Changing the integral activity



NESA exemplar question solutionsC3.1 Areas and volumesSolutions for questions from the NESA topic guidance related to areas and volumes.Sketch the region bounded by the curve y=x2 and the lines y=4 and y=9. Evaluate the area of this region. Note: The region in desmos is shaded using y≥x2{4≤y≤9}Method 1: The shaded area is twice the area between the curve and the y-axis where x≥0Area=249y12dy=22y33249=229332-24332=22?273-2?83=2543-163=2383=763=2513units2Method 2: The shaded area is twice the area between x=0 and x=3 Shaded area between x=0 and x=2 is 2?5=10 units2Shaded area between x=2 and x=3 is:A=9-23x2dx=9-x3323=9-333-233=83units2Total area =210+83=2513units2The graphs of the curves y=x2 and y=12-2x2 are shown in the diagram.Find the points of intersection of the two curves.Solve y=x2and y=12-2x2 simultaneously.x2=12-2x23x2=12x2=4x=±2When x=2, y=22=4When x=-2, y=-22=4Points of intersection: (-2,4) and (2,4)The shaded region between the curves and the y-axis is rotated about the y-axis. By splitting the shaded region into two parts, or otherwise, find the volume of the solid formed.V=πabfy2.dyVolume 1: The volume of the curve rotated about the y-axis between 4 and 12y=x2x=±y=±y12 V1=π412y12 2.dy= π412y.dy=πy22412= π1222-422=64π units3Volume 2: The volume of the curve rotated about the y-axis between 0 and 4y=12-2x2x=±6-y2=±6-y212V2=π046-y212 2.dy= π046-y2.dy=π6y-y2404= π6?4-424-6?0-024= π24-4-0=20π units3Total volume=64π+20π=84π units3The region bounded by the curve y=x-1 3-x and the x-axis is rotated about the line x=3 to form a solid. When the region is rotated, the horizontal line segment at height y sweeps out an annulus.Find the area of the annulus as a function of y.Area=πR2-r2r=3-x2 and R=3-x1Where x1 is the smaller x value and x2 is the larger x value for a given height y.y=x-1 3-xy=-x2+4x-3x2-4x=-3-yComplete the square.x2-4x+4=-3-y+4x-22=1-yx=2±1-yx1=2-1-y and x2=2+1-yr=3-2+1-y=1-1-yR=3-2-1-y=1+1-yArea=π1+1-y2-1-1-y2=π1+21-y+(1-y)-1-21-y+(1-y)=π2-y+21-y-2-y-21-y=4π1-yFind the volume of the solid. Sum of all annuluses from y=0 to y=1V=014π1-ydy=4π01(1-y)12dy=4π-21-y32301=4π-21-1323--21-0323=4π0--23=8π3 units3The region enclosed by the curve y=4x and the x-axis between x=0 and x=4 is rotated about the x-axis. Find the volume of the solid of revolution.Note: The region in desmos is shaded using 0≤y≤4x {0≤x≤4}V=πabfx2.dx=π044x2.dx=π0416x.dx=π8x204=π8?42-8?02=π128-0=128π units3A curved funnel has a shape formed by rotating part of the parabola y=2x about the y-axis, where x and y are given in cm. The funnel is 4 cm deep. Find the volume of liquid which the funnel will hold if it is sealed at the bottom.Note: The diagram shows an example of the region to be rotated about the y-axis. If the lower bound is y, then the upper bound is y+4 as the funnel is 4 cm deep.y=2xx=y24V=πabfy2.dy=πyy+4y242.dy=πyy+4y416.dy=πy580yy+4=πy+4580-y580=y+45-y580π units3If the solution is expanded:=π1?y5?40+5?y4?41+10?y3?42+10?y2?43+5?y1?44+1?y0?4580-y580=π20y4+160y3+640y2+1280y+102480 =y44+2y3+8y2+16y+645π units3Sketch the region bounded by the curve y=sinx+cosx and the coordinate axes in the first quadrant, taking the upper limit of x as 3π4. Note: The region in desmos is shaded using 0≤y≤sinx+cosx {0≤x≤3π4}Show the intercepts on the axes, and calculate the area of the region. y-intercept, substitute x=0y=sin0+cos0=0+1=1x-intercept, substitute y=00=sinx+cosx-cosx=sinx-cosxcosx=sinxcosx-1=tanxx=3π4 (limiting the x value to the upper bound of 3π4)Area=0 3π4sinx+cosxdx=-cosx+sinx03π4=-cos3π4+sin3π4--cos0+sin0=22+22--1+0=1+2 units2Find the volume of the solid formed if the region is rotated about the x-axis to form a solid of revolution.V=π0 3π4sinx+cosx2.dx=π0 3π4sin2x+2sinxcosx+cos2x.dx=π0 3π41+sin2x.dx=πx-12cos2x03π4=π3π4-12cos2?3π4-0-12cos2?0=π3π4-12?0-0-12=π3π4+12=3π24+π2 units3Note: This volume can be check using online calculators such as symbolab. It is entered as volume y=sinx+cosx,x=0,x=3π4C3.2 Differential equations part 1Solutions for questions from NESA’s topic guidance related to differential equations.If a product which has just been launched is judged by the market to be of poor quality, sales will decline as people try the product but do not continue to buy it. For a certain product, the rate of weekly sales is modelled by S't=400t+13-200t+12, where S is the number of sales in millions and t is the number of weeks since the launch of the product.Find the function that describes the weekly sales.S't=400t+13-200t+12S(t)=400t+13-200t+12dtS(t)=400t+1-3-200t+1-2dtSt=400t+1-2-2-200t+1-1-1+CSt=-200t+12+200(t+1)+CS0=00=-2000+12+200(0+1)+C0=-200+200+CC=0St=-200t+12+200(t+1)Find the number of sales for the first week and for the tenth week.S1=-2001+12+2001+1=-2004+2002=-50+100=50Number of sales in week 1 = 50 000 000S10=-20010+12+20010+1=-200121+20011=-200121+2200121=2000121 or 16.528…Number of sales in week 10=16 528 925.619…Comment on your results in the context of the given information.Based on the assumption given, this is a poor quality product as sales have declined as people try the product but do not continue to buy. The model only approximates products sold as only a whole number of products can be sold.Which of the following direction fields best represents the differential equation dydx=x-y?Answer: C. When x=y, dydx=x-y=0, i.e. the slope of the line segments will be zero.C3.2 Differential equations part 2This contains solutions for the questions from the NESA sample unit for the content: model and solve differential equations including but not limited to the logistic equation that will arise in situations where rates are involved, for example in chemistry, biology and economics (ACMSM132) AAMGardeners are concerned about the spread of a particular pest. All specimens detected so far lie within a circular region with radius 25 km and it is suggested that the increase of the radius rkm might be modelled by a differential equation drdt=16r, where t denotes the time in months. What does this model predict for the radius of the region affected by the pest after t months?drdt=16r6drr=1×dt 6r12dr=1?dt6r12dr=1?dt4r32=t+Cr=25 when t=04?2532=0+C4?125=CC=5004r32=t+500r32=t+5004r=t+500423Water is slowly leaking from a tank. The depth of the water after t hours is h metres and the variables are related by the equation dhdt=-ae-0.1t. Initially the depth of water is 6 metres and after 2 hours it has fallen to 5 metres. At what depth will the level eventually settle?dhdt=-ae-0.1th=-ae-0.1tdth=10ae-0.1t+C where C is the height the water will eventually settle as limt→∞10ae-0.1t=0h=6 when t=06=10ae-0.1?0+C6=10a+Ca=6-c10h=5 when t=25=10ae-0.1?2+C5=10?6-c10?e-0.2+C5=6-ce0.2+C5=6e0.2-ce0.2+C5-6e0.2=C-1e0.2+15e0.2-6e0.2=Ce0.2-1e0.2C=5e0.2-6e0.2e0.2-1e0.2C=5e0.2-6e0.2-1 or 1.0175… mThe water will settle at a depth of approximately 1.02 mWhen a ball is dropped from the roof of a tall building the greatest speed that it can reach is u ms-1. One model for its speed v ms-1 when it has fallen a distance x m is given by the differential equation dvdx=cu2-v2v where c is a positive constant. Find an expression for v in terms of x.dvdx=cu2-v2vvu2-v2dv=c?dxvu2-v2dv=c?dx-12-2vu2-v2dv=c?dx-12lnu2-v2=cx+C-12lnu2-v2=cx+C as u≥vv=0, x=0-12lnu2-02=c?0+C-12lnu2=C-12?2lnu=CC=-lnuSubstitute C into the function -12lnu2-v2=cx+C-12lnu2-v2=cx-lnulnu2-v2=-2cx+2lnulnu2-v2=-2cx+lnu2u2-v2=e-2cx+lnu2u2-v2=e-2cx?elnu2u2-v2=e-2cx?u2u2-v2=u2e-2cxv2=u2-u2e-2cxv=±u2-u2e-2cxv=u2-u2e-2cx as v speedis positive.Note: The solution could be expressed in a range of alternate forms such as:v=u1-e-2cx12. The solution can be checked by finding the derivative.The number of cane toads in a colony after t months can be modelled by t=40 0001+3e-t .Sketch the function Ct=40 0001+3e-t.What is the initial cane toad population?Substitute t=0C0=40 0001+3e-0=400001+3=10000The initial cane toad population is 10000.What is the population after 2 months, correct to 3 significant figures?C2=40 0001+3e-2=28449.383…≈28400 (correct to 3 significant figures)The cane toad population after two months is approximately 28400, correct to 3 significant figures.When does the population reach 30 000?Substitute Ct=3000030000=40 0001+3e-t300001+3e-t=400001+3e-t=433e-t=43-13e-t=13e-t=19-t=ln19t=-ln19=ln9=2.1972…monthsThe population will reach 30000 after approximately 2.2 months or 2 months and 6 daysShow that C't=120 000etet+32Ct=40 0001+3e-t=400001+3e-t-1C't=-1?-1?3e-t?400001+3e-t-2=120000et1+3e-t2=120000et1+3et 2=120000etet+3et 2=120000etet2et+32=1200001etet+32=120000etet+32 as required.Find the maximum growth rate of the cane toad population.The maximum growth rate of the population is at the time corresponding to the maximum value of the derivative. i.e. Solve C''t=0This corresponds to the point of inflexion in the graph of the original function, which is the Ct value halfway between the asymptotes. i.e. Solve Ct=20000Find the time of the maximum growth rate of the population:Option 1: Solve C''t=0C'(t)=120000etet+32C'(t)=120000etet+3-2Apply the product rule:u=120000et, u'=120000etv=et+3-2, v'=-2etet+3-3C''t=u'v+v'uC''t=120000etet+3-2-2etet+3-3?120000et=120000et1et+32-2etet+33=120000etet+3-2etet+33=120000et3-etet+33Solve C''t=00=120000et3-etet+330=3-etet=3t=ln3=1.098…Check for change in concavity either side of t.C''1=491.466…C''1.2=-505.205…∴ t=ln3=1.098… is time of the maximum growth rate in the cane toad population.Option 2: Solve Ct=2000020000=40 0001+3e-t200001+3e-t=400001+3e-t=23e-t=2-13e-t=1e-t=13-t=ln13t=-ln13=ln3=1.098…Find the maximum growth rate of the population:C'ln3 is the maximum value of the derivative and the maximum growth rate in the cane toad population.C'ln3=120 000etet+32=120 000eln3eln3+32=120 000?33+32=360 00036=10000The maximum growth rate in the cane toad population is 10000 toads per month.The population of a town is decreasing at a rate proportional to the square root of the population at that time.Write a differential equation to describe the situation.Let P'(t) be the rate of change in the population at time t.P't∝t P't=kt where k is a constant.If the population was initially 2500 and decreased to 2025 after 10 years, find the population after 20 years.P't=ktPt=kt?dtPt=kt12?dtPt=2kt323+CUse the initial conditions to find C and k: P0=25002500=2k?0323+C2500=CPt=2kt323+2500P10=20252025=2k?10323+2500-475=2k?10323-1425=2k?1032-14252?1032=kk=-142521000Pt=2?-142521000?t323+2500Pt=-1425?t3231000+2500Find the population after 20 years:P20=-1425?203231000+2500=1156.497…The population of the town after 20 years is 1156.497… ................
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