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School

Ho Chi Minh City University of Technology**We aren't endorsed by this school

Course

IM 3063

Subject

Mathematics

Date

Dec 24, 2024

Pages

Uploaded by PresidentBravery15955

1. The force exerted on the dam by integralThe Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curvey=0.7x2and below the liney=220. (Here distances are measured in meters). The water level is 36 meters below the top of the dam. Find the force (in Newtons) exerted on the dam by water pressure. (Water has a density of 1000kgm3, and the acceleration of gravity is 9.8msec2.) SolutionDensity of water:ρwater=1000kgm3Depth of water: h=220−36=184mFunction of height of the dam: h(y)=184−y ,(0≤ y≤184)According to Pascal's principle, the force exerted by the water on the dam is given by:F=∫abρgh(y)∆xdy(1)The shape of the dam wall is determined by y=0.7x2. Then, solving for x:x=±√y0.7So, the length of a thin horizontal strip is given by:∆ x=2√y0.7Then, from (1), we have:

F=ρwaterg∫abh(y)∆ xdy(ρwater, gare constant)→F=1000∗(9.8)∫0184(184−y)2√y0.7dy→F=1000∗(9.8)∗2√0.7∫0184(184−y)√y dy→F=19600√0.7∫0184(184y12−y32)dyUse calculator:¿>F=2.87∗109NNet access profitEx1: Suppose that tyears from now, one investment will be generating profit at the rate of P'1(t)=100+t2hundred dollars per year, while a second investment will be generating profit at the rate of P'2(t)=220+2thundred dollars per year.a. For how many years does the rate of profitability of the second investment exceed that of the first?b. Compute the net excess profit assuming that you invest in the second plan for the time period determined in part (a) Solutiona. The rate of profitability of the second investment exceeds that of the first until P'1(t)=P'2(t)↔100+t2=220+2t↔t2−2t−120=0↔t=12,−10¿>t=12yearsb. The excess profit of plan 2 over plan 1 isE(t)=P2(t)– P1(t), and the net excess profit NEover the time period 0≤t ≤12determined in part (a) is given by the definite integral.NE=E(12)−E(0)=∫012E'(t)dt¿∫012[P'2(t)−P'1(t)]dt¿∫012[(220+2t)−(100+t2)]dt¿∫012[−t2+2t+120¿]dt Usecalculator = 1008 hundred dollarsThus, the net profit is $1008

Ex2: Suppose that tyears from now, one investment plan will be generating profit atthe rate of P'1(t)=50e0.1tthousand dollars per year, while a second investment will be generating P'2(t)=110e0.06tthousand dollars per year. a. For how many years does the rate of profitability of the second investment exceed that of the first? b. Compute the net excess profit, in thousands of dollars, assuming that you invest in the second plan for the time period determined in part (a)Solutiona. The rate of profitability of the second investment exceeds that of the first untilP'1(t)=P'2(t)↔50e0.1t=110e0.06t↔e0.1t=115e0.06t↔e0.04t=115↔t ≈20yearsb. The excess profit of plan 2 over plan 1 isE(t)=P2(t)– P1(t), and the net excess profit NEover the time period 0≤t ≤20determined in part (a) is given by the definite integral.NE=E(20)−E(0)=∫020E'(t)dt¿∫020[P'2(t)−P'1(t)]dt¿∫020[110e0.06t−50e0.1t]dt Usecalculator ≈1059 hundred dollarsThus, the net profit is $1059Lorenz CurvesFind the Gini index for the given Lorenz curveEx1: L(x)=x3Ans:GI=2∫01[x−L(x)]dx¿2∫01[x−x3]dxUse calculator, we have:¿12

Ex2: L(x)=23x3.7+13xAns:GI=2∫01[x−L(x)]dxGI=2∫01[x−(23x3.7+13x)]dxUse calculator, we have: ≈0.383Gini index Form: Distribution of incomeEx1: In a certain state, it is found that the distribution of income for lawyers is givenby the Lorenz curve y=L1(x),whereL1(x)=45x2+15xwhile that of surgeons is given by y=L2(x), whereL2(x)=58x4+38xCompute the Gini index for each Lorenz curve. Which profession has the more equitable income distribution? Solution● L1(x)=45x2+15xG1=2∫01[x−L(x)]dx¿2∫01[x−(45x2+15x)]dxUse calculator, we have:G1=415● L2(x)=58x4+38xG2=2∫01[x−L(x)]dx¿2∫01[x−(58x4+38x)]dxUse calculator, we have:G2=524Since the Gini index for surgeons is smaller, it follows that in this state, the incomes of surgeons are more equitably distributed.

Ex2: In a certain state, it is found that the distribution of income for lawyers is givenby the Lorenz curve y=L1(x),whereL1(x)=0.3x2+0.7xwhile that of surgeons is given by y=L2(x), whereL2(x)=0.75x2+0.25xCompute the Gini index for each Lorenz curve. Which profession has the more equitable income distribution?Solution●L1(x)=0.3x2+0.7xG1=2∫01[x−L(x)]dx¿2∫01[x−0.3x2−0.7x]dx¿2∫01[0.3x−0.3x2]dxUse calculator, we have:G1=110● L2(x)=0.75x2+0.25xG2=2∫01[x−L(x)]dx¿2∫01[x−0.75x2−0.25x]dx¿2∫01[0.75x−0.75x2]dxUse calculator, we have:G2=14Since the Gini index for lawyers is smaller, it follows that in this state, the incomes of lawyers are more equitably distributed.Average valueEx1: A manufacturer determines that tmonths after introducing a new product, thecompany’s sale will be S(t) thousand dollars, where S(t)=144t√16t2+25What are the average monthly sales of the company over the first three months afterthe introduction of the new product?

SolutionThe average monthly sales V over the time period 0≤t ≤3is given by the integral:V=1b−a∫abf(x)dxV=13−0∫03144t√16t2+25dtTo evaluate this integral, make the substitution:u=16t2+25limits of integrationdu=32tdtif t=0, then u=25132du=tdtif t=3, then u=169You obtain:V=13∫25169144√u(132du¿)=32∫251691√udu¿Use calculator, we have:V=24Thus, for the 3-month period immediately after the introduction of new product, the company’s sales average $24000 per month.Ex2: A researcher models the temperature T(in degrees Celsius) during the time period from 6AM to 6PM in a certain city by the functionT(t)=−136t3+18t2+73t−2for 0≤t ≤12where t is the number of hours after 6PMa. What is the average temperature in the city during the workday, from 8AM to 5PMb. At what tduring the workday is the temperature in the city the same as the average temperature found in part (a).Solutiona. Since 8AM to 5PM, respectively, are t=2andt=11after 6AM, we want to compute theaverage of the temperature T(t)for 2≤t ≤11,which is given by the definite integral

Tave=111−2∫211(−136t3+18t2+73t−2)dtUse calculator, we have:V=1153144≈8℃Thus, the average temperature in the city during the workday is approximately 8℃.b. We want to find a time, t=tawith 2≤ta≤11such that T(ta)=1153144. Solving this equation,we find that:−136ta3+18ta2+73ta−2=1153144→t=−9.09∨9.36∨4.23Since t=−9.09is outside the time interval 2≤t ≤11(8AM to 5PM), it follows that the temperature in the city is the same as the temperature only when t=9.36∨4.23Two interpretationsCalculate the average rate of change of a function, f(x) = 3x + 12 as x changes from 5to 8SolutionWe have formula V=F(b)−F(a)b−aF(5)=15+12=27F(8)=24+12=36→V=F(8)−F(5)8−5=36−278−5=3Therefore, V(x)=3 Calculate the average rate of change of the function f(x) = x2 – 9x in the interval 2 ≤ x ≤ 7SolutionWe have formula V=F(b)−F(a)b−aF(2)=4−18=−14

F(7)=49−63=−14→V=F(7)−F(2)7−2=−14+148−5=0Therefore, V(x)=0