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Chabot Mathematics. §5.6 Int Apps Life&Soc Sci. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] 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. 5.5. Review §.

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Chabot Mathematics §5.6 Int AppsLife&SocSci Bruce Mayer, PE Licensed Electrical & Mechanical [email protected] 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 5.5 Review § Any QUESTIONS About §5.5 → Biz & Econ Integral Apps Any QUESTIONS About HomeWork §5.5 → HW-26 §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 days τ ≡ The “Time Constant” in % of MaxAge In this case the time constant is 21.715% of MaxAge Survival and Renewal The Survival fcn for the SYS % 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 = [xminxmax]; zyh = [0 0]; zxv = [0 0]; zyv = [yminymax]; % % 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([xminxmaxyminymax]),... 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]) MATLAB Code 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 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 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: % of those born on day 440 that survive to day 730 No. added on day 440 thatSurvive to 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 Survival & Renewal P(T) model Development Using Rewrite P(T) as Recognize that sum is in the Riemann form; Thus as ∆t→0, the Sum→Integral Survival & Renewal P(T) model Development 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 Survival & Renewal Running the Numbers on MuPAD find P(730) = 9483Spotted 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 Example RainBow Trout S&R Thus: Now Engage the substitution Find Example RainBow Trout S&R Running the Numbers There are a projected 76,383 rainbow trout fingerlings in the river after 5 years Incremental area, dA = [Length]·[Width]→ Population Density Calculate Population density (people divided by area) usingConcentric ring Integration People Living in the Ring = [Pop-Density at Location r]·[Area] Incremental area, dA = [Length]·[Width]→ Population Density Then Add-up, or Integrate, all the dp’s to obtain the total no. of people living in the area 2πR People Living in the Ring = [Pop-Density at Location r]·[Area Population Density Or Or in condensed terms 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? Example Urban Population SOLUTION: The population in the 2-3 mile ring: Use Substitution Example Urban Population Making theSubstitutions STATE: The population between 2&3 miles away from the center is approximately 13,065 People Volumes of Revolution Rotate y = f(x) about x-axis form solid 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 Volumes of Revolution Rotate y = f(x) about x-axis form solid Then the increment volume dV is the [DiskArea]·[DiskWidth] =dV = {π[f(xj )] 2} ·{dx} Adding up all the Incremental Disk Volumes Example Solid of Revolution Find the volume of the solid created by rotating about the x-axis over x = [0,1] the Graph of Example Solid of Revolution The solid after Rotation Example Solid of Revolution SOLUTION: Using the volume Formula: The volume of the solid is approximately 0.239 cubic units WhiteBoard Work Problems From §5.6 P24 → Arterial Blood Flow P36 → LifeExpectancy P40 → HumanRespiration 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 = [xminxmax]; zyh = [0 0]; zxv = [0 0]; zyv = [yminymax]; % % 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([xminxmaxyminymax]),... 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') Xint = 2.1833 All Done for Today A Rotated“Logistic”Curve Chabot Mathematics Appendix Do On Wht/BlkBorad Bruce Mayer, PE Licensed Electrical & Mechanical [email protected] – 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 Max @ (1.26, 4.807) Vtot = Area-Under-Curve Xint = 2.1833 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)

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