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Learning Activity #1 (to be posted)
Please read the following fact pattern and address all three of
the questions. Only discuss relevant issues! I am not as
interested in your outcome (who will end up winning) as I am in
how you use the law to support your points. Base your answer
only on facts given. If additional information is needed, tell me
that. Your answers should be about 200 words
McBain Truck Company had an established rule that its truck
drivers were not allowed to carry passengers in the trucks.
Further, this rule was included in the contract between the union
and the company. Balsom, a driver for McBain, invited a friend
to ride in his truck on a day’s delivery trip. As a result of
Balsom’s carelessness during the trip, the truck was involved in
a collision and his friend was seriously injured.
a. Is the employer responsible to third parties for injuries
caused by an employee in the normal performance of
employment?
b. Would Balsom be held responsible for the injuries?
c. Is an employer liable to third parties for all actions of an
employee?
Learning Activity #2 (to be posted)
Instructions: Please choose ONE of the three questions to
answer. Do not give yourself extra work by answering all three
questions.
1) Helena, the purchasing agent for Cassadine Inc., contacts
five different suppliers to obtain price quotes for office supplies
she needs to order. The price for the identical office supplies
ranges from a high of $855 to a low of $500, including shipping
and handling charges. As purchasing agent, Helena has
complete autonomy on selecting suppliers and on placing orders
up to $5,000; before placing orders in excess of $5,000, she
must get signed authorization from the company’s comptroller.
A. If Helena purchases the supplies from the most expensive
supplier because she is angry at her employer for turning down
her recent request for a raise, what duty of an agent has she
breached? What are Cassadine’s remedies?
B. Helena purchases a $20,000 computer system from Alpha
Inc., a dealer she has never ordered from, but who knows
Helena is the purchasing agent for Cassadine. If Cassadine
refuses to accept the system when it arrives, what argument will
Alpha make for Cassadine’s liability? What will be Cassadine’s
defense?
2)Brandon is a traveling salesperson for Best Publishing Co.
His job is to drive around to bookstores and persuade them to
stock Best publications. One day, while on his way to his next
appointment in College Park, Brandon accidentally rear-ends
Emily.
A. Has Brandon committed a tort? What will Emily have to
prove to hold Best liable?
B. Assume Brandon uses his own car, sets his own hours,
decides without direction from Best what bookstores to visit,
and is paid according to how many books he places in
bookstores. What kind of agent is Brandon? How does this
affect Best’s liability?
C. Assume Brandon takes directions from Best on most aspects
of his job. While stopping to get gas, Brandon gets into a
fistfight with the station attendant, who makes a negative
comment about Brandon’s favorite baseball team. Brandon
injures the attendant. If the attendant sues Best, what will Best
argue to avoid liability?
3) A buyer, wishing to buy a Christmas gift for her husband,
visits a computer store and asks the salesperson to recommend a
laser printer. She tells the salesperson that the printer must print
at a minimum of 12 pages per minute at a minimum resolution
of 1200 dots per inch (dpi) and must have a duty cycle of at
least 10,000 copies per month. The salesperson recommends a
low-end printer on sale that week but fails to mention that it
prints at a speed of only 8 pages per minute and is rated for not
more than 5,000 copies per month.
A. Has the salesperson breached any express warranty?
B. Has the salesperson breached any implied warranty? Explain.
C. If the salesperson demonstrates a color printer that meets all
of the seller’s requirement but then sells her a cheaper
monochrome laser printer instead, has he breached any
warranty?
MATLAB sessions: Laboratory 4
MAT 275 Laboratory 4
MATLAB solvers for First-Order IVP
In this laboratory session we will learn how to
1. Use MATLAB solvers for solving scalar IVP
2. Use MATLAB solvers for solving higher order ODEs and
systems of ODES.
First-Order Scalar IVP
Consider the IVP {
y′ = t− y,
y(0) = 1.
(L4.1)
The exact solution is y(t) = t− 1+ 2e−t. A numerical solution
can be obtained using various MATLAB
solvers. The standard MATLAB ODE solver is ode45. The
function ode45 implements 4/5th order
Runge-Kutta method. Type help ode45 to learn more about it.
Basic ode45 Usage
The basic usage of ode45 requires a function (the right-hand
side of the ODE), a time interval on which
to solve the IVP, and an initial condition. For scalar first-order
ODEs the function may often be specified
using the inline MATLAB command. A complete MATLAB
solution would read:
1 f = inline(’t-y’,’t’,’y’);
2 [t,y] = ode45(f,[0,3],1);
3 plot(t,y)
(line numbers are not part of the commands!)
• Line 1 defines the function f as a function of t and y, i.e., f(t,
y) = t − y. This is the right-hand
side of the ODE (L4.1).
• Line 2 solves the IVP numerically using the ode45 solver. The
first argument is the function f,
the second one determines the time interval on which to solve
the IVP in the form
[initial time, final time], and the last one specifies the initial
value of y. The output of ode45
consists of two arrays: an array t of discrete times at which the
solution has been approximated,
and an array y with the corresponding values of y. These values
can be listed in the Command
Window as
[t,y]
ans =
0 1.0000
0.0502 0.9522
0.1005 0.9093
0.1507 0.8709
0.2010 0.8369
........
2.9010 2.0109
2.9257 2.0330
2.9505 2.0551
2.9752 2.0773
3.0000 2.0996
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 4
Since the output is quite long we printed only some selected
values.
For example the approximate solution at t ≃ 2.9257 is y ≃
2.0330. Unless specific values of y are
needed it is better in practice to simply plot the solution to get a
sense of the behavior of the
solution.
• Line 3 thus plots y as a function of t in a figure window. The
plot is shown in Figure L4a.
0 0.5 1 1.5 2 2.5 3
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
Figure L4a:
Solution
of (L4.1).
Error Plot, Improving the Accuracy
Error plots are commonly used to show the accuracy in the
numerical solution. Here the error is the
difference between the exact solution y(t) = t−1+2e−t and the
numerical approximation obtained from
ode45. Since this approximation is only given at specified time
values (contained in the array t) we only
evaluate this error at these values of t:
err = t-1+2*exp(-t)-y
err =
1.0e-005 *
0
0.0278
0.0407
0.0162
-0.0042
......
-0.0329
-0.0321
-0.0313
-0.0305
-0.0298
(in practice the exact solution is unknown and this error is
estimated, for example by comparing the
solutions obtained by different methods). Again, since the error
vector is quite long we printed only a
few selected values. Note the 1.0e-005 at the top of the error
column. This means that each component
of the vector err is less than 10−5 in absolute value.
A plot of err versus t is more revealing. To do this note that
errors are usually small so it is best
to use a logarithmic scale in the direction corresponding to err
in the plot. To avoid problems with
negative numbers we plot the absolute value of the error (values
equal to 0, e.g. at the initial time, are
not plotted):
semilogy(t,abs(err)); grid on;
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 4
0 0.5 1 1.5 2 2.5 3
10
−8
10
−7
10
−6
10
−5
Figure L4b: Error in the solution of (L4.1) computed by ode45.
See Figure L4b. Note that the error level is about 10−6. It is
sometimes important to reset the default
accuracy ode45 uses to determine the approximation. To do this
use the MATLAB odeset command
prior to calling ode45, and include the result in the list of
arguments of ode45:
f = inline(’t-y’,’t’,’y’);
options = odeset(’RelTol’,1e-10,’AbsTol’,1e-10);
[t,y] = ode45(f,[0,3],1,options);
err = t-1+2*exp(-t)-y;
semilogy(t,abs(err))
See Figure L4c.
0 0.5 1 1.5 2 2.5 3
10
−16
10
−15
10
−14
10
−13
10
−12
10
−11
10
−10
Figure L4c: Error in the solution of (L4.1) computed by ode45
with a better accuracy.
Parameter-Dependent ODE
When the function defining the ODE is complicated, rather than
entering it as inline, it is more
convenient to enter it as a separate function file, included in the
same file as the calling sequence of
ode45 (as below) or saved in a separate m-file.
In this Section we will look at an example where the ODE
depends on a parameter.
Consider the IVP {
y′ = −a (y − e−t)− e−t,
y(0) = 1.
(L4.2)
with exact solution y(t) = e−t (independent of the parameter a!).
An implementation of the MATLAB
solution in the interval [0, 3] follows.
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 4
1 function ex_with_param
2 t0 = 0; tf = 3; y0 = 1;
3 a = 1;
4 [t,y] = ode45(@f,[t0,tf],y0,[],a);
5 disp([’y(’ num2str(t(end)) ’) = ’ num2str(y(end))])
6 disp([’length of y = ’ num2str(length(y))])
7 %-------------------------------------------------
8 function dydt = f(t,y,a)
9 dydt = -a*(y-exp(-t))-exp(-t);
• Line 1 must start with function, since the file contains at least
two functions (a driver + a
function).
• Line 2 sets the initial data and the final time.
• Line 3 sets a particular value for the parameter a.
• In line 4 the parameter is passed to ode45 as the 5th argument
(the 4th argument is reserved for
setting options such as the accuracy using odeset, see page 3,
and the placeholder [] must be
used if default options are used).
Correspondingly the function f defined in lines 8-9 must include
a 3rd argument corresponding to
the value of the parameter. Note the @f in the argument of
ode45. This is necessary when the
function is not defined as inline. See the help on ode45 for more
information.
• On line 5 the value of y(3) computed by ode45 is then
displayed in a somewhat fancier form than
the one obtained by simply entering y(end). The command
num2string converts a number to a
string so that it can be displayed by the disp command.
The m-file ex with param.m is executed by entering ex with
param at the MATLAB prompt. The
output is
>> ex_with_param
y(3) = 0.049787
length of y = 45
• The additional line 6 in the file lists the length of the array y
computed by ode45. It is interesting
to check the size of y obtained for larger values of a. For
example for a = 1000 we obtain
>> ex_with_param
y(3) = 0.049792
length of y = 3621
This means that ode45 needed to take smaller step sizes to cover
the same time interval compared
to the case a = 1, even though the exact solution is the same!
Not all problems with a common solution are the same! Some
are easier to solve than others.
When a is large the ODE in (L4.2) is said to be stiff. Stiffness
has to do with how fast nearby
solutions approach the solution of (L4.2), see Figure L4d.
⋆ Other MATLAB ODE solvers are designed to better handle
stiff problems. For example replace ode45
with ode15s in ex with param.m (without changing anything
else) and set a = 1000:
4 [t,y] = ode15s(@f,[t0,tf],y0,[],a);
>> ex_with_param
y(3) = 0.049787
length of y = 18
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 4
Figure L4d: Direction field and sample solutions in the t-y
window [0, 0.1] × [0.9, 1] as obtained using
DFIELD8: a = 1 (left) and a = 1000 (right).
A solution exhibiting blow up in finite time
Consider the Differential Equation
dy
dt
= 1 + y2, y(0) = 0
The exact solution of this IVP is y = tan t and the domain of
validity is [0, π2 ). Let’s see what happens
when we try to implement this IVP using ode45 in the interval
[0, 3].
>> f=inline(’1+y^2’,’t’,’y’);
>> [t,y]=ode45(f,[0,3],0);
Warning: Failure at t=1.570781e+000. Unable to meet
integration
tolerances without reducing the step size below the smallest
value
allowed (3.552714e-015) at time t.
> In ode45 at 371
The MATLAB ode solver gives a warning message when the
value of t = 1.570781 is reached. This is
extremely close to the value of π/2 where the vertical asymptote
is located.
If we enter plot(t,y) we obtain Figure L4e on the left (note the
scale on the y-axis), however, if we use
xlim([0,1.5]), we can recognize the graph of y = tan t.
0 0.5 1 1.5 2
0
2
4
6
8
10
12
14
16
18
x 10
13
0 0.5 1 1.5
0
5
10
15
Figure L4e:

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Learning Activity #1 (to be posted)Please read the following fac.docx

  • 1. Learning Activity #1 (to be posted) Please read the following fact pattern and address all three of the questions. Only discuss relevant issues! I am not as interested in your outcome (who will end up winning) as I am in how you use the law to support your points. Base your answer only on facts given. If additional information is needed, tell me that. Your answers should be about 200 words McBain Truck Company had an established rule that its truck drivers were not allowed to carry passengers in the trucks. Further, this rule was included in the contract between the union and the company. Balsom, a driver for McBain, invited a friend to ride in his truck on a day’s delivery trip. As a result of Balsom’s carelessness during the trip, the truck was involved in a collision and his friend was seriously injured. a. Is the employer responsible to third parties for injuries caused by an employee in the normal performance of employment? b. Would Balsom be held responsible for the injuries? c. Is an employer liable to third parties for all actions of an employee? Learning Activity #2 (to be posted) Instructions: Please choose ONE of the three questions to answer. Do not give yourself extra work by answering all three questions. 1) Helena, the purchasing agent for Cassadine Inc., contacts five different suppliers to obtain price quotes for office supplies she needs to order. The price for the identical office supplies ranges from a high of $855 to a low of $500, including shipping and handling charges. As purchasing agent, Helena has complete autonomy on selecting suppliers and on placing orders up to $5,000; before placing orders in excess of $5,000, she must get signed authorization from the company’s comptroller.
  • 2. A. If Helena purchases the supplies from the most expensive supplier because she is angry at her employer for turning down her recent request for a raise, what duty of an agent has she breached? What are Cassadine’s remedies? B. Helena purchases a $20,000 computer system from Alpha Inc., a dealer she has never ordered from, but who knows Helena is the purchasing agent for Cassadine. If Cassadine refuses to accept the system when it arrives, what argument will Alpha make for Cassadine’s liability? What will be Cassadine’s defense? 2)Brandon is a traveling salesperson for Best Publishing Co. His job is to drive around to bookstores and persuade them to stock Best publications. One day, while on his way to his next appointment in College Park, Brandon accidentally rear-ends Emily. A. Has Brandon committed a tort? What will Emily have to prove to hold Best liable? B. Assume Brandon uses his own car, sets his own hours, decides without direction from Best what bookstores to visit, and is paid according to how many books he places in bookstores. What kind of agent is Brandon? How does this affect Best’s liability? C. Assume Brandon takes directions from Best on most aspects of his job. While stopping to get gas, Brandon gets into a fistfight with the station attendant, who makes a negative comment about Brandon’s favorite baseball team. Brandon injures the attendant. If the attendant sues Best, what will Best argue to avoid liability? 3) A buyer, wishing to buy a Christmas gift for her husband, visits a computer store and asks the salesperson to recommend a laser printer. She tells the salesperson that the printer must print at a minimum of 12 pages per minute at a minimum resolution of 1200 dots per inch (dpi) and must have a duty cycle of at least 10,000 copies per month. The salesperson recommends a
  • 3. low-end printer on sale that week but fails to mention that it prints at a speed of only 8 pages per minute and is rated for not more than 5,000 copies per month. A. Has the salesperson breached any express warranty? B. Has the salesperson breached any implied warranty? Explain. C. If the salesperson demonstrates a color printer that meets all of the seller’s requirement but then sells her a cheaper monochrome laser printer instead, has he breached any warranty? MATLAB sessions: Laboratory 4 MAT 275 Laboratory 4 MATLAB solvers for First-Order IVP In this laboratory session we will learn how to 1. Use MATLAB solvers for solving scalar IVP 2. Use MATLAB solvers for solving higher order ODEs and systems of ODES. First-Order Scalar IVP Consider the IVP { y′ = t− y, y(0) = 1. (L4.1) The exact solution is y(t) = t− 1+ 2e−t. A numerical solution can be obtained using various MATLAB
  • 4. solvers. The standard MATLAB ODE solver is ode45. The function ode45 implements 4/5th order Runge-Kutta method. Type help ode45 to learn more about it. Basic ode45 Usage The basic usage of ode45 requires a function (the right-hand side of the ODE), a time interval on which to solve the IVP, and an initial condition. For scalar first-order ODEs the function may often be specified using the inline MATLAB command. A complete MATLAB solution would read: 1 f = inline(’t-y’,’t’,’y’); 2 [t,y] = ode45(f,[0,3],1); 3 plot(t,y) (line numbers are not part of the commands!) • Line 1 defines the function f as a function of t and y, i.e., f(t, y) = t − y. This is the right-hand side of the ODE (L4.1). • Line 2 solves the IVP numerically using the ode45 solver. The first argument is the function f, the second one determines the time interval on which to solve the IVP in the form [initial time, final time], and the last one specifies the initial value of y. The output of ode45 consists of two arrays: an array t of discrete times at which the solution has been approximated, and an array y with the corresponding values of y. These values can be listed in the Command Window as
  • 5. [t,y] ans = 0 1.0000 0.0502 0.9522 0.1005 0.9093 0.1507 0.8709 0.2010 0.8369 ........ 2.9010 2.0109 2.9257 2.0330 2.9505 2.0551 2.9752 2.0773 3.0000 2.0996 c⃝2011 Stefania Tracogna, SoMSS, ASU MATLAB sessions: Laboratory 4 Since the output is quite long we printed only some selected values. For example the approximate solution at t ≃ 2.9257 is y ≃
  • 6. 2.0330. Unless specific values of y are needed it is better in practice to simply plot the solution to get a sense of the behavior of the solution. • Line 3 thus plots y as a function of t in a figure window. The plot is shown in Figure L4a. 0 0.5 1 1.5 2 2.5 3 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 Figure L4a: Solution of (L4.1).
  • 7. Error Plot, Improving the Accuracy Error plots are commonly used to show the accuracy in the numerical solution. Here the error is the difference between the exact solution y(t) = t−1+2e−t and the numerical approximation obtained from ode45. Since this approximation is only given at specified time values (contained in the array t) we only evaluate this error at these values of t: err = t-1+2*exp(-t)-y err = 1.0e-005 * 0 0.0278 0.0407 0.0162 -0.0042
  • 8. ...... -0.0329 -0.0321 -0.0313 -0.0305 -0.0298 (in practice the exact solution is unknown and this error is estimated, for example by comparing the solutions obtained by different methods). Again, since the error vector is quite long we printed only a few selected values. Note the 1.0e-005 at the top of the error column. This means that each component of the vector err is less than 10−5 in absolute value. A plot of err versus t is more revealing. To do this note that errors are usually small so it is best to use a logarithmic scale in the direction corresponding to err in the plot. To avoid problems with negative numbers we plot the absolute value of the error (values
  • 9. equal to 0, e.g. at the initial time, are not plotted): semilogy(t,abs(err)); grid on; c⃝2011 Stefania Tracogna, SoMSS, ASU MATLAB sessions: Laboratory 4 0 0.5 1 1.5 2 2.5 3 10 −8 10 −7 10 −6 10 −5
  • 10. Figure L4b: Error in the solution of (L4.1) computed by ode45. See Figure L4b. Note that the error level is about 10−6. It is sometimes important to reset the default accuracy ode45 uses to determine the approximation. To do this use the MATLAB odeset command prior to calling ode45, and include the result in the list of arguments of ode45: f = inline(’t-y’,’t’,’y’); options = odeset(’RelTol’,1e-10,’AbsTol’,1e-10); [t,y] = ode45(f,[0,3],1,options); err = t-1+2*exp(-t)-y; semilogy(t,abs(err)) See Figure L4c. 0 0.5 1 1.5 2 2.5 3 10 −16
  • 11. 10 −15 10 −14 10 −13 10 −12 10 −11 10 −10 Figure L4c: Error in the solution of (L4.1) computed by ode45 with a better accuracy. Parameter-Dependent ODE When the function defining the ODE is complicated, rather than
  • 12. entering it as inline, it is more convenient to enter it as a separate function file, included in the same file as the calling sequence of ode45 (as below) or saved in a separate m-file. In this Section we will look at an example where the ODE depends on a parameter. Consider the IVP { y′ = −a (y − e−t)− e−t, y(0) = 1. (L4.2) with exact solution y(t) = e−t (independent of the parameter a!). An implementation of the MATLAB solution in the interval [0, 3] follows. c⃝2011 Stefania Tracogna, SoMSS, ASU MATLAB sessions: Laboratory 4 1 function ex_with_param
  • 13. 2 t0 = 0; tf = 3; y0 = 1; 3 a = 1; 4 [t,y] = ode45(@f,[t0,tf],y0,[],a); 5 disp([’y(’ num2str(t(end)) ’) = ’ num2str(y(end))]) 6 disp([’length of y = ’ num2str(length(y))]) 7 %------------------------------------------------- 8 function dydt = f(t,y,a) 9 dydt = -a*(y-exp(-t))-exp(-t); • Line 1 must start with function, since the file contains at least two functions (a driver + a function). • Line 2 sets the initial data and the final time. • Line 3 sets a particular value for the parameter a. • In line 4 the parameter is passed to ode45 as the 5th argument
  • 14. (the 4th argument is reserved for setting options such as the accuracy using odeset, see page 3, and the placeholder [] must be used if default options are used). Correspondingly the function f defined in lines 8-9 must include a 3rd argument corresponding to the value of the parameter. Note the @f in the argument of ode45. This is necessary when the function is not defined as inline. See the help on ode45 for more information. • On line 5 the value of y(3) computed by ode45 is then displayed in a somewhat fancier form than the one obtained by simply entering y(end). The command num2string converts a number to a string so that it can be displayed by the disp command. The m-file ex with param.m is executed by entering ex with param at the MATLAB prompt. The output is >> ex_with_param y(3) = 0.049787 length of y = 45
  • 15. • The additional line 6 in the file lists the length of the array y computed by ode45. It is interesting to check the size of y obtained for larger values of a. For example for a = 1000 we obtain >> ex_with_param y(3) = 0.049792 length of y = 3621 This means that ode45 needed to take smaller step sizes to cover the same time interval compared to the case a = 1, even though the exact solution is the same! Not all problems with a common solution are the same! Some are easier to solve than others. When a is large the ODE in (L4.2) is said to be stiff. Stiffness has to do with how fast nearby solutions approach the solution of (L4.2), see Figure L4d. ⋆ Other MATLAB ODE solvers are designed to better handle stiff problems. For example replace ode45
  • 16. with ode15s in ex with param.m (without changing anything else) and set a = 1000: 4 [t,y] = ode15s(@f,[t0,tf],y0,[],a); >> ex_with_param y(3) = 0.049787 length of y = 18 c⃝2011 Stefania Tracogna, SoMSS, ASU MATLAB sessions: Laboratory 4 Figure L4d: Direction field and sample solutions in the t-y window [0, 0.1] × [0.9, 1] as obtained using DFIELD8: a = 1 (left) and a = 1000 (right). A solution exhibiting blow up in finite time Consider the Differential Equation dy
  • 17. dt = 1 + y2, y(0) = 0 The exact solution of this IVP is y = tan t and the domain of validity is [0, π2 ). Let’s see what happens when we try to implement this IVP using ode45 in the interval [0, 3]. >> f=inline(’1+y^2’,’t’,’y’); >> [t,y]=ode45(f,[0,3],0); Warning: Failure at t=1.570781e+000. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-015) at time t. > In ode45 at 371 The MATLAB ode solver gives a warning message when the value of t = 1.570781 is reached. This is
  • 18. extremely close to the value of π/2 where the vertical asymptote is located. If we enter plot(t,y) we obtain Figure L4e on the left (note the scale on the y-axis), however, if we use xlim([0,1.5]), we can recognize the graph of y = tan t. 0 0.5 1 1.5 2 0 2 4 6 8 10 12 14 16
  • 19. 18 x 10 13 0 0.5 1 1.5 0 5 10 15 Figure L4e: