Chabot Mathematics
§5.6 Int Apps
Life&Soc Sci
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Final Exam Ref Document
 Students may HAND WRITE at 3x5
CARD as open Reference for the Final
Exam
• Final Exam
Tu/17Dec13
at 6:30pm
in Rm1613
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Review §
5.5
 Any QUESTIONS About
• §5.5 → Biz & Econ Integral Apps
 Any QUESTIONS About HomeWork
• §5.5 → HW-26
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
§5.6 Learning Goals
 Examine survival and renewal functions
 Use definite integration to compute
• population-totals from Population Density
• explore the flow of blood through an artery
 Derive an integration formula for the
volume of a solid of revolution, and use
it to estimate the size of a tumor
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Survival & Renewal
 Consider a Population of
Spotted Yellow Squirrels
(SYS) confined to a game preserve
 The SYS are carefully counted every 5
years by Dept of Fish & Game
Biologists. The Last “census” ended
today with a total, P0, of 7500 SYS
 The Biologists need a method of
Estimating the change in Population
before the next Census
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Survival & Renewal
 After Researching the SYS the
biologists have found
• That the SYS have a LifeSpan (maximum
age, Amax) of about 2200 days (≈ 6 yrs)
• The SYS have a
“Survival” function:
– Where
S t   100% e
 t ≡ time in days
 τ ≡ The “Time Constant” in % of MaxAge
 In this case the time constant is 21.715% of MaxAge
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
t 
Survival and Renewal
 The Survival fcn for the SYS
MTH15 • Spotted Yellow Squirel
100
S t   100% e
  21.715%
 t  Amax  
90
Survival %, S(A)
80
70
60
50
40
30
20
10
0
Bruce May er, PE
0
Chabot College Mathematics
7
10
20
30
40
50
60
70
Age, A (% of Max)
80
90
100
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 23Jun13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
%
clear; clc; clf; % clf clears figure window
%
% The Limits
xmin = 0; xmax = 100;
ymin = 0; ymax = 100; % in PerCent of
MaxAge
% The FUNCTION
tau = 21.715 % in PerCent of MaxAge
x = linspace(xmin,xmax,500); y = 100*exp(-x/tau);
%
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
plot(x,y, 'LineWidth', 4),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14} Age, A (% of Max)'),
ylabel('\fontsize{14}Survival %, S(A)'),...
title(['\fontsize{16}MTH15 • Spotted Yellow Squirel',]),...
annotation('textbox',[.21 .05 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
27Jul113','FontSize',7)
hold on
plot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)
set(gca,'XTick',[xmin:10:xmax]); set(gca,'YTick',[ymin:10:ymax])
Survival & Renewal
 After Researching the SYS the biologists
find That the SYS have a roughly
constant Birth, or Replacement, Rate:
Rt   40 SYS day
 It is now 2 yrs after the Last census, so
the “Term” of Projection, T, is 730 days
 The Biologist can now develop a model
for the Population, P(T) at T = 730 days
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Survival & Renewal
 P(T) model Development
• Multiply starting population by S(730) to
determine how many
N 0 s  S T  P0
of the original 7500
are alive 2 years later:
 
• To N0S add the births over some short time
period, ∆t, that survive until the end of the
term
– For example, it is much more likely that a SYS
born on day 701 will survive as compared with a
SYS born on day 49
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Survival and Renewal
 P(T) model Development
• In this case it is convenient to take
∆t = 1 day.
• Thus the number of SYS born on, say, day
440 that make it to day 730 must survive a
total of 730−440 = 290 days; a math
expression:
N AS 440  R440 1 day S 730  440
No. added on
day 440 that
Survive to 730
Chabot College Mathematics
11
No. Born on day 440 =
Rate·∆time
% of those born on
day 440 that survive to
day 730
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Survival and Renewal
 P(T) model Development
• The No. Added by births in variable form
N AS 440  R440  1 day S 730  440

N AS t   Rt   t  S T  t 
• Then the Total SYS 2 years later
P730   N 0 S 730   N AS 1  N AS 2  N AS 728  N AS 729 
– Recall that
Chabot College Mathematics
12
N 0 S 730   P 0 S 730 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Survival & Renewal
 P(T) model Development
• Using
N AS t   Rt   t  S T  t   Rt  S T  t  t
• Rewrite P(T) as
Max
PT   P0  S T    Rt   S T  t   t
k 1
• Recognize that sum is in the Riemann form;
Thus as ∆t→0, the Sum→Integral
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Survival & Renewal
 P(T) model Development
• So the final Math model if S(t) and R(t) are
known:
PT   P0  S T    Rt  S T  t  dt
T
0
 Now can calc the SYS Population 2
years (730 d) after the last SYS-Count
P730   7500  e
 730 d 2200 d 100% 
21.715%

730
0
40  e
• Note: Times expressed as the
% of Term-Time of 730days
Chabot College Mathematics
14
 730 t  22 %
21.715%
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
 dt
Survival & Renewal
 Running the
Numbers on
MuPAD find
• P(730) = 9483
Spotted Yellow
Squirells
• 5 Years Later
expect a
Population of
about 10,032
SYS
Chabot College Mathematics
15
MTH15_Spotted_Yellow_Squirel_S-n-R_1307.mn
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Example  RainBow Trout S&R
 About 48% of rainbow trout stocked as
fingerlings in the Clinch River die each
year, so that the fraction surviving out to
t years is e−0.65t
 The stocking rate of new fish is about
50,000 per year.
 If there are initially 63,000 fingerlings,
how many are projected to remain after
five years?
Chabot College Mathematics
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Example  RainBow Trout S&R
 SOLUTION:
0.65t


S
t

e
 This is a survival and
renewal application, with Rt   50k
 Then the number of fish present (in
k-fish) after five years is given by
P5  P0 S 5   Rt S 5  t  dt
5
0
number of
initial pop'n
remaining
after 5 years
Chabot College Mathematics
17
number of fish added
during the five years
which remain after 5
years
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Example  RainBow Trout S&R
 Thus:
P(5) = 63× e
-0.65(5)
5
+ ò 50e
-0.65(5-t )
dt
0
 Now Engage the u  0.65(5  t )
substitution
 Find
æ 1
ö
P(5) » 2.442775 + 50 ò e ç
du ÷
è 0.65 ø
t=0
t=5
u
50 é u ùt=5
= 2.442775 +
ëe ût=0
0.65
Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Example  RainBow Trout S&R
 Running the Numbers
50 é -0.65(5-t ) ù5
P(5) » 2.442775+
ëe
û0
0.65
50 é -0.65(5-5) -0.65(5-0) ù
= 2.442775+
-e
ëe
û
0.65
 76.383
 There are a projected 76,383 rainbow
trout fingerlings in the river after 5 years
Chabot College Mathematics
19
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Population Density
Incremental area,
dA = [Length]·[Width]→
dA  2r  dr
 Calculate
Population
density
(people
divided by
area) using
Concentric
ring Integration
People Living in the Ring =
[Pop-Density at Location r]·[Area]
dp  pr  dA
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Population Density
Incremental area,
dA = [Length]·[Width]→
dA  2r  dr
 Then Add-up,
or Integrate, all
the dp’s to
obtain the total
no. of people
living in the
area 2πR
P R   
PR 
0
dp   pr  dA
A
0
People Living in the Ring =
[Pop-Density at Location r]·[Area
dp  pr  dA
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Population Density
 Or PR    pr  dA   pr  2rdr 
A
R
0
0
 Or in condensed
terms
PR   2  pr  r  dr
R
0
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Example  Urban Population
 A town’s population is centralized and
drops off dramatically toward the
outskirts of the city.
 Census results suggest
6
a model for the population
2
1

r
2
density in k-People/mi
 How many people are between two and
three miles from the center of the city?
Chabot College Mathematics
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Example  Urban Population
 SOLUTION:
3
6
 The population
P = ò 2p r ×
dr
2
1+ r
2
in the 2-3 mile ring:
length of a width of a
small band small band
of population of population
r
=12p ò
dr
2
2 1+ r
3
 Use Substitution
u  1 r 2
du
du
 2r  dr 
dr
2r
Chabot College Mathematics
24
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Example  Urban Population
3
 Making the
Substitutions
u  1 r
2
du
dr 
2r
r
P  12 
dr
2
1 r
2
P =12p
= 6p
r æ1 ö
ò u çè 2r du÷ø
r=2
r=3
r=3
1
ò u du
r=2
 STATE: The
r=3
é
ù
population between = 6p ëln u ûr=2
2&3 miles away
P = 6p éëln (1+ r )ùû
from the center is
= 6p éëln (1+ (3)2 ) - ln (1+ (2)2 )ùû
approximately
 13.0655
13,065 People
2
3
2
Chabot College Mathematics
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Volumes of Revolution
 Rotate y = f(x) about x-axis form solid
y  f x 
xj
a
 At position xj the height of
the disk r = y = f(xj )
 The Area of the disk at xj
is the area of a circle,
A = πr2 = π[f(xj )]2
b
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Volumes of Revolution
 Rotate y = f(x) about x-axis form solid
y  f x 
xj
a
 Then the increment volume
dV is the
[DiskArea]·[DiskWidth] =
dV = {π[f(xj )] 2} ·{dx}
 Adding up all the
Incremental Disk Volumes
V     f  x  dx
b
b
Chabot College Mathematics
27
a
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
2
Example  Solid of Revolution
 Find the volume of the solid created by
rotating about the x-axis
3
over x = [0,1] the Graph of f (x) = x - x
0.3
0.2
y = f(x)
0.1
0
f (x) = x 3 - x
-0.1
-0.2
-0.3
0
0.1
0.2
Chabot College Mathematics
28
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
1
Example  Solid of Revolution
 The solid after Rotation
0.3
0.2
Z
0.1
0
-0.1
-0.2
-0.3
0.2
0
-0.2
0
0.1
0.2
0.4
0.3
0.5
0.6
0.7
0.8
Y
X
Chabot College Mathematics
29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
0.9
1
Example  Solid of Revolution
V     f ( x) dx
 SOLUTION:
0
1
2
3
 Using the
   x  x dx
0
volume Formula:
1
1
2
6
4
2


x

2
x

x
dx
• The volume of the
0
solid is
1
é1 7 2 5 1 3ù
approximately V = p ê x - x + x ú
ë7
5
3 û0
0.239 cubic units
1 7 2 5 1 3
1 7 2 5 1 3
V    (1)  (1)  (1)     (0)  (0)  (0) 
5
3
5
3
7

7

8
V
  0.239
105
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
WhiteBoard Work
 Problems From §5.6
• P24 → Arterial
Blood Flow
• P36 → Life
Expectancy
• P40 → Human
Respiration
MTH15 • Human Respiration
Respiration Flow, R (liter/sec)
5
4
3
2
1
0
Bruce May er, PE • 27Jul113
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
Time, t (sec)
Chabot College Mathematics
31
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 23Jun13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
%
clear; clc; clf; % clf clears figure window
%
% The Limits
xmin = 0; xmax = 2.25;
ymin = 0; ymax = 5; % in PerCent of MaxAge
% The FUNCTION
tau = 21.715 % in PerCent of MaxAge
x = linspace(xmin,xmax,500); y = x.*(-1.2*x.^2 +5.72);
Xint = roots([-1.2 0 5.72])
%
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
area(x,y, 'LineWidth', 4, 'FaceColor',[0.6 0.8 1]),axis([xmin xmax ymin
ymax]),...
grid, xlabel('\fontsize{14} Time, t (sec)'), ylabel('\fontsize{14}
Respiration Flow, R (liter/sec)'),...
title(['\fontsize{16}MTH15 • Human Respiration',]),...
annotation('textbox',[.21 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor',
'none', 'String', 'Bruce Mayer, PE • 27Jul113','FontSize',7)
hold on
plot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)
set(gca,'XTick',[xmin:.25:xmax]); set(gca,'YTick',[ymin:1:ymax])
set(gca,'Layer','top')
MTH15 • Human Respiration
Respiration Flow, R (liter/sec)
5
4
3
2
1
0
0
0.25
Chabot College Mathematics
33
Xint = 2.1833
Bruce May er, PE • 27Jul113
0.5
0.75
1
1.25
Time, t (sec)
1.5
1.75
2
2.25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
All Done for Today
A Rotated
“Logistic”
Curve
Chabot College Mathematics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
42
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
43
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
44
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
45
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Bruce Mayer, PE
MTH15 • 27Jul13
P5.6-36
L := 41.6*(1+1.07*t)^0.13
Lavg := (1/60)*int(L, t=10..70)
T80 := subs(L, t = 80)
T := 73.4916
Le := (1/T)*int(L, t=0..T)
plot(L, t =0..100, GridVisible = TRUE,
LineWidth = 0.04*unit::inch)
y = Life Expectancy, L * t = Current Age
From MATLAB
>> Tz = @(T) 41.6*(1+1.07*T).^0.13 - T
Tz =
@(T)41.6*(1+1.07*T).^0.13-T
>> LL = fzero(Tz, 50)
LL =
73.4916
Chabot College Mathematics
47
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
48
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
49
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
MTH15 • Human Respiration
Respiration Flow, R (liter/sec)
5
Max @ (1.26, 4.807)
4
3
Vtot =
Area-Under-Curve
2
1
0
0
0.25
Chabot College Mathematics
50
Xint = 2.1833
Bruce May er, PE • 27Jul113
0.5
0.75
1
1.25
Time, t (sec)
1.5
1.75
2
2.25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
52
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
53
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Bruce Mayer, PE
MTH15 • 27Jul13
P5.6-40
R := -1.2*t^3 + 5.72*t
t0 := numeric::solve(R,t)
tph := max(t0)
Vtot := int(R, t=0..tph)
Vavg := (1/tph)*int(R, t=0..tph)
Rmax := subs(R, t = 1.26)
plot(R, t =0..tph, GridVisible = TRUE,
LineWidth = 0.04*unit::inch)
y = Respiration Rate (liters/se) * t = time (sec)
Chabot College Mathematics
54
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx
Chabot College Mathematics
55
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx