### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# WesternCivilizationI HISM101

GPA 3.75

### View Full Document

## 23

## 0

## Popular in Course

## Popular in History

This 73 page Class Notes was uploaded by Erick Adams on Monday October 5, 2015. The Class Notes belongs to HISM101 at Capital Community College taught by Staff in Fall. Since its upload, it has received 23 views. For similar materials see /class/218878/hism101-capital-community-college in History at Capital Community College.

## Similar to HISM101 at

## Popular in History

## Reviews for WesternCivilizationI

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/05/15

FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES SPINOFF 11A Newton s law of Cooling and Heating Vaporization of Liquid Helium Solution We start with the following general lnction that relates the temperature T of a liquid to the time t that it has been cooling down or heating up T TS 0 43 1 We are given that the ambient temperature is 800 F This is equivalent to 26660 C as shown in the conversion below 80 F 32 F 266 C The ambient temperature TS 26660 C and the temperature at time 0 To 727150 C Substituting into equation 1 gives T 2666 2715 2666 2 Since the temperature is 7271 C when t 10 minutes we can use equation 2 to write the following 271 2666 2715 2666e k391 3 We now solve equation 3 for k to obtain 0000168 29766 00016783596 10k 1n 29816 10 Thus the equation that describes the relationship between the temperature of the helium and the time is T 2666 29816e 00001682 We are now able to nd the time when the temperature reaches 7269 5 C 29616 2695 2666 29816600001682 2 0000168t 1n 29816 3 t 4006 minutes NASA AMATYC N SF 11 34 FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES LTA 15 Population Size Matters Formulating a Mathematical Model for Scrub Jay Population at the Kennedy Space Center Level of Presentation College Algebra Precalculus Mathematical Prerequisites Linear Equations and Graphing Technology Requirement Scienti c calculator with twovariable statistics or a graphing calculator such as TI83TM or computer software capable of regression analysis such as ExcelTM Time Required 2 Class Hours Recommended Working Format Short lecture small groups and student use of graphing calculator or appropriate software Comments The instructor may need to elaborate on some concepts that might be unfamiliar to students such as the concept of a regression line and the coef cient of determination Your Turn 1 Sketch a scatterplot 0f the data from Table 1 using an X y coordinate system with year the independent variable and population size the dependent variable Does the pattern of the plotted points appear to t any of above models Ifwe plot the data in the above table we obtain the scatterplot shown below NASA AMATYC NSF 1531 KSC Scrub Jay Population Estimated Population N O O O O O O 1975 1980 1985 1990 1995 2000 Year Notice how the points form a downward trend indicating a decreasing population size over time The pattem of the plotted points suggest a linear exponential or power lnction model Also a decreasing part of the quadratic model would seem to t the data quite well Your Turn 2 Find the following regression equations for the coded data in Table 2 a Quadratic regression model b Exponential regression model c Power regression model Following are the screen displays from the T183TM graphing calculator EuadReg 3911538346 E325393125 The resulting regression equations for the diiTerent models are summarized below where N represents the population size and t represents time t 1 corresponds to 1980 Linear regression equation N 35377 11963t Quadratic regression equation N 25125t2 171538t 37251 Exponential regression equation N 383364094754t NASA AMATYC NSF 15 32 Power regression equation N 406618t390393148397 Your Turn 3 Sketch graphs of the quadratic regression exponential regression and power regression equations superimposed on the scatterplot of the scrub jay data Visually inspect the graphs and describe how well each regression equation ts the data To see how closely the different models t the scatterplot we use the TI 83TM calculator to display graphs of the different regression equations superimposed on the scatterplot From the following graphs we see that all of the regression equations appear to t the data reasonably well but the power regression model appears to provide the weakest t This would suggest that we eliminate the power regression equation as a candidate for the best mathematical model mg 391 quotH Linear Regression Quadratic Regression aka Exponential Regression Power Regression Your Turn 4 Using your calculator or computer software retrieve the values of the coef cient of determination for each of the models considered in Step 2 Use the coef cients of detennination to decide which model best ts the data NASA AMATYC NSF 1533 The R2 values are shown on theTI 83TM screens below L1nReg DuadEeg Habx Hax3bxc a353F21254 a251259 54 119EEB41 b11533346 P394SB4B598 c3F25598125 39921323691 R39631355925 I ExPReg PHPREQ Hab x Hax b a38336416 a BEE18345 b945413351 b39314339F44E P 9453HEHB4 r33331335953 r924EESBEE r896163264 The regression equations and their corresponding coef cients of determination are listed below Linear N 35377 11963t r2 09450 Quadratic N 25125t2 l7l538t 37251 R2 09601 Exponential N 383364094754 r2 09457 Power N 406618t390393148397 r2 08031 Based on a comparison of the coef cients of determination we should eliminate the power model as a candidate for the best model since the other values are closer to 1 In comparing the three remaining models linear quadratic exponential it might be tempting to simply select one with the largest 12 value but the values of 09601 and 09457 are not that far apart and we should not exaggerate the importance of their dilTerence The differences among the linear quadratic and exponential models are relatively small we should not simply select the model with the largest value of r2 NASA AMATYC NSF 1534 Your Turn 5 Predict the population size for the year 2100 using each of the regression equations you determined in Step 2 Determine which equations best t the data on the basis of the reasonableness of the answers Quadratic N 251251212 171538121 37251 19755 Exponential N 383364094754121 6 Power N 40661812139 393148397 898 The quadratic model yields a population size of 19755 and this also lacls realism Based on the data in our original table a population size of only 6 given by the exponential model does seem feasible whereas the population size of 898 given by the power model seems too large Exercises 1 a 2222 b 2142 c 2119 d 1911 Both the quadratic model and the exponential model give reasonable predicted values when compared to the population values for 1989 and 1991 given in Table 1 Either of these models could be considered to provide good t However the exponential model provides a more accurate t when based on other criteria a 1025 b 1231 c 1236 d 1559 The quadratic model exponential model and power model all give predicted population values that are comparable with the data in Table 1 However the exponential model was better based on other criteria so the result of 1236 is best a y 384 250286x with r2 0998 b y 0160714x2 261536x 369 with R2 0998 c y 557504123921X with i2 0972 d y 604502x039510757 with r2 0989 NASA AMATYC NSF 1535 The linear and quadratic models both have r2 values close to 1 so both models are very good When making a prediction for year 7 the linear model results in 214 and the quadratic model results in 212 so both estimates are close 4 a y 846733 4364x with i2 0728 b y 169429x2 7496x 7346 with R2 0962 c y 998671250224X with i2 0999 d y 1574350851242 withi2 0970 The exponential model appears best because it has the highest r2 and because of the same reasoning used in the ScrubJay example discussed in Step 2 of the LTA The estimated value of y when x 7 is 6134 5 In the following equations x 1 represents 1980 a Use y 933185 470315x with r2 0621 to get an estimate of39587 b Use y 899986x2 299318x 876186 with R2 0626 to get 17413 c Use y 12030140886814X with r2 0726 to get 122769 d Use y 1538310x 5859quot 4 with r2 0534 to get 204110 The exponential model appears best because it has the highest r2 and because of the same reasoning used in the ScrubJay example discussed in Step 2 of the LTA 6 Answers will vary depending on the data obtained from the Intemet NASA AMATYC NSF 1536 FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES LTA 14 Space Vehicle Hold Down and Release Mechanism Design General Remarks Typical story problems in mathematics require students to formulate an equation to model a situation This LTA requires students to develop a qualitative Mm model the physical situation and to interpret graphs representing a physical event Section I Restraining Force vs Displacement Graphs Hand out the rst few pages of LTA 14 Space Vehicle Hold Down and Release Mechanism Design through Section I This will provide the students with a general description of what a so release hold down system should do Students should work in groups of 2 4 for approximately 20 minutes to determine from the information handed out what the graph of Force vs Displacement should look like Then have the groups critique each other s graph to see if the graphs satisfy the description of a so release mechanism Results should be a decreasing function in the first quadrant which ends on or near the x axis where x is displacement As an example see the restraining force graph in Figure 3 Section II It is suf cient to accept oral discussion for the explanations of Sectiont I a written explanation is not required Ifa written report is desired students tend to need a good deal of direction as to the structure of the report Since Section I is to be used to elicit student ideas and stimulate discussion a written report is not recommended Section II Characteristics of a Restraining Force vs Displacemerl Graphs Hand out the rest of the LTA The graph of the CRM restraining force in Figure 3 has several important features a The force starts out high and decreases to either a zero or to a small positive value When the launch vehicle is released any net upward force imparts a kick to the launch vehicle Such an impulse or kick must be kept small b A near zero slope at beginning and end makes the change in the restraining force in those regions very small and leads to a smoother li off However not all so release mechanisms have small initial and ending slopes The graph of the CRM restraining force shown in Figure 3 is based on the release mechanism for the Satum V used during the Apollo program NASA AMATYC N SF 14 38 Solutions 1 The graph should show that as the displacement increases the restraining force decreases The graph should be smooth and it should decrease to zero 2 Mechanism A a Characteristics exhibited Energy absorbed by the CRM is within the acceptable range 125000 250000 in lb No sharp comers on graph Characteristics lacking Flat or nearly at graph at initial and nal displacements Decreasing graph throughout displacement interval Low restraining force at release b Energy Absorbed is approximately 175000 in lb The result is based on one triangle with height 50000 lbs and base 7 inches General comments Mechanism A is a bad design since the Force vs Displacement curve increases throughout the displacement interval We want a high restraining force initially and a low one at release 3 Mechanism B a Characteristics exhibited Flat or nearly at graph at initial and nal displacements Decreasing graph throughout displacement interval Low restraining force at release No sharp comers on graph Characteristics lacking Energy absorbed by the CRM is within the acceptable range 125000 250000 in lb b Energy Absorbed is approximately 25000 inlb The result is based on 3 trapezoids and one triangle General comments Except for the small amount of energy that Mechanism B absorbs it is a good design The Force vs Displacement curve is decreasing smooth and relatively at at both ends The restraining force is initially high and near zero at release This curve has the best overall shape but the mechanism does not absorb enough energy NASA AMATYC N SF 14 39 4 Mechanism C a Characteristics exhibited Low restraining force at release No sharp corners on graph Energy absorbed by the CRM is within the acceptable range 125000 250000 in lb Characteristics lacking Flat or nearly at graph at initial and nal displacements Decreasing graph throughout displacement interval b Energy Absorbed is approximately 227000 in lb The result is obtained by doubling the area for the le half of the region The area for the le half is based on 7 trapezoids and one triangle General comments Mechanism C is a bad design since the Force vs Displacement curve is increasing during a part of the displacement The curve is smooth but is not at at both ends One good feature is that the force at release is near zero The initial restraining force is zero leading to a big jerk at the beginning The CRM represented by this curve would be good if it were preloaded to about 35 in of displacement 5 Mechanism D a Characteristics exhibited Decreasing graph throughout displacement interval Low restraining force at release No sharp corners on graph Energy absorbed by the CRM is within the acceptable range 125000 250000 inlb Characteristics lacking Flat or nearly at graph at initial and nal displacements b Energy Absorbed by the CRM is approximately 248000 inlb The result is based on 7 trapezoids General comments Mechanism D is a good design The Force vs Displacement curve is decreasing smooth and at at the beginning The curve is not at at release but the force at release is near zero It would be better if the curve attened a bit more at the end 6 Mechanism E a Characteristics exhibited Decreasing graph throughout displacement interval Low restraining force at release No sharp corners on graph Energy absorbed by the CRM is within the acceptable range 125000 250000 in lb Characteristics lacking Flat or nearly at graph at initial and nal displacements NASA AMATYC N SF 14 40 b Energy Absorbed by the CRM is approximately 175000 inlb The result is based on one triangle with height 50000 lbs and base 7 inches General comments Ifthe curve for Mechanism E were attened at the ends it would be a good force vs displacement pro le for a CRM The following table summarizes the answers for Exercises 2 6 Summary table for Exercises 2 6 Ex 2 Ex 3 Ex 4 Ex 5 Ex 6 Characteristics Mech A Mech B Mech C Mech D Mech E Flat or nearly at at both ends L E L L L Decreasing on entire interval L E L E E Low force at release L E E E E No shaip comers E E E E E Energy absorption within range E L E E E Energy absorbed inlbs 175000 25000 227000 248000 175000 E Exhibits characteristic L Lacks characteristic 7 The only candidates for good CRM Mechanisms are B D and E because they are the only ones with a decreasing restraining force throughout the entire displacement However each of these could be improved Mechanism B would be better if it absorbed more energy to bring it within the acceptable range 125000250000 inlb Mechanism E would be better if its graph were attened at the beginning and end of the displacement Mechansim D would be better if its graph were attened at the end of the displacement 8 In order from least to most energy absorbed we have Mechanisms B A or E C D 9 Mechanism B would be the best one if it absorbed enough more energy to bring it within the Acceptable range 125000 250000 inlb while its graph retained the same general shape On the other hand Mechanisms D and E would be very good if their graphs were attened at both ends Section III Controlled Release Mechanism for the Evolved Expendable launch Vehicle 10 a Graph 1 Tensile Bar Mechanism b The description of the Tensile Bar Mechanism indicates that preloading is required Because Graph 1 increases for the rst part of its displacement it represents a CRM that must be pre loaded to the length at which the restraining force is largest That is the graph represents a retraining force that must be preloaded to the top of the curve where it is relatively at A second reason for making this selection is that the description of the Tensile Bar Mechanism NASA AMATYC N SF 14 41 states that the material s properties may not allow for a at or nearly at curve just before release Certainly Graph 1 is not at just before release The force versus displacement graph a er preloading is obtained by shi ing the original graph to the le so that its maximum value intersects the vertical axis Only the rst quadrant part of the shi ed graph is retained O V 11 a Graph 2 Friction Rod and Washers Mechanism b The description of the Friction Rod and Washers Mechanism as well as its sketch indicates that as the central rod clears each level of washers the restraining force decreases in steps This is because fewer washers remain to exert a force on the central rod Another reason for selecting the Friction Rod and Washers Mechanism is that it exerts a maximum force at zero displacement Graph 2 shows a maximum force at zero displacement No preloading is required 0 V 12 a Graph 3 Mandrel Mechanism b The description of the Mandrel Mechanism states that preloading is required Since Graph 3 increases during the rst part of its displacement it represents a mechanism that requires pre loading to the top of the curve where it is relatively at Also the Mandrel Mechanism indicates that its restraining force can be made relatively at just before release and we see that Graph 3 is nearly at just before release The force versus displacement graph a er preloading is obtained by shi ing the original graph to the left so that its maximum value intersects the vertical axis Only the rst quadrant part of the shi ed graph is retained O V Section IV Data Analysis and Curve Fitting MathematicaTM ExcelTM or a TI83 85 86 TM will fit a polynomial to data The TL 838586 TM only go up to degree 4 For MathematicaTM or ExcelTM you can set the order you wish 13 T39I838586TM gives Fx 87523 x4 14982 x3 73879 x2 12162 x 49964 ExcelTM for order 6 gives Fx 2542 x6 51104 x5 29319 x4 21424 x3 54927 x2 27848 x 49989 Observe that the coef cients in these polynomials have been rounded to ve signi cant digits When calculating the answers for the following exercises digits of the coef cients produced by the calculators or spreadsheets were used Ifyou use the polynomials om this exercise whose coef cients have been rounded to ve signifcant digits your answers will be somewhat dilTerent 14 531 lbs is from the TT83 85 86M ExcelTM order 6 gives 443 lbs 15 6412 inches is from the TI 83 85 86M ExcelTM order 6 gives 6221 inches NASA AMATYC NSF 1442 16 50 is 7 inches 47 elongation is 658 inches and the corresponding restraining force is 863 lbs from the TT83 85 86TM and 619 lbs from ExcelTM order 6 17 53 elongation is 742 inches and the corresponding restraining force is 1003 lbs on the TI 83 85 86TM and 723 lbs for the ExcelTM degree 6 Each of these don t make much sense The result from TT83 8 5 86TM implies that the restraining force is pushing the rocket olT the pad instead of restraining it The result from ExcelTM implies that the restraining force for a displacement of 742 inches is larger than the restraining force for a displacement of 658 inches That is according to the ExcelTM polynomial the restraining force would increase as the displacement increased beyond 7 inches This situation points out the problem with polynomial extrapolations The graphs of the polynomials of order 4 and order 6 are shown below in Figures 1 and 2 respectively 60000 50000 40000 30000 Force lb 20000 10000 10000 Displacement in Figure 1 graph of polynomial of order 4 NASA AMATYC N SF 14 43 Force lb 60000 50000 40000 30000 20000 10000 2 3 4 5 6 7 Displacement in Figure 2 graph of polynomial of order 6 NASA AMATYC N SF 14 44 FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES SPINOFF 14C Analyzing Graphs SpinolT 14C mostly concems creating and interpreting graphs The SpinolT might best be used as a homework assignment It addresses not only the mathematics of linear cubic exponential and trigonometric functions but also provides a reading assignment that puts the mathematics into context Along with having to read about a real world design problem there is a writing assignment The latter involves interpreting graphs based on design criteria for a working controlled release mechanism Solutions 1 Concept A 14285 lbs Concept B 31775 lbs Concept C 9413 lbs 2 Graphing on a TI85TM and using ISECT x 35 in and the force is 25000 lbs 3 FrD0 50000 lbs FrD7 456 lbs The graph of FrD is shown below Force Release Mechanism D 60000 50000 40000 30000 20000 10000 Force lbs x in 4 Writing Assignment Each CRM has an initial restraining force of 50000 lbs and decreases to less than 100 lbs at x 7 in Thus all Concepts meet the basic requirements All are smooth decreasing functions Concepts A and D are not at at the beginning but D does become at at the end Not explicitly mentioned but of interest is that the area under the restraining curve represents the amount of work done on the CRM by the rocket That is the area under the restraining curve represents the amount of the rocket s energy that is absorbed by the CRM NASA AMATYC N SF 14 51 5 a Concept A Cost 375000 17258 125 2100000 b Concept B Cost 1025000 9758 125 2000000 Choose B Cost functions are CAX 375000 13800x and CBX 1025000 7800x Set these two lnctions equal to each other and solve for x to obtain 10833 However we cannot have a fractional number of launches Ifthere are 108 or fewer launches then Concept A is less expensive if the number of launches is 109 or more then Concept B is less expensive Solve graphically by nding the point of intersection of the graphs of the two cost functions NASA AMATYC N SF 14 52 FACULTYNO TES Ta LTAs andSpmstm dzslgmdsa um eachpmfessaxcmxm zmzm nasmaaaawayuaaa as manna as aaaaaatsxass pmjec39s wa mmm amass guldance m dnmg masaamas wuka class m LTAs andSpmn smz amsaams a ma mnpcaapm ve vmxkandiypw y bem n mm u use a mm Izammg techmlngy Smce ms abjechve amas LTAs andSpmst n yml setfaxyunrsmdzmsthz masadsaaabs passan Th2 pllrpuse amas Fundinmzs as m pmwdz mfannanan um assists ymna mkz ran advantage amas LTAs andSpmst Thls mcludzssuggesmns raa msmwonn as we as answers raa ms sxsasasss um mm am an FACULTY NOTES LTA 4 Just Say NO to Cracks in the Space Shuttle Background on the Kennedy Space Center The 140000acre John F Kennedy Space Center KSC is the major launch site for the National Aeronautics and Space Administration NASA Prior to the Space Shuttle era manned launches from KSC included the Apollo missions which landed men on the Moon three ights to Skylab the first United States Space Station and the Apollo Soyuz Test Project Mission NASA also launched a wide variety of unmanned spacecraft using expendable rockets li ing off from Cape Canaveral in Florida and from Vandenberg Air Force Base in California These have included weather and communications satellites orbiting scienti c satellites lunar explorers and landers and interplanetary probes to Mercury Venus Mars Jupiter Saturn Uranus and Neptune Except for operational areas the KSC reservation is designated as a national wildlife re lge Background on the Space Shuttle A Space Shuttle consists of an Orbiter an external tank and two solid rocket boosters SRBs The Shuttle is 1842 feet in height with a gross li o weight of about 4500000 pounds and a total li olT thrust of 7700000 pounds This thrust decreases during the two minutes the solids burn to about 3000000 pounds during the last 10 seconds A er the SRBs are discarded the Orbiter engines burn for 6 12 more minutes and take the vehicle to nearorbital velocity The deltawinged Orbiter resembles an airplane and is about the size of a DC9 jetliner The four Orbiters in service are Columbia Discovery Atlantis and Endeavour Only the Orbiters have names A Space Shuttle Orbiter is 12217 feet in length with a wingspan of 7806 feet The three main engines each produce 37500 pounds of thrust at sea level and 470000 pounds in the vacuum of space The cargo bay is 60 feet in length and 15 feet in diameter The Space Shuttle could launch a bus into orbit A er the boosters separate from the Shuttle they descend by parachute to a watery landing They are retrieved by two speciallydesigned ships and returned to Kennedy Space Center for re lrbishment and eventual reuse A er the Orbiter s three main engines shut down the external tank is jettisoned breaks up in the atmosphere and falls into the Indian Ocean It is the only major Shuttle component that is not reused NASA AMATYC NSF 4 21 The Orbiter uses its smaller maneuvering system engines supplied by onboard propellants to enter orbit After performing its mission the Orbiter uses these engines again to slow the vehicle and return through the atmosphere to land as an unpowered glider More than 200 signi cant modi cations have been made to the Orbiter eet since 1986 One of these was a crew escape system It allows the crew members to bail out if the Orbiter is in level ight within the atmosphere but cannot reach a safe runway The Space Shuttle will be the primary launch vehicle for the United States for years to come Background on Fracture Mechanics The eld of Fracture Mechanics has only existed since approximately the time of World War II Naval ships were manufactured in the US and then shipped for use in the North Sea At least one of these ships actually snapped in half as a result of cracks and the reaction metal has to extreme cold Once the cracks reached the critical stage they literally grew at the speed of sound and snapped the ship in half The equations and scenarios given in this LTA are a simpli cation of what Fracture Mechanical Engineers really use and experience For example deciding what the allowable K Kg Stress Intensity Factor should be is an extensive process The critical K KC must rst be determined and it is not as simple as looking it up on a chart There are many factors which contribute to KC Initially you consider the type of material used for example high strength steel and then other factors such as temperature are considered Once a KC is determined then the allowable Stress Intensity Factor Ka is calculated This is not simply a matter of some sort of ratio or rule of thumb but the material s end use must be considered If the material is used for the Space Station then you want a big buffer between critical and allowable On the other hand if the material is for a less demanding application or if it can be readily repaired the allowable can push the envelope more Another item which has been simpli ed is that of measuring the length of a crack In the LTA it is expected that students will simply measure the crack from end to end Actually the crack should be measured normal to the direction of stress which caused the crack as indicated below Stress 079 a 810 6121 NASA AMATYC NSF 4 22 Curricular Information General It is expected that this will be a culminating activity after students have solved simple equations with square roots Students should also have had experience with dimensional analysis basic graphing and solving literal equations Depending upon a college s curriculum students at the Intermediate Algebra level should be capable of completing this activity However because of an increase in maturity level it might be better suited for students at the College Algebra level It is best done in groups and therefore suggested that the students have some previous experience in group work The activity is designed to increase the student s ability to work with multiple equations a variety of unfamiliar units and with technology Assessment Although assessment will vary with the instructor assessment techniques could include the following 9 O O O O Require an oral presentation on part of the project or on the whole project Require a written report for the project a part of the project or one of the exercises Give students a problem similar to the project and have them write a recommendation to the project director about whether or not corrective action is needed On exams be sure to include problems where students encounter smalllarge numbers and use formulas that are not already solved for the appropriate variable Expect students to write or give an oral report on the relationships between the variables in this unit based on what they have seen numerically and graphically in the problems Group Work Group work can be incorporated in many ways here are a few suggestions 00 O 9 Allow students to work on the exercises together Have students work together on the project and present a group report Have diiTerent groups assigned to diiTerent problems in the project and then have each group do a presentation about the portion they solved Make sure students have already worked in groups before this activity Technological Information This activity was written to be calculatorindependent However it is expected that all students will have access to a scienti c calculator at a minimum Ideally students will be able to use computer algebra so ware such as DeriveTM or a graphics calculator such as the TI83TM this is especially advisable for parts 3 through 5 in The Project section With the domain of 0 S a S 05 once students determine the appropriate range the graphs characteristic features are obvious Students are then easily able to determine the needed values of K Since we are dealing with a combination of very large and very small numbers students need to use the increased accuracy of a calculator that can store or carry answers When using two equations students need to use the calculator39s results from the rst equation not rounded for the second equation NASA AMATYC NSF 4 23 Timeline for Classroom Use This activity is intended for approximately two class periods with students doing some work outside of class The following is one suggestion as to how to handle the workload This is in no way the only choice since everyone has diiTerent schedules and demands on classroom time Class period 1 Allow students to become familiar with the problem physics background and formulas Work some of the exercises in class and assign the rest as homework Class period 2 7 One week later to allow for this to all sink in have students work in groups to do the project If time allows have them do oral presentations or assign a written report to be handed in at a later date Solutions Your Job 1 Pressure If you increase the pressure of the gas the crack will be more likely to rupture the pipe 2 Thickness of pipe wall If you decrease the thickness of the pipe wall the crack will be more likely to rupture the pipe 3 Radius of pipe If you increase the radius of the pipe but keep the wall thickness constant the crack will be more likely to rupture the pipe 4 Stress on pipe The vulnerability of a pipe to being ruptured by a crack referred to as stress on the pipe increases if the pressure is increased increases if the wall thickness is decreased and increases if the radius of the pipe is increased 5 Length of crack If you increase the length of the crack the crack will be more likely to rupture the pipe 6 Stress Intensity Factor The likelihood that a crack will rupture a pipe referred to as the stress intensity factor increases if the length of the crack is increased and increases if the stress on the pipe is increased Exercises Note The length of the crack in the tank measures approximately 1 inch That is 2a m 1 inch 1 Length ofcrack 28mmso a 14mm 7 112 MPa K ixa 112Mpam 743 MPamm 2 Crack 1 a 01 in P 50000 psi K 28025 psi in Crack 2 a 005 in P 100000 psi K 39633 psi in NASA AMATYC NSF 4 24 12 600 144 3 P600psr r 12 11 144 m Setup atab1e usmg Uf The tab1e below uses Tble 0 and ATbl 0 1 where xt and Y1 c r vanes mversely vmht Astrhereases a de 4 K 14000psh 710125 1r 8773 p51 erhx The 1dea15 that any crack we eahhot to the halfrcracklengths from 0 to 0 125 as shown m the graph ovas 11 K 8773 0 125 Backtn the Emblem Criterinn 1 e se our deersror on the gweh allowable value ofkacvame Note that for the average radms to be 48 hr the 8 dAmEnsxon gweh rh Frgure 1 must be consxdered an average dwmeter Kg 100000 p51 em 10 375 hr p 110 psr 48 1h 11 05m W14080psx 0 375 K 140804 7 05 17647ps1em Smee17647 lt 100000 the crackxs safe andreqmres no acuon NASA 7 AMATYC 7 Ni 425 Criterion 2 As an alternative approach we compare the length of the crack with the thickness of the tank Length of crack in tank 2a z 1 in gt 2 in gin Therefore by this criterion the crack is unsafe and it would require corrective action The Lam Equation 5 6 7 r0 2375 in rt 207 in average radius 22225 in wall thickness 0305 in radius 7287 gt 6 wall th1ckness P 6000 psi Thin Wall 7 W 43721 psi 0305 1n 39 Z 39 2 Lame a 6000 psi w 43927 psi 2375 in2 207 in2 average radius 355 in wall thickness l in amp 355 gt 6 wall th1ckness 120000 psi 2 P 3380 psi 1n r30mm 7 100MPa P10MPa 100MPa Mst3mm 250MPa thl2mm t NASA AMATYC NSF 4 26 8 a0125 in P600 psi t 15 in Va 24in 2325 in 24 in 225 in r f a 600 psi 2325 in 15 in K 9300psix7270125 in 5828 psi in It is safe to operate To nd the highest allowable stress use Km21X cm W 9300 psi 75000psi in 119683 si n0125 in p Pm X r am X I 119 683 3915 39 Pmax 1 51 m 7721 psi 2325 in The highest allowable pressure for the pipe is approximately 7721 psi The Project 1 r0 61 feet 732 inches P 900 psi 7 130000 psi rr r702 and tr0 n 3n r0 t 2r7 t 2 900 psi27321n t 130000psi 2 3130000t 65880 450t t 130450t 65880 I 0505 in NOTE The use of Va as r results in a thickness of 0507 in rather than 0505 in 2 a 532 in Va 61 feet 732 in ri 727in 732 in2 727 in2 732 in2 727 in2 K 131312 psi in 92000 psi in 92000 psi in lt Maximum Allowable K of 100000 psi in a900psi 30131312 psi Thus no action is needed NASA AMATYC NSF 4 27 3 Let x length ofthe crack then a Graph K y 1313121 727 MIHDUM Hm1n39BS Hmax1 H511 V551SBBBB Hr951 4 For a 01 inch crack K 52043 psi in For a 025 inch crack K 82288 psi in For a 03 inch crack K 90142 psi in 5 The bold entries in the table show those K values that indicate needed corrective action for the associated cracks K y 1313121 727 Crack K Length psi in NASA AMATYC NSF 4 28 FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES SPINOFF 12A Finding the Rate of Return for Energy Savings Investment Comments The contractor invests 12000000 in work and is paid back 23000000 The 11000000 can be considered retum earned on the contractor s investment The money comes from savings created by reduced energy consumption and deferred maintenance Deferred maintenance means that by installing new equipment replacement of the old equipment is delayed for the life of the new equipment The nance option on some calculators permits nding interest rates or present values Amortization formulas for ordinary annuities include V 1wa VVVV where n is the number of compounding periods per year p is the payment every lnth of a year for t years r is the annual rate of interest andP is the principal present value I onP where I is the interest paid over the term oft years Solutions 3000000107A2 8000000107Al 1000000 12994270 The rate of return on the investment is 9361 66666667 8333333 thion 1 Solution using a calculator Value of rst 12 investments on September 30 1998 3133231382 Value of second 12 investments on September 30 1999 8355283685 Value of third 12 investments on September 30 2000 104441046 Total value of investments on September 30 2000 1376810785 thion 2 Solution using a spreadsheet The answer using a spreadsheet is on the next page Table Spreadsheet Solution NASA AMATYC N SF 12 39 Monthly Investment Value at end of Value of all pn39or Accumulated Value investments month investments at end of Investments at end 1 3 6 of month of month 0 250000 111111 85 778 149 724825 1492 435 797 102 729 014 OOONONUIAUJN 715 111145 19591 111145 732244 111145 762 111145 756 111145 6711111145 111145 49 111145 447369 111145 46065 58 897 107 25 Answer 1376810785 NASA AMATYC N SF 12 40 3 The rate of return on the investment is 9725 Obtained using the nance option on a TI 83TM graphing calculator with present value of 1274824801 on September 30 99 NASA AMATYC N SF 12 41 FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES SPINOFF 14B Force Versus Displacement Solutions 1 The graph should show that as the displacement increases the restraining force decreases The graph should be smooth and it should decrease to zero Students should have no trouble graphing the functions in Exercises 2 5 They may need to be reminded to label the axes In Exercises 25 Simpson s Rule and its associated error formula are needed With n 6 Simpson s Formula is Ax 7 y0 4y1 2y2 432y4 4y5 y6whereAxg Thefunction values can be found using a calculator A calculator with a TABLE feature yields these values easily N 3 U V O V E 3 O V Q V The graph of F x 50000 7143x is a straight line that passes through the points P0 50000 and Q7 0 Simpson s Formula yields l749965 inlb 5 Error Bound S KEZTZ if Ff xl S K for all x in the interval 0 7 Since the n fourth Derivative of E equals 0 it follows that the error in Simpson s approximation is zero J 50000 7143xdx 174996 5 inlb Hence Simpson s Rule gives the exact value of the definite integral The graph of E x 50000 1458x3 is a concave down curve that passes through the points P0 50000 and Q7 0 Simpson s Formula yields 26248355 in lb 5 Error Bound S KEZTZ if Ff xl S K for all x in the interval 0 7 Since the n fourth Derivative of E equals 0 it follows that the error in Simpson s approximation is zero J 50000 1458x3 dx 26248355 inlb The answers for b and d are equal Thus Simpson s Rule gives the exact value of the definite integral NASA AMATYC N SF 14 48 4 a The graph of E x 250001 cos is concave down during the first part of the interval 0 7 concave up over the second part of the interval and passes through the points P0 50000 and Q7 0 Simpson s Formula yields 175000 inlb U V Ff x S K for all x in the interval 0 7 Since 0 V 5 Error Bound IE l S KEZT if n 4 Ff X 25000cos D it follows that all X in the interval 0 7 IE lt Kb a5 10142557 05 180124 18064 4 EM g 25000 1014255 for 73074 Q V J250001 cosdx 175000 inlb The answers forb and d are equal Thus Simpson s Rule gives the exact value of the integral V 8 The graph of Ex 500006 03x 87469x is concave up on the interval 0 7 and passes through the points P0 50000 and Q7 0 b Simpson s Formula yields 12483937 inlbs c Since F4X 50000 0346 03 it follows that Kb a5 4057 05 180114 18064 d 500008 0 3 87469x1 x 12482736 inlb The estimate by Simpson s Rule differs E x s 50000 03quot 405 Thus E1S 2918 from the exact value by 1201 in lb Students especially those without a physics background may need help in understanding that the area under a curve is measured in inchpounds and thus represents work In this case the area represents the work done by the rocket on a CRM which is equal to the energy absorbed by the CRM during liftoff 6 Ifthe mechanism fails the restraining force drops immediately to zero The value of each function at X 58 represents the kick imparted to the launch vehicle These tum out to be about 8571 lb 21553 lb 3539 lb and 3703 lb for the functions ofExercises 2 3 4 and 5 respectively NASA AMATYC N SF 14 49 FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES LTA 12 lVIission Control We Have Energy Savings Background Information Math Prerequisites Arithmetic Skills Averages Reading and reasoning Gmphing Leaming Technologies Suggested Scienti c Calculator Graphing Calculator Approximate Class Time required for LTA 12 2 hours Comments Students will gain experience with mathematical reasoning and problem solving as they meet the following objectives Collect numerical data Estimate data from personal life Calculate sums of data Measure rectangles and calculate areas Calculate ratios Calculate total cost given the unit cost and the number of units Find to what scal year a given month belongs Find the average of a set of numbers Draw conclusions from data Graph data Draw conclusions from graphs NASA AMATYC N SF 12 29 Section I Utilities and Energy Use Comments Typical home and of ce treatments to protect against bad weather include Note installing wall ceiling and pipe insulation replacing windows or adding storm windows insulating pipes wrapping water heaters providing tint or shade for windows awnings buying more energy ef cient air conditioners water heaters refrigerators and clothes washers and dryers installing motion detectors as light switches changing habits for using water using low or reduced water ow plumbing xtures In completing the tables students will need to multiply watts by hours per day to get watt hours per day While the units are watthours per day in this Section larger amounts of electricity used in later parts of the LTA will require kilowatthours kWh A brief discussion of the meaning of this unit of measurement may help avoid confusion later It may be helpful to review the formulas for the area of a rectangle triangle and circle Part A 1 7 812 Solutions Re ecting on Energy Use Answers will vary It is anticipated that students will suggest a number of types of buildings and related energy savings items Answers will vary NASA AMATYC N SF 12 30 Part B Light Fixtures and Lamps 13 The answeIs in columns 1 and 2 of the following table should be the same for all students The estimated daily usage in column 3 will diiTer from student to student Table 1 Light Fixtures and Lamp Information Light Fixture or Lamp Watts HouIs Day Watt HouIs Per Day Kitchen Fluorescent Light 160 4 640 4 bulbs 40 W Kitchen Ceiling Fixture 240 5 1200 4 bulbs 60 W Living Room LR Ceiling Fixture 240 3 720 4 bulbs 60 W Living Room 2 Floor Lamps 300 5 1500 1 bulb 150 W First oor Bathroom Recessed Light 300 15 450 4 bulbs 75 W First oor Hallway Table Lamp 180 25 450 3 bulbs 67 60 W Family Room 2 Reading Lamps 150 6 900 1 bulb 75 W Bedroom 1 Reading Lamp 75 3 225 1 bulb 75 W Bedroom 1 Ceiling Fixture 240 25 600 4 bulbs 67 60 W Bedroom 2 2 Reading Lamps 150 1 150 1 bulb 75 W Bedroom 2 Ceiling Fixture 240 1 240 4 bulbs 60 W Second oor Hall Reading Lamp 75 3 225 1 bulb 75 W Second oor Hall Ceiling Fixture 240 3 720 4 bulbs 60 W Second oor Bathroom Fluorescent 160 15 240 Light 4 bulbs 40 W Bedroom 3 Table Lamp 180 4 720 3 bulbs 60 W Bedroom 3 Floor Lamp 150 3 450 1 bulb 150 W Bedroom 3 Ceiling Fixture 240 25 600 4 bulbs 60 W TOTAL 10030 watthours NASA AMATYC N SF 12 31 The pre x kilo means one thousand 1000 Divide watthours by 1000 Note 1000 watthours 1 kilowatthour With reference to Exercise 13 10030 watthours 10030 kilowatthours Multiply the number of kilowatt hours by 005 With reference to Exercise 13 the total cost is equal to 005 per kWh10030 kWh 05015 Answers vary They should include the cost of electricity as well as a comparison to 005 used by NASA Answers will vary f Answers will vary An example of an assumption that may be used is that the year is not a leap year and therefore the year has 365 days Part C Window to Wall Ratio 15 Students follow the example of Table 2 to complete a table of window and wall dimensions Students answers may vary slightly Window and wall information for House A House A Front Window Dimension Window Area Exterior Wall Exterior Wall Wall Dimensions Area windows 5 8 mm by 20 mm 5160 sq mm 113 mm by 33 3729 sq mm 800 sq mm mm door 13 mm by 29 mm Totals 800 sq m 3729 sq mm Window and wall information for House B House B Front Window Dimension Window Area Exterior Wall Exterior Wall Wall Dimensions Area Le double 12 mm by 8mm 96 sq m Rectangular window part of front Rectangular 8 mm by 21 mm 4168 mm W311 118 mm 2950 sq mm windows 4 672 sq m by 25 mm Triangular Triangle 2 77 2 Triangular windows 2 b 7 m h 11 mm 77 sq m gable end Trapezoidal altitude 8 mm 28462 base 60 1350 sq mm windows 2 side 1 17 mm 368 sq mm mm altitude side 2 29 mm 45 mm Door 10mmby 21 mm Totals 1213 sq m 4300 sq mm NASA AMATYC N SF 1232 Window and wall information for House C House C Front Window Window Area Exterior Wall Exterior Wall Wall Dimensions Dimensions Area Rectangular 32 mm by 20 mm 640 sq m Rectangular 4290 sq mm part of triple part of front xrindmxr xmll39 Halfdisk Radius 6mm 12p6 mm2 130 mm by above triple 57 sq m 33 mm window Rectangular 19mmby11mm 418 sqm Triangularpart 125118 part of right of front wall 459 sq m side windows Base 51 2 mm altitude 18 mm Half disc part Radius 6 mm 212 p6 mm2 ofright side 113 sq m windows 2 Half disc Radius 7 m 12 p397 mm2 window above 77 sq m door Door 13mmby 23mm Totals 1305 sq mm 4749 sq mm 16House A WWR 800s 0215215 3729 sq mm House B WWR 1 213 Sq mm 0280 280 4300 sq mm House C WWR 1 305 sq m 0275 275 4749 sq m Part D Energy Savings Features Building Types Climate 17 a Both might need re ecting tints for snow or bright sun Both locations use double thicknesses to reduce heat transfer but in opposite directions b Answers will vary c Answers will vary d Answers will vary NASA AMATYC N SF 1233 18 a The units of area cancelled out b The WWR would not have changed c Ifk is the scaling factor for inches and if c is the conversion factor from square millimeters to square inches then we have the following The true area of the windows k2 area of the windows in the diagram in square inches k2 c area of the windows in the diagram in square millimeters The true area of the wall k2 area of the wall in the diagram in square inches k2 c area of the wall in the diagram in square millimeters Since constant factors cancel when we form the WWR ratio it is not necessary to know the actual dimensions of the house Any drawing done to scale is suf cient 19 a Answers will vary b Answers will vary c Answers will vary d Wall W Approximately 50 Wall X Approximately 20 Wall Y Approximately 30 Part E Follow Up Exercises Related to NASA 20 3010 hours per year which is obtained by solving the equation 2089kWx 62878kWh 21 0071kWh calculation 447162878 kWh 00711kWh 22 4400 square feet calculation 88000 sq ft005 4400 sq it 23 00481 per kWh calculation 96001995kWh 00481kWh 24 31 years calculation 3000009600 per year 3125 years Section 2 Establishing a Baseline for Energy Use Comments The contractor invests 12000000 in work and is paid back 23000000 The difference of 11000000 can be considered a return on the contractor s investment The government obtained the money to pay the contractor from savings created by reduced energy consumption and deferred maintenance Deferred maintenance means that by installing new equipment replacement of the old equipment is delayed for the life of the new equipment The finance option on some calculators permits nding interest rates or present values For a more detailed discussion about rate of retum on investment see the Faculty Notes section for SpinoiT 12 A NASA AMATYC N SF 12 34 Part B Fiscal Year and Calendar Year Baseline For Electricity Use 1 January 1 and December 31 2 Expenses such as instructor contracts are between August and June Also most tuition is paid at the beginning of the fall and winter terms Thus the major nancial expenses for schools occur in the same scal year if the scal year is chosen to be July 1 to June 30 3 September 30 1998 4 FY 1998 5 1995 6 1997 7 Answers are found in the far right column of each of the following tables Table 3 E amp 0 En ineerin and O erations Buildin Electrici Use in kWh Month FY96 FY97 FY98 Baseline Monthly Average kWh kWh kWh kWh rounded to nearest hundred Oct 79130 70320 87120 78900 NOV 67910 61920 60720 63500 Dec 58010 56280 60650 58300 Jan 51870 80720 56280 63000 Feb 50990 52560 47880 50500 Mar 55570 69000 47040 57200 Apr 64500 58560 59020 60700 May 75600 71760 69210 72200 Jun 75600 77520 80400 77800 Jul 86140 90240 89870 88800 Aug 93550 93240 95290 94000 Sep 96010 97920 95330 96400 NASA AMATYC N SF 12 35 Baseline Monthly Average rounded to nearest Baseline Monthly Average rounded to nearest 1 149 760 730 299 780 730 299 780 1 NASA AMATYC N SF 12 36 Part C 8 Exploring Electricity Use Patterns A graphing calculator or spreadsheet can be used to construct a connected line graph for the data in each table Coded values 1 through 12 for the months should be recorded on the x axis with corresponding electricity use in kilowatt hours on the y axis Although the EampO building uses more electricity than Hangar S the electricity use pattern for the two buildings is very similar Both buildings use considerably less electricity from November through April than they do from May through September For both buildings the electricity use for February is less than for any other month Hangar AF uses more electricity the other two buildings and its electricity use pattern is quite different The electricity use for Hangar AF for the six month period from August through January is considerably lower than for the six month period from February through July The lowest electricity use occurs in November No the summer months may have higher air conditioning costs Seasonal dilTerences a ect energy use Building usage is another cause for variability in energy use For example more electricity is used in Hangar S and in Hangar AF for the months October through March FY97 than for the same months in FY98 This dilTerence would not be attributable to seasonal di erences It may be due to higher building usage in FY97 10 October through March FY98 132690 kWh April through September FY98 262400 kWh Usage for April through September is indeed higher than for October through March FY98 1 1 Total electricity use for each scal year by building is shown in the following table FY96 FY97 FY98 420 2976000 2456880 2370540 NASA AMATYC N SF 12 37 FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises V FACULTY NOTES LTA 11 Helium Usage at Kennedy Space Center Solutions First use Boyle s Law to determine the volume at 3500 psia P1V1 P2V2 147psia70000 scf 3500 psiaV2 294 3 v2 Since we know that each railcar holds 1050 3 1050 113 3 357 days 294 ft per day Thus the gaseous helium in one railcar will last 357 days or they will use approximately 2 railcars of helium per week Here we use Boyle s Law once more to determine the volume at 3500 psia P1V1 P2V2 147 psia1000000 scf 3500 psiaV2 4 0 3 7 V2 KSC will use 4200 3 of gaseous helium from a railcar Each railcar holds 1050 3 so 4200 113 4 rallcars 1050 it per railcar There are several steps conversions we have to perform a First convert all temperatures to the Kelvin scale C59F732 KC273 3C5970732 3K72715273 3C211 C 3K15 and K C 273 3 K 211 273 3 K 2941 So 700 F 2941 K and72715O C 15 K NASA AMATYC N SF 11 28 Now use the appropriate gas laws to convert 70000 scf to liters at 47 psia and 15 K ie 70000 113 147 psia and 2941 K 3 X liters 47 psia and 15 K b First change 70000 3 to an equivalent volume of liters 3 3 3 39 7000 002833m 10003dm llrtesr 219810001item 1 ft m dm Thus 70000 3 of gaseous helium 147 psia and 2941 K is equivalent to 1981000 liters of liquid helium 147 psia and 2941 K c Ifwe keep the temperature constant at 2941 K and since the mass remains unchanged we can use Boyle s Law to nd the volume of helium needed at 47 psia P1 147 psia P2 47 psia V1 1981000 liters V2 unknown P1V1 P2V2 147 psia1981000 113 47V2 619589 liters V2 Thus the volume of helium heeded at 47 psia is 619589 liters d Ifwe now leave the pressure constant at 47 psia mass remains unchanged we can use Charles Law to determine the new volume V1 619589 liters V2 T1 2941 K T2 15 K 1712 T T 1 2 619589 2 17L 2941 15 3 V2 3160 liters Thus KSC uses approximately 3160 liters of liquid helium for normal operations 3 Since one liquid helium trailer holds approximately 42000 liters each tanker will last 1 day 42 000 liters 133 days 3160 liters 4 70000 scf of gaseous helium 0075 per scf results in 5250 3160 liters of liquid helium 200 per liter results in 6320 The gaseous helium is more cost effective than the liquid helium NASA AMATYC N SF 11 29 Mathematical Aside Direct and Inverse Proportions Solutions Questions 1 k 04 for eaeh temperaturevolume pair 2 Ln The shape of the graph is a straight line Siuw the quot quot quot 39 04 for eaeh pair ofdala the equation can be written as follows V 04T or X 04 T 4 k 140 for eaeh volumepressure pair NASAVAMATYCVNSF 1130 Pressure approaehes zero as volume becomes very large I J n39 t l t it 6 I 0f uie variable r of data the equation can be written as follows VPl40 or Pm V 140 for eaeh pair Exercises 1 a directproportion b k 200 c P200T 2 a inverse proportion b k 00 w M c T 3 a direct proportion b k 3 c y 3X 4 a inverseproportion b c yl2X NASAVAMATYCVNSF 1131 Mathematical Aside The Gas Laws Questions 1 The graph is linear and goes through the point 0 0 so it is a direct proportion 2 k 6 3 Notice that the pressure is very large when the volume approaches zero and similarly the pressure is nearly zero when the volume is very large This indicates that there is an inverse relationship between volume and pressure NASA AMATYC N SF 11 32 Mathematical Aside Unit Analysis Solutions Exercises 1 a 2236mph b 117h0u1s 2 a 441psia b 30385 kPA 3 14524 z NASA AMATYC N SF 11 33 FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES LTA 16 The Space Shuttle Landing Learning Objective The students will determine a piecewise lnction that models the nal landing approach of the Space Shuttle Topics Version I deals with setting up a coordinate system deriving graphing and using linear functions deriving graphing and using exponential functions and basic right triangle trigonometry Version II has two parts Part A deals with setting up a coordinate system and deriving graphing and using linear functions Part B involves deriving graphing and using piecewise functions Level Version I is suitable for Precalculus students a er they have studied exponential functions Ifthey are not familiar with piecewise functions they can also do Part B of Version II Version II is suitable for Intermediate Algebra Technical Mathematics and College Algebra students Mathematical Prerequisites All students should be able to construct a suitable coordinate system and write the equation of a line Students doing Version I should be able to use right triangle trigonometry and be familiar with exponential functions Technology A graphing calculator is strongly recommended since students can store values and repeat calculations easily Suggested Strategies Groups are strongly recommended for this LTA You can have each group tum in one nal report or you can have each group member write his or her own report while using the group for problem solving We have found that students responded very well to this LTA when working in groups and were therefore able to complete much of the work outside of class For both versions it would be use ll to read the introduction with the students in class before actually doing any of the work This will set the stage and get the students thinking about the problem Discuss the rst part of the assignment 7 setting up the coordinate system and plotting the three data points This is not as easy as it seems since the points cover large distances and yet all must be plotted so that they can be clearly seen Encourage the groups to discuss their methods for constructing this graph including pasting several sheets of graph paper together We have had some students tape four sheets together while others have preferred poster board Have them come to the next class prepared to discuss any problems they had with the coordinate system Intermediate Algebra Technical Mathematics and College Algebra students could do the beginning of Version II modeling the rst phase of the landing in class in their groups The remaining two phases of the landing could be assigned for groups to do outside of class Part B piecewise lnctions should probably be started with the groups in class it could be completed by groups or individuals outside of class NASA AMATYC N SF 16 34 Precalculus stude can for the most part complete the LTA as a lab cumide of class with pa haps a linle class time allowed for questions and clan39fication Time Needed For Intemediate Algebra Technical Matha39mn39cs and College Algebra smdmts allow 23 hours spread over sevaal class pen39ods This assumes smde are Cu39npldjng pan afthe sc s quotat Vias may could be givm 12 weeks to complete the projecL Solu ons Version I The Final Approach to the Runwa Using Linear Func ons Exponmu39al Funcu39ms and Trigonometry 1 nn u i i i i i y u u i i 39 H quot 39 I 39 39 I 39 H I unealmuinlm the runwa 39 I I I I 39 39 39 I smdenus i i ii i i I m A axes Rarely a student group will chooseto scale in miles instead offeet and even more rmx y as P u i i i here only for this construction y slt376uu 13355 14 39 iz i suuu suuu damn Rlt75uu175 i zuuu Kiwava gzsmaai issuu ssuu 13EEB 175nm zsuuu 325nm avuuu NAsAeAMATYceNsF m Note on calculations When a numerical answer requires more than one calculation the full decimal place accuracy of the calculator should be used to obtain the answers given in the Solutions Thus if students round off intermediate answers and use them in subsequent calculations their answers may diiTer somewhat from the ones given in the Solutions 2 The data points are rising from le to right That is the farther the Shuttle is from the touchdown point the higher it is above the ground This is equivalent to saying that the nearer the Shuttle is to the touchdown point the closer it is to the ground Relative to the coordinate system the Shuttle travels from right to le and moves down 13365 1750 11615 39600 7500 32100 The units of slope are feet of altitude per foot of directed distance from the runway threshold The slope indicates how much the Shuttle descends for each foot that it nears the touchdown point Possible answers For every 32100 it measured horizontally that the Shuttle is closer to the runway its altitude decreases by 11615 ft L V Slope 036184 OR For every foot measured horizontally that the Shuttle is closer to the runway its altitude decreases by about 036 it 4 y 036184x 96378505 This Jnction is avalid model for 7500 S x S 39600 5 Connect the points R7500 1750 and S39600 13365 in Exercise 1 with a line segment 6 Ifx 10000y 03618410000 96378505 2655 7 Glide Slope tan391 w 199 32100 8 The linear model for the transitional phase is y 033381x 75360825 O V Glide Slope for a linear transitional function tan391 m 185 10 A linear function has constant slope so if a linear model is used the glide slope will not change during this phase The line will not give a shallow glide slope as the Shuttle exits this phase NASA AMATYC N SF 16 36 11 An approximate path shape for the transitional phase is shown below 12 1750 abmo 131 abmo 1 7500 13 ab 54850 3 b 50 1000534612 131 abmo 131 131 a 10005346122650 Thus the exponential transition function is y 31780571000534612X The domain ofthe transitional model is 2650 S x S 7500 3178057 14 The calculator answer is y 31780565921000534612x 15 Connect the data points Q2650 131 and R7500 1750 in Exercise 1 with the exponential lnction you obtained in Exercise 13 Students should evaluate the function at a few points between 2650 and 7500 and draw a smooth curve through them 16 y 002701x 5942268 The domain ofthe nal phase is 2200 S x S 2650 17 Connect the data points P 2200 0 and Q2650 131 in Exercise 1 with a straight line segment 18 Glide Slope tan391 i m 15 4850 19 tan3 2 d 255019 ft m 48 miles It would take the commercial aircra 255019 it or about 48 miles For the Shuttle the total distance is 41800 it or about 8 miles Thus the commercial aircraft would take about 6 times as much distance to land 13365 d NASA AMATYC N SF 16 37 20 exercises R M on TTA an L y361 4x763735 5 317E 571 534612Ax swan 75mm yuan 45mm u27mX auuu 7375 Version 11 The Final Approach 1 use Runway Using Linear Fundims s quot Thm 39 39 quot Relative to due left and moves down coordinau systan the Shuttle goes from right to NASAVAMATYCVNSF 163 3 Distance from Altitude in ft runway in ft y x 39600 13365 7500 1750 4 Slope 13365 1750 11615 39600 7500 32100 m 036184 The units of slope are feet of altitude per foot of directed distance from the runway threshold The slope indicates how much the Shuttle descends for each foot that it nears the touchdown point Possible answers For every 32100 it measured horizontally that the Shuttle is closer to the runway its altitude decreases by 11615 ft OR For every foot measured horizontally that the Shuttle is closer to the runway its altitude decreases by about 036 it 5 y 036184x 96378505 ifx is between 7500 and 39600 6 Connect the points R75001750 and S39600 13365 in Exercise 1 with a line segment 7 Ifx 10000 theny 03618410000 96378505 m 2655 ft 8 Distance from Altitude in ft runway in ft y x 7 500 1750 2 65 0 l3 1 1619 9 Slope m 033381 4850 10 033381x 75360825 where x is between 2650 and 7500 11 Connect the points Q2650 131 and R7500 1750 in Exercise 1 with a line segment NASA AMATYC N SF 16 39 12 A linear lnction has constant slope so if a linear model is used the slope will not change during this phase The model will not describe a steep slope on entry and a shallow slope as the Shuttle exits this phase An approximate path shape that might accomplish the transitional phase is shown below 13 Distance from Altitude in ft runway in ft y x 26 5 0 l3 1 2200 0 14 Slope of the nal linear phase L i 2650 2200 4850 m 002701 15 y 002701x 5942268 ifx is between 2200 and 2650 16 Connect the data points P 2200 0 and Q2650 131 in Exercise 1 with a straight line segment PaltB 17 3000 lt x lt 12000 18 120SxS765 NASA AMATYC N SF 16 40 19 The line y 3 2 is used only when x is between 0 and 3 including both and the line y s x 1 is used only when x is between 3 and 7 not including either number 20 21Ifx1y3125 22Ifx6y76148 23Equa 0n 1 y 036184xr96378505 for 7500 S x S39600 24Equa 0n 2 y 033381xr7535965 for 2650 lt x 7500 25Equa 0n 3 y 002701x5942268 for 72200 S x S 2650 26 5 miles 26400 R Using Equation 1 y 03618426400 796378505 s 8589 R 27 Using Equation 3y 0027011000 5942268 s 86 28Using Equation 2 y 0333815000 7 75360825 s 915 R 29Using Equation 2 1450 033381 7 75360825 x s 6601 R 30 Using Equation 3 100 002701 5942268 NAsAeAMATvceNsF 1541 31 the exercises 39 39 39 39 39 39 L c pawn w m mum u w mm to include Shes mm lll quot 39 hm chat H19 L L y 36134x 763735 5 33331x 7 7535755 an y 27 1x 742253 715uu 15mm auuu 45mm suuu 75mm yuan 7375 NASAVANMJYCVNSF M FACULTYNOTES The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with hisher teaching style and course objectives This may range from using the materials as out of class projects with minimal in class guidance to doing most of the work in class The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some leaming technology Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students the Faculty Notes are not intended to be prescriptive The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs This includes suggestions for instruction as well as answers for the exercises FACULTY NOTES SPD IOFF 16A Mndeiing The Space Shuttle Landing The Cubic Spline V Fir V but m from the LTA The puipose odeis Spmoffls to examine the exponenuai pieee that models the in with new th F exponenuai pieee eauses the funeuon to be nonrddfferenuable at those two points The nail task m w denvanve eondiuons This results in a system of four equauons in four unknowns Wth very unWieidy coef cients The Spinotrbneiy inuodiees students to matnx methods for solving a system F mm W t t i T V a TTegzma and Denvem Upon eompieuon ofthe Spinoff students eouidbe askedto suggest othe functions whieh might also Work Snllltinns 1 and 2 Here is the nal gaph 3 The glide slope dinng Lhe rst linear phase is atetanw 3618A19 9 The glide slope the slope othe line 4 The dmvanve ofthe exponenuai pieeeis y39 e 0 0169857341050 00053461 NASA 7 Ammo NSF IE 9 The glide slope at x is the arctangent of the derivative at x That is glide slope arctan f x Whenx 7500 the glide slope is arctan0935323459841 m 431 Notice that the slope is 0935323459841 but the slope of the linear piece of the rst phase is 036184 The glide slope of the exponential is 431 whereas the glide slope of the linear is 199 This means that the piecewise function is not differentiable at x 7500 Whenx 2650 the glide slope is arctan0070015528847 m 4 0 Notice that the slope is 0070015528847 but the slope of the linear piece of the nal phase is 002701 The glide slope of the exponential is 40quot whereas the glide slope of the linear is 15quot This means that the piecewise function is not diiTerentiable at x 2650 The third degree polynomial is y ax3 bx2 cx 61 Substituting 7500 1750 we get 75003a 75002b 7500c d 1750 Substituting 2650 131 we get 26503a 26502b 26500d 131 The derivative is y 361x2 2bx c 10 f 2650 should be 002701 and f 7500 should be 036184 11 If f 2650 002701 then 326502a 2 2650bc 002701 If f397500 036184 then 375002a27500bc 036184 12 The solution as given in the Spino is 2 3 75003 75002 7500 1 1750 26503 26502 2650 1 131 13 The augmented matnx 1s 375002 27500 1 0 036184 326502 22650 1 0 002701 14 The reduced row echelon form of this matrix is 1 0 0 0 11851583206610 8 0 1 0 0 0000214958911022 0 0 1 0 0862588999214 i0 0 0 1 112786541439 J 15 The solutionto the system is a 11851583206610 8 b 0000214958911022 c 0862588999214 and d 1127 86541439 NASA AMATYC N SF 16 45 15 The polynomial model is y 71185158232066403 000021495891102212 7 0362588999214 112786541439 e daiva ve is y 73555474961983 0000429917822044x70362588999214 f 2650 0270099999975 which agees with due slope ofche linear function in four decimal places f 7500 351340000002 which agees wich d1 slope ofthe linear in ve decimal p aces A gap ofche new piecewise mc on is 45175 y 3611 x 7 953715 7111515132uss 1u1mmmmzuvsnmzzwza 27113 55525 13431 11125 137 15511 19459 NASA 7 AMATYC r N SF 115 as FACULTYNO TES Ta LTAs andSpmstm dzslgmdsa um eachpmfessaxcmxm zmzm nasmaaaawayuaaa as manna as aaaaaatsxass pmjec39s wa mmm amass guldance m dnmg masaamas wuka class m LTAs andSpmn smz amsaams a ma mnpcaapm ve vmxkandiypw y bem n mm u use a mm Izammg techmlngy Smce ms abjechve amas LTAs andSpmst n yml setfaxyunrsmdzmsthz masadsaaabs passan Th2 pllrpuse amas Fundinmzs as m pmwdz mfannanan um assists ymna mkz ran advantage amas LTAs andSpmst Thls mcludzssuggesmns raa msmwonn as we as answers raa ms sxsasasss um mm am an FACULTY NOTES LTA 2 NASA Aquatics Lab Background Information This LTA is written so that in Part III Exercise 2 students will develop a function for the load of the form Lx 12 where x2 is the radius of the pallet In Part III Exercise 3 the students x should get a linear function of the height Mathematical topics covered include Relations involving Measurement and Geometry Basic Level Rational Equations Basic Level Graphs Geometry Rational Functions and Limits intuitively Timeline This Laboratory Technical Activity LTA is intended as a two hour laboratory activity where students interact with peers and instructors It is suggested that students begin work on this project during one class period and take it home for further work either individually or collaboratively They can then nish the project either outside of class or during the following class period Required Technology Any graphing calculator or computer algebra system may be employed The LTA assumes students have a basic knowledge of how to graph functions and perform elementary operations with the technology the instructor uses in class Level Precalculus or College Algebra Instructional Methodology This LTA is intended for group activity either in class or take home or both Instructors are better able to meet individual needs in this learning environment because the capable students may be able to proceed with little or no assistance Possible Modi cations The LTA could be easily modi ed to provide a more dif cult exercise for students when constructing their initial load function The rst two sections could be replaced by a single paragraph which instructs the students to identify appropriate variables constants and units needed to solve the load problem One variation of this would be to have the students start the identi cation in small groups share their results with a larger group and successively larger groups until the entire class has discussed appropriate variables and so forth It might be possible to have the class split into two competitive groups to come up with the best list with more points awarded to those with the best list Another variation would be to have the students start on the identi cation of variables etc and then hand out Parts I and II after the students have worked on the problem for a while NASA AMATYC NSF The rst problem in Part III could be rewritten so that students would be allowed or even encouraged to choose the area of the pallet as the independent variable The remaining questions would not need to be changed Part V Exercise 1 could be omitted without losing the value of the LTA as a learning tool The question is intended to help put the mathematics into context Part VI could be eliminated also if time is a problem It would make the LTA shorter and the instructor may have writing activities in another part of the course which are suf cient Assessment Speci c leaming objectives are listed in Part VI Problem Point Value Objectives Part I 10 12 point each Mathematical Modeling Unit Analysis Part II 8 2 points each Mathematical Modeling Unit Analysis one for constantvariable Part III 20 4 points for 1 Mathematical Modeling Critical 8 each for 2 amp 3 Thinking Communicating Mathematics Part IV 36 4 points for 5 Interpreting Graphs Unit Analysis 8 each for l4 Communicating Mathematics Function Values Critical Thinking Part V 10 5 points each Communicating Mathematics Critical Thinking Interpreting Graphs Part VI 16 Communicating Mathematics Critical Thinking Total 1 00 Leaming Objectives As a result of doing this LTA students should improve their skills in the following areas 1 Mathematical Modeling understand connections between a technical application and an algebraic representation 2 Interpreting Graphs nd points of intersection interpret practical meaning of the coordinates and recognize asymptotic behavior 3 Unit Analysis convert units and determine the units of each variable involved 4 Communication communicate mathematics to others both written and verbal V39 Critical Thinking deduce conclusions and explain the practical meaning of mathematical results 6 Function Values evaluate rational functions at various points in the domain nd a value in the domain corresponding to a given value in the range NASA AMATYC NSF Solutions Note Answers for LTA 2 have been rounded to the three most signi cant digits to be consistent with the accuracy of the data provided Part I Relevant Quantities Table 1 Quantity Constant or Variable Constant Value or Appropriate Units Variable Symbol number of gallons per cubic foot constant 748 i gallonscubic foot volume of entire tank in cubic feet constant 754 ii cubic feet volume of entire tank in qallons constant 564 iii gallons radius of the tank constant 2 feet radius of the pallet variable r feet height of the tank constant 6 feet water level in the tank variable h feet lbs 6247 lb 1 z S a a 1 Number of gallons per cub1c foot 2 748201 3 g 748201 g 3 834 S lbs gal ii Volume oftank in cubic feet V 1 2 ft26 ft 753982 ff 1 g i753982 ff 56413gal iii Volume of tank in gallons V 748201 Part II Relationships aA 2 aV 3 aW b Am2 b V4nh b W784h lbs 4 We1ght offull tank ofwater 834 l56413gal 4700 lbs rounded to 3 s1gn1 ant d1g1ts ga NASA AMATYC NSF rm m r Madeling 47mm mtg 1 mmmdmmmmm ZW 2 ludzpndzmwmhlz rum21 09pmka Lmpmmdspxsqunfam Fm L ludzpndzmwmhlz 11mfee 09pmka Lmpmmdspxsqunfam 11m 1 W m mmmmmmwmgm 1 3 1 1 Th ndnlsafthz pllztcanbe mmme 11mm feetbecmlse 11 mm rm 11 centzxaf 11 11am an 1 7 z 7 u my gmhpramr 4 21m 511 2 Name 111111111111 fallnwmg graph we have 111211111 damn m be 11 mter um T1115 w 111W 1 shriythz e 39ecmfmakmg 111 ndnlsafthz p11 anynumhex gnatznhn rm 1111115511 arequl 5 rm um mm 11 1 n A 624h2EIEI 2 h321 Tnhehexgxasmewenmyemesmemdexmreemmemkeepmexmmaennn Frsq 5 7 42m 253 21 am 722 The number afgallnns um embe serexysmxea m me m canmt exceed 3m gallnns rm v onquot Cumiduxlimu 1 Vmaus wn en xespunses m fnss nlz 2 The Lead dzcnases andeven u ets alas m um ssme ndnlsafthz enseunweam Increase wnhnmbannd y F rmvxe Cumluxinm A wnnen xepm um mm m 1 n

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### 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.