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# APPLIED CALCULUS MATH 115

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This 34 page Class Notes was uploaded by Margarita Tillman on Monday October 19, 2015. The Class Notes belongs to MATH 115 at Rhodes College taught by Christopher Seaton in Fall. Since its upload, it has received 34 views. For similar materials see /class/224922/math-115-rhodes-college in Mathematics (M) at Rhodes College.

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Date Created: 10/19/15

Applied Calculus Fall 2008 Module 3 Curve Fitting Problem 1 Root Growth a One Parameter Linear Model Problem 2 Development Test Results Two Parameter Linear Model Problem 3 Concentration of Chemical Two Parameter Non linear Model Introduction In Module 1 we dealt with situation where a model could be tted exactly to the data It is always possible to nd a model that ts the data exactlyi This is not however the most appropiate way of modeling the data especially if want to use the model to make predictions about future events It is also inef cient since there are as many parameters to calculate as there are data points An efficient model is one with the fewest parameters that still manages to describe the behaviour of the data The disadvantage of using a model with fewer parameters than data points is that it will not pass through all the data points and so we nd the model that is the best t7 for the data The rst problem shows you have to t the best7 model of the form y mm to the data Within this problem we look at various ways of selecting the best7 model In the second problem we t the best7 model of the form y mm 12 In the nal problem and in the projects that your group will be assigned you will rst t nd and exact t model then a two parameter nonlinear model and nally t a model with more than two parameters We will then compare the properties of these models In the process of completing these problems we will see how to nd minimum and maximum values of functions and see how to nd derivatives of functions in several variables Schedule for Module 3 Date Lecture Assignment Mon 6th Oct Problem 1 Selecting and FoyeWPFP 1 see MOOdle Fitting a Model or escnptlon Prepare for the Module 2 Test Wed 8 Oct Problem 1 Optimizing the Model Prepare for the Module 2 Test Fri 10 h Oct Module 2 Test Mon 13ch Oct Finding and classifying critical points Homework 2 Sec 23 1 5 7 8 Sec 43 27 28 Supplimentary Problems See page 6 below Wed 15ch Oct Problem 2 Finding the best mm 12 model for data Homework 3 See page 11 below Fri 17th Oct Problem 3 Selecting the Model and Objective Function Read your project description look up the research paper read the abstract and nd the table from which your data wa taken Wed 22nd Oct Problem 3 Optimizing the Model Work on your projects Fri 24th Oct Multiple Variable Functions and Partial Derivatives Homework 4 Sec 72 2 3 5 7 12 13 14 21 22 Mon 27m Oct Problem 4 Working with other multiple parameter Homework 5 See page 17 mo e s belowi Mon 3rd NOV Presentation of Module 3 Projects Wed 5 h NOV Presentation of Module 3 Projects 1 Problem 1 11 Statement of the Problem A biologist is studying how fast the root of a particular specises of plant growsi The results of one experiment are belowi The biologist believes that the law7 of root growth is Growth m X Days7 and has asked us to determine the value of m from the experimental data Growth mm 0 mmewmwog ltlt 4 M Instructions 1 Plot these points on the graph paper provided 2 Draw a straight line that goes through the origin7 and is as Close to the other points as possib e 3 Calculate the slope of the line 12 Calculate the Distance of the Line from the Points Fill out the table below for your line 771102 AAAAAAA H Assignment Task 1 0 Complete this calculation of 0 Type your answer into the subject line of an email to your professorl Your answers must be braced together7 for example7 if your value for m is 148 and the value for is 1887 type 148188 0 Send the emaill Do not type anything else 0 You will lose your homework grade if you do not follow these directions 13 Solution The best possible Value of m is the Value that minimizes the function V2m If we plot the function in Mathematica we ge 6 which 7 7 039 We ceh use Mathematica to nd the derivative of mm by typing 77L D V2 in m The output is Then to solve 7 0 we can type Solve D V2 in m oh and if this fails to work m we ceh use FlndRoot D V2 in m0m1315 Homework 2 Find all the location of the critical poihts within the given domeih of the following fuhctiOhs 1 r 7 r1hr for a 0 2 r 7 3325 for all real as ie 700 lt a lt oo 339 fa 7 3t5 7184 7 5 7715 for all reelt Hint use Methemetice 4 HQ 7 9cos59 for 9 e 77 Questions 1 and 4 will be graded 2 Alternative Optimization Methods In Problem 1 to nd the best tting line of the form y mz we found the value of m that minimized the function V2021 12 7 m2 38 7 2m2 42 7 3m2 63 7 4m2 72 7 5m2 73 7 M2 It was claimed that represented the total distance between the data points and the model the y mz line so the lower this distance the better the t There are however alternative ways to measure the distance between a data set and a model 21 Alternative Residuals A residual is a distance from a single data point and the curve of the model The total distance function is a combination of all the residuals When deriving the function V2 we used Vertical residuals iiei residuals which are parallel to the yaxis We could however use a horizontal residual ie one parallel to the zaxis or a perpendicular residual ie one at rightangles to the model line These three alternatives are illustrated below Vertical Residual Horizontal Residual Perpendicular Residual To calculate the total distance functions from these three residuals we can square each residual distance sum all these and nd the square root of this sumi The values of the seven residuals for these three possibilities are given below Data Point Vertical Horizontal Perpendicular 07 0 112 127m 17 13 238 38 7 2m 2 7 351quot 342 42 7 3m 3 7 4 51quot 463 6374721 47 g 5 72 72 7 5m 5 7 g 6 7 3 7 3 7 6m 6 7 g 731quot From this table we can easily write down the different total distance functions V2071 1 2 7m2 3 8 7 2m2 4 2 7 37212 637 4m2 72 75m2 7376721 127721 2 3872721 2 7275721 2 7376721 2 P2m 7 7 7 7 m21 m21 m21 m21 These functions are de ned in the Mathematica le named M310ther0Fsnbi We nd the critical point of these functions and use a test to con rm that the point is a local minimumi Note that each function gives a different optimal value for the slope of the of the straight line modeli 22 Different ways of totaling the distance Once the type of residual has been selected we have to have a scheme for combining the individual residuals into a total distanceli The scheme we have been using is to square add and square root the sum we can call this as pythagorean scheme7i Other schemes are 0 Just sum the residuals o Cube sum and cube root the sumi 0 Take the fourth power sum and take the fourth root of the sumi 0 Take the fth power sum and take the fth root of the sumi 0 etc In other words we can de ne the total distance function for this set of data by any one of the formulae below for any whole value 0 p Wm 1i27mp3 872mpn7 275mp7376mp 1 atlt2gtplt57gt lt6gtT P 127721 gt17 3872721 lt7i275mgt17lt7i376mgt r m 7 7 7 7 p m21 m2l m21 m2l Note that for odd values ofp the functions Vp H17 and P17 do not have a minimumi They either increase or decrease They do however pass through the zaxis meaning that there is a value of m for which the total distance7 between data and model is zero this is when the positive residuals cancel out the negative residualsi So for odd values of p we nd the optimal value of m by nding the zero of the distance function 3 Problem 2 31 Statement of the Problem The following data shows the age of seven children7 in months7 and their score on a particular developmental testi Age Months Score 20 95 27 90 30 98 35 103 41 112 44 109 46 115 32 Fitting y mm To try to t y mz model to this data7 we can set the objective function to be 95 7 20m2 90 7 27m2 109 7 44m2 115 7 46m27 nding the minimum of this function is attained when m l l l l l l l H ll in this answer from the lecture When we plot the data together with y z the Mathematica output is 10 20 3O 4O 50 Clearly this is a poor t7 so all models of the form y mm are invalid We need to use a different type of model 33 Fitting y mm b For this data no value of m can make y mz t the data well The next simplest model is one of the form y mm 12 The gure below shows the vertical7 horizontal and perpendicular residualsi Vatical Residual Houzoml Residual Pelpendiculal Residual Hence the list of objective functions we have to choice from is Vpm b 7 95 7 20m 7 b 90 7 27m 7 b 109 7 44m 7 b 115 7 46m 7 b i l HAW 20 95747gt 27 9077b 11lt44 w 46 my m m m m P b 7 95720m7bgtp 907271 rtil7gtpni 109744m7bgtp lt115746mibgt1 p m 7 W W W W Their values will Change with both m and 12 Their graphs are surfaces Below is the graph of V2mb Ms 9 s s EH s s s mmwgaary W It is hard to see that this function does have a minimum value However we can try to solve the two of simultaneous equations 72 0 and 72 01 This will give us 6m 61 10 points7 if any exist7 where both rate of change in the m direction is zero7 and the rate of change in the b direction is zero Homework 3 0 Using the commands in the Mathematica workbook M32AffineModel nb7 an swer the questions belowi o The answers should by written in a worddocument7 named quotyour last namequotJV13JIw3 doc 0 You may work together with your fellow group members if you wish7 but should submit individual copies of your work 1 Using the total distance function H4m7 b7 nd the optimal model for the test data7 iiei Age Months Score 20 95 27 90 30 98 35 103 41 112 44 109 46 115 Provide a graphic evidence that the model is valid 2 Using the P2 distance function7 t the best model of the form y mm b to the data about the growth of plant roots from Problem 17 ie Growth mm 0 mmewmwog ltlt 4 M Comment on whether a model of the form y mm or a model of the form y mm b is most suitable for this data 4 Problem 3 41 Statement of the Problem A factory needs to use a certain toxic chemical in its processing There is a steady ow of water through the machinary that washes the chemical into a settlement tank The factory runs its process for eleven hours and then stops for one hour to allow the equipment to be washed clear of the chemical The table below shows the concentration of chemical in the outlet pipe7 at different times during a typical twelve hour shifti Time Hours Concentration mgL 0 000 2 000 4 0120 6 142 8 4230 9 20500 10 69000 105 101000 11 103000 115 83000 12 0 00 42 Selecting the Model When plotting the data Mathematica gives a graphic of the form 1000 39 2 4 6 8 10 12 The model that is suggested by this data is one of the form y a12 7 z2e b12 To nd the best values of a and 127 we will set up a total distance funtion in a manner similar to how we proceeded for straight line models 12 43 De nition of Residuals to curves The next three diagrams show how the different residuals are de ned Vertical Residual Horizontal Residual Perpendicular Residual The value of i is the solution of yi 7 f i 7 mi and the value of Q 1s Notice that in order to calculate either horizontal or perpendicular residuals one must rst solve an equation Consequently7 these often involve practical dif culties7 and vertical residuals are often the only ones that can be calculated 44 Misleading Residuals There are situations Where using horizontal or vertical residuals can give inaccurate values to the distance from point to curve7i These are illustrated belowi 21312 Vertical residual gives an erroneously large distancei w Horizontal residual gives an erroneously large distance Note that in both of these situations the perpendicular residual gives a more rea sonable value to distance from point to curve77 however nding the i which satisfies yi 7 i 7 could be dif cult 45 Useful Tricks for Minimizing Total Distance Functions 451 Simplifying the total distance function The Pythagorean total distance for vertical residuals is the function V2a717 91 fab112 y2 fabr22 yn fab1n27 where a and b are the parameters of the model The values of a and b which minimize this function will also minimize the function V220397b 91 fabati2 y2 fabr22 yn 7 fab1n2 If this seems unlikely to you7 just think about a list ofpositive numbers7 eg 5728798114 7 take note of the position of the minimum number7 now square every number in the list eg 235476478171217 l67 and the minimum of this list is in the same position as the minimum of the original list Notice that this trick only works when the function or list of numbers is positive 452 Estimating the parameters by nding an exact t7 Often we can find reasonable estimates for the parameters by finding a model that is an exact fit to a few data points If your model has 3 parameters7 say7 then you should pick three of the data points that are representative of the spread of the data7 then find the parameter values that force the model to pass exactly through these points These parameter values will not be the best fit for all the data but very often they are close to the optimal values For this data and model7 we would pick two points eg 6142 and 111030 avoid the points with y value zero and solve the following equations simultaneously 142 a X 62 gtlt 5 1030 a X12 gtlt e bxl 5 Problem 4 51 Statement The most critical stage of a 100m race is as the athlete accelerates out of the blocks The athlete will be able to maintain the speed they manage to reach in this phase Below are the values of an athlete7s acceleration in the rst 08 seconds of a practice runi What is the athletes speed 08 seconds into the run Time sec Acceleration ms2 204 Oil 203 0 2 194 0 3 173 0 4 140 0 5 9 8 0 6 5 3 0 7 1 6 0 8 0 52 Fitting a four parameter Suppose we want to t a cubic model to the data7 ie a model of the form at at3 1272 ct d then we will follow the same proceedure as before We de ne the total distance function7 and nd where this function is a minimumiln this example if used the V2 total distance function it would be de ned as V2abc d 7 204 7 a02 203 7 a0i12 194 7 a0i22 173 7 a0i32 14 7 a042 7 9 8 7 a0i52 5 3 7 a0i62 7 1 6 7 am 2 7 0 7 a0i82 To minimize this function we nd all four partial derivatives of V27 and nd the locations when all four are simultaneously equal to zero That is so ve 6V2 8V2 8V2 6V2 E W W W W 7 8b 7 BC 7 8d We now need to run some type of test to classify the type of critical points we have found Unfortunately the equivalent of the second derivative test is too complex for this course We can t even examine the graph of V2 to see if the location looks like a minimum7 because the graph of V2 lies in ve dimensional space The only check we can perform is to plot the data and the new model and see if it is a reasonable ti 53 Solution When each of the stages is done on Mathematica using le M3P4acce1erationnb we nd that the model that minimizes the V2 function is at 80 8923t37120i209t24r 16 Module 4 Areas under curves and Anti olerivatives Problem 1 Velocity from Acceleration Problem 2 Mass from Concentration in the outlet pipe Problem 3 Mass from Concentration in the settlement tank In this module we Will see how to calculate total accumulations of quantities from the rate of accumulation This includes calculating total amounts from densities and concentrations This type of calculation can also be used to nd the area With a curved boundary The formal mathematical techniques we develope here Will be studied in Module 5 Schedule Date Lecture Assignment Wed 29 11 Oct Problem 1 Speed from None acceleration Fri 31th Oct Finding anti derivatives Homework 1 Sec 61 1 2 3 4 13 16 37 40 Sec 83 35 36 39 40 Fri 7th Nov Areas under curves Homework 2 Exercise on page 9 Mon 10ch Nov Problem 2 Finding total mass owing into the settle ment tan Homework 3 Exercise on page 14 Wed 12th Nov Problem 3 Finding total mass in a settlement tank Homework 4 Exercise on page 19 Mon 24th Nov Present posters of Projects in MSC 1 Problem 1 11 Statement The most critical stage of a 100m race is as the athlete accelerates out of the blocks The athlete will be able to maintain the speed they manage to reach in this phase Below are the values of an athlete7s acceleration in the rst 03 seconds of a practice runi What is the athletes speed 03 seconds into the run Time sec Acceleration ms2 204 0 1 203 0 2 194 03 173 04 140 0 5 9 8 0 6 53 0 7 1 6 0 8 0 12 Estimation from the Data If we assume that the athlete maintains each of these accelerations for the full interval7 then we could calculate their speed as follows Acceleration Length of interval Increase in speed 1 204 X 0 1 2 04 203 0 1 203 X 0 1 2 03 194 0 1 194 X 01194 173 0 1 173 X 01173 140 0 1 140 X 0 1 140 9 8 0 1 9 8 X 0 1 0 98 53 0 1 53 X 0 1 0 53 16 0 1 1 6 X 0 1 0 16 3 Zai X 011 10181 i0 Each of these multiplications can be represented as an area of a rectangle and together can be draw as o o 15 o 10 g 5 O o r j o 2 04 o 6 08 Total area is 1081 13 Improved approximations using a tted model The athlete does not however maintain their acceleration over the whole 01 second interval their acceleration is continually decreasing If we had their acceleration every 005 seconds and repeated the calculation above we would obtain a more accurate approximation To nd the athletes acceleration at these intermediate times we rst t an interpolation function to the data If we nd the best tting model of the form y at4 bt3 c using the techniques of Module 3 we would nd that this data is approximated by with the function at 150514 7160287 2042 5 So the speed would be Za0i05 gtlt X 005 10299 This can be calculated 390 in Mathematica with the commands acc t150 51 t 416028t 320 42 Sumacc005i 005i0080051 As before we can represent this a the sum of the areas of rectangles each rectangle has a height of a0i05 gtlt and a width 005 These are pictured 0 2 0 4 0 6 0 8 It would be even more accurate if only assumed constant acceleration for 001 79 second time intervals7that is7 calculate Za0i01 gtlt gtlt 00L The Mathemat i0 ica command is SumEacc 001i 001i00 80 011 and the result is 989 The illustration of this is 20 02 04 06 08 Here the bars are so narrow that sum in approximately equal to the area under the curve 14 Best possible approximation The shorter the time interval over which we assume that the acceleration is constant the more accurate the approximation is So if we take nd the limit of these approximations as the interval width tended to zero we would have best possible approximation of the athletes speed7 and the exact value of the 030124 area under the curve That is lim 2 adt gtlt gtlt dt in Mathematica this is dtgt0 i0 calculated with Limit Sum acc dti dt i 0 0 8dt1 dtgtO and the answer we get is 978853 So the athletes speed to 3 signi cant gures at the end of this spurt of acceleration is 979 m s 15 Another method Acceleration is by de nition the instantaneous rate of change of velocity speedi So if at represents the acceleration of a body at time t and vt represents the speed of that body at time t then d1 3 at In this problem the body is the athlete we know their acceleration at 1495474 7159i51t3 2042 and wish to know the athletes speed at t 08 In other words if we differentiated the function we want to know about speed we would get 1495474 7 li li t3 2042 acceleration so we have to undo the differentiation that led to 1495474 7 li li t3 2042 Another way to say this is that to nd the speed we need to solve the differential equation d1 4 3 a 150517 7160287 2042 This equation is also called the model of the problemi Mathematica will solve this differential equation in response to the comman DSolve v t 15051 t 4 16028t 32042vt t The solution it gives is 2042t 7 400774 30175 C1 where C1 denotes the arbitrary constant We do not have to worry about the value of this constant if we think of our problem as nding the increase in speed from t 0 to t 08 then we are nding 1108 7 110 and this is 2042 X 0 8 7 4007 X 084 301 X 085 Cl 7 0 Cl and the value of this is 979 to 3 signi cant gures 2 Antiderivatives and Areas 21 Summary of antiderivat ives The table below summarizes the antiderivatives that should be memorized fltzgt fltzgtdz z forn7 il llzn1C l lnz C 6am 16am C 1 a cosaz 7 sinaz C a sinaz fl cosaz C a For any other use the Mathematica command DSolve F x f x F x x 22 Calculating areas under curves 221 Theoretical summary We have seen that we can nd the area under a curve inbetween two values of I we can sum the areas of rectangles under the curve7 and take the limit as the Width of the rectangles tends to zero In other words calculate lbad1l 011111110 2 fa 2dr dz ln Mathematica the command for this is Limit Sum f aidx i0 badx1 dxgtO 222 Example For the function 7213 81 calculate the following areas a The sum of the areas of rectangles under the curve of Width 02 from I 06 to z 16 Provide a graphical illustration 9 The sum of the areas of rectangles under the curve of Width 005 from I 06 to z 16 Provide a graphical illustration The sum of the areas of rectangles under the curve of Width 001 from I 06 to z 16 Provide a graphical illustration 0 FL The area under the curve from I 06 to z 16 e The sum of the areas of rectangles under the curve of width 02 from I 06 to z 36 Provide a graphical illustration f The sum of the areas of rectangles under the curve of width 001 from I 06 to z 36 Provide a graphical illustrationl g The area under the curve from I 06 to z 36 Note Areas below the zaxis are negative See le aresnb for the solution 223 Exercise for Homework 2 For the function sinz7 calculate the following a The sum of the areas of rectangles under the curve of width 02 from I 0 to z Provide a graphical illustrationl b The sum of the areas of rectangles under the curve of width 005 from I 0 to z Provide a graphical illustrationl c The sum of the areas of rectangles under the curve of width 001 from I 0 to z Provide a graphical illustrationl d The area under the curve from I 0 to z e The sum of the areas of rectangles under the curve of width 02 from g to Provide a graphical illustrationl f The sum of the areas of rectangles under the curve of width 001 from g to Provide a graphical illustrationl g The area under the curve from g to Copy your solutions into a word document and submit within WebCTi 3 Problem 2 31 Statement A factory needs to use a certain toxic chemical in its processing There is a steady ow7 of 500 litres of water per hour7 through the machinary that washes the chemical into a settlement tank The factory runs its process for eleven hours and then stops for one hour to allow the equipment to be washed clear of the chemical The table below shows the concentration of chemical in the outlet pipe7 at different times during a typical twelve hour shifti Our task is to calculate the total mass of chemical in the tank when it is full The settlement tanks measure 18m by 5m and are 2m deepi Time Hours Concentration mgL 0 0100 2 0100 4 0128 6 3133 8 100 9 481 10 1618 1015 2368 11 2415 1115 1946 12 0 00 32 Estimation from the Data If we assume that the concentration is constant over the time intervals Concentration Length of interval Amount of chemical i0 000 X 4 0 X 500 0 00 028 20 028 X 2 0 X 500 0 56 333 20 333 X 2 0 X 500 6 66 100 1 0 100 gtlt1i0 X 500 100 481 1 0 481x10 X 500 481 1618 05 1618 X 0 5 X 500 809 2368 05 2368 X 0 5 X 500 1184 2415 05 2415 X 0 5 X 500 12075 1946 05 1946 X 0 5 X 500 973 9 Zai gtlt dti X 500 2 380860 i0 Each of these multiplications can be represented as an area of a rectangle times 5007 and together can be draw as 2000 O O 1500 1000 500 O O a g n i a 6 8 10 12 Total area is 476172 So the total amount of chemical owing into the tank in one shift is 476172 X 500 ie 2380 860 33 Improved approximations using a tted model The concentration does not however remain constant for the whole time inter val it changes continuously within the intervals If we had a measurement of the concentration every 30 minutes we could assume that it remained constant over these intervals and obtain a more accurate estimate The closest we can get to 30 minute measurements is to t an interpolation function to the data In Module 3 we saw a very similar problem and using that process we see that the concentrationc at time t may be written as Ct 1830212 773261929102quot 23 So the 30 minute7 estimate would be 500 Z COiSi 05 2 540 896 To calculate i0 this in Mathematica we can type con t 18302 12t 2E 1 929112t 500Sumcon0 5i 0 5i 0 23 The graphical representaion of this sum is 2500 2000 1500 1000 2 4 6 8 10 12 It would be more accurate still to assume that the concentration was constant for only 6 minutes and calculate 500 0115011239 The Mathematica command that does this is 500Sumcon01i01i 0 120 11 and the result is 2 549 429 The illustration of this is 2500 2000 1500 1000 34 Best possible approximation Taking this process to its limit the best approximation we could make by this process is 120114 01111310 500 2 C2 dt dt 1 The Mathematica command that would calculate this is Limit 500Sum conidt dt i 0 12dt 1dtgtO Unfortunately Mathematica can have problems calculating limits for transenden tal functions so the best we can do is to set dt to a very small value eg 10 10 and calculate the sumi The answer is 2154944 X 105 This is in mg so is equal to 2154944kgi 35 Differential Model If the amount of a chemical is evenly distributed through out a volume of water then the concentration is given by the word formula Amount Concentratlon 7 Volume In this problem however the amount in the water changes continuously So if we focus on a very short time interval from t to t dt say and think about the water that leaves the outlet pipe in this time interval we can say Average Concentration of water that passes between time t and time t dt 7 Amount of Chemical to pass in dt hours 7 Volume of water to pass in dt hours 7 At dt 7 At 500 dt where we are using At to represent the total mass of chemical to have passed through the pipe in t hours The instantaneous concentration at time It will therefore be given by Atdt7At Lg 500 dt 7 500 dt t l39 6lt gt 190 This allows to say that the differential model of the problem is dA E 5004 500 lt1830212 7 t2e19291lt12 gtgt To solve this equation in Mathematica we type DSolveEA t500bestA1211 2E bestB12tAft t The solution that Mathematica gives is t 19290836612579962 C1 E 2 t 0010500525421595833 t 000041940429675416064 006583748987344133 If we rephase the main question to be nd the increase in the amount to ow into the tank in a 12 hour shift then we are asking for A12 7 A0 and so will not have to worry about the value of the arbitary constant C1 ln Mathematica we type A12 A 0 and we get 254944 X 105 3 6 Final calculation The solution of 254944 gtlt 105mg is the mass of chemical to ow into the tank in one shift To calculate the total amount to ow into a full tank we must nd out how many shifts it takes to ll a tank Volume of a tank is 18 X 5 X 2 180mg The total amount of water in a full tank is therefore 180 OOOL a volume of lm3 holds 1 OOOL of water The ow rate is 500L per hour so it takes W 360 hours to ll the tank There are 12 hours in a shift so it takes 30 shifts to ll a tank Hence the total mass of chemical to ow into the tank is 30 X 254944 X 106 764833 gtlt 107mg or 76 4833kgl 37 Exercise for Homework 3 1 Another factory runs on on 4 hour shifts and the concentration function is Ct 3l2t4 7 te 23954quot l The ow rate is 100 Lhourl a What is the total amount of chemical to ow into the tank in the rst 2 hours of the shift b What is the total amount of chemical to ow into the tank in a shift c A full tank holds 1600Ll What is the mass of the chemical in a full tank 2 At another factory the concentration function for a 6 hour shift is know to be Ct 36 7 736700103quot The ow rate is 200 Lhourl a State the differential model for At the total amount of chemical to ow out in the rst t hours of the shift b Find the increase in the amount of chemical to ow into the tank between the second and fth hour of the shift Copy your solutions into a word document and submit within Moodlel 4 Problem 3 41 Statement Ten days after a settlement tank is full7 concentration measuments are taken various depths in the tank The results are belowi Our task is to estimate the total amount of chemical in the tank Remember that the tank measures 18m by 5m and is 2m deepi Depth m Concentration mgL 0 0 63 5 0125 103 0 5 152 0175 178 1 0 281 1125 303 1 5 387 1175 892 210 3120 42 Estimation from the Data If we assume that the concentration is constant for the 25cm inbtween measure ments Concentration Length of interval Amount of chemical i i 6315 X 0125 X 5 X 18 X 1000 1143 gtlt106 103 0125 103 X 0125 X 5 X 18 X1000 2 32 gtlt106 152 0125 152 X 0125 X 5 X 18 X1000 3 42 gtlt106 178 0125 178 X 0125 X 5 X 18 X1000 401x106 281 0125 281gtlt 0125 X 5 X 18 X1000 6 32 gtlt106 303 0125 303 X 0125 X 5 X 18 X 1000 6182 X 106 387 0125 387 X 0125 X 5 X 18 X1000 871x106 892 0125 892 X 0125 X 5 X 18 X 1000 2011gtlt 106 9 Zai gtlt dhi gtlt5 X 18 X100053088800 i0 Each of these multiplications can be represented as an area of a rectangle times 18 X 5 X 10007 and together can be draw as v 3000 2500 2000 1500 1000 500 o 0 0 0 7 0 5 1 15 27 Total area is 58918751 So the total amount is 5891875 X 18 X 5 X 10007 iiei7 53 088 8001 43 Improved approximations using a tted model The concentration does not7 however7 remain constant for the whole 25cm7 it changes continuously within the depths If we had a measurement of the concentration every 10cm we could assume that it remained constant over these intervals7 and obtain a more accurate estimate The closest we can get to 10cm measurements is to t an interpolation function to the data Tutorialal groups from previous semesters have tted a model of the form a c h 7 7 W to this data and found that the optimal model was 1055 6001145 7 W 19 So the 10cm7 estimate would be 90 000 Z COili 011 64 328 2001 To calculate i0 this in Mathematica we can type con h105 51381 44801Sqrt h 90000Sumcon01i01i019 The graphical representaion of this sum is 3000 2500 2000 1500 1000 05 1 15 2 It would be more accurate still to assume that the concentration was constant for only 1cm7 and calculate 90 000 211 0015001239 The Mathematica com mand that does this is 90000Sum con001i 001i020011 and the result is 75127900 The illustration of this is 3000 2500 2000 1500 1000 500 44 Best possible approximation Taking this process to its limit7 the best approximation we could make by this process is hil 113090 000 Cz dh dh The Mathematica command that would calculate this is Limit 90000Sum conidh dh i 0 2dh1 dhgtO Unfortunately Mathematica can have problems calculating limits for transenden tal functions7 so the best we can do is to set dh to a very small value A further 17 problem arises because Mathematica refuses to calculate the sum for values of dh below 000001 So the best approximation we can calculate is 764722 X 107 This is in mg so is equal to 764722kg 45 Differential Model If the amount of a chemical is evenly distributed through out a volume of water then the concentration is given by the word formu a Amount Average Concentration Volume In this problem however the amount in the water changes continuously as we go deeper down in the tank So if we focus on the thin sheet of water illustrated below A h hdh dh between the depths of h and h dh we can say Average Concentration between depths of h and h dh 7 Amount of Chemical within the sheet Volume of water with in the sheet 7 AU dh 7 Ah 18x5x1000xdh where we are using Ah to represent the total mass of chemical that is contained in the top hm of the tank The instantaneous concentration at depth h will therefore be given by 7 Ahdh 7Ah 7 1 dA 50 115310 90 000 dh 90 000 E This allows to say that the differential model of the problem is dA 90000a f 90000 h 7 dh Cl ViaE To solve this equation in Mathematica we type DSolve A h 90000bestASqrt bestB Sqrt h AEh h 18

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#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.