UBC Wiki - User contributions [en]

Science:Math Exam Resources/Courses/MATH104/December 2013/Question 01 (l)/Solution 1

2016-12-15T05:51:33Z

Konradbe: Adds an extended explanation.

None of these have to be true! The fact that the limit exists does not imply anything about the value at that point. In fact, <math>f(x)</math> need not even be defined at <math>a</math>. However, if <math>f(x)</math> is not defined at <math>a</math> (C), then writing <math>f(a)</math> does not make sense because this value does not exist (D). <math>f(x)</math> can not be continuous at a point where it's not defined (B), and a function can only be differentiable at those points where it is defined and continuous (A). Therefore, the following is a counter example

:[[File:MATH104 Dec2013 Q01l.jpg|400px]]

Hence, the correct answer is (E).

Science:Math Exam Resources/Courses/MATH104/December 2015/Question 08 (a)/Solution 1

2016-12-14T08:13:01Z

Konradbe: Adds alternative formulation as suggested by type report form.

The two key pieces of information required for this question are:
* "when the unit ''price'' is <math>${\color{OrangeRed} 16}</math> per toaster, then the weekly ''demand'' is <math>{\color{OliveGreen} 20}</math> toasters"
* "for every <math>${\color{WildStrawberry} 2}</math> ''decrease'' in the unit ''price'', the weekly ''demand increases'' by <math>{\color{PineGreen} 10}</math> toasters"
from which it is clear that the unit price <math>p</math> and the weekly demand <math>q</math> are linearly related.

Using the 'slope-point' form of the equation of a line and considering <math>q</math> as a function of <math>p</math>, we can write
:<math>
q - q_0 = m(p - p_0),
</math>
where <math>(p_0, q_0)</math> is a 'coordinate' consisting of a price and the demand ''at that price'', and <math>m</math> is the slope of the line in the <math>pq</math>-plane.

Hence we can take <math>(p_0, q_0) = ({\color{OrangeRed} 16}, {\color{OliveGreen} 20})</math> and <math>m = \frac{{\color{PineGreen} 10}}{-{\color{WildStrawberry} 2}}</math> (when <math>p</math> '''''decreases''''' by 2, <math>q</math> ''increases'' by 10). Thus
:<math>
\begin{align}
q - {\color{OliveGreen} 20} &= \frac{{\color{PineGreen} 10}}{-{\color{WildStrawberry} 2}} (p - {\color{OrangeRed} 16}) \\
q &= {\color{blue} -5(p - 16) + 20} \,.
\end{align}
</math>

Note: Considering <math>p</math> as a function of <math>q</math> is equally correct and leads to the equivalent answer <math>p = -\frac{q}5 + 20</math>.

Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (o)/Solution 1

2016-12-14T07:35:17Z

Konradbe: Fixes typo as reported in typo report form.

Define the function <math>f(x)</math> as
:<math>
f(x) = 5^{x} - 10x - 7.
</math>
Notice that this function is continuous for all values of <math>x</math>. Hence, we need only find a closed interval where the sign of <math>f(x)</math> changes and invoke the intermediate value theorem to get the desired conclusion.

Consider <math>x = 2</math> and <math>x = 3</math>. If we take these two points and plug them into <math>f(x)</math>, we get
:<math>
\begin{align}
f(2) &= 5^{2} - 10(2) - 7 = -2 < 0\\
f(3) &= 5^{3} - 10(3) - 7 = 88 > 0
\end{align}
</math>
and so the sign of <math>f(x)</math> changes over the closed interval <math>[2,3]</math>.

 Therefore, there exists a value <math>c</math> in <math>[2,3]</math> such that 
:<math>
{\color{blue} f(c) = 5^{c} - 10c - 7 = 0}
</math>
by the Intermediate Value Theorem.

Science:Math Exam Resources/Courses/MATH100/December 2013/Question 05/Solution 1

2016-12-14T07:25:03Z

Konradbe: Fixes typo as reported in typo report form.

The solution of the differential equation <math>\frac{dT}{dt} = k(T-T_s)</math> is <math>\displaystyle T - T_s = (T_0 - T_s)e^{kt}</math>, where <math>T</math> is the temperature of an object, <math>T_s</math> is the temperature of its surroundings, and <math>k</math> is some constant.

 We are given that <math>T_s = 40,\, T_0 = 70,\, T(0.5) = 60</math>, where <math>t</math> is in hours. Thus we begin by solving for <math>k</math>:

<math>
\begin{align}
T(t) - T_s &= (T_0 - T_s)e^{kt}
\\ 60 - 40 &= (70 - 40)e^{k\cdot0.5}
\\ e^{0.5k} &= \frac{2}{3}
\\ \frac{1}{2}k &= \ln\frac{2}{3}
\\ k &= \ln\frac{4}{3}
\\
\\ \therefore T(t) &= 30e^{\ln\left(\frac{4}{3}\right)\cdot t} + 40
\\ &= 30\left(\frac{4}{3}\right)^t + 40
\end{align}
</math>

 Finally, solving for <math>t</math> when <math>\displaystyle T(t) = 50</math> yields:

<math>
\begin{align}
50 &= 30\left(\frac{4}{3}\right)^t + 40
\\ \frac{1}{3} &= \left(\frac{4}{3}\right)^t
\\ \ln\frac{1}{3} &= t\ln\frac{4}{3}
\\ t &= \frac{\ln\frac{1}{3}}{\ln\frac{4}{3}} = {\color{blue} \frac{\ln3}{\ln\frac{3}{4}} \,\,\mathrm{hours}}
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH104/December 2013/Question 05/Solution 1

2016-12-11T21:53:06Z

Konradbe: Cla

To use the elasticity of demand, we evaluate <math>\epsilon</math> at <math>p = 4</math>. If <math>\epsilon < -1</math>, then demand is elastic and the price should be lowered to increase revenue. Otherwise, if <math>\epsilon > -1</math>, then demand is inelastic and the price should be raised.

A quick calculation using the given price-demand relationship confirms that if <math>q = 1800</math>, then <math>p = 4</math>.

To evaluate <math>\frac{dq}{dp}</math>, we opt to use the chain rule on the price-demand relationship. Differentiating it with respect to <math>p</math>, gives

:<math>\begin{align}
\frac{d}{dp}p &= \frac{d}{dp}\left(\frac{q-3000}{600}\right)^2 \\
1 &= 2\left(\frac{q - 3000}{600}\right)\frac{1}{600}\frac{dq}{dp} \\
1 &= \left(\frac{q - 3000}{600}\right)\frac{1}{300}\frac{dq}{dp}
\end{align}</math>

Solving for <math>\frac{dq}{dp}</math> we obtain
:<math>
\frac{dq}{dp} = 300\left(\frac{600}{q - 3000}\right)
</math>
Putting together all the pieces to evaluate <math>\epsilon</math> at <math>p = 4</math>, we get
:<math>
\begin{align}
\epsilon &= \frac{p}{q}\frac{dq}{dp} \\
&= \frac{1}{q} \left(\frac{q-3000}{600}\right)^2 300\left(\frac{600}{q - 3000}\right) \\
&= \frac{300}{q}\left(\frac{q - 3000}{600}\right)
\end{align}
</math>

Hence, when <math>q = 1800</math>, then <math>\epsilon = -6/18</math>.

Thus the price should be raised in order to increase revenue.

Note:
If we weren't told that the current number of passengers is 1800, we would have to consider the alternate solution <math>p = 4, q = 4200</math>. This gives us a positive price elasticity of <math>3/21</math>, which again tells us that increasing the price will increase revenue.

Note that while it may happen that a good has positive elasticity, and thus it does not follow the law of demand (e.g. Do expensive handbags become more or less desirable to wealthy shoppers if the price drops?), this is unlikely to be the case in our scenario. Clearly the demand equation we are using cannot hold for all <math>q</math> and <math>p</math> (otherwise it would suggest that the price should be increased to infinity!), reminding us that mathematical models are only valid within certain bounds.

Science:Math Exam Resources/Courses/MATH104/December 2014/Question 01 (h)/Solution 1

2016-12-10T08:10:01Z

Konradbe: Clarified solution as it was pointed out in the typo report form.

By the chain rule we get that <math>h'(x) = f'(g(x))g'(x)</math>, in particular, <math>h'(4) = f'(g(4))g'(4)</math>.

From the fact that the point <math>(4,7)</math> is on the graph <math>g</math> we know that <math>g(4) = 7</math>.
For a differentiable function the slope of the tangent line at a point equals the derivative of the function at that point. The slope of the tangent line of <math>g(x)</math> at <math>x=4</math> is <math>(3x-5)' = 3</math>, hence <math>g'(4) = 3</math>. Similarly, <math>f'(7) = (-2x+23)' = -2</math>.

So, evaluating <math>h'</math> at <math>x = 4</math> gives:

<math>
\begin{align}
h'(4) &= f'(g(4))g'(4) \\ &= f'(7)g'(4) \\ &= (-2)(3) \\ &= -6
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH100/December 2011/Question 02 (a)/Solution 1

2016-12-10T07:06:35Z

Konradbe: Fixes typo as reported in typo report form.

Newton's Law of Cooling which states that
:<math>\displaystyle
T(t)=T_a+(T_0-T_a)e^{-kt}
</math>.
Here, the initial temperature of the body ''T''0 is 33°C; the temperature of the environment ''T''a is given to be 21°C; and the temperature of the body after 1 hour (''T''(1)) is 31°C.
Plugging in these numbers give us
:<math>\displaystyle
31=21+(33-21)e^{-k}
</math>
Using this equation, we can solve for ''k'':
:<math>
\begin{align}
e^{-k} & =\frac{31-21}{33-21}=\frac{5}{6} \\
-k & =\ln(5/6) \\
k & =-\ln(5/6) \\
\end{align}
</math>

And so now we have found the equation that gives us the temperature of the body at any time ''t'':
:<math>\displaystyle
T(t) = 21 + 12e^{t\ln(5/6)}
</math>
and we would like to know the time of death, which is the time at which the body temperature was 37°C. For this, we simply solve
:<math>\displaystyle
T(t) = 37
</math>
that is
:<math>\begin{align}
21 + 12e^{t\ln(5/6)} &= 37 \\
12e^{t\ln(5/6)} &= 16 \\
e^{t\ln(5/6)} &= 16/12 \\
t\ln(5/6) &= \ln(4/3) \\
t &= \frac{\ln(4/3)}{\ln(5/6)} \\
t &\approx -1.578
\end{align}</math>

Therefore, the police arrived <math>\ln(4/3)/\ln(5/6)</math> hours after the murder, that is, just over an hour and a half.

Science:Math Exam Resources/Courses/MATH104/December 2015

2016-11-27T05:23:02Z

Konradbe: Removes IP flag

{{MER Exam page|creation_box=true}}
[[Category:MER Exam QG flag]]
[[Category:MER Exam Page]]

Science:MER/Committee

2016-09-05T21:28:53Z

Konradbe:

'''Contact email''': [[File:Mer-wiki-email.png]]

'''Contributors''' (After September 2015):
* [[User:NicholasHu|Nicholas Hu]]
* [[User:WilsonHsu|Wilson Hsu]]

Science:MER/Credit

2016-09-05T21:26:29Z

Konradbe: /* History of the MER wiki */

==Citing the MER wiki in a CV or Resume==

[[User:ChristinaKoch|Christina]] says: I list my work on the wiki under either "Relevant Experience" or "Other Educational Experience" (depending on the context), with a little blurb that reads something like:
* Mathematics Wiki Manager
** Co-managing an online, open educational resource wiki aimed at first-year students in math courses at the University of British Columbia.
** Developing resource architecture, appearance, and documentation to improve usability.
** Designing surveys and coordinating a study to measure resource effectiveness.
** Successfully authored a grant application for evaluation and development funding.

[[User:CarmenBruni|Carmen]] says: Though I've never used my CV - I have listed this under professional service. My CV is available on my webpage for those interested to see it.

[[User:Konradbe|Bernhard]] says: I list MER under "Side Projects" and mention the skills MediaWiki, JavaScript, CSS, and my tasks as Founding member, Head of organizing committee, Project manager, Web developer. My short summary is
** Reaches >80% of the several thousand first year UBC students that take math courses.
** Over a million pageviews and students cummulatively spent 4.5 years on the resource in the first 2 years after inception.
** Crashed the UBC server during the April 2014 exam period because too many users tried to access the MER webpage.

Other ideas for citation may be the number of solutions you've written (if you know!) or other ways you've contributed to the wiki. Please add ideas/what you currently use!

==History of the MER wiki==


{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Jan - Aug 2015 (Winter Term 2, 2014-2015 school year, summer 2015)
|-
|style="padding:1em"|
* Committee:
** Committee Heads: [[User:CarmenBruni|Carmen Bruni]] and [[User:Konradbe|Bernhard Konrad]]
** Data Analysis: [[User:WilliamThompson|Will Thompson]] and [[User:Konradbe|Bernhard Konrad]]
** Research Advisor: [[User:MikeL|Michael Lindstrom]]
** Android Developer: [[User:Konradbe|Bernhard Konrad]]
** Technical Advisor: [[User:IainMoyles|Iain Moyles]]
** Other Contributors: [[User:VincentChan|Vincent Chan]], [[User:IsabellGraf|Isabell Graf]], [[User:NicholasHu|Nicholas Hu]], [[User:Hambrook|Kyle Hambrook]]

* Wrote up FL-TLEF final report.
* Published paper on history and purpose of MER in PRIMUS: http://www.tandfonline.com/doi/full/10.1080/10511970.2015.1127301
* All major contributors finished up their degree and left UBC.
* We have completed the following exams

{{MER Completed exams in timespan|start=20150101|end=20150831}}
|}


{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Sep - Dec 2014 (Winter Term 1, 2014-2015 school year)
|-
|style="padding:1em"|
* Committee
** Heads: [[User:CarmenBruni|Carmen Bruni]] and [[User:Konradbe|Bernhard Konrad]]
** Data Analysis: [[User:WilliamThompson|Will Thompson]] and [[User:Konradbe|Bernhard Konrad]]
** Research Advisor: [[User:MikeL|Michael Lindstrom]]
** Lead FL-TLEF study: [[User:JiaoJi|Jiao Ji]]
** Android Developer: [[User:Konradbe|Bernhard Konrad]]
** Technical Advisor: [[User:IainMoyles|Iain Moyles]]
** Senior Advisory Member: [[User:ChristinaKoch|Christina Koch]]
** Other Contributors: [[User:VincentChan|Vincent Chan]], [[User:IsabellGraf|Isabell Graf]], [[User:NicholasHu|Nicholas Hu]], [[User:Hambrook|Kyle Hambrook]].

* On 2014-09-25 we put a more modern design on our [[Science:Math_Exam_Resources|front page]]. Mainly developed by [[User:ChristinaKoch|Christina Koch]]. Later [[User:WilliamThompson|Will Thompson]] gave the layout further refining touches.
* At the [http://cms.math.ca/Events/winter14/ 2014 Canadian Mathematical Society meeting] [[User:CarmenBruni|Carmen Bruni]] and [[User:Konradbe|Bernhard Konrad]] presented a poster about MER. [[User:CarmenBruni|Carmen Bruni]] also [http://cms.math.ca/Events/winter14/abs/ume#cb2 gave a talk] in the session ''Undergraduate Mathematics Education in 21st Century: Rethinking Curriculum'' and [https://twitter.com/BernhardKonrad/status/542944369695657985 looked sharp!]
* [[User:Konradbe|Bernhard Konrad]] wrote [https://github.com/MER-wiki developed a Python Script] to extract the MediaWiki source code for all question content. The content is compiled to LaTeX using [https://pypi.python.org/pypi/pypandoc/ Pypandoc].
* [[User:IsabellGraf|Isabell Graf]] refined the Python Script and developed an algorithm to extract only the final answer from the full solution. This feature was requested by several of the FL-TLEF interviewees. The resulting pdfs are available [http://mer-wiki.github.io/pdf_version/ for download here], and were downloaded more than 8500 times during the exam period!
* Several waves of email to undergraduates were sent to advertise for MER and to recruit students to take our online survey and be interviewed. Thanks to [[User:JiaoJi|Jiao Ji]], Warren Code and the Math Secretaries.
* As part of our FL-TLEF study [[User:JiaoJi|Jiao Ji]] interviewed 15 students, with help from [[User:Konradbe|Bernhard Konrad]], [[User:CarmenBruni|Carmen Bruni]], [[User:MikeL|Michael Lindstrom]] and [[User:IainMoyles|Iain Moyles]]. [[User:CarmenBruni|Carmen Bruni]] and [[User:Konradbe|Bernhard Konrad]] also helped to transcribe the recorded audio.
* Building on previous work, [[User:Konradbe|Bernhard Konrad]] started to organize the tags into a natural hierarchy.
* Sadly, [[User:ChristinaKoch|Christina Koch]] is leaving the MER committee for her new job in the US. Good luck Christina, and thank you for your tremendous work and commitment to pushing this project forward!
* The following '''5''' exams were completed:
** MATH100 - December 2013
** MATH102 - December 2013
** MATH104 - December 2013
** MATH110 - December 2013
** MATH307 - December 2010
|}

{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Summer 2014
|-
|style="padding:1em"|
* Committee:
** Heads: [[User:CarmenBruni|Carmen Bruni]] and [[User:Konradbe|Bernhard Konrad]]
** Data Analysis: [[User:WilliamThompson|Will Thompson]] and [[User:Konradbe|Bernhard Konrad]]
** Research Advisor: [[User:MikeL|Michael Lindstrom]]
** Android Developer: [[User:Konradbe|Bernhard Konrad]]
** Technical Advisor: [[User:IainMoyles|Iain Moyles]]
** Senior Advisory Member: [[User:ChristinaKoch|Christina Koch]]
**Other Contributors: [[User:VincentChan|Vincent Chan]], [[User:IsabellGraf|Isabell Graf]]

* The MER wiki is mentioned as a prime example of user-driven innovation in higher education. Read the full article [http://www.educause.edu/visuals/shared/er/extras/2014/ReclaimingInnovation/default.html#extra here].
* We presented a poster at the [http://events.ctlt.ubc.ca/events/view/3405 Flexible Learning Open House at UBC] on June 10th: [[User:ChristinaKoch|Christina Koch]], [[User:Konradbe|Bernhard Konrad]], [[User:IainMoyles|Iain Moyles]], [[User:MikeL|Michael Lindstrom]].
* We successfully interview several candidates for the MER FL-TLEF RA research position and decide to hire [[User:JiaoJi|Jiao Ji]]! Congratulations and let the studies begin! We are very proud of our first ''employee''.
* Our Ethics application is approved, we can start to roll out our research study in the coming term. Thank you to [[User:JiaoJi|Jiao Ji]], [[User:MikeL|Michael Lindstrom]], [[User:Konradbe|Bernhard Konrad]], [[User:CarmenBruni|Carmen Bruni]], [[User:IainMoyles|Iain Moyles]], [[User:ChristinaKoch|Christina Koch]] and [[User:WilliamThompson|Will Thompson]].
* No more guessing on which topics you need to master: Course pages [https://twitter.com/MERWiki/status/487392947113979904 now] display all the topics of the questions in that course. This idea was brought up by [[User:CarmenBruni|Carmen Bruni]] and implemented using dpl and JavaScript by [[User:Konradbe|Bernhard Konrad]].
* The new version of the [https://play.google.com/store/apps/details?id=mer.study.helper Android App] automatically recognizes newly solved questions and exams by connecting with an online SQL database, implemented by [[User:Konradbe|Bernhard Konrad]] and [[User:IsabellGraf|Isabell Graf]].
* [[User:WilliamThompson|Will Thompson]] made the [http://mer-wiki.github.io/MER_satisfaction_survey/MER_Satisfaction_Survey_2014.html results] of our April 2014 Satisfaction Evaluation available.
* [[User:Konradbe|Bernhard Konrad]] and [[User:WilliamThompson|Will Thompson]] made the data analysis of our google analytic click statistics available [[Science:Math_Exam_Resources/Usage|on our Usage page]].

* The following '''5''' exams were completed
** MATH312 - December 2009 & December 2012
** MATH307 - December 2008
** MATH200 - December 2011 & April 2012
|}

{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Jan - Apr 2014 (Winter Term 2, 2013-2014 school year)
|-
|style="padding:1em"|
* Committee
** Heads: [[User:CarmenBruni|Carmen Bruni]] and [[User:Konradbe|Bernhard Konrad]]
** Webmaster: [[User:Konradbe|Bernhard Konrad]]
** Designer: [[User:WilliamThompson|Will Thompson]]
** Workshop instructor: [[User:DavidKohler|David Kohler]]
** Android Developers: [[User:Konradbe|Bernhard Konrad]] and [[User:MikeL|Michael Lindstrom]]
** Technical Advisor: [[User:IainMoyles|Iain Moyles]]
** Senior Advisory Member: [[User:ChristinaKoch|Christina Koch]]
** Other Contributors: [[User:VincentChan|Vincent Chan]], [[User:PamS|Pam Sargent]], [[User:KaterynaMelnykova|Kateryna Melnykova]], [[User:CindyBlois|Cindy Blois]], [[User:KristinaB|Kristina]], [[User:IsabellGraf|Isabell Graf]]

* We met with the AVP of the AMS to discuss common interests of this wiki with the [http://www.ams.ubc.ca/2013/03/vp-academic-office-succeeds-in-getting-exam-database-passed-by-ubc-senate/ AMS Exam Database]. Thanks to [[User:ChristinaKoch|Christina Koch]] and [[User:Konradbe|Bernhard Konrad]].
* To help students prepare for their midterm exams, the syllabus for Math101, Math103, Math105, and Math110 were added. Each syllabus includes links to the category pages of the given topics. Thanks to [[User:CarmenBruni|Carmen Bruni]], [[User:konradbe|Bernhard Konrad]], [[User:IainMoyles|Iain Moyles]], [[User:ChristinaKoch|Christina Koch]] and [[User:PamS|Pam Sargent]].
* Apparently, the first midterm was on [https://twitter.com/_christinaLK/status/444327028414619648 Riemann sums and the Fundamental theorem of Calculus]. Thank you [[User:ChristinaKoch|Christina Koch]] for this screenshot on Feb 6th 2014.
* To prettify the list of questions on category pages, [[User:Konradbe|Bernhard Konrad]] and [[User:WilliamThompson|Will Thompson]] created a new [[Template:MER_list_questions_in_category|template]].
* [[User:WilliamThompson|Will Thompson]] and David M. Thompson made a great banner to celebrate our '''2 year anniversary''' on Feb 20th 2014 [[File:DavidWebster2ndAnniv.png||Celebrating two years of MER.|thumb]]
* Using the [http://www.mediawiki.org/wiki/Extension:Variables Variables extension], [[User:Konradbe|Bernhard Konrad]] refined some of the architecture so that general information about courses, exams, and layout, can be stored in one place only. With [[User:CarmenBruni|Carmen Bruni]] this was used to great effect on the course pages, with [[User:WilliamThompson|Will Thompson]] this was exploited to make changing the colour scheme easier and more convenient.
* An additional feedback option using [https://docs.google.com/forms/d/1n6Pf-fegzLB4n1QYTHmxvIcIWr4swidLaTm1kekZk38/viewform google forms] was included by [[User:ChristinaKoch|Christina Koch]] to inspire more feedback. [[User:WilliamThompson|Will Thompson]] created a suitable home for the form at the bottom of each solution box.
* [[User:IainMoyles|Iain Moyles]] and [[User:konradbe|Bernhard Konrad]] built on that idea and added a [https://docs.google.com/forms/d/1Y08giPVAqZXyQFthQ-76Ocs6FQ9CNLQP9yodDtNqNbI/viewform google form] to allow anyone to suggest a new or alternative solution.
* [[User:WilliamThompson|Will Thompson]] moved the rating bar out of the navigation bar to a better home at the bottom of the question page.
* After a redesign of the navigation bar the question pages have a more modern and slick look. They also display the Easiness of each question. Thank you [[User:Konradbe|Bernhard Konrad]]. [[User:WilliamThompson|Will Thompson]] further refined the design.
* Apparently, [https://twitter.com/_christinaLK/status/444332217360072704 voting] on the easiness of a question is very popular among Math 103 students. [[Science:Math_Exam_Resources/Courses/MATH103/April_2013/Question_02_(d)|This Math 103 question]] is the first question to receive more than 10 easiness votes.
* On the question pages, student looking for additional help are now reminded of the [http://www.math.ubc.ca/Ugrad/ugradTutorials.shtml Math Learning Centre]. We also offer help [https://docs.google.com/forms/d/1nV32JI4R3Mp96lYtLIpNEE-LB_T-fEHz-8lfAZXd430/viewform finding a private tutor]. Thanks to [[User:Konradbe|Bernhard Konrad]] and [[User:IainMoyles|Iain Moyles]].
* A slick Google analytics script records the time that students spend on a question page ''before'' opening the solution box. We hope to use this insight to better understand how our resource is used. Thanks to [[User:IainMoyles|Iain Moyles]] and [[User:Konradbe|Bernhard Konrad]] for this idea and its technical implementation.
* Our '''FL-TLEF grant application''' is successful! With these funds from [http://flexible.learning.ubc.ca/ flexible learning] we will hire a RA to study the effectiveness of our resource!
* No joke: On April 1st 2014 the MER wiki team solved its [https://twitter.com/MERWiki/statuses/451098004754149378 1000th good quality solution]!
* We are now on the google play store! Download the MER study helper app [https://play.google.com/store/apps/details?id=mer.study.helper here]. Programmed by [[User:Konradbe|Bernhard Konrad]] with help from [[User:MikeL|Michael Lindstrom]] and [[User:WilliamThompson|Will Thompson]].
* The MER team prepared and presented a poster at the [http://www.cwsei.ubc.ca/ Carl Wiemann Science Education Initiative] End of Year event. Thank you to [[User:CarmenBruni|Carmen Bruni]], [[User:ChristinaKoch|Christina Koch]], [[User:Konradbe|Bernhard Konrad]], [[User:MikeL|Michael Lindstrom]], [[User:WilliamThompson|Will Thompson]], and [[User:IainMoyles|Iain Moyles]].
* In four workshops with a total of '''32''' participants [[User:DavidKohler|David Kohler]] gives valuable general study tips. Thank you also to the helpers [[User:Konradbe|Bernhard Konrad]], [[User:WilliamThompson|Will Thompson]], and [[User:IainMoyles|Iain Moyles]].
* On Wednesday 24th 2014, the day before the Math 103, 105, and 110 exams, MER creates so much traffic that the UBC wiki server crashes. This is the first time since the inception of the UBC wiki that it crashed due to overuse. After an hour of hectic email exchanges with system admins, as well as forum posts, emails, and phone calls from panicking students, the service is finally restored when UBC adds additional cores to its wiki servers.
* Students used the rating bar feature almost '''3000''' times to rate the easiness of previous exam questions.
* During the exam period, our team answered '''75''' typo or question reports on existing solutions. Several typos were removed and solutions were clarified. Thank you to [[User:Konradbe|Bernhard Konrad]], [[User:IainMoyles|Iain Moyles]], [[User:CarmenBruni|Carmen Bruni]], and [[User:ChristinaKoch|Christina Koch]] for responding so quickly.
* The '''private tutor matching''' brought together several students and private tutors (not necessarily MER contributors). This new feature was brought up and realized by [[User:Konradbe|Bernhard Konrad]] and [[User:IainMoyles|Iain Moyles]].
* In the [[Science:Math_Exam_Resources/Contest|'''Solution writing contest''']] students were invited to contribute their own solutions to the wiki. We received '''22''' good submissions which were reviewed and are now fully integrated in the MER. Three lucky winners were drawn at random to receive $20 UBC food gift cards. Sponsored and initiated by [[User:IainMoyles|Iain Moyles]].
<center>
<gallery heights=75px widths=75px mode="nolines">
Image:MERContest Winter2014 ALim.jpg|Anne Lim (Right)
Image:MERContest Winter2014 NHuynh.jpg|Nghi Huynh (Middle)
Image:MERContest Winter2014 TLe.jpg|Toan Le (Right)
</gallery>
</center>
* The following '''7''' exams were completed
** MATH101 - April 2013
** MATH103 - April 2013
** MATH105 - April 2013
** MATH110 - April 2013
** MATH152 - April 2012 & April 2013
** MATH215 - December 2013
|}

{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Sep - Dec 2013 (Winter Term 1, 2013-2014 school year)
|-
|style="padding:1em"|
* Committee:
** Head: [[User:ChristinaKoch|Christina Koch]] and [[User:konradbe|Bernhard Konrad]]
** Math Grad Committee Rep: [[User:IainMoyles|Iain Moyles]]
** Webmaster: [[User:konradbe|Bernhard Konrad]]
** Senior Advisory Member: [[User:CarmenBruni|Carmen Bruni]]
** Other Contributors: [[User:KaterynaMelnykova|Kateryna Melnykova]], [[User:CindyBlois|Cindy Blois]], [[User:MikeL|Michael Lindstrom]], [[User:JeromeLefebvre|Jerome Lebefvre]], [[User:WilliamThompson|Will Thompson]], [[User:ThomasWong|Thomas Wong]], [[User:VincentChan|Vincent Chan]], [[User:Mstykow|Maxim Stykow]], [[User:Hambrook|Kyle Hambrook]], Edward Kroc
* Committee wrote and submitted a grant proposal for the [http://tlef.ubc.ca/ Teaching and Learning Extension Fund]. Thanks to [[User:ChristinaKoch|Christina Koch]], [[User:konradbe|Bernhard Konrad]], [[User:IainMoyles|Iain Moyles]], [[User:CarmenBruni|Carmen Bruni]], [[User:Hambrook|Kyle Hambrook]], Edward Kroc.
* Committee wrote and submitted an application for the [[Library:Scholarly_Communications/Innovative_Dissemination_of_Research_Award | Innovative Dissemination of Research Award]]. Thanks to [[User:ChristinaKoch|Christina Koch]], [[User:konradbe|Bernhard Konrad]], [[User:IainMoyles|Iain Moyles]], [[User:CarmenBruni|Carmen Bruni]], [[User:VincentChan|Vincent Chan]], [[User:Hambrook|Kyle Hambrook]], Edward Kroc.
* Math 110 pilot student project: Instructor [[User:ThomasWong|Tom Wong]] agreed to make his final exam review session for MATH 110, section 1, into a "solution-writing" session. We posted hints, but refrained from posting solutions to last year's Math 110 December exam on the wiki. Students (in groups) were then given 3-4 problems from the Math 110 December 2012 exam and asked to write full and complete solutions. These solutions later served as reference for the solutions that were put on the wiki. Thanks to [[User:ThomasWong|Thomas Wong]] and [[User:ChristinaKoch|Christina Koch]].
* The tagging system was revisited. Several new tags were added, many questions were tagged, additional content and videos were added to some tag main pages. Thanks to [[User:CarmenBruni|Carmen Bruni]] and [[User:konradbe|Bernhard Konrad]].
* The [[Science:MER/Contributors_Guide|Contributors Guide]] was completely restructured and partially rewritten. Thanks to [[User:ChristinaKoch|Christina Koch]] and [[User:konradbe|Bernhard Konrad]].
* Math Exam Resources is mentioned at [http://openedconference.org/2013/ openEd].
* Various advertisement emails and posters were sent to instructors and undergrads and distributed in the MLC.
* The exam progress counter was redone to more accurately reflect how much work is already completed. Thanks to [[User:CarmenBruni|Carmen Bruni]], [[User:IainMoyles|Iain Moyles]] and [[User:konradbe|Bernhard Konrad]].
* We get our own email alias [[File:Mer-wiki-email.png]]
* Organization is moved to [http://www.hojoki.com hojoki].
* We created a [https://github.com/organizations/MER-wiki github] account for any non-wiki code.
* The following '''7''' exams were completed
** MATH100 - December 2010 & December 2012
** MATH102 - December 2012
** MATH104 - December 2012
** MATH110 - December 2012
** MATH215 - December 2011
** MATH221 - April 2013
|}

{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Summer 2013
|-
|style="padding:1em"|
* Discussions with CTLT wiki staff as to new possibilities
* Formation of a leadership/governing committee
|}

{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Jan - Apr 2013 (Winter Term 2, 2012-2013 school year)
|-
|style="padding:1em"|
* The following '''6''' exams were completed
** MATH101 - April 2012
** MATH103 - April 2012
** MATH105 - April 2012
** MATH110 - April 2011 & April 2012
** MATH152 - April 2010
|}

{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Sep - Dec 2012 (Winter Term 1, 2012-2013 school year)
|-
|style="padding:1em"|
* The following '''9''' exams were completed
** MATH100 - December 2011
** MATH102 - December 2011
** MATH104 - December 2010 & December 2011
** MATH110 - December 2010 & December 2011
** MATH220 - December 2011
** MATH221 - December 2011
** MATH257 - December 2011
|}

{| class="collapsible wikitable collapsed" width="100%" style="background: #FFFFFF;"
! style="padding-left:1em; text-align: left;" | Jan-Apr 2012 (Winter Term 2, 2011-2012 school year)
|-
|style="padding:1em"|
* Wiki created.
* [[Help:Dynamic_Page_Lists|Dynamic Page Lists]] (dpl) are heavily used to organize content and for peer-review quality control.
* Content focuses on Term 2 courses: MATH 101, 103, 105.
* Within a few weeks the [[Science:Math Exam Resources|main page]] gets almost 10000 clicks, 1500 of which are in the 12 hours before big final exam (one click every 30 seconds). [[File:Wiki-Klicks-2011-2012-Term2.png|thumb|Top x refers to [[Special:PopularPages]].]]
* Main architect of page design and architecture: [[User:DavidKohler|David Kohler]] with help from Steve Bennoun, [[User:AdrianKeet|Adrian Keet]] and [[User:Konradbe|Bernhard Konrad]].
* Main contributors of content: [[User:CindyBlois|Cindy Blois]], [[User:CarmenBruni|Carmen Bruni]], [[User:CameronChristou|Cameron Christou]], [[User:RobertFraser|Robert Fraser]], [[User:IsabellGraf|Isabell Graf]], Kyle Hambrook, Galo Higuera Rojo, Robert Klinzmann, [[User:DavidKohler|David Kohler]], [[User:Konradbe|Bernhard Konrad]], Eva Koo, Michael Lindstrom, [[User:IainMoyles|Iain Moyles]], Simon Rose, William Thompson, Simon Tse, Erick Wong, [[User:ErinMoulding|Erin Moulding]]
* The following '''15''' exams were completed
** MATH101 - April 2005 & April 2006 & April 2007 & April 2008 & April 2009 & April 2010 & April 2011
** MATH103 - April 2009 & April 2010 & April 2011
** MATH105 - April 2010 & April 2011
** MATH152 - April 2011
** MATH220 - April 2005 & April 2011
|}

[[Category:MER]]

Thread:Science talk:Math Exam Resources/Courses/MATH100/broken link to course main page/reply

2016-09-05T20:26:01Z

Konradbe: Reply to broken link to course main page

The link this year is http://www.math.ubc.ca/~andrewr/maths100180/2016/home.html, but I removed it so that it defaults back to http://www.math.ubc.ca/Ugrad/index.shtml

Science:Math Exam Resources/Courses/MATH100

2016-09-05T20:25:19Z

Konradbe:

{{#vardefine:course_description|
Differential Calculus with Applications to Physical Sciences and Engineering

}}{{#vardefine:course_name|
Math 100/180

}}{{#vardefine:course_url|

}}{{#vardefine:midterm1|

}}{{#vardefine:midterm2|

}}{{#vardefine:midterm3|

}}{{#vardefine:midterm4|

}}{{#vardefine:exam_date|

}}{{#vardefine:message|
== MATH 100/180 ==
Since both MATH 100 and 180 have the same curriculum and same exams, we are listing everything here under the title MATH 100 only but all the content is intended for MATH 180 students as well.

}}<noinclude>{{MER Course page|message={{#var:message}}
|course_url={{#var:course_url}}
|midterm1={{#var:midterm1}}
|midterm2={{#var:midterm2}}
|midterm3={{#var:midterm3}}
|midterm4={{#var:midterm4}}
|exam_date={{#var:exam_date}}
}}</noinclude>

Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (m)/Solution 1

2016-09-05T02:48:18Z

Konradbe: Fixes typo as mentioned in the typo report form

To determine the values of <math>t</math> where <math>f(t)</math> is increasing, we must solve for the critical points of <math>f(t)</math> and examine the sign of the derivative in between the critical points, solving <math>f'(t) = 0</math> gives
:<math>
\begin{align}
\displaystyle f'(t) &= 0\\
2t\exp(t) + t^{2}\exp(t) &= 0\\
(t^{2} + 2t)\exp(t) &= 0.
\end{align}
</math>
Since the exponential function is never zero for any value of <math>t</math>, we can solve the above equation by solving
:<math>
\begin{align}
\displaystyle t^{2} + 2t &= 0\\
t(t + 2) &= 0, \quad \rightarrow \quad t = 0,\,-2
\end{align}
</math>
Now we need to evaluate the sign of <math>f'(t)</math> in the intervals <math>(-\infty,-2), (-2,0), (0,\infty)</math>. Taking test points <math>t = -3,-1,1</math> and plugging them into <math>f'(t)</math> gives:
:<math>
\displaystyle f'(-3) = 3e^{-3} > 0, \quad f'(-1) = -e^{-1} < 0, \quad f'(1) = 3e > 0.
</math>
So <math>f'(t)</math> is positive on <math>(-\infty,-2),(0,\infty)</math> and negative on <math>\displaystyle(-2,0)</math>.

Therefore, <math>f(t)</math> is increasing for <math>t < -2</math>, <math>t > 0</math> and decreasing for <math>-2 < t < 0</math>.

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 25/Solution 1

2016-07-27T01:33:22Z

Konradbe:

Notice that

<math>\det(A-I\lambda)
= \det\left( \begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
-\begin{bmatrix}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda
\end{bmatrix}
\right)
= -\lambda^{3}
</math>

and the roots of this polynomial are <math>\lambda =0 </math>. This is a triple root and so this is the only eigenvalue. Next, for any vector ''v'', we have

<math> Av = 0 = 0v </math>

and hence every '''nonzero''' vector ''v'' is an eigenvector (the zero vector is excluded by definition).

Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 05 (b)/Solution 1

2016-07-27T01:24:03Z

Konradbe: Fixes a calculation error as pointed out by the typo report form

We begin by computing the eigenvalues.

:<math>
\begin{align}
\det(P-\lambda I) &=
\det \left(\begin{bmatrix}
0 & 1/2 \\ 1 & 1/2
\end{bmatrix}
-
\begin{bmatrix}
\lambda & 0 \\ 0 & \lambda
\end{bmatrix}
\right)\\
&= \det \left(\begin{bmatrix}
-\lambda & 1/2 \\ 1 & 1/2 -\lambda
\end{bmatrix}\right)\\
&= (-\lambda)(1/2-\lambda) - 1/2 \\
&= \lambda^{2} - \frac{1}{2}\lambda - \frac{1}{2} \\
&= \frac{1}{2} (2\lambda^{2} - \lambda - 1) \\
&= \frac{1}{2} (2\lambda + 1)(\lambda-1) \\
\end{align}
</math>

and thus the roots (and hence eigenvalues) are <math>\displaystyle \lambda = -1/2,1 </math>. To compute the eigenvectors, we look at the nullspace of <math>\displaystyle P-\lambda I </math> for each eigenvalue. When <math>\displaystyle \lambda = -1/2 </math>, we have

:<math>
\begin{bmatrix}
0 & 1/2 \\ 1 & 1/2
\end{bmatrix}
-
\begin{bmatrix}
-1/2 & 0 \\ 0 & -1/2
\end{bmatrix} =
\begin{bmatrix}
1/2 & 1/2 \\ 1 & 1
\end{bmatrix}
</math>

and a vector (hence an eigenvector) in the kernel of this matrix is given by <math>
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
</math>

Similarly, for <math>\lambda = 1 </math>, we have

:<math>
\begin{bmatrix}
0 & 1/2 \\ 1 & 1/2
\end{bmatrix}
-
\begin{bmatrix}
1 & 0 \\ 0 & 1
\end{bmatrix} =
\begin{bmatrix}
-1 & 1/2 \\ 1 & -1/2
\end{bmatrix}
</math>

and a vector (hence an eigenvector) in the kernel of this matrix is given by <math>
\begin{bmatrix}
1 \\ 2
\end{bmatrix}
</math>

Adjoin these eigenvectors to make a matrix

:<math>
M = \begin{bmatrix}
1 & 1 \\ -1 & 2
\end{bmatrix}
</math>

Then, our theory of diagonalizability gives us that

<math>
D = M^{-1}PM \qquad \text{or, equivalently,} \qquad P = MDM^{-1}
</math>

where

:<math>
D = \begin{bmatrix}
-1/2 & 0 \\ 0 & 1
\end{bmatrix}
</math>

is the diagonal matrix consisting of the eigenvalues. Next, notice that

:<math>
P^{20} = (MDM^{-1})^{20} = \underbrace{(MDM^{-1})(MDM^{-1})...(MDM^{-1})}_{20 \text{ times}} = MD^{20}M^{-1}
</math>

Finally, our work will pay off: Taking the 20th power of a diagonal matrix is easy and convenient. Having to inverting a 2x2 matrix is a small price to pay for this convenience.

<math>
\begin{align}
P^{20} = MD^{20}M^{-1} &= \begin{bmatrix}
1 & 1 \\ -1 & 2
\end{bmatrix}\begin{bmatrix}
-1/2 & 0 \\ 0 & 1
\end{bmatrix}^{20} \begin{bmatrix}
1 & 1 \\ -1 & 2
\end{bmatrix}^{-1}
\\
&= \begin{bmatrix}
1 & 1 \\ -1 & 2
\end{bmatrix}\begin{bmatrix}
(-1/2)^{20} & 0 \\ 0 & (1)^{20}
\end{bmatrix} \left(\frac{1}{3}\begin{bmatrix}
2 & -1 \\ 1 & 1
\end{bmatrix}\right)\\
&= \frac{1}{3}\begin{bmatrix}
1/(2^{20}) & 1 \\ -1/(2^{20}) & 2
\end{bmatrix} \begin{bmatrix}
2 & -1 \\ 1 & 1
\end{bmatrix}\\
&=
\frac{1}{3}\begin{bmatrix}
-1/(2^{19}) + 1 & 1 + 1/(2^{20}) \\ 2 + 1/(2^{19}) & -1/(2^{20}) + 2
\end{bmatrix}
\end{align}
</math>

Applying the vector <math>\begin{bmatrix}1 \\ 0 \end{bmatrix}</math> to this matrix gives

<math>
P^{20}\begin{bmatrix}1 \\ 0 \end{bmatrix} = \begin{bmatrix}\frac{-1/(2^{19}) + 1}{3}\\ \frac{1/(2^{19}) + 2}{3} \end{bmatrix}
</math>

and this completes the problem.

Science:Math Exam Resources/Courses/MATH200/April 2012/Question 01 (a)/Hint 3

2016-04-27T06:11:18Z

Konradbe: Created page with "(Alternative solution) The vector that we're looking for is perpendicular to two vectors. Which? What is a quick way to calculate such a vector, given the other two?"

(Alternative solution) The vector that we're looking for is perpendicular to two vectors. Which? What is a quick way to calculate such a vector, given the other two?

Science:Math Exam Resources/Courses/MATH200/April 2012/Question 01 (a)/Solution 2

2016-04-27T06:11:15Z

Konradbe: Created page with "This quicker alternative solution relies on the cross product: Given two vectors <math>v_1, v_2</math>, their cross product <math>v_3 = v_1 \times v_2</math> is perpendicular..."

This quicker alternative solution relies on the cross product: Given two vectors <math>v_1, v_2</math>, their cross product <math>v_3 = v_1 \times v_2</math> is perpendicular to both, <math>v_1</math> and <math>v_2</math>.

We are asked to find a vector that is parallel to a plane, and perpendicular to a line. By the definition of the normal vector of a plane, all vector parallel to a plane are perpendicular to the normal vector. Further, vectors are perpendicular to a line if they are perpendicular to the direction vector of that line. Hence, choose <math>v_1</math> as the normal vector of the plane, <math>v_1 = [2, 1, -1]</math>, and choose <math>v_2</math> as the direction vector of the line, <math>v_2 = [-1, -2, 3]</math>. Then

<math> \displaystyle
v_3 = v_1 \times v_2 = [2, 1, -1] \times [-1, -2, 3] = [1, -5, -3]
</math>

is parallel to the plane, and perpendicular to the line.

Science:Math Exam Resources/Courses/MATH200/April 2012/Question 01 (a)/Solution 1

2016-04-27T05:50:03Z

Konradbe: Fixes typo, as pointed about by the typo form

Let the line L be written in vector form as
:<math> \displaystyle
[a_{0},b_{0},c_{0}] + [a_{1},b_{1},c_{1}]t.
</math>
If L is parallel to the plane 2x + y - z = 5, then L must be perpendicular to the normal vector of said plane, n = [2,1,-1]. i.e. The dot product between n and the direction of L must be equal to zero:
:<math>
[2,1,-1]\cdot[a_{1},b_{1},c_{1}] = 0.
</math>
Evaluating the dot product gives:
:<math>
\mathbf{(1)} \quad [2,1,-1]\cdot[a_{1},b_{1},c_{1}] = 2a_{1} + b_{1} - c_{1} = 0
</math>
Remember that we also need the vector is perpendicular to the line,<math> [3 - t, 1 - 2t, 3t] = [3,1,0] + [-1,-2,3]t.</math>
Hence, the following equation must also be satisfied:
:<math> \displaystyle
\mathbf{(2)} \quad [-1,-2,3] \cdot [a_{1},b_{1},c_{1}] = -a_{1} - 2b_{1} + 3c_{1} = 0
</math>
So we have two equations and three unknowns. This means there are infinitely many vectors that will satisfy the given conditions. We only need to find one. (Note: Clearly <math>a_{1} = b_{1} = c_{1} = 0</math> is a solution to the above equations (1), (2), but it is the trivial solution and is perpendicular to every vector, including the direction of the line L.)

Solving (1) for <math>c_{1}</math> gives <math>c_{1} = 2a_{1} + b_{1}</math>. Subbing this into (2) gives
:<math>\displaystyle
-a_{1} - 2b_{1} + 3(2a_{1} + b_{1}) = 0 \quad \rightarrow \quad b_{1} = -5a_{1}.
</math>
From this we get that <math>c_{1} = 2a_{1} - 5a_{1} = -3a_{1}</math>.

Thus, any vector of the form k[1,-5,-3] where k is a constant (not equal to zero) will be parallel to L. For example, the vectors [1,-5,-3] and [-2,10,6] are acceptable solutions.

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 04 (b)/Solution 3

2016-04-20T03:45:50Z

Konradbe: Created page with "We apply integration by parts with <math>u = \frac{x^2}2</math> and <math>dv = 2x(1+x^2)^{\frac12}\,dx</math>. Then <math>du = x\,dx</math> and <math>v = \frac23(1+x^2)^{\frac..."

We apply integration by parts with <math>u = \frac{x^2}2</math> and <math>dv = 2x(1+x^2)^{\frac12}\,dx</math>. Then <math>du = x\,dx</math> and <math>v = \frac23(1+x^2)^{\frac32}</math>. Thus

:<math>
\begin{align}
\int x^3 \sqrt{1+x^2} dx &= \int \frac{x^2}{2} \cdot 2x(1+x^2)^{\frac12}dx\\
&=\frac{x^2}{2}\frac23(1+x^2)^\frac32 - \int x\frac23(1+x^2)^\frac32 dx\\
&=\frac{x^2}{2}\frac23(1+x^2)^\frac32 - \frac13\frac25(1+x^2)^\frac52 + C\\
&=\frac{x^2}{3}(1+x^2)^\frac32 - \frac{2}{15}(1+x^2)^\frac52 + C
\end{align}
</math>

''Note:'' The fastest way to confirm that this is the same answer as we get in the solutions above, check that they both simplify to <math>\sqrt{1+x^2}\left(-\frac{2}{15} + \frac{1}{15}x^2 + \frac15x^4\right) + C</math>.

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 04 (b)/Hint 3

2016-04-20T03:14:56Z

Konradbe: Created page with "As another alternative solution, rewrite the integrand as <math>\frac{x^2}2 \cdot 2x(1+x^2)^{\frac12}</math> and apply integration by parts."

As another alternative solution, rewrite the integrand as <math>\frac{x^2}2 \cdot 2x(1+x^2)^{\frac12}</math> and apply integration by parts.

Science:Math Exam Resources/Courses/MATH152/April 2012/Question 01 (b)/Solution 1

2016-04-19T06:30:56Z

Konradbe:

From [[Science:Math_Exam_Resources/Courses/MATH152/April_2012/Question_1_(a)|1(a)]] we saw that the equations for the planes <math>S_1</math> and <math>S_2</math> are

:<math>
\begin{align}
2x_1+x_2-x_3&=1\\
x_1+x_2&=1.
\end{align}
</math>

These two equations taken together are precisely the equation form of the line since any <math>x_1,x_2</math> and <math>x_3</math> that satisfy this will be on both planes and therefore be on the line of intersection.

Science:Math Exam Resources/Courses/MATH110/April 2014/Question 05/Solution 1

2016-04-17T16:57:29Z

Konradbe: Fixes a small error as pointed out by typo form

We can proceed by using the standard template for solving optimisation questions. The exact steps might vary between course and section, but the overall steps will be similar.

'''1.''' Let's define the coordinates of <math>P = (x,y)</math>.

'''2.''' The area of the triangle is therefore: <math>A = \frac{1}{2} (1+x)\sqrt{1-x^2}</math>

'''3.''' The area equation is already in one variable. The domain is <math>x \in [-1,1]</math>. In that we can have <math>x</math> between <math>-1</math> and <math>1</math> (the diameter of the circle) and we can include the endpoints since this will give an area of <math>0</math>.

'''4.''' Differentiate:

<math>\begin{align}
A'(x) &= \frac{1}{2} \left[ (1+x) \cdot \frac{-2x}{2\sqrt{1-x^2}} + 1\cdot \sqrt{1-x^2}\right]\\
&= \frac{1}{2} \left[ \frac{-x -x^2 + (1-x^2)}{\sqrt{1-x^2}}\right]\\
&= \frac{1}{2} \left[ \frac{1-x -2x^2}{\sqrt{1-x^2}}\right]
\end{align}</math>

In this case, the critical points are when the numerator is equal to zero (denominator is zero when <math>x=\pm1</math>).

<math>\begin{align}
x &= \frac{1 \pm \sqrt{1 - 4(-2)(1)}}{-4}\\
&= \frac{1 \pm \sqrt{9}}{-4} = \frac{1}{2}, -1\end{align}</math>

We can then use the closed interval method:

<math>\begin{align}
A(1) &= 0\\
A(-1)&=0\\
A\left(\frac{1}{2}\right) &= \frac{1}{2}\left(1+\frac{1}{2}\right)\sqrt{1-\left(\frac{1}{2}\right)^2} = \frac{3\sqrt{3}}{8} > 0\end{align}</math>

'''5.''' The maximum area happens at point <math>P = \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)</math>.

Science:Math Exam Resources/Courses/MATH101/April 2005/Question 01 (e)/Solution 1

2016-04-16T20:43:49Z

Konradbe: Fixes a typing error as pointed out by typo form

To find the general solution for this problem, we need to find both the particular solution, <math>y_{p}</math>, and homogeneous solution, <math>y_{h}</math>.

The homogeneous solution satisfies

:<math>
\displaystyle{}y_{h}'' + 2y_{h}' + y_{h} = 0,
</math>

which can be solved using the technique from [[Science:Math_Exam_Resources/Courses/MATH101/April_2005/Question_1_(d)|part (d)]] of this exam to obtain the solution

:<math> \displaystyle
y_{h}(x) = c_{1}e^{-x} + c_{2}xe^{-x}.
</math>

To find the particular solution, we look at the right hand side of the equation and notice that there is a term that is linear in <math>x</math> (as opposed to quadratic, cubic, etc.). So we can assume a trial solution of <math>y_p = ax + b</math> and plug this into the differential equation to find the constants <math>a,b</math> such that the original differential equation is satisfied.

:<math>
\begin{align}
y_{p}'' + 2y_{p}' + y_{p} &= x \\
0 + 2a + ax + b &= x
\end{align}
</math>

By comparing the linear and constant terms on both sides of the equation, we can see that for the differential equation to be satisfied, we must satisfy <math>a = 1</math> and <math>2a + b = 0</math>. Thus, <math> a = 1</math>, <math>b = -2</math>. So the general solution to the differential equation is the sum of the homogeneous and particular solutions:

:<math> \displaystyle
y(x) = x - 2 + c_{1}e^{-x} + c_{2}xe^{-x}
</math>

where <math>c_{1},\,c_{2}</math> are arbitrary constants.

Science:Math Exam Resources/Courses/MATH105/April 2013/Question 04 (b)/Solution 1

2016-04-16T18:20:41Z

Konradbe: Fixes a typing error as pointed out by typo form

From part [[Science:Math Exam Resources/Courses/MATH105/April 2013/Question 04 (a)|a)]], the critical points are <math>
(x,y) = (0,0),\,\left(\frac{1}{2},1\right)
</math>
and the first partial derivatives are
:<math> \frac{\partial f}{\partial x} = y(1 - 2x)e^{-2x - y} = 0 \quad \frac{\partial f}{\partial y} = x(1 - y)e^{-2x - y} = 0 </math>
The second partial derivatives are given by
:<math>
\begin{align}
\frac{\partial^{2} f}{\partial x^{2}} &= y(- 2)e^{-2x - y}+y(1 - 2x)(-2)e^{-2x - y}\\
& = -2ye^{-2x-y}(1+1-2x) \\
&= -4ye^{-2x-y}(1-x) \\
&= 4ye^{-2x-y}(x-1)\\
\\
\frac{\partial f}{\partial x\partial y} &= y(1 - 2x)e^{-2x - y}\\
&= (1 - 2x)e^{-2x - y} +y(1-2x)(-1)e^{-2x-y}\\
&= (1 - 2x)e^{-2x - y}(1 +y)\\
&= (1-y-2x+2xy)e^{-2x - y}\\
\\
\frac{\partial^2 f}{\partial y^2} &= x(-1)e^{-2x - y} + x(1 - y)(-1)e^{-2x - y}\\
&= xe^{-2x - y}(-1-1+y)\\
&= xe^{-2x - y}(y-2)
\end{align}
</math>

To classify the critical points, we need to compute the Hessian matrix, <math>H</math>, of the function <math>f(x,y)</math>:
:<math>
H(x,y) = \left[\begin{array}{cc} \frac{\partial^{2}f}{\partial x^{2}} & \frac{\partial^{2}f}{\partial x \partial y} \\ \frac{\partial^{2}f}{\partial x \partial y} & \frac{\partial^{2}f}{\partial y^{2}} \end{array}\right] = \left[\begin{array}{cc} 4y(x - 1)e^{-2x-y} & (1-y-2x+2xy)e^{-2x-y} \\ (1-y-2x+2xy)e^{-2x-y} & x(y-2)e^{-2x-y} \end{array}\right].
</math>

Evaluating <math>H</math> at the critical point <math>(x,y) = (0,0)</math> gives
:<math>
H(0,0) = \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right].
</math>
The determinant of <math>H</math> at the point <math>(0,0)</math> is equal to <math>(0 \cdot 0) - (1 \cdot 1) = -1</math>, which is less than zero. Hence the point <math>(0,0)</math> is a saddle point.

Evaluating <math>H</math> at the critical point <math>(x,y) = (1/2,1)</math> gives
:<math>
H\left(\frac{1}{2},1\right) = \left[\begin{array}{cc} -2e^{-2} & 0 \\ 0 & -\frac{1}{2}e^{-2} \end{array}\right].
</math>
The determinant of <math>H</math> at the point <math>(1/2,1)</math> is equal to <math>(-2e^{-2}) \cdot \left(-\frac{1}{2}e^{-2}\right) - 0 \cdot 0 = e^{-4}</math> which is greater than zero and so <math>(1/2,1)</math> is not a saddle point of <math>f</math> and must be either a local max or min. Since <math>\frac{\partial^{2}f}{\partial x^{2}}\left(\frac{1}{2},1\right) = -2e^{-2}</math> is less than zero, the point <math>(1/2,1)</math> is a local maximum of <math>f</math>.

 Therefore, (x,y) = (0,0) is a saddle point and (x,y) = (1/2,1) is a local maximum.

Science:Math Exam Resources/Courses/MATH221/December 2009/Question 02 (b)/Solution 1

2016-04-15T05:09:56Z

Konradbe: Fixes error resulting from solving for the wrong matrix, as pointed about by the typo form

Strategy:

Use the following determinant properties.

The determinant of upper triangular matrix (i.e a matrix in row echelon form in this case) is equal to the product of its diagonal entries. If i<math>\neq</math>j , then subtracting some constant time row i from row j does not change the determinant.

Step 1:

Perform the following row operations on B:

row 2 = row 2 - 2 row 1

row 3 = row 3 - 2 row 1

and we get

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& -1\\
0& 2& -5& 0\\
0& 3& -1& 1
\end{bmatrix}</math>

Since the above matrix is still not in upper triangular form, perform the following row operations to it:

row 3 = row 3 - 2 row 2

row 4 = row 4 - 3 row 2

Then we get:

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 2\\
0& 0& 5& 4
\end{bmatrix}</math>

After performing one more row operation,

row 4 = row 4 + 5 row 3,

we get the following upper triangular matrix:

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 0\\
0& 0& 0& 14
\end{bmatrix}</math>

Step 2 :

Let <math>U' = \begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 0\\
0& 0& 0& 14
\end{bmatrix}</math>

Then <math>det(A) = det(U') = (1)(1)(-1)(14) = -14</math> and thus <math>det(A) = -14</math>.

Science:Math Exam Resources/Courses/MATH221/December 2009/Question 02 (b)/Statement

2016-04-15T05:02:12Z

Konradbe: Fixes typo, as pointed about by the typo form

Find the determinant of the matrix

<math>B = \begin{bmatrix}
1& 0& 3& 0\\
2& 1& 4& -1\\
3& 2& 4& 0\\
0& 3& -1& 1
\end{bmatrix}</math>

Science:Math Exam Resources/Courses/MATH101/April 2014/Question 01 (d)/Solution 1

2016-04-14T06:25:11Z

Konradbe: Clarifies solution in response to user comment

Let <math>x=2\sin \theta</math>. Then, <math>dx=2\cos \theta d\theta</math>. By substitution,

<math>\begin{align}
\int \sqrt{4-x^2} dx
&= \int \sqrt{4-4\sin^2 \theta}\cdot 2\cos \theta d\theta
=\int \sqrt{4\cos^2\theta}\cdot 2\cos \theta d\theta\\
&= \int 2\cos\theta\cdot 2\cos \theta d\theta
=\int 4\cos^2\theta d\theta.
\end{align}</math>

We now use another trigonometric identity,

<math>\begin{align}
\cos^2 \theta=\frac{1+\cos 2\theta}{2},
\end{align}</math>

to obtain

<math>\begin{align}
\int 4\cos^2 \theta d\theta &=\int 2 (1+\cos 2\theta) d\theta \\
&= 2\int d\theta + 2\int \cos 2\theta d\theta \\
&= 2\theta + 2\left(\frac{\sin 2\theta}{2}\right) + C \\
&= 2\theta+2\sin\theta\cos\theta+C.
\end{align}</math>

In the last step we used that <math>\sin 2\theta=2\sin\theta\cos\theta</math>. All that's left to do is to express this result in terms of <math>x</math>.

First, notice that <math>x=2\sin\theta</math>, <math>\theta=\arcsin\left(\frac{x}{2}\right)</math> and hence <math>\sin\theta=\frac{x}{2}</math>. Also, by using the below picture, <math>\cos \theta=\frac{\sqrt{4-x^2}}{2}</math>. Thus,

[[File:Math101Exam2014Aprilpicture1q1d.jpg|Math101Exam2014AprilQ1d]]

<math>\begin{align}
\int \sqrt{4-x^2} dx
=2\theta+2\sin\theta\cos\theta+C
=2\arcsin\frac{x}{2}+2\cdot\frac{x}{2}\cdot\frac{\sqrt{4-x^2}}{2}+C
=2\arcsin\frac{x}{2}+\frac{x\sqrt{4-x^2}}{2}+C.\end{align}</math>

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 07/Solution 1

2016-04-14T05:57:07Z

Konradbe: Fixes a typing error as pointed out by typo form

Since <math>f(a+i(b-a)/N) = f(-1+3i/N) = -1+3i/N - 1</math> we follow the hints, to obtain

:<math>\begin{align}\int_{-1}^2 (x-1) \,dx &= \lim_{N \to \infty} \sum_{i=1}^N f(-1+3i/N)\frac{3}{N}\\
&= \lim_{N \to \infty} \sum_{i=1}^N (-1+3i/N - 1)\frac{3}{N}\\
&= \lim_{N \to \infty} \frac{3}{N} \sum_{i=1}^N \left(-2+ \frac{3i}{N}\right)\\
&= \lim_{N \to \infty} \left( -\frac{6}{N} \sum_{i=1}^N 1 + \frac{9}{N^2} \sum_{i=1}^N i \right)\end{align}</math>

Since

:<math>\sum_{i=1}^N 1 = N \quad \text{ and }\quad \sum_{i=1}^N i = \frac{N(N+1)}{2},</math>

we can continue our computation

:<math>\begin{align}\int_{-1}^2 (x-1) \,dx &= \lim_{N \to \infty} \left( -\frac{6}{N}N + \frac{9}{N^2} \frac{N(N+1)}{2} \right)\\
&= -6 + \frac{9}{2}\lim_{N \to \infty}\frac{N+1}{N}\\
&= -6 + \frac{9}{2}\\
&= -\frac{3}{2}.\end{align}</math>

''You can always compare your answer to the computation of the integral directly and see that the answers match:''
:<math> \begin{align}
\int_{-1}^2 (x-1)\,dx &= \left. \frac{x^2}2-x\right|_{-1}^2 \\
&= \frac42 - 2 - \left(\frac12 + 1\right) \\
&= -\frac32.
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 06 (e)/Solution 3

2016-04-11T00:57:13Z

Konradbe: New solution as suggested by feedback form

In this particular example there is an even faster way to see that <math>D</math> must be outside of the triangle <math>T</math>, because <math>D</math> is even outside the cuboid spanned by the three corners of <math>T</math>. Indeed, for all points <math>(x, y, z)</math> in <math>T</math> it must be true that

:<math>\begin{align}
\min(0, 1, -1) &\leq x \leq \max(0, 1, -1) \\
\min(1, 1, 2) &\leq y \leq \max(1, 1, 2) \\
\min(2, 5, 2) &\leq z \leq \max(2, 5, 2)
\end{align}</math>

However, since the <math>y</math> and <math>z</math> coordinates of <math>D</math> do not satisfy the inequalities above, <math>D</math> is outside the cuboid, and hence outside of <math>T</math>.

Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 06 (e)/Solution 2

2016-04-11T00:28:32Z

Konradbe:

The answer will be that ''(0, 0, -1)'' is outside of ''T''. There are many different possible ways to show this, but they all take a correct formulation of a ''decisive'' criterion and then patience and carefulness to work till the end.

For this solution, let's explore some methods using basic geometry. We will hardly use the concept of vectors, and so for the subsequent paragraphs, let us denote <math> A = (0,1,2), </math> <math>B = (-1,2,2), </math> <math>C = (1,1,5) </math>, and <math> D = (0,0,-1) </math> for brevity.

This are a few ways to formulate a criterion to determine whether ''D = (0, 0, -1)'' is inside or outside ''T''. Here we suggest 2 possible criteria.

In both cases, we consider the triangles <math> \triangle ABD, \triangle BCD, \triangle CAD </math>. There are two qualitatively different pictures for ''D'' inside or outside ''T''.

'''Criterion 1 - Comparing Lengths'''

If at least one of the lengths ''AD'', ''BD'' or ''CD'' is strictly greater than the largest of ''AB'', ''BC'', ''CA'', then <math>D</math> is outside <math>T</math>.

Conversely if ''D'' is inside ''T'', then all the lengths ''AD'', ''BD'' or ''CD'' is less than or equal to the largest of ''AB'', ''BC'', ''CA''.

Now, we calculate:

:<math> AB = |(-1,1,0)| = \sqrt{2} </math>
:<math> BC = |(2,-1,3)| = \sqrt{14} </math>
:<math> CA = |(1,0,3)| = \sqrt{10} </math>

But since
:<math> DC = |(1,1,6)| = \sqrt{38} </math>,
''D'' is outside of ''T''.

'''Criterion 2 - Compare Areas'''

If ''D'' is outside ''T'', then the sum of areas of <math> \triangle ABD, \triangle BCD, \triangle CAD </math> is strictly greater than that of <math> \triangle ABC </math>

Conversely if ''D'' is inside ''T'', then equality holds instead.

We already knew that area of <math> T = \triangle ABC </math> is <math> \sqrt{19}/2 </math>. Let's calculate the area of <math> \triangle BCD </math>
:<math>\begin{align}
\text{Area}(\triangle BCD)&= \frac{1}{2} \left|\vec{BD} \times \vec{CD} \right|\\
&= \frac{1}{2} \left|(-1,2,3) \times (1,1,6) \right|\\
&= \frac{1}{2} |(9,9,-3)|\\
&= \frac{3\sqrt{19}}{2}
\end{align}</math>

So, without calculating the other areas, we can be sure that ''D'' is outside of ''T''.

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 09/Solution 1

2016-04-10T23:41:29Z

Konradbe: Clarifies solution in response to user comment

Because <math>e^{x}</math> is its own derivative, our anti-derivative for the function <math>e^{3-x}</math> will probably be very similar to the original function. To check our first anti-derivate candidate <math>e^{3-x}</math> we have to use the chain rule, <math>(G(H(x)))' = G'(H(x))H'(x)</math>, where <math>G(x) = e^x</math> and <math>H(x) = 3-x</math>. Then <math>G'(x) = e^x</math> and <math>H'(x) = -1</math>, so we calculate:

:<math> \displaystyle
(e^{3-x})' = (G(H(x)))' = G'(H(x))H'(x) = (e^{3-x})(-1) = -e^{3-x}.
</math>

Hence our candidate almost works, we just have to bring the minus sign over: <math>F(x) = - e^{3-x}</math>. Using the chain rule to double check, we find that, indeed,

:<math> \displaystyle
F'(x) = (-e^{3-x})' = (-e^{3-x})(-1) = e^{3-x} = f(x).
</math>

This is one anti-derivative. To find another one, we can simply add a constant to the anti-derivative shown above, say <math>F_2(x) = -e^{3-x} + 5</math>. When we differentiate, the constant will disappear, giving us the same derivative as before.

Science:Math Exam Resources/Courses/MATH101/April 2011/Question 06 (a)/Solution 1

2016-04-10T21:27:13Z

Konradbe: Clarifies solution in response to user comment

We have the series expansion

:<math>\begin{align}
\frac{1}{1 + x^3} &= \frac{1}{1 - (-x^3)} \\
&= 1 + (-x^3) + (-x^3)^2 + (-x^3)^3 + \cdots \\
&= 1 + (-1)x^3 + (-1)^2 x^6 + (-1)^3 x^9 + \cdots \\
&= \sum_{n=0}^\infty (-1)^n x^{3n}
\end{align}</math>

and so, integrating term by term we find that

:<math>\begin{align}
\int \frac{1}{1+x^3}dx = \sum_{n=0}^\infty (-1)^n \int x^{3n} dx = \sum_{n=0}^\infty (-1)^n\frac{x^{3n+1}}{3n+1} + C
\end{align}</math>

It only remains to find the radius of convergence. Computing the limit in the ratio test yields

<math>
\begin{align}
\lim_{n\to\infty} \left|\frac{c_{n+1}}{c_{n}}\right|
&= \lim_{n\to\infty} \left|\frac{(-1)^{n+1}\frac{x^{3(n+1)+1}}{3(n+1)+1}}{(-1)^n\frac{x^{3n+1}}{3n+1}}\right| \\
&= \lim_{n\to\infty} \left|\frac{\frac{x^{3n+4}}{3n+4}}{\frac{x^{3n+1}}{3n+1}}\right| \\
&= \lim_{n\to\infty} \frac{|x|^{3n+4}}{|x|^{3n+1}}\frac{3n+1}{3n+4}\\
&= \lim_{n\to\infty} |x|^{3}\frac{n(3+1/n)}{n(3+4/n)}\\
&= \lim_{n\to\infty} |x|^{3}\frac{3+1/n}{3+4/n}\\
&= |x|^{3}
\end{align}
</math>

Now the ratio test tells us that the series converges whenever <math> \displaystyle |x|^{3} < 1 </math> and thus our radius of convergence is 1.

Science:Math Exam Resources/Courses/MATH215/December 2013/Question 02 (a)/Solution 1

2016-04-10T02:08:13Z

Konradbe: Clarifies solution in response to user comment

We begin with <math>y'+x^3 y= 3 x^3, \quad y(0)=8</math>. We can use an integrating factor here, taking <math>\mu(x) = e^{\int x^3 dx} = e^{\frac{x^4}{4}}</math> where we chose the arbitrary constant to be 0. Multiplying both sides of the equation by <math>\mu(x)</math> gives

:<math>e^{\frac{x^4}{4}} y' + e^{\frac{x^4}{4}}x^3y = 3 x^3 e^{\frac{x^4}{4}}</math> or

:<math>(e^{\frac{x^4}{4}} y)' = 3x^3 e^{\frac{x^4}{4}}</math>.

We can integrate both sides of the equation (doing a simple substitution on the right-hand side) giving us

:<math>\int \frac{d}{dx} (e^{\frac{x^4}{4}} y(x)) dx = \int 3x^3 e^{\frac{x^4}{4}} dx</math>

:<math>e^{\frac{x^4}{4}} y = \underbrace{\int 3 e^u du}_{u=x^4/4,\, du = x^3dx} = 3 e^u + C = 3 e^{\frac{x^4}{4}} + C</math>

:<math>y = 3 + C e^{\frac{-x^4}{4}}</math>.

If <math>y(0) = 8</math> then <math>8 = 3 + C</math> so <math>C=5</math> and the final answer is

<math>y(x) = 3 + 5 e^{\frac{-x^4}{4}}</math>.

Science:Math Exam Resources/Courses/MATH221/April 2010/Question 06/Solution 1

2015-12-16T18:13:57Z

Konradbe:

The answer to this statement is '''false''': No solution to <math>A\mathbf{x} = \mathbf{c}</math> must exist (however, if a solution exists, it must be unique).

If <math>A\mathbf{x} = \mathbf{c}</math> then <math>\mathbf{c}</math> is a linear combination of the columns of <math>A</math>. However, the vector <math>\mathbf{c}</math> is in <math>\mathbb{R}^5</math> and this space requires 5 vectors to form a basis. We can therefore not guarantee that the 4 vectors we have from the columns of <math>A</math> could properly represent any vector <math>\mathbf{c}</math> and therefore we are not guaranteed that a solution exists to <math>A\mathbf{x}=\mathbf{c}</math>.

We can illustrate this with an example.

Let <math>A = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>, <math>\mathbf{b} = \begin{bmatrix}
1 \\ 0 \\ 0 \\ 0 \\ 0
\end{bmatrix}</math> and <math>\mathbf{c} = \begin{bmatrix}
0 \\ 0 \\ 0 \\ 0 \\ 1
\end{bmatrix}</math>.

Then <math>A\mathbf{x} = \mathbf{b}</math> has the unique solution <math>\mathbf{x} = \begin{bmatrix}
1 \\ 0 \\ 0 \\ 0
\end{bmatrix}</math>, but <math>A\mathbf{x} = \mathbf{c}</math> has no solution.

However, '''if''' a solution to <math>A\mathbf{x} = \mathbf{c}</math> exists, then it must be unique: If <math>A\mathbf{x}=\mathbf{b}</math> has a '''unique''' solution then it must have a trivial nullspace, i.e. there must not be a vector <math>\mathbf{y} \not= \mathbf{0}</math> such that <math>A\mathbf{y}=\mathbf{0}</math> as otherwise

:<math>
A(\mathbf{x}+\mathbf{y})=A\mathbf{x}+A\mathbf{y}=\mathbf{b}+\mathbf{0}=\mathbf{b}
</math>

and <math>\mathbf{x}+\mathbf{y}</math> is a solution violating the uniqueness assumption of <math>A\mathbf{x}=\mathbf{b}</math>. So assume that <math>\mathbf{r}</math> and <math>\mathbf{s}</math> are both solutions of <math>A\mathbf{x} = \mathbf{c}</math>. We need to show that the solution is unique, that is, <math>\mathbf{r} = \mathbf{s}</math>. Subtracting the two equations we obtain <math>A\mathbf{r} - A\mathbf{s} = \mathbf{c} - \mathbf{c} = \mathbf{0}</math> and thus <math>A(\mathbf{r} - \mathbf{s}) = \mathbf{0}</math>, that is, <math>\mathbf{r} - \mathbf{s}</math> is in the nullspace of <math>A</math>. However, this nullspace only consists of the null vector, which implies that indeed <math>\mathbf{r} = \mathbf{s}</math>.

Science:Math Exam Resources/Courses/MATH100/December 2014/Question 03 (c)/Solution 1

2015-12-16T10:57:45Z

Konradbe: Clarifies solution as request by feedback form

Convert the base to base <math>e</math>. Since <math>a^b = e^{\log(a^b)} = e^{b\log(a)}</math> we obtain

<math>
y = x^{\log x} = x^{\log(x)\log(x)} = e^{\left((\log(x))^2\right)}.
</math>

To find the derivative we use the chain rule <math>(f(g(h(x))))' = f'(g(h(x)))g'(h(x))h'(x)</math> with <math>f(x) = e^x</math>, <math>g(x) = x^2</math> and <math>h(x) = \log(x)</math>. With this we calculate

<math>\begin{align}
y &=e^{\left((\log (x))^2\right)} \\
\frac {dy}{dx} &= (f(g(h(x))))' \\
&= f'(g(h(x)))g'(h(x))h'(x) \\
&= e^{\left((\log(x))^2\right)}(2\log (x))(1/x) \\
&= 2\log (x) \cdot\frac 1 {x}\cdot x^{\log (x)}\\&= 2\log (x) \cdot x^{\log (x) - 1}\end{align}</math>

Science:Math Exam Resources/Courses/MATH221/April 2010/Question 06/Solution 1

2015-12-15T22:49:29Z

Konradbe:

The answer to this statement is '''false'''.

To begin with, notice that <math>A</math> must have full rank, that is, the rank of <math>A</math> is <math>4</math>. Otherwise <math>Ax=b</math> never has a unique solution.

'''Proof''': Assume <math>A</math> does not have full rank. Then there exists a vector <math>\vec{0} \neq y \in \mathbb{R}^4</math> such that <math>Ay = 0</math>. Hence <math>A(x+y) = Ax + Ay = Ax + \vec{0} = b</math> has two distinct solutions, <math>x</math> and <math>x + y</math>.

Hence, if <math>Ax = c</math> has a solution, it must be unique. However, <math>A</math> has only rank 4, and thus can not span the whole of <math>\mathbb{R}^5</math>. That is, there is a vector <math>c \in \mathbb{R}^5</math> that is not in the image of <math>A</math> and <math>Ax = c</math> has no solution.

Alternative:

The statement is '''false''' Let <math>A = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}</math>, <math>b = \begin{bmatrix}
1 \\ 0 \\ 0 \\ 0 \\ 0
\end{bmatrix}</math> and <math>c = \begin{bmatrix}
0 \\ 0 \\ 0 \\ 0 \\ 1
\end{bmatrix}</math>.

Then <math>Ax = b</math> has the unique solution <math>x = \begin{bmatrix}
1 \\ 0 \\ 0 \\ 0
\end{bmatrix}</math>, but <math>Ax = c</math> has no solution.

Science:Math Exam Resources/Courses/MATH221/April 2010/Question 06/Solution 1

2015-12-15T22:43:28Z

Konradbe:

Science:Math Exam Resources/Courses/MATH221/April 2010/Question 06/Hint 1

2015-12-15T21:15:23Z

Konradbe: Created page with "Think about what properties of <math>A</math> are necessary in order for a solution to exist for <math>Ax = c</math>, for every vector <math>c</math> in <math>\mathbb{R}^5</ma..."

Think about what properties of <math>A</math> are necessary in order for a solution to exist for <math>Ax = c</math>, for every vector <math>c</math> in <math>\mathbb{R}^5</math>.

Science:Math Exam Resources/Courses/MATH100/December 2014/Question 06 (b)/Solution 1

2015-12-15T20:24:48Z

Konradbe: Adds some details as requested by feedback form

Multiply by the conjugate and then simplify:

<math>\begin{align}
\sqrt{x^2+2x}-\sqrt{x^2-2x}&=(\sqrt{x^2+2x}-\sqrt{x^2-2x}) \cdot\frac{\sqrt{x^2+2x}+\sqrt{x^2-2x}}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\\
&= \frac {x^2+2x-(x^2-2x)}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\\
&= \frac {4x}{\sqrt{x^2(1+2/x)}+\sqrt{x^2(1-2/x)}}\\
&= \frac {4x}{x(\sqrt{1+2/x}+\sqrt{1-2/x})}\\
&= \frac {4}{\sqrt{1+2/x}+\sqrt{1-2/x}}
\end{align}</math>

assuming <math>x>0</math>. Hence

<math>\begin{align}
\lim_{x\rightarrow +\infty}(\sqrt{x^2+2x} - \sqrt{x^2-2x}) &= \lim_{x\rightarrow +\infty}\frac 4 {\sqrt{1 + 2/x}+\sqrt{1-2/x}}\\&= \frac 4 2 = 2\end{align}</math>

Template:MER Exam page

2015-11-29T12:07:25Z

Konradbe: Removes "April 2014" because it's outdated

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| MATH100 = 100+180
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Science:Math Exam Resources/Courses/MATH104/December 2013/Question 01 (i)/Solution 1

2015-11-29T12:05:17Z

Konradbe: Clarifies solution as request by feedback form

We use the point slope formula

:<math>
\frac{y-f(a)}{x-a} = f'(a).
</math>

Since <math>a = 1</math> we obtain <math>f(a) = \ln(1) = 0</math>, while <math>f'(x) = \frac1{x}</math> implies <math>f'(a) = 1</math>. Hence

<math> \displaystyle \frac{y-0}{x-1} = 1.</math>

Solving gives

<math> \displaystyle y = x-1</math>

Thus, the approximation we seek is given by

<math> \displaystyle y = 1.25-1 = 0.25</math>.

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (l)/Solution 1

2015-11-24T11:07:52Z

Konradbe:

We first take the derivative to see that

<math> \displaystyle f'(x) = e^{x}(\sin x - \cos x ) + e^{x}(\cos x + \sin x) = 2e^{x}\sin x </math>

Setting this to zero, we see that the function is zero whenever <math> \displaystyle \sin x =0</math> (the exponential function is always bigger than 0) and on our interval <math> [\tfrac{\pi}{2},2\pi] </math>, this occurs at <math> \displaystyle \pi </math> and at <math> \displaystyle 2\pi </math>.

Checking the endpoints and values at the critical points, we have

<math> \displaystyle f(\tfrac{\pi}{2}) = e^{\tfrac{\pi}{2}}(\sin (\tfrac{\pi}{2}) - \cos (\tfrac{\pi}{2}) ) = e^{\tfrac{\pi}{2}} </math>

<math> \displaystyle f(\pi) = e^{\pi}(\sin (\pi) - \cos (\pi) ) = e^{\pi}</math>

<math> \displaystyle f(2\pi) = e^{2\pi}(\sin (2\pi) - \cos (2\pi) ) = -e^{2\pi} </math>

Hence, our function obtains its absolute maximum at <math> \displaystyle \pi </math> and the value is <math> \displaystyle e^\pi </math>.

Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (j)/Solution 1

2015-10-24T19:55:10Z

Konradbe:

Recall the formula for continuously compounded interest,
:<math>
A = Pe^{rt}
</math>
where <math>A,P,r</math> are the future value, principal and growth rate respectively and <math>t</math> is the time since the investment was made. The rate of growth of the investment is simply the time derivative of the future value (i.e. <math>dA/dt</math>). Evaluating the time derivative of <math>A</math> gives:
:<math>
\frac{d}{dt}A = \frac{d}{dt}\left(Pe^{rt}\right) = rPe^{rt}.
</math>
By the information given in the question, we want to solve for the time <math>t</math> when the <math>dA/dt = 10,000</math> dollars/year, given <math>P = 100,000</math> dollars and <math>r = 0.07</math>/year.
:<math>
\begin{align}
10000 &= (0.07)(100000)e^{0.07t}\\
10000 &= 7000e^{\frac{7}{100}t} \\
\frac{10}{7} &= e^{\frac{7}{100}t}\\
\ln\left(\frac{10}{7}\right) &= \frac{7}{100}t\\
\rightarrow \quad t &= \frac{100}{7}\ln\left(\frac{10}{7}\right) \quad (\approx 5.10)
\end{align}
</math>
Therefore, we would need to wait (100/7)ln(10/7) years for our initial investment to be growing at a rate of $10,000 per year so we can retire.

Science:Math Exam Resources/Courses/MATH220/December 2009/Question 07 (b)/Solution 1

2015-08-18T18:22:48Z

Konradbe: Fixes minor error as reported in typo form

We show that the sequence converges to 3. To see this, it suffices to show that for any <math> \displaystyle \epsilon > 0</math> there exsists an <math>n_0</math> such that

:<math> \displaystyle \left|\frac{3n^{2} - 4n}{n^{2} + 5n + 2} - 3 \right| < \epsilon</math>

for all <math>n \geq n_0</math>. Let <math> \displaystyle \epsilon > 0</math>. Simplifying the left hand side above yields:

:<math>
\begin{align}
\left|\frac{3n^{2} - 4n}{n^{2} + 5n + 2} - 3 \right| &= \left|\frac{3n^{2} - 4n}{n^{2} + 5n + 2} - \frac{3(n^{2} + 5n + 2)}{n^{2} + 5n + 2} \right| \\
&= \left|\frac{- 19n-6}{n^{2} + 5n + 2} \right|\\
&= \frac{19n+6}{n^{2} + 5n + 2} \\
&< \frac{19n + 19}{n^2 + n} \\
&= \frac{19(n+1)}{n(n+1)} \\
&= \frac{19}{n}
\end{align}
</math>

Now, choose <math>n_0</math> large enough so that <math> \displaystyle \frac{19}{n_0} < \epsilon</math> which can be done by choosing <math>n_0</math> larger than <math> \displaystyle \frac{19}{\epsilon}</math>. Then, for each <math>n \geq n_0</math>, we have that

:<math>
\begin{align}
\left|\frac{3n^{2} - 4n}{n^{2} + 5n + 2} - 3 \right| &< \frac{19}{n} < \epsilon
\end{align}
</math>

which shows that our original sequence converges to <math>3</math>.

Science:Math Exam Resources/Courses/MATH104/December 2014/Question 04/Solution 1

2015-06-26T01:12:03Z

Konradbe: Clarify solution as reported in typo form

We know that if we sell <math>x</math> bus tour tickets, we sell <math>y = 100 - x</math> train tour tickets. Let us now make the revenue function:

<math>\begin{align}
R & = x(30 - \frac{x}{4}) + y(70 - \frac{y}{2}) \\
& = x(30 - \frac{x}{4}) + (100 - x)(70 - \frac{100 - x}{2}) \\
& = 30x - \frac{x^2}{4} + 70(100-x) - \frac{(100-x)^2}{2}\\\end{align}</math>

The maximum of <math>R</math> can either be at the boundary of the interval or a local maximum. For <math>x = 0</math> we find <math>R(0) = 2000</math> and for <math>x = 100</math> we obtain <math>R(100) = 2500</math>. To find the critical point we take the derivative of <math>R</math> with respect to <math>x</math>. Being careful with the chain rule in the last summand we obtain
<math>\begin{align}
R'(x) & = 30 - \frac{x}{2} - 70 - (100 - x)(-1) \\
& = 60 - \frac{3x}{2}\end{align}</math>

Now we want to maximize the revenue with respect to the number of bus tickets we sell; hence we let <math>R'(x) = 0</math> and determine <math>x</math>. We obtain that <math>x = 40</math> and <math>R(40) = 3200</math>. This means that if we sell 40 bus tour tickets and 60 train tour tickets, we maximize the revenue.

Science:Math Exam Resources/Courses/MATH104/December 2013/Question 05/Solution 1

2015-06-26T00:57:09Z

Konradbe: Clarify solution as reported in typo form

To use the elasticity of demand, we evaluate <math>\epsilon</math> at <math>p = 4</math>. If <math>\epsilon < -1</math>, then demand is elastic and the price should be lowered to increase revenue. Otherwise, if <math>\epsilon > -1</math>, then demand is inelastic and the price should be raised.

A quick calculation using the given price-demand relationship shows that if <math>p = 4</math>, then either <math>q = 4200</math> or <math>q = 1800</math>. Hence, we will have to consider both scenarios.

To evaluate <math>\frac{dq}{dp}</math>, we opt to use the chain rule on the price-demand relationship. Differentiating it with respect to <math>p</math>, gives

:<math>\begin{align}
\frac{d}{dp}p &= \frac{d}{dp}\left(\frac{q-3000}{600}\right)^2 \\
1 &= 2\left(\frac{q - 3000}{600}\right)\frac{1}{600}\frac{dq}{dp} \\
1 &= \left(\frac{q - 3000}{600}\right)\frac{1}{300}\frac{dq}{dp}
\end{align}</math>

Solving for <math>\frac{dq}{dp}</math> we obtain
:<math>
\frac{dq}{dp} = 300\left(\frac{600}{q - 3000}\right)
</math>
Putting together all the pieces to evaluate <math>\epsilon</math> at <math>p = 4</math>, we get
:<math>
\begin{align}
\epsilon &= \frac{p}{q}\frac{dq}{dp} \\
&= \frac{1}{q} \left(\frac{q-3000}{600}\right)^2 300\left(\frac{600}{q - 3000}\right) \\
&= \frac{300}{q}\left(\frac{q - 3000}{600}\right)
\end{align}
</math>
Depending on the value of <math>q</math>, the value of <math>\epsilon</math> will be different.

If <math>q = 4200</math>, then <math>\epsilon = 3/21</math>.

If <math>q = 1800</math>, then <math>\epsilon = -6/18</math>.

Thus, in either case, the price should be raised in order to increase revenue.

Note: The case where <math>q = 4200</math> is may seem somewhat counter-intuitive since the elasticity of demand relationship suggests here that raising the price actually increases demand (in stark contrast to the general rule that increasing price decreases demand). However, this phenomenon is not actually very unusual. You may want to ponder what this means... (e.g. Do expensive handbags become more or less desirable to wealthy shoppers if the price drops?)

Science:Math Exam Resources/Courses/MATH215/December 2013/Question 05 (b)/Solution 1

2015-06-22T04:49:27Z

Konradbe: Fixes typo as reported in typo form

We need a particular solution. To obtain it, we will use the method of undetermined coefficients.

If <math>\vec{x}_p = \begin{bmatrix} at e^{3t} + b e^{3t} \\ ct e^{3t} + d e^{3t} \end{bmatrix}</math> then <math>\vec{x_p}' = \begin{bmatrix} (a + 3b) e^{3t} + 3 a t e^{3t} \\ (c + 3d) e^{3t} + 3c t e^{3t} \end{bmatrix} </math> and

:<math>A x_p + \begin{bmatrix} 25 e^{3t} \\ 0 \end{bmatrix} = \begin{bmatrix} 3 ct e^{3t} + (3d + 25)e^{3t} \\ (2a + c) t e^{3t} + (2b + d) e^{3t} \end{bmatrix}</math>

Therefore if <math>\vec{x_p}' = A \vec{x_p} + \begin{bmatrix} 25 e^{3t} \\ 0 \end{bmatrix}</math> we get

:<math>\begin{bmatrix} (a + 3b) e^{3t} + 3 a t e^{3t} \\ (c + 3d) e^{3t} + 3c t e^{3t} \end{bmatrix} = \begin{bmatrix} 3 ct e^{3t} + (3d + 25)e^{3t} \\ (2a + c) t e^{3t} + (2b + d) e^{3t} \end{bmatrix}</math>.

In comparing the coefficients of <math>e^{3t}</math> from row 1, the coefficients of <math>t e^{3t}</math> from row 1, the coefficients of <math>e^{3t}</math> from row 2, and the coefficients of <math>t e^{3t}</math> from row 2, the following equations must hold:

:<math>\begin{align}
a + 3b &= 3d + 25 \\
3a &= 3c \\
c + 3d &= 2b + d \\
3c &= 2a + c
\end{align}</math>

We can turn this into a matrix equation and row-reduce:

:<math> \begin{bmatrix} 1 & 3 & 0 & -3 \\ 1 & 0 & -1 & 0 \\ 0 & 2 & -1 & -2 \\ 1 & 0 & -1 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\d \end{bmatrix} = \begin{bmatrix} 25 \\ 0 \\ 0 \\ 0 \end{bmatrix}</math>

Subtracting row 2 from row 4 and row 1 from row 2 yields:

:<math> \begin{bmatrix} 1 & 1 & 1 & -1 \\ 0 & -3 & -1 & 3 \\ 0 & 2 & -1 & -2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\d \end{bmatrix} = \begin{bmatrix} 25 \\ -25 \\ 0 \\ 0 \end{bmatrix}</math>

We next multiply the third row by 3 and then add the second row twice:

:<math> \begin{bmatrix} 1 & 1 & 1 & -1 \\ 0 & -3 & -1 & 3 \\ 0 & 0 & -5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\d \end{bmatrix} = \begin{bmatrix} 25 \\ -25 \\ -50 \\ 0 \end{bmatrix}</math>

Next, divide the third row by -5 and then add the result to the second row, and subtract it from the first row:

:<math> \begin{bmatrix} 1 & 1 & 0 & -1 \\ 0 & -3 & 0 & 3 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\d \end{bmatrix} = \begin{bmatrix} 15 \\ -15 \\ 10 \\ 0 \end{bmatrix}</math>

In our last step we divide the second row by -3 and then subtract the result from the first row:

:<math> \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\d \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \\ 10 \\ 0 \end{bmatrix}</math>

We see that ''d'' is a free parameter, and since we only need one particular we choose ''d=0'' for simplicity. Then we read off ''a = 10'', ''b - d = 5'', ''c = 10''. Putting this together we find a particular solution

<math>\vec{x_p} = \begin{bmatrix} 10 t e^{3t} + 5 e^{3t} \\ 10 t e^{3t} \end{bmatrix} </math>.

The general solution is the homogeneous solution (which we found in [[Science:Math_Exam_Resources/Courses/MATH215/December_2013/Question_05_(a)|part (a)]]) plus a particular solution:

<math>x(t) = C_1 \begin{bmatrix}1 \\ 1 \end{bmatrix} e^{3t} + C_2 \begin{bmatrix}3 \\ -2 \end{bmatrix} e^{-2t} + \begin{bmatrix} 10 t e^{3t} + 5 e^{3t} \\ 10 t e^{3t} \end{bmatrix} </math>.

Science:Math Exam Resources/Courses/MATH221/December 2009/Question 02 (b)/Solution 1

2015-06-20T06:41:00Z

Konradbe:

Strategy:

Use the following determinant properties.

The determinant of upper triangular matrix (i.e a matrix in row echelon form in this case) is equal to the product of its diagonal entries. If i<math>\neq</math>j , then subtracting some constant time row i from row j does not change the determinant.

Step 1:

Perform the following row operations on B:

row 2 = row 2 - 2 row 1

row 3 = row 3 - 2 row 1

and we get

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 2& -5& 0\\
0& 3& -1& 1
\end{bmatrix}</math>

Since the above matrix is still not in upper triangular form, perform the following row operations to it:

row 3 = row 3 - 2 row 2

row 4 = row 4 - 3 row 2

Then we get:

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 0\\
0& 0& 5& 1
\end{bmatrix}</math>

After performing one more row operation,

row 4 = row 4 + 5 row 3,

we get the following upper triangular matrix:

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 0\\
0& 0& 0& 1
\end{bmatrix}</math>

Step 2 :

Let <math>U' = \begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 0\\
0& 0& 0& 1
\end{bmatrix}</math>

Then <math>det(A) = det(U') = (1)(1)(-1)(1) = -1</math> and thus <math>det(A) = -1</math>.

Science:Math Exam Resources/Courses/MATH221/December 2009/Question 02 (b)/Solution 1

2015-06-20T06:40:06Z

Konradbe: Fixes typo as reported in typo form

Strategy:

Use the following determinant properties.

The determinant of upper triangular matrix (i.e a matrix in row echelon form in this case) is equal to the product of its diagonal entries. If i<math>\neq</math>j , then subtracting some constant time row i from row j does not change the determinant.

Step 1:

Perform the following row operations on B:

row 2 = row 2 - 2 row 1

row 3 = row 3 - 2 row 1

and we get

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 2& -5& 0\\
0& 3& -1& 1
\end{bmatrix}</math>

Since the above matrix is still not in upper triangular form, perform the following row operations to it:

row 3 = row 3 - 2 row 2

row 4 = row 4 - 2 row 2

Then we get:

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 0\\
0& 0& 5& 1
\end{bmatrix}</math>

After performing one more row operation,

row 4 = row 4 + 5 row 3,

we get the following upper triangular matrix:

<math>\begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 0\\
0& 0& 0& 1
\end{bmatrix}</math>

Step 2 :

Let <math>U' = \begin{bmatrix}
1& 0& 3& 0\\
0& 1& -2& 0\\
0& 0& -1& 0\\
0& 0& 0& 1
\end{bmatrix}</math>

Then <math>det(A) = det(U') = (1)(1)(-1)(1) = -1</math> and thus <math>det(A) = -1</math>.

Science:Math Exam Resources/Courses/MATH100/December 2014/Question 01 (h)

2015-05-23T04:35:01Z

Konradbe:

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