MATH6040: Winter 2019, Week 12

Mathematics Exam Advice

The first piece of advice is to read questions carefully. Don’t glance at a question and go off writing: take a moment to understand what you have been asked to do.
Don’t use tippex; instead draw a simple line(s) through work that you think is incorrect.
For equations, check your solution by substituting your solution into the original equation. If your answer is wrong and you know it is wrong: write that on your script.

If you do have time at the end of the exam, go through each of your answers and ask yourself:

have I answered the question that was asked?
does my answer make sense?
check your answer (e.g. differentiate/antidifferentiate an antiderivative/derivative, substitute your solution into equations, check your answer against a rough estimate, or what a picture is telling you, etc)

Student Feedback

You are invited to give your feedback on my teaching and this module here.

Test 2 – Results

Have been emailed to you along with final CA results.

Week 12

We looked at centroids of laminas and centres of gravity of solids of revolution.

We had one and a half classes of tutorial time.

Week 13

There is an exam paper at the back of your notes — I will go through this on the board in the lecture times (in the usual venues):

Monday 16:00
Tuesday 09:00
Thursday 09:00

We will also have tutorial time in the tutorial slots. You can come to as many tutorials as you like.

Monday at 09:00 in B180
Monday at 17:00 in B189
Wednesday at 10:00 in F1. 3

The exam is on Friday 13 December. Past exam papers (MATH6040 runs in Semester 1 and Semester 2) may be found here.

Academic Learning Centre

If you are a little worried about your maths this semester, perhaps after the Quick Test or in general, I would just like to remind you about the Academic Learning Centre. Most students received slips detailing areas of maths that they should brush up on. The timetable is here.

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage.

CIT Mathematics Exam Papers

These are not always found in your programme selection — most of the time you will have to look here.

Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.

6 comments

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December 10, 2019 at 11:21 am

Student

Hi J.P.

Just regards Autumn 2018, Q. 1 ii? It says to show the vectors are perpendicular but their dot product does not equal zero?

December 10, 2019 at 11:24 am

J.P. McCarthy

I find

$\mathbf{u}+\mathbf{v}=4\mathbf{i}-4\mathbf{j}+4\mathbf{k}$

and

$\mathbf{u}-\mathbf{v}=-2\mathbf{i}+2\mathbf{k}$ .

$(\mathbf{u}+\mathbf{v})\bullet(\mathbf{u}-\mathbf{v})=(4)(-2)+(-4)(0)+4(2)=0$ .

Regards,
J.P.

December 12, 2019 at 8:21 am

Evening,

Sorry to disturb you so late, I’m a little confused here on Q 1 (b) i. from the Winter 2017 paper.

How does
$\mathbf{F}=(260)(\widehat{\mathbf{AB}})$
give
$\mathbf{F}=(240, 60, 80)$ .

Clearly each coefficient of $\mathbf{i}, \, \mathbf{j},\,\mathbf{k}$ is multipiled by two, but here they are multiplied by 260?

Regards.

December 12, 2019 at 8:25 am

Let us note that $\mathbf{AB}=(120,30,40)$ .

We know that $\mathbf{F}$ is of magnitude 260 and in the direction of $\mathbf{AB}$ .

$|\mathbf{AB}|=130$ , and so $\mathbf{F}=2\mathbf{AB}=(240,60,80)$ .

If we cannot see that 260 is twice 130, or if these magnitudes are not comparable, instead we find a unit vector in the direction of $\mathbf{AB}$ :

$\displaystyle\widehat{\mathbf{AB}}=\frac{\mathbf{AB}}{|\mathbf{AB}|}=\frac{(120,30,40)}{130}$ ,

$\displaystyle\mathbf{F}=260\widehat{\mathbf{AB}}=260\frac{(120,30,40)}{130}=(240,60,80)$ .

December 12, 2019 at 12:11 pm

Hi J.P.,

Just to clarify, sorry I know you explained already, could you show me how to answer part c.

December 12, 2019 at 12:19 pm

Part i is

$\displaystyle \underbrace{\left(\begin{array}{ccc}1 & -1 & -1 \\ 2 & 0 & 5 \\ 2 & 4 & 0\end{array}\right)}_{=A}\underbrace{\left(\begin{array}{c}x\\ y \\ z\end{array}\right)}_{\mathbf{b}}=\underbrace{\left(\begin{array}{c}0\\ 6 \\ 4\end{array}\right)}_{\mathbf{b}}$

Part ii, we want to get rid of the $A$ to get $\mathbf{x}$ on its own. To do this we multiply both sides (on the left) by $A^{-1}$ :

$A\mathbf{x}=\mathbf{b}\Rightarrow A^{-1}A\mathbf{x}=A^{-1}\mathbf{b}$
$\Rightarrow \mathbf{x}=A^{-1}\mathbf{b}$ .

$A^{-1}$ is given in part iii. and so

$\displaystyle\mathbf{x}=A^{-1}\mathbf{b}=\frac{1}{38}\left(\begin{array}{ccc}20 & 4 & 5 \\ -10 & -2 & 7 \\ -8 & 6 & -2\end{array}\right)\left(\begin{array}{c}0\\6\\4\end{array}\right)$

Fix the constant $\frac{1}{38}$ .Multiply the $3\times 3$ matrix by the $3\times 1$ matrix to get a $3\times1$ matrix. Multiply (all entries) by $\frac{1}{38}$ .

This is the vector

$\displaystyle \mathbf{x}=\left(\begin{array}{c}x \\ y \\ z\end{array}\right)$

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MATH6040: Winter 2019, Week 12

Mathematics Exam Advice

Student Feedback

Test 2 – Results

Week 12

Week 13

Academic Learning Centre

Study

CIT Mathematics Exam Papers

Student Resources

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J.P. McCarthy Maths

6 comments

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MATH6040: Winter 2019, Week 12

Mathematics Exam Advice

Student Feedback

Test 2 – Results

Week 12

Week 13

Academic Learning Centre

Study

CIT Mathematics Exam Papers

Student Resources

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J.P. McCarthy Maths

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