Effective
George Miller
Dosage (Unit 72)
Nassau CommunityZollege
.:
tpidemies (Unit 73)
Fredeeick,Mosteljer
Harvard University
Tracer Methods inPermeability
Walter E. -.6dtrs
(Unit
4)"
Uniwersitylbf MicAigan Press
.
George Spi-f4er
Indiana University
+)
Arnold A. Strassenburg
SUNY at Stony Brook
Alfred,A Willcox
Mathematical Association of America
This material was nena
red with the
support of National
Science Foundatjon Grant No;
SED76-19615 A02.
Recommendations
are'
expressed
those of tin a
uthors and di) not necessarily reflect
the views of the NSF,; nor of
the NetiOnai Steering Committee.
,
?roject
Qf
19q9 EDC
UMAP
Ali righttrreserved.
o
4
Our question is
How may we use the empirical data
MEASURING CARDIArOUTPUT
'given in Table 1 to'determine the
cardiac output?
.
.
TABLE 1
THE TECHNIQUE OF DYE DILUTION
1.
f
Typical Data for the Dye Dilution Technique
1
.
The volume'of blood a
pexson s heart pqmpb per unit
Time (seconds):
0
time (that Fs, the rate at which it icumps
2
5
4
3
7
8
6
1
brood) is
'
0.6 0.9,1.4 .1.9
Concentratibn:
0.1
called- the peson's cardi'ac
0
0
0
2.7
oufrput.
Normally itl a per-
(mg /liter)
son at rest this rate is about '5 liters pet minute.
But
A
IS
afte,
strenuous exercise It can rise to more then 30
10
Time (seconds):
9
12
13
14
11
15
17.
16
6
.
liei's per minute.
can also be raised or,,,lowered-
3.0 3.7
I
4.0
Cincentraion:
4.0
3.8
2.9
2.2
3.7
4.1
(mg/litek)
'Signiticantly, by
certain
*
diseases of the biooa vessels,
.
heart, and nervous system.
,
Time (seconds) :
18
20
19
'
23
22
21
24.
,
r.
V(
In this unit we shall discuss
a technique fOr
1.5
0.9
1.1
0.8
0.9
Concentration:
0.9
d.9'
.
-
..-
(mg/1 i ter)
measuringcaidiac output known as
dye dilution.
The
technique works as follows.. At time
t = 0.a known amount
$
Dof a dye is injected into
o
a main Vein near the heart.
The dyed blood circulates"through'the
s
right side of Ole
.
heart, the lungs: then the left side of
the, heart, and
/
finally appears ifi;the 'arterial systel.
The concentration
.%.
.
I
of the-dye is monitdred at fixed
time intervals,At at %
.),
some convenient point Ed the arterial
system.
Typically,
At might equal one secOnd.... For purposes of
L.
0
the mathe;
'
, matical development
we shall assume the
.
.
.
monitoring 1s done in the
,aorta
rta near the )lgirt..
In
.
.
4
A.Exercise 1 it will be assumed that*the
dye concentration
c
2
-is Anitored in t branch artery
_instead, and you will be
nr
A
asked to make appropriate changes
in the analysis that
-
*
c
1
_
fallows
v
1^
0
C
.
U
o:
6
'Normally it will take only a few seconds fdr
the dye,
td1Tasi throue the heart and
lungs once and begin to
,
a
'appear in the aorta.
, A typical set of readings will be
as in Table. 1., where-we see one result of injecting
5
a'c
25
10
20
15
D = 5 mg of dye in q4aIn'vein'ne,Ar the heart
at time A
Time (seconds)
= 0" seconds. , we plot these readings-.on,graph R4er
.,
Figure 1.
Typical readings in the dye dilution techniqqe.
.
we ,get tlie points shownftin Figure 1.-
= 5
when D
mg of dye 'are injected,at time t = 0 seconds'.
2
6 4.
v
Even if we 4ere to
We cannot achieve this situation.
THE FORMULAFOR CARD/AC OUTPUT
*!
2.,
wait a very extended period of time to achieve a constant
concentration of dye in the bloodstream, this would be
Preliminary Illustration
2.1
.
.
.
useless since we would have no vay of knowing how'long
,
Let, us-set the stage by considering a S.omewhat`.
the dye to pass the monitoring' point
it
Suppose
trtificial simplified version of the question.
once.'
of
h_e dye
it were possible to set things ,up so eilZ
How can we modify this simple algebraic comRutation
interval
flowed through the heart exactly once in a 'time
a
to analyze the data of Figure 1, where the dye concentra-
Of length T seconds,tat a constant concentration of C mg/k.
tion is not constant?
(A record of our observations would look,something like
Then we could express the amount Of dye by
Figure 2.)
r
Rectangular AporoxiM'ation
2.2
ldtei-s) of,
the formula D = CV, where V is the volume (in
blood flowing through the heart:in this time interval, or
-There are two essential differIncesbetween the
The cardiac output R (the rate) would then be
V = D/C.
idealized observations of Figure 2 and the'realistic
b;heformula V
= 1%1*(volume.= rate x time), which
"given
One is that in Figure 2 the
observations of Figure 1.
can be written in the fprm
dye conce tration is cons'ant.
\'the other is less striking
important in Figure4.2 we can identify a time
but-equall
R = V/T = D/CT,
(11
2
infervaL during which we kIlOw exactly how much dye has
the,i
where D, C, and T are\all known. Notice teat CT is
passed by our monitoring point.
area-of thexectangle in Figure 2.
Let us consider this second difference first.
Iii
k
Figure 1.we see that the dye concentration rises sharply,
then falls sharply, and then, just-when we think it is
This second
going to fall off tp zero, it rises again.
C.
20seconds.
Now, physi%logists
rise occurs at about t-=
I
know that 20 seconds is just about long enough for some
1
blood passing through the aorta to make a round trip of
it
ts
the body and the lungs, and reappear in°the aorta.
Apparently what is happening is that most of the dye
passes through the aorta in the first 14 or 15 seconds.
e
The dye concentration then falls off rapidly (from t = 15,
to t = 21).asathe rest of the dye trickles through.
Then,
0
at about't = 21; a little dye, having completed its round
1
trip, appears fox...the second time and mingles with what
Time (seconds)
is left Of the "first-time-through" dyne to cause the jump
in the graph.
..
,
Idealized observations of dye concentration in
I
Figure 2.
8
the aorta.
We mUst attempt,to pick out what part of the dye
6,
!
C
concentration after t,= 21 is due to "first-time-thraugli
3
of
.
*
rr
\\:
.
i
A
.11
dye.
Notice that just before't = 21 (and especially from
section.
In Figure '3 we have drawn a fuceesion
of
t = 19 t6 t = 21) the.iye concentration is decreasing
rectangle37
at
Each rectangle has base tt (the interval
a pretty steady` rate.
Let us assume that nfirstztime-
between observations) and height c(ti) (the
observed
through" concentration continues to acrease at,this rate:
'Concentration at time ti).
,In our example tt = 1.
In
Then the graph of "first-time-through" concentration,
Figure 3 we have illustrated, for i
=,t6 = 6
=
iinstead of rising at t =.21",- will
pass through the points
and e(ti)
c(6) = 1.4.
A., B, and C as shown1in Figure 3.
Now let us consider the time interval from
ti to
/
.In Figure 3 we simply drew in A, B, and C by
At the beginning'of
of 'length At.
t
eye.
that interval
i+1'
They are approximately:
nd
A (22,0.5), B.(23,0.3)
the dye concentration i,s observed
to be c(ti).
The volume
C (24,0)..
Therjepresentat best, a.shrewd gu
of blood flowing past our
observation.point during the
there is no point in agonizing
over their Axact location.,
time.interval:is-R.At.
Recall that R is s: rate.
If the
By the end of this section we shall
see that the portion
dye concentration were .constant for this
time interval,
of the g
ph after t = 2/ has only a small effect
the tOtalamoubt of dye flowirig"th.rough
on our
in this interval
-
result.
would bei6(tORAt, of R times the.
area of rectangle num-
ber.i4n Figue 3.
Now w
are ready to confront the first of the two
°
esstential differences mentioned
at the beginning of this
:Me time interval At is rather small
cowered to t b e
total time, involved, aid' the dye concentration
-
newer
changes abruptly, so the error introduced
by making this
approximation is not great.
f-i
i
...,
...--
1
S lnce we have assumed that
the monitoring is done in
I
.
the aorta near the heart, 'all the dye
I
must flow by our
I
I
I
0
monitoring point between t = 0 and t
4-
= T0.
If we add all
the approximations Corregpondilig
to the rectangles from
I
. I.
I
1
I
I I
I
1
I
s-
= 0 to t = T
we mult account approximately for the
I
i
0
I
4,,
I
I'
total amount. of dye D:
I
1
0
.
I
II
I.
I
I
'
i
D =
(2)
E c(t4)RAt = R ii.c(tdAt,
I
I
C
ft--,
/
I 1
I
i=1
i=1
-1
:
1 :
1
0
r
1
*I
where n is the number of rectangles.
4--,
!
In our example,.
1,-1
1 ,
.
.
1
n = 23 if we count the first two "rectangles," from
7- -1
IA
t = 1
1 I
'
.
I
I
1
I
_ItiltIT-ig
'
I
I
I
-
I
to t = 3, each of which has "height"
;111111y-,
zero.
Thus,
1
1
I
'C
I I
I
I I
I I
D.
"15 i
*
R =
(3)
10
20
25
c(t..)At
40
Time (seconds),.
i=1
where the denominator is the-total area'of
Figure 3.
Rectangular approximatUoil of the dye flow:
the rectangle
5
in Figure 3.
4
.6
.
t
X10
)
*
,
write
We can no
The Definite
integral-.
,
D
slifiCtly._ upon the
o
R -
(4)
s.er,Irld'Values-c(t.),-aaa-tWobserver'-determined'Nalues
Ic(t)dt
Now let us note that underlying_ the empirical'
74d: At.
th4ie is a function
°
values plotted, in Figures 1 and 3
We must use an approximation sign because our curve c(t)
c(tYdefined, (but not obierved) for all t between t.= 0
We have no
is at best a,curve which fits the data well.
This function may be approxi-
(see Figure 4).
and t =,'T
0
way of knowing if it is exact.
mated .by fitting. a Smooth curve, to the observed points
jti-,c(ti)1, 'nstag' the :es 'Suied poilitS A, B, and C at the
COMPUTATION OF CARDIAC OUTPUT
3.
enth'. The -dinoliiitator' in (3) 'is 'ten an! estimate Of the
In fact it is one of the approxi-
area under this cUrve.
Antidifferentiation.
3.1
kiting- sums' used in lefinlirg tbe- definite. integral'
-How we use Equation (4) depends on the nature of the
lim
c(t.
c(t)dt
It may be that a curve can be fitted to
3.)
function c(t)-.
. 11+0, 1=1
the data points in Figure 1 which is the graph of a func-
To/n.
To as given, and set At
which we think of n and
In that
tion c(t) whose antiderivative C(t) is known.
use/ the° fundamental theorem of calculus to
case we would
compute
T
0
°
C(T0) - C(0)\
. c(t)dt =
,
.
J
.
,
t
and then
41
C(T)
R'
- C(0)
:,
, -
: .
--
,
:
Numerical' Methods
. ,3 . 2
.
.
.
.
there will be no explicit
however
.MO-re likely,
.
2
In
let- alone for its antiderivative .
rmuld for- c(t) ,
use one, of, a iariety, Of ways' to estimate
:this case we
0-
,
the denominator
*46
of Equation (4), and thus obtain .an
.
list several, and illustrate
14 shall
approximation of R.
that these
with the data of Table I.
Recall
somelaf them
data were.ohtained wi,tb
tag.
a dye dosage of D = 5
r
. -
-
.
can use the
,
denominatqr ,of EqUation (3).
".*
"-
,-
the
This
rectangles in Figure 3, rather
uses the areas of
the 'area
"
tinder the curve in Figure 4.
,curve-c(t
1.3
In our example,
In our example,
°
.$1
a
24'
23
0
.421%
.+....
c(t-.)at =
+ 0.1 +
c(t)dt =
c(*ti)-.= 0
... +2.2
(040404 0.2 + 1.2+ 1.81-
10
i=1
i=1
0.8.+ 0.5'+ 0%3 = 44.1.
+ 1;8 + 1.6 + 1:0 + 0.6 + 0)
"h 1.,1
.
Then
'
0
= 44.1.
4-(88.2)
,
o
=
:4
.
/
0:.113 liters/second
44.1
--,
44--.---
R.
tg.
As in part '(a)', R = 6.8 Liters/minute.
.
-
= 6.8 liters/minute.
N.
-..,
-;.--
......
If the interval [a,b] is'divided into n equal
(d)
.-
.
'(Although the, concentration nleasUrements were taken by the
...
parts (a = to, < t1_ <
< tn-1 < tn =4
where n .is an
b],
.
second, outpu't is
measured in liters per minute.)
r
usually
number, then the
,'al,so known as
even
parabolic rule
°
Notice that in making this computation ,wereplaced_ the
..
_____.
Simpson's rube, says:
,
°
'; lasithiee experimental points of Table 1 wits the points
..............__
\
A "(22,0.5), B (23,0.3.), and b,(24,0).
the reason for
lb
b - a
4
c(t)dt.
doing thi,s was discussed in Section /.2.
.
+ 4y1 + 2y2 + 4y3 + 2y4 +....,
(.yo
.
a
1
More laboriously, but
,
more accurately, we
(b)
also,
.
+ 2yn:2 + 4yn.., + yn),
'"_
could sketch_Figure 4 on a large sheet of graph paper and
.
count the number of squares' that fall between 'c(t) and
.--
with the, notation of part (c)
with
.
.
the horizontal axis.
We -would then multipljrthis toter
by the unit of area tepresenteds;by a single tquare?
In our example,
,
(
There ,are also mechanical devices, called planimeterg,
..
,
24
with which it is possible to trace the boundary,ofa
.
t
+ 2(0)
0 (0
4(0)
c(t)dt
the area of
region and then read an estimate
.
21D.6)
I- 400.) +
region
+
'4.
3(24)
the
ofi
from a meter.
Weicould use one of these instead of coUnt:
;.
+ :...
+ 2(0.9) + 4(0.8) + 2(0.5)
!ti
4(0.9)
+
ing
.
...
squares.
.
+ 4(0.3) + 0)
If the interval ra,b) if. divided into n'equal
(c)
;
*
a.
<fro
[a = t
< t
parts
< ..... < tn-1 < to = LI), then the
=-1-(132.2)
.
....
0
1
says!: .
(
4
trapezoidal' zule
.
= 44.1 (to the. nearest -tenth).
.
b
Again; R-= 6.8, liters/minute.
+ 2y1 + 2y2 + ...:,+ 2yn'_1 +yn);
c(t)dt = -b---721-1
-a.(yo
"
a
t'-
-
c(t1) for_ i= 0,1,2,...,n.
where we have written y
9
10
rTh
15
-1
.
EXUCI
SES'
Find R in terms of To, Co, and,thetotal amount" of dye
I
4.
a:
injected.
,,r/
i.,Assume the dye monitoring takei7Alace at a 8rinch artery which
How isitaffected if (1) T
is doubled and CO is kept
b.
0
What
receives only 1/10 of the .blood cominTirom the heart.
is'
halved and CO is doubled?
(2) To
-constant?
-changes are necessary In the analysis contaihed In* Sect4ons 2.2
and 2.32
How does this affect Equations (3) and (4)?'
Find R in terms of T0, CO, and D '(the total amount of dye
5.
3.
injected) if c(t)
is as'shown is Figure -6.
t in secnild, and
Suppose c(t) --is measured -in
D in milligrams.
In what units should
ti
c(t)cit
be expressed.
Suppose that at time t the dye,concentration is
3.
.
=
-bt2 +
c(t) = -bt(t - To)
2T'/3
/3
"
whereb and T
'are positive constants.
Time (seConds)
0
s.
Graph c(0101
I
A hypothetical concentration curve.
Figure 6.
b. .Find R in terms of b, To,.and the total amount D of dye
Injected., .
liven attempt to determine cardiac output, 10 milligrams of dye
6.
are Jnjected into a main vein near the heart.
The dye.concen-
.
Suppose c(t) is'as'shown in Figure 5.
4.
trationis monitored at the aorta.
The following observations
are made:
,
6
4.-
Time (seconds):
7
8
0 2
5
3'
1
.
Concentration:
1.2.02.0
0.2 0.6
4.2
3.0
5.5
0,1
0.
(mg/literi4L
0
iv
16
Time (seconds):
9
12 13
15 17
10
14
Concentration:46.3
7.8
7.8
6.9
7.9
7.0 7.5 7.9
6.1
(mg/liter)
Time4(seconds):
25 26
20
24
22 23
18 19
21
'
T
/2
0
Concentration:
5.4 '4.7
2.2
2.8
3.5
2.2
I
4.1
2.1
'2.1
(mg/liter)
Time (seconds)
A hypothetical concentration
curve.
Figure 5.
- -
.1
-
,
-
-
a:
Plot thesq observations on graph paper.
6.
a.
8
b., At wyt'time does recirculation begin?
7
c.- What points would yog6lidd to the graph corresponding -to A,
6
6, and C in Figure 3?
o
5
...,
"
$.
0
7.
Calculate the cardiac output R from the data in Exercise 6:
'-'
4
c
23
a.
using Equation (3) directly.
o
-
2
b.
using the trapezoidal rule.
.
using the parabolic Nle.
c.
1
0
15
10 (
'20
5
25
30
ANSWERS TQ EXERCISES
5.
=
Time (seconds)
1.
Throughout Section 2.2, D and'R must be replaced by 0/10 and
b,
just after 23 seconds.
R/10, respectively.
Equations (3) and (4)
are unaffected.
c. A (24,1.2),
B (25,0.7), C (26,0) is one possible
answer.
2.
Milligram-seconds per liter.
ti
,7.
Did you Kemgmber to replace the last three
day points i6.the
3 a.
table-bl three 'Points approaching the t-axis?
A (24:1.2),
B125,
, C(26,0) will do.
a.
The denominator of EqUetion'(1) is
26
26
rc(ti)At =
I cal)
i=1
i=1
o + 0.1 + 0.2 +
+ 1.8 + 2:1 + 1.4 + 0.7 + 0
=
106.7.
6D
D
.Therefore,
b.-.
-3
bT
0.
-47=
(btT0 - bt2)dt'
R=
0.094 liters/second
0
`-s
5.6 liters/minute.
$.
2D
D .where A is the area )
4.
a.
2/sec L
cl
min
k /
T
26
C
b.
C T
A' of the triangle
,
,'
0 0
0
0
c(4)dt
2
° (0
+ 0.2 +
+ T.2 + +
+
02:7
0)
2
(26)
b.
(1) halved.
r,
(2) unchanged.
1
=
-(213.4)
2
where A is the area
D
(
5. 2I:2 /sec = 112-2 /Min
2C-T
C T
of the trapezoid
= 106.7.
0 0
1
.
14
19
13
'
.,;