bruce mayer, pe licensed electrical & mechanical engineer [email protected]

[email protected] • MTH15_Lec-27_sec_5-6_Integral_Apps_Life-n-Soc_Sci.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§5.6 Int Apps

Life&Soc Sci



Final Exam Ref Document Students may HAND WRITE at 3x5

CARD as open Reference for the Final Exam• Final Exam

Tu/17Dec13at 6:30pmin Rm1613

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Review §

Any QUESTIONS About• §5.5 → Biz & Econ Integral Apps

Any QUESTIONS About HomeWork• §5.5 → HW-26

5.5

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§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



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




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

t ≡ time in daysτ ≡ The “Time Constant” in % of MaxAge

In this case the time constant is 21.715% of MaxAge

tetS %100



Survival and Renewal The Survival fcn for the SYS

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

60

70

80

90

100

Age, A (% of Max)

Sur

viva

l %, S

(A)

MTH15 • Spotted Yellow Squirel

Bruce May er, PE

%715.21%100 max

AtetS



MATLA

B C

ode% Bruce Mayer, PE% MTH-15 • 23Jun13% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m%clear; clc; clf; % clf clears figure window%% The Limitsxmin = 0; xmax = 100; ymin = 0; ymax = 100; % in PerCent of MaxAge% The FUNCTIONtau = 21.715 % in PerCent of MaxAgex = linspace(xmin,xmax,500); y = 100*exp(-x/tau);% % The ZERO Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];%% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenplot(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 onplot(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:

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

daySYS 40tR



Survival & Renewal P(T) model Development

• Multiply starting population by S(730) to determine how many 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

00 PTSN s



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:

440730day 1440440 SRN ASNo. added on day 440 that

Survive to 730

% of those born on day 440 that survive to

day 730

No. Born on day 440 = Rate·∆time



Survival and Renewal P(T) model Development

• The No. Added by births in variable form

• Then the Total SYS 2 years later

– Recall that

tTSttRtN

SRN

AS

AS

440730day 1440440

72972821730730 0 ASASASASS NNNNNP

730730 00 SPN S




• Using

• Rewrite P(T) as

• Recognize that sum is in the Riemann form; Thus as ∆t→0, the Sum→Integral

ttTStRtTSttRtN AS

Max

10

k

ttTStRTSPTP




• So the final Math model if S(t) and R(t) are known:

Now can calc the SYS Population 2 years (730 d) after the last SYS-Count

• Note: Times expressed as the % of Term-Time of 730days

T

dttTStRTSPTP00

730

0%715.21

%22730%715.21

%1002200730

407500730 dteePtdd



Survival & Renewal Running the

Numbers on MuPAD find• P(730) = 9483

Spotted YellowSquirells

• 5 Years Laterexpect aPopulation of about 10,032SYS MTH15_Spotted_Yellow_Squirel_S-n-R_1307.mn



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?



Example RainBow Trout S&R SOLUTION: This is a survival and

renewal application, with Then the number of fish present (in

k-fish) after five years is given by

ktR

etS t

50

65.0

5

00 555 dttStRSPP



Example RainBow Trout S&R Thus:

Now Engage the substitution

Find

)5(65.0 tu



Example RainBow Trout S&R Running the Numbers

There are a projected 76,383 rainbow trout fingerlings in the river after 5 years

383.76



Population Density Calculate

Population density (people divided by area) usingConcentric ring Integration

Incremental area, dA = [Length]·[Width]→

drrdA 2

People Living in the Ring = [Pop-Density at Location r]·[Area]

dArpdp

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Population Density Then Add-up,

or Integrate, all the dp’s to obtain the total no. of people living in the area 2πR

Incremental area, dA = [Length]·[Width]→

drrdA 2

People Living in the Ring = [Pop-Density at Location r]·[Area

dArpdp

ARP

dArpdpRP00

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Population Density Or

Or in condensed terms

RA

rdrrpdArpRP00

2

R

drrrpRP0

2

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Example Urban Population A town’s population is centralized and

drops off dramatically toward the outskirts of the city.

Census results suggest a model for the population density in k-People/mi2

How many people are between two and three miles from the center of the city?

216r



Example Urban Population SOLUTION: The population

in the 2-3 mile ring:

Use Substitution21 ru

rdudrr

drdu

22



Example Urban Population Making the

Substitutions

STATE: The population between 2&3 miles away from the center is approximately 13,065 People

3

22

112 dr

rrP

rdudrru2

1 2

0655.13



Volumes of Revolution Rotate y = f(x) about x-axis form solid

jx

xfy 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

a



Volumes of Revolution Rotate y = f(x) about x-axis form solid

jx

xfy Then the increment volume dV is the [DiskArea]·[DiskWidth] = dV = {π[f(xj )] 2} ·{dx}

Adding up all the Incremental Disk Volumes

dxxfVb

a

2

b

a



Example Solid of Revolution Find the volume of the solid created by

rotating about the x-axis over x = [0,1] the Graph of

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

x

y =

f(x)



Example Solid of Revolution The solid after Rotation

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.2

0

0.2

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

XY

Z



Example Solid of Revolution SOLUTION: Using the

volume Formula:• The volume of the

solid is approximately 0.239 cubic units

1

0

2 )( dxxfV

1

0

23 dxxx

1

0

246 2 dxxxx

357357 )0(

31)0(

52)0(

71)1(

31)1(

52)1(

71 V

239.0105

8 V



WhiteBoard Work Problems From §5.6

• P24 → Arterial Blood Flow

• P36 → LifeExpectancy

• P40 → HumanRespiration

Time, t (sec)

Res

pira

tion

Flow

, R (l

iter/s

ec)

MTH15 • Human Respiration

0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.250

1

2

3

4

5

Bruce May er, PE • 27Jul113

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MATLA

B C

ode% Bruce Mayer, PE% MTH-15 • 23Jun13% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m%clear; clc; clf; % clf clears figure window%% The Limitsxmin = 0; xmax = 2.25; ymin = 0; ymax = 5; % in PerCent of MaxAge% The FUNCTIONtau = 21.715 % in PerCent of MaxAgex = linspace(xmin,xmax,500); y = x.*(-1.2*x.^2 +5.72);Xint = roots([-1.2 0 5.72])% % The ZERO Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];%% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenarea(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 onplot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)set(gca,'XTick',[xmin:.25:xmax]); set(gca,'YTick',[ymin:1:ymax])set(gca,'Layer','top')


Bruce Mayer, PE Chabot College Mathematics Time, t (sec)

Res

pira

tion

Flow

, R (l

iter/s

ec)


0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.250

1

2

3

4

5

Bruce May er, PE • 27Jul113Xint = 2.1833



All Done for Today

A Rotated“Logistic”

Curve



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22



Bruce Mayer, PEMTH15 • 27Jul13P5.6-36

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 - TTz = @(T)41.6*(1+1.07*T).^0.13-T

>> LL = fzero(Tz, 50)LL =73.4916


Bruce Mayer, PE Chabot College Mathematics Time, t (sec)

Res

pira

tion

Flow

, R (l

iter/s

ec)


0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.250

1

2

3

4

5

Bruce May er, PE • 27Jul113Xint = 2.1833

Vtot = Area-Under-Curve

Max @ (1.26, 4.807)



Bruce Mayer, PEMTH15 • 27Jul13P5.6-40

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)



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