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Question (1): (a) Write Mathematica NoteBook which generates a table of error function (erf) and its derivatives for real arguments (z) between --3 and 3 in steps of 0.25. The error function is defined by the equation below (but is rarely evaluated by performing the integration).

(see http://mathworld.wolfram.com/Erf.html for information the error function ) The values in the table should be given with 5 decimal places. The table should have headers explaining what the columns are. Explain how you designed the NoteBook and give an example of the output. (b) How would you change this NoteBook if 10 significant digits were required? Mathematica NoteBook should also be supplied Question (2): Write a NoteBook that reads your name in the form <first name> <middle name> <last name> and outputs the last name first and adds a comma after the name, the first name, and initial of your middle name with a period after the middle initial. If the names start with lower case letters, then these should be capitalized. The NoteBook should not be specific to the lengths of your name (ie., the NoteBook should work with anyone’s name.

As an example. An input of thomas abram herring would generate: Herring, Thomas A. Question (3): Write a Mathematica NoteBook that will compute the motion of a bicyclist and the energy used cycling along an oscillating, sloped straight--line path. The path followed will be expressed as

H (x)  Sx  A sin(2 x / )  B cos(2 x / )

where H(x) is the height of the path above the starting height, S is a slope in m/m, A and B are amplitudes of sinusoidal oscillations in the path. The wavelength 2of the oscillations is . The forces acting on the bicycle are: Wind Drag Rolling Drag

Fd  1/2 Ar Cd V Fr  M r gC r

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where Ar is the cross--sectional area of the rider, Cd is the drag coefficient, r is the density of air and V is the velocity of the bike. For the rolling drag, Mr is the mass of the rider and bike, g is gravitation acceleration and Cr is rolling drag coefficient.

The bicyclist puts power into the bike by pedaling. The force generated by this power is given by Rider force Fr  Pr /V where Fr is the force produced by the rider, Pr is power used by the rider and V is velocity that the bike is traveling (the force is assumed to act along the velocity vector of the bike). Your NoteBook can assume that the power can be used at different rates along the path. The energy used will be the integrated power supplied by the rider.

Assume that there is maximum value to the rider force.

Your code should allow for input of the constants above (path and force coefficients). The NoteBook can assume a constant power scenario and constant force at low velocities.

As a test of your NoteBook use the following constants to compute: (a)Time to travel and energy used to travel 10 km along a path specified by S=0.01, A=5.0 m, B=0.0 m and = 2km, with constant power use of Pr =100Watts and a maximum force available of 20N. (b)The position and velocity of the bike tabulated at a 100--second interval. (c)Add graphics to your NoteBook which plots the velocity of the bike as a function of time and position along the path.

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Assume the following values Cd = 0.9 Cr = 0.007 Ar = 0.67 m2 r = 1.226 km/m3 g = 9.8 m/s2 Mr = 80 kg In this case, the Mathematica NoteBook will not be of the type used for fortran and C/C++. Look at the documentation on NDSolve for this problem.

Your answer to this question should include: (a)The algorithms used and the design of your NoteBook (b)The Mathematica NoteBook with your code and solution (I run your NoteBook). (c)The results from the test case above.

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This latest version of Mathematica has some issues with copying and pasting reproduce subscripts and superscripts from word, which previously was able to equations and graphics from Microsoft Word. It also seems not to be able to do. It is not clear if the incompatibility is from the Microsoft Word or the Mathematica side. This is another indication of how changes in versions of software can have unintended consequences.

(Question 1): (a) Write Mathematica NoteBook which generates a table of error function (erf) and its derivatives for real arguments (z) between -3 and 3 in steps of 0.25. The error function is defined by the equation below (but is rarely evaluated by performing the integration). (see http://mathworld.wolfram.com/Erf.html for information the error function ) The values in the table should be given with 5 decimal places. The table should have headers explaining what the columns are. Explain how you designed the NoteBook and give an example of the output. (b) How would you change this NoteBook if 10 significant digits were required?

Mathematica NoteBook should also be supplied

● Solution: Given that the error function is defined in Mathematica themost tricky part of the this problem is formarting the ourput. There are many ways to approach the output problem.

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Off@General::spell1D; H* Stops message about head looking like a command*L head = 8"Argument", "Erf@zD", "dErf@zDêdz"<; derf@arg_D := 2 ê Sqrt@PiD * Exp@-x^2D; ents = Table@8N@SetAccuracy@x, 10D, 3D, In[1]:= N@SetAccuracy@Erf@xD, 10D, 6D, N@SetAccuracy@derf@xD, 6D, 6D<, 8x, -3, 3, 0.25<D; full = Insert@ents, head, 1D; Out[6]//TableForm= TableForm@full, TableAlignments Ø RightD Argument

Erf@zD

dErf@zDêdz

-3.00 -2.75 -2.50 -2.25 -2.00 -1.75 -1.50 -1.25 -1.00 -0.750 -0.500 -0.250 0. µ 10-10 0.250 0.500 0.750 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00

-0.999978 -0.999899 -0.999593 -0.998537 -0.995322 -0.986672 -0.966105 -0.922900 -0.842701 -0.711156 -0.520500 -0.276326 0. µ 10-10 0.276326 0.520500 0.711156 0.842701 0.922900 0.966105 0.986672 0.995322 0.998537 0.999593 0.999899 0.999978

0.00014 0.00059 0.00218 0.00714 0.02067 0.05277 0.118930 0.23652 0.41511 0.64293 0.87878 1.06001 1.12838 1.06001 0.87878 0.64293 0.41511 0.23652 0.118930 0.05277 0.02067 0.00714 0.00218 0.00059 0.00014

(b) Output of table to 10 deciminal places.

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Off@General::spell1D; H* Stops message about head looking like a command*L head = 8"Argument", "Erf@zD", "dErf@zDêdz"<; In[22]:= derf@arg_D := 2 ê Sqrt@PiD * Exp@- x^2D; ents = Table@8N@SetAccuracy@x, 10D, 3D, N@SetAccuracy@Erf@xD, 10D, 10D, Out[26]//TableForm= N@SetAccuracy@derf@xD, 10D, 10D<, 8x, -3, 3, 0.25<D; full = Insert@ents, head, 1D; TableForm@full, TableAlignments Ø Erf@zD dErf@zDêdz RightD Argument -3.00 -2.75 -2.50 -2.25 -2.00 -1.75 -1.50 -1.25 -1.00 -0.750 -0.500 -0.250 0. µ 10-10 0.250 0.500 0.750 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00

-0.999977910 -0.999899378 -0.999593048 -0.998537283 -0.995322265 -0.986671671 -0.9661051465 -0.9229001283 -0.8427007929 -0.7111556337 -0.5204998778 -0.2763263902 0. µ 10-10 0.2763263902 0.5204998778 0.7111556337 0.8427007929 0.9229001283 0.9661051465 0.986671671 0.995322265 0.998537283 0.999593048 0.999899378 0.999977910

0.000139253 0.0005862772 0.002178284 0.0071423190 0.0206669854 0.0527749959 0.1189302892 0.2365211224 0.415107497 0.642931069 0.878782579 1.060014129 1.128379167 1.060014129 0.878782579 0.642931069 0.415107497 0.2365211224 0.1189302892 0.0527749959 0.0206669854 0.0071423190 0.002178284 0.0005862772 0.000139253

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period after the middle initial. If the names start with lower case letters, then these should be capitalized. The NoteBook should not be specific to the lengths of your name (ie., the NoteBook should work with anyone’s name.

As an example. An input of thomas abram herring would generate: Herring, Thomas A.

● This problem is not too bad to solve. This solution works in >5.0 Mathematica and does explicitly some things such as splitting a string apart that are now Mathemat- ica commands (StringSplit).

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In[27]:=

H* Define a function that will convert chararacter of a string to upper case *L confirst@a_D := StringReplacePart@a, ToUpperCase@StringTake@a, 81<DD, 81, 1<D; H* Get the name from the user*L inname = InputString@"Enter Name Hfirst, middle, lastL "D; H* Convert whole string to lower case*L fullname = ToLowerCase@innameD; H* Now get the list of blanks in the string*L posblanks = StringPosition@fullname, " "D; H* Get the position of first blank*L posfirst = Extract@Extract@posblanks, 1D, 1D; firstname = StringTake@fullname, posfirst - 1D; firstname = confirst@firstnameD; H* Use our confirst routine *L H* Get Middle Name *L posmid = Extract@Extract@posblanks, 2D, 1D; midinit = StringTake@fullname, 8posfirst + 1, posfirst + 1<D; midinit = confirst@midinitD; H* Get Last Name *L lastname = StringTake@fullname, 8posmid + 1, StringLength@fullnameD<D; lastname = confirst@lastnameD; Input name thomas abram herring converted to: Herring, Thomas A. H* Output the string *L finalname = lastname <> ", " <> firstname <> " " <> midinit <> "."; outline = "Input name " <> inname <> " converted to: " <> finalname; H* Output the results*L Print@outlineD;

Write a Mathematica NoteBook that will compute the motion of a bicyclist and the energy used cycling along an oscillating, sloped straight--line path. The path followed will be expressed as

H ( x)  Sx  A sin(2x / )  B cos(2x / )

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where H(x) is the height of the path above the starting height, S is a slope in m/m, A and B are amplitudes of sinusoidal oscillations in the path. The wavelength of the oscillations is . The forces acting on the bicycle are: Wind Drag Rolling Fd  1/2Ar C d V 2 Drag

Fr  M r gC r where Ar is the cross--sectional area of the rider, Cd is the drag coefficient, r is the density of air and V is the velocity

of the bike. For the rolling drag, Mr is the mass of the rider and bike, g is gravitation acceleration and Cr is rolling drag coefficient.

The bicyclist puts power into the bike by pedaling. The force generated by this power is given by

Rider force

Fr  Pr /V

where Fr is the force produced by the rider, Pr is power used by the rider and V is velocity that the bike is traveling (the force is assumed to act along the velocity vector of the bike). Your NoteBook can assume that the power can be used at different rates along the path. The energy used will be the integrated power supplied by the rider.

Assume that there is maximum value to the rider force.

Your code should allow for input of the constants above (path and force coefficients). The NoteBook can assume a constant power scenario and constant force at low velocities.

As a test of your NoteBook use the following constants to compute: (a)

Time to travel and energy used to travel 10 km along a path specified by S=0.001, A=5.0 m, B=0.0 m and =

2km, with constant power use of Pr =100Watts

programminghomeworkhelp.com and a maximum force available of 20N. The position and velocity of the bike tabulated at a 100--second interval. Add (b) graphics to your NoteBook which plots the velocity of the bike as a (c) function of time and position along the path.

Assume the following values Cd = 0.9 Cr = 0.007 Ar = 0.67 m2  = 1.226 kg/m3 g = 9.8 m/s2 Mr = 80 kg In this case, the Mathematica NoteBook will not be of the type used for fortran and C/C++. Look at the documentation on NDSolve for this problem.

Your answer to this question should include: (a)

The algorithms used and the design of your NoteBook

(b)The Mathematica NoteBook with your code and solution (I run your NoteBook). (c) The results from the test case above. ● There are several ways of approaching this problem and two solution are presented below. The problem itself is dvided into a number of cells that allow parts of prob- lem to be re-execuated. There two basic steps to the solution: (1)Use NDSolve to solve the second order differential equation that describes the problem. The solution effectively becomes an equation that returns values at any specified time.

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(2) Use FindRoot to determine when the bike has reached the end of track. ● Set up the defaults first. This cell should be evaluated Clear@t, x, xd, hx, vx, xp, xd, th, daccD; cd = cr =; 0.007; 0.90 = 0.67; H* m^2 *L area H*Print@"Parameter cd ",cdD*L = 80; H* kg *L mass prider = 100; H* Power Watts*L slope = 0.001; fmax = 20; as = 5.0; bs = 0.0; lambda = 2000; tracklen = 10 000; outint = 100.0; H* Define the acceleration functions we will need*L 9.8 ; rhoair = 1.226; Print@"Default Values set"D

grav =

Default Values set

● Now allow the user to enter values of there own. This cell does not need to be evaluated if the default values are desired.

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In[82]:=

Input@"Drag Coefficient ", cdD; Input@"Rolling friction coefficient ", crD; tracklen = Input@"Length of track HkmL ", tracklen ê 1000D * 1000.0; slope = Input@"Track slope ", slopeD; as = Input@"Sin Amplitude HmL ", asD; bs = Input@"Cos Ampliude Hm L", bsD; lambda = Input@"Periodic wavelength HkmL ", lambda ê 1000D * 1000.0; mass = Input@"Rider+Bike Mass HkgL ", massD; area = Input@"Rider Area Hm^2L ", areaD; cd = cr = outint = Input@"Output interval HsL", outintD;

o Now set up the force model equations theta@xp_D := ArcTan@slope + as * Cos@2 * Pi * xp ê lambdaD * 2 * Pi ê lambda bs * Sin@2 * Pi * xp ê lambdaD * 2 * Pi ê lambdaD htrack@xp_D := slope In[92]:= * xp + as * Sin@2 * Pi * xp ê lambdaD + bs * Cos@2 * Pi * xp ê lambdaD; H* Second derivative of surface for computing cenripical acceleration *L dy2dx2@xp_D := - as * Sin@2 * Pi * xp ê lambdaD * H2 * Pi ê lambdaL^2 bs * Cos@2 * Pi * xp ê lambdaD * H2 * Pi ê lambdaL^2; vmag@xd_, zd_D := Sqrt@xd^2 + zd^2D;

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H* Acceleration due to drag *L dragx@xd_, zd_D :=- rhoair * vmag@xd, zdD * area * cd * xd ê H2 * massL; dragz@xd_, zd_D := zd ê H2 * massL; - rhoair * vmag@xd, zdD * area * cd * H* Rolling force.This act along surface *L rollx@xp_D := - grav * cr * Cos@theta@xpDD; rollz@xp_D := - grav * cr * Sin@theta@xpDD; H* Gravity force acting perpendicular to surface *L gravx@xp_D := -grav * Cos@theta@xpDD * Sin@theta@xpDD; gravz@xp_D := +grav * Cos@theta@xpDD^2;

H* Centripetal acceleration *L H* Could divide by Sqrt@H1+Tan@theta@xpDD^2L^3DD factor to curvature term. This change does not seem to make solution closer to track shape*L

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centx@xp_, xd_, zd_D := Hxd^2 + zd^2L * Sin@theta@xpDD * dy2dx2@xpD; centz@xp_, xd_, zd_D := Hxd^2 + zd^2L * Cos@theta@xpDD * dy2dx2@xpD; H* This commented code has expression for radius of curvature centx@xp_,xd_,zd_D := Hxd^2+zd^2L*Sin@theta@xpDD* dy2dx2@xpDêSqrt@H1+Tan@theta@xpDD^2L^3D; centz@xp_,xd_,zd_D := Hxd^2+zd^2L*Cos@theta@xpDD* dy2dx2@xpDêSqrt@H1+Tan@theta@xpDD^2L^3D; *L H* To see the effects of the centripetal force, remove the comments below centx@xp_,xd_,zd_D :=0; centz@xp_,xd_,zd_D :=0; *L H* Now the rider force *L ridrx@xp_, xd_, zd_D := HMin@If@ vmag@xd, zdD > 0, prider ê vmag@xd, zdD, fmaxD, fmaxD ê massL * Cos@theta@xpDD; ridrz@xp_, xd_, zd_D := HMin@If@ vmag@xd, zdD > 0, prider ê vmag@xd, zdD, fmaxD, fmaxD ê massL * Sin@theta@xpDD; Print@Acceleration functions SetD Acceleration functions Set

● The following cell can be used to test that the acceleration functions above gener- ate numeric results when called with distance and velocity. This test is useful if NDSolve does not return answer.

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In[121]:=

dx = 1500.; txv = 2.0; tzv = 0.01; Print@"Slope ", theta@dxD, " 2nd derivative ", dy2dx2@dxD, " Height ", htrack@dxDD Print@"Drag X ", dragx@txv, tzvD, " Z ", dragz@txv, tzvDD Print@"Roll X ", rollx@dxD, " Z ", rollz@dxDD Print@"Gravity X ", gravx@dxD, " Z ", gravz@dxDD Print@"Centripetal X ", centx@dx, txv, tzvD, " Z ", centz@dx, txv, tzvDD Print@"Rider X ", ridrx@dx, txv, tzvD, " Z ", ridrz@dx, txv, tzvDD Slope 0.001 2nd derivative 0.000049348 Height -3.5 Drag X -0.0184822 Z -0.0000924109 -0.0000686

Roll X -0.0686 Z

Gravity X -0.00979999 Z 9.79999

Centripetal X 1.97397 µ 10- 7 Z 0.000197397 Rider X 0.25 Z 0.00025

● Now set up the solution for NDSolve. There are two equations in the x and z accel- erations and initial conditions for x and z and x' and z' at time zero. The solution is set to solve for a maximum of 10000 seconds. This is OK for the standard case but may need to be modified for other longer running cases.

The evaluaton of the solution is saved for x and z positions and z and z velocities.

(NDSolve contains examples of setting up these solutions.

We can then find the length of the time need by solving the equation xp[t]-tracklen == 0. This is not with FindRoot.

In[177]: solution =

= NDSolve@

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8x''@tD ã ridrx@x@tD, x'@tD, z'@tDD + dragx@x'@tD, z'@tDD + rollx@x@tDD + gravx@x@tDD + centx@x@tD, x'@tD, z'@tDD, z''@tD ã ridrz@x@tD, x'@tD, z'@tDD + dragz@x'@tD, z'@tDD + rollz@x@tDD + gravz@x@tDD - grav + centz@x@tD, x'@tD, z'@tDD, x@0D == 0, z@0D ã 0, x'@0D ã 0, z'@0D ã 0<, 8x, z<, 8t, 0, 10 000<D; xp@t_D := First@Evaluate@x@tD ê. solutionDD; zp@t_D := First@ Evaluate@z@tD ê. solutionDD; xv@t_D := First@Evaluate@x'@tD ê. solutionDD; zv@t_D := First@Evaluate@z'@tD ê. solutionDD; H* Compute the error in the distance *L endt = t ê. FindRoot@xp@tD - tracklen ã 0 , 8t, 1, 3000.0<D; H* Now integrate to get work done *L work = NIntegrate@mass * Sqrt@xv@tD^2 + zv@tD^2D * Sqrt@ridrx@xp@tD, xv@tD, zv@tDD^2 + ridrz@xp@tD, xv@tD, zv@tDD^2D, 8t, 0, endt<, AccuracyGoal Ø 4D; Print@"Time to reach end of track ", endt, " sec, Speed ", xv@endtD, " mês"D; Print@"Work done ", work, " Joules, ", work ê 4.1868 ê 10^3, " kcal" D; NIntegrate::ncvb : NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in t near 8t< = 8678.279<.

NIntegrate obtained 174398.48726962117` and 0.39321119853291564` for the integral and error estimates. à Time to reach end of track 2255.17 sec, Speed 2.65009 mês Work done 174 398. Joules, 41.6544 kcal

● The case above gives an example of how tricky Mathematica can be in telling you what it is actually doing. In the integration above for work a numerical rounding error message is printed. The message suggests there is some problem around 678 seconds. The cell below divides the calculation into 2 parts normally split at around 678 seconds and in this case no numerical error warning is printed. It’s not at all clear why the program is behaving this way.

In[166]:=

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tt = 678.279; worksplit = NIntegrate@mass * Sqrt@xv@tD^2 + zv@tD^2D * Sqrt@ridrx@xp@tD, xv@tD, zv@tDD^2 + ridrz@xp@tD, xv@tD, zv@tDD^2D, 8t, 0, tt<D + NIntegrate@mass * Sqrt@xv@tD^2 + zv@tD^2D * Sqrt@ridrx@xp@tD, xv@tD, zv@tDD^2 + ridrz@xp@tD, xv@tD, zv@tDD^2D, 8t, tt, endt<D; Print@"Work computed in one step ", work, " and split at time ", tt, " " , worksplit, ". Difference in results ", work - worksplitD; In[185]:= Work computed in one step 174 399. and split at time 678.279 174 399.. Difference in results 0.

● Final output for version A (full solutions) Print@"12.010 HW 4: Mathematica Bike Problem"D; Print@"Solution Parameters"D; Print@"Track Length ", tracklen ê 1000.0, " km"D; Print@"Track Slope ", slope, " Sin Cos ", as, " ", bs, " m, Lambda ", lambda ê 1000., " km"D; Print@"Rider Power ", prider, " Watts, Max Force ", fmax, " N"D; Print@"Time to reach end of track ", endt, " sec, Speed ", xv@endtD, " mês"D Print@"Work done by rider ", work, " Joules, ", work ê 4.1868 ê 10^3, " kcal" D

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12.010 HW 4: Mathematica Bike Problem Solution Parameters

Track Length 10. km

Track Slope 0.001 Sin Cos 5. 0. m, Lambda 2. km

Rider Power 100

Watts, Max Force 20 N Time to reach end of track 2255.17 sec, Speed 2.65009 mês rider 174 398. Joules, 41.6544 kcal

Work done by

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In[241]:=

● Add some graphics to the results: Velocity versus time, Z position versus X posi- tion, and difference from track shape. The latter plot shows how well our integra- tion matched the actual shape of the track.

Plot@ xv@tD, 8t, 0, endt<, AxesLabel Ø 8"Time HsecL", "Velocity HmêsL"<D ParametricPlot@ 8xp@tD ê 1000 , zp@tD<, 8t, 0, endt<, AxesLabel Ø 8"X position HkmL", "Z positionHmL"<, AspectRatio Ø 1 ê 1D ParametricPlot@ 8xp@tD ê 1000 , Hzp@tD - htrack@xp@tDDL * 1000.<, 8t, 0, endt<, AxesLabel Ø 8"X position HkmL", "Error in Z HmmL"<, AspectRatio Ø 1 ê 1D Out[241]=

Velocity HmêsL 7 6 4 5 3 2 1

500

1000

1500

2000

Time

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Z positionHmL

10

Out[242]=

5

2

4

6

8

10

X position

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Error in Z HmmL 0.2 0.1

2

Out[243]=

4

6

8

10

X position

- 0.1

- 0.2 - 0.3 - 0.4

-0.5

● We now use table and table form to have to put the positions and velocities as a function of the interval selected by the user. (The default interval is 100 seconds). To generate the table with 100s output, and to add the final value, we generate the table in two parts and save the table in a list called outlst. We use the append function to join the two tables together.

programminghomeworkhelp.com In[274]:=

outlst = Table@8t, xp@tD, zp@tD, xv@tD, zv@tD<, 8t, 0, endt, outint<D; H* Now add the final entry to the tabe; *L outlst = Append@outlst, Transpose@Table@8t, xp@tD, zp@tD, xv@tD, zv@tD<, Out[276]//TableForm= 8t, endt, endt, outint<DDD; TableForm@outlst, TableHeadings Ø 8None, 8"Time HsL", "X Pos HmL", "Z pos HmL", "X Vel HmêsL", "Z Vel HmêsL"<<D Time HsL 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2255.17

X Pos HmL 0.

79.9901 306.763 826.797 1513.81 1951.68 2202.42 2608.52 3292.9 3834.83 4117.37 4428.12 5046.56 5681.7 6035.25 6294.17 6803.7 7493.54 7941.44 8193.43 8588.77 9269.22 9821.37 10 000.

Z pos HmL 0.

1.32329 4.41341 3.41531 -3.48135 1.19559 5.17194 7.32075 -0.685554 1.35511 5.91931 9.30101 4.31778 1.47426 6.58758 10.2845 9.69517 2.49435 7.0264 11.0479 13.3952 5.52648 7.16001 9.99951

X

Vel HmêsL 0.

1.48869 3.36696 6.79761 5.87698 2.95947 2.65422 5.8297 6.88819 3.88251 2.35353 4.53331 7.24973 4.97372 2.48085 3.32052 6.71443 5.98646 3.03341 2.60451 5.72489 6.96317 3.99206 2.65009

Z Vel HmêsL - 5.29396 µ 10-23 0.0241386 0.0335387 -0.0845564 0.00988077 0.0489105 0.0361969 -0.0247871 -0.0586485 0.0568391 0.0368374 0.0204786 -0.105412 0.0471862 0.0412101 0.0347469 -0.0793262 0.00407604 0.0498764 0.0361917 -0.0190286 -0.0655712 0.0570797 0.0442763

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● Alternative solution, Here we solve the one dimensional problem which is basically the roller coaster solution that keeps the bike on the ground. In the Fortran and C-versions we computed the forces along the sloped surface and integrated horizontal motions from the motion along the slope. If a similar approach is followed here, then we need expressions for height and slope as functions of the distance along the sirface. These could be derived given that we have the eqautions. A simpler solution to solve for the horizontal motion and compute the z-motion consistent wth staying on the track. The z-motion is needed because drag and rider force depend on the total velocity not just the horizontal velocity.

In[284]:= We use the same constants above but we re-define the accerations and NDSolve here.

theta@xp_D := ArcTan@slope + as * Cos@2 * Pi * xp ê lambdaD * 2 * Pi ê lambda bs * Sin@2 * Pi * xp ê lambdaD * 2 * Pi ê lambdaD H* Get total velocity given running on track *L vtot@xd_, xp_D := Sqrt@xd^2 + Hxd * Tan@theta@xpDDL^2D; dacc@xd_, xp_D := -cd * rhoair * vtot@xd, xpD^2 * area ê H2 * massL; racc@th_D := - grav * cr * Cos@thD; facc@xd_, xp_D1-D accelerations set := Min@If@ vtot@xd, xpD > 0, prider ê vtot@xd, xpD, fmaxD, fmaxD ê mass gacc@th_D := - grav * Sin@thD; Print@"1-D accelerations set"D

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H* Set up differential equations to be be solved, x@tD is horizontal position, There is still a problem here with along track versus horizontal distance Here we use y for the dependent variable*L solnb = NDSolve@ 8y''@tD ã Hdacc@y'@tD, y@tDD + gacc@theta@y@tDDD + racc@theta@y@tDDD + facc@y'@tD, y@tDDL * Cos@theta@y@tDDD, y@0D ã 0, y'@0D ã 0.0<, y, 8t, 0.0, 10 000.0<, AccuracyGoal Ø 10D; py@t_D := Evaluate@y@tD ê. solnbD; vy@t_D := Evaluate@y'@tD ê. solnbD; H* Compute the error in the distance *L endb = t ê. FindRoot@py@tD - tracklen ã 0 , 8t, 1, 3000.0<D; work = NIntegrate@mass * facc@First@vy@tDD, First@py@tDDD * In[311]:= vtot@First@vy@tDD, First@py@tDDD, 8t, 0, endt<D; Print@"Time to reach end of track ", endb, " sec, Speed ", First@vy@endtDD, " mês"D; Print@"Work done ", work, " Joules, ", work ê 4.1868 ê 10^3, " kcal" D; Time to reach end of track 2255.05 sec, Speed 2.64848 mês done 174 405. Joules, 41.6558 kcal Plot@8vy@tD, Sqrt@xv@tD^2 + zv@tD^2D<, 8t, 0, endb<, PlotRange Ø All, AxesLabel Ø 8"Time HsecL", "Velocity HmêsL"<D

Work

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Velocity HmêsL 7 6 Out[311]=

45 3 2 1 500

1000

1500

2000

Diff Vel HmmêsL 6 4

Out[312]=

2

500 -2 -4

● Now output table of values

1000

1500

2000

Time

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In[308]:=

Out[310]//TableForm=

outl1d = Table@8t, py@tD, vy@tD<, 8t, 0, endb, outint<D; H* Now add the final entry to the tabe; *L outl1d = Append@outl1d, Transpose@Table@8t, py@tD, vy@tD<, 8t, endb, endb, outint<DDD; TableForm@outl1d, TableHeadings Ø 8None, 8"Time HsL", "X Pos HmL", "X Vel HmêsL"<<D Time HsL 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2255.05

X Pos HmL 0.

80.0296 306.907 827.102 1514.08 1951.82 2202.58 2608.89 3293.34 3835.08 4117.54 4428.5 5047.16 5682.12 6035.47 6294.5 6804.39 7494.16 7941.75 8193.74 8589.47 9270.08 9821.86 10 000.

X

Vel HmêsL 0.

1.48938 3.36857 6.79856 5.87554 2.95863 2.65548 5.83165 6.88692 3.8804 2.35409 4.537 7.25006 4.97145 2.48028 3.32362 6.7169 5.98318 3.03121 2.6065 5.72868 6.96082 3.98801 2.65031