MATH6037: Week 2

I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.

Note: We are supposed to be C212 rather than B212 as advertised by Conor.

Missing Manual

One of you left your manual somewhere… email me if you want it before Wednesday (I have it).

Quiz 1 Results and Solutions

Below find the results. You are identified by the last four digits of your student number unless you are winning the league. The marks are out of 2.5 percentage points. Your best eight quizzes go to the 20% mark for quizzes. The R % column is your running percentage (for best eight quizzes) and the GPP is your Gross Percentage Points (for best eight quizzes).

S/N	Q1	Q2	Q3	Q4	Q5	Q6	Q7	Q8	Q9	Q10	R %	GPP
Kelliher	2.5										100	2.5
3281	2.4										96	2.4
2859	2.4										96	2.4
8335	2.375										95	2.4
8416	2.3										92	2.3
7878	2.15										86	2.2
5527	1.9										76	1.9
7209	1.9										76	1.9
8478	1.9										76	1.9
8403	1.85										74	1.9
2567	1.55										62	1.6
6548	1.55										62	1.6
8556	1.35										54	1.4
1864	1.3										52	1.3
9464	1.2										48	1.2
1852	0.7										28	0.7
****	0.6										24	0.6
****	0.5										20	0.5
5553	0.4										16	0.4
8455	0.2										8	0.2
5546	0										0	0
4198	0										0	0

Quiz 2 Question Bank

Tuestion bank for Quiz 2 (in Week 3) is as follows:

P.23, Q. 1-3, 6
P.31, Q.1-5, 7, 10

Here you can find the tables that will be allowed. The big tables will not be allowed. If you have any specific difficulties with these questions, please email me. Your Quiz 2 Questions will be taken from these. The final answers will not be given on the quiz paper and neither is there any value in writing down the final answers alone — you will receive marks for full and correct solutions — but nothing for final answers without justification or skipping important steps. No hints will appear either. Please don’t learn off model solutions — you need to understand the material not just on a superficial level to do well later on.

Quiz 2 runs from 19:15 to 19:30 sharp on Wednesday 18 February.

Week 6

As briefly mentioned in class, I will be away in Week 6 (11 March). I am proposing that we have a class over the Easter Break. In particular, I am thinking 1 April. Please email me if this not possible for you.

Maple Labs

Due to the fact that there are more than 20 students attending the module MATH6037 Mathematics for Science 2.1, we will have to introduce a lab split starting this week, on Wednesday 18 February.

The following is the proposed schedule for Weeks 3, 5, 6′, 7, 9, 11:

Group 1 – Starts at 18:00 and Finishes at 20:50:

Wednesdays 18:00-19:05 – Maple Lab in room C219

Wednesdays 19:15-19:30 – Weekly Quiz in C212

Wednesdays 19:30-20.50 – Theory class in room C212

Group 2 – Starts at 19:15 and Finishes at 22:00:

Wednesdays 19:15-19:30 – Weekly Quiz in C212

Wednesdays 19:30-20:50 – Theory class in room C212

Wednesdays 20.55-22:00 – Maple Lab in room C219

Spaces in each group will be allocated on a first come first served basis.

I will be starting the labs next Wednesday 18 February, so please respond to this ASAP outlining your preference for Group 1 or Group 2.

In weeks where there is no Maple Lab, the quiz will take place from 19:00-19:15 sharp.

Week 2

In Week 2 we looked at $u$ -substitutions. Then we studied Integration by Parts: this is the start of the new material (i.e. not MATH6019 material).

Week 3

In Week 3 we will look at Partial Fractions. They provide us with a way of integrating rational functions and will be very important later for when we study Laplace Methods.

In Maple we will do some basic plotting, differentiation and integration.

Notes

I have given out 20 sets of the notes (and still have three). I received the €11 from everyone who got a manual. Please bring €11 for next week if you need a manual.

A student was asking did I have MATH6019-type notes that revision could be done with. You could look at an old higher level maths book, look in the library for anything with “Calculus” in the title.

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage. Anyone can give me exercises they have done and I will correct them.

I also advise that you visit the Academic Learning Centre.

Continuous Assessment

The Continuous Assessment is broken into Weekly Quizzes (20%) and Maple (10%).

There will be ten weekly quizzes and your eight best results will count (so 2.5% per quiz from eight quizzes). You will receive an email (i.e. this one) on Thursday/Friday detailing the examinable exercises.

Maple consists of five labs and a Maple Test in the sixth lab. Satisfactory participation in labs gives you 1.5% and the Maple Test is worth 2.5%. More on this in the coming days.

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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February 15, 2015 at 11:29 am

David whitty

Put me down for group one of there still room JP

February 16, 2015 at 11:26 am

J.P. McCarthy

David,

Yes, no problem.

Regards,
J.P.

February 16, 2015 at 11:14 am

Student

Hi J.P.,

Would you a look at my Q.10 and tell me if its solved correctly and I have two queries on Q4 and Q5.

Thanks.

February 16, 2015 at 11:24 am

Q. 10: perfect.

Q.4: Usually in calculus\pure maths when we write $\log x$ we mean base $e$ :

$\log x\equiv \log_ex\equiv \ln x$

The calculator takes

$\log x\equiv \log_{10}x$ ,

and you see it behind the $10^{\cdot}$ button like $\ln x$ is behind the $e^{\cdot}$ button.

To be honest logs base 10 are a relic (http://math.stackexchange.com/questions/552038/are-base-ten-logarithms-relics)

Q.5: This is the Chain Rule (http://en.wikipedia.org/wiki/Chain_rule)

$\displaystyle\frac{d}{dx}\ln (2x)=\frac{1}{2x}\cdot \frac{d}{dx}(2x)=\frac{1}{2x}\cdot 2=\frac{1}{x}$ .

In this particular example, there is another way of showing that

$\displaystyle\frac{d}{dx}\ln(2x)=\frac{1}{x}$ .

We have that logs transform multiplication into addition as follows (here the $\log$ can have any base at all):

$\log (x\cdot y)=\log(x)+\log(y)$ .

Applied here we have

$\ln(2x)=\ln(2\cdot x)=\ln(2)+\ln(x)$ .

Now $\ln(2)\approx 0.6931$ is a constant so its derivative is zero:

$\displaystyle\frac{d}{dx}\ln(2x)=\frac{d}{dx}(\ln 2+\ln x)=0+\frac{1}{x}=\frac{1}{x}$ .

February 16, 2015 at 6:07 pm

Having a little trouble with one part of the questions on p. 31 and p. 32, it’s how you come up with your $v$ value?

In Q.1 you anti-differentiate $\cos(2x)\,dx$ and get $\frac12 \sin{2x}$ and this is fine to me as the $dx$ is cancelled by the anti differentiation.

But in Q.4 when you anti-differentiate $dx$ and you come up with $x$ as an answer? To me if I am to follow the format of the other questions my answer would have been $1$ as the $dx$ would be cancelled again? Am I looking at this in the wrong way?

February 16, 2015 at 6:17 pm

Yes, slightly.

Firstly $v$ isn’t a “value”. The $v$ -function would be better.

We have, in the first example

$dv=\cos (2x)\,dx\Rightarrow \int dv=\int \cos(2x)\,dx$ .

To say the $dx$ is cancelled isn’t quite correct. When we write

$\int f(x)\,dx$ ,

we are looking for all the functions, that when differentiated WITH RESPECT TO $x$ , give $f(x)$ .

For example,

$\displaystyle \int \cos(2x)\,dx=\frac{1}{2}\sin(2x)+C$

because when we differentiate $\frac12 \sin(2x)+C$ with respect to $x$ we get:

$\displaystyle\frac{d}{dx}\left(\frac{1}{2}\sin(2x)+C\right)$
$=\frac12 \cos(2x)\cdot\frac{d}{dx}(2x)+0=\frac12\cos(2x)\cdot 2=\cos(2x).$

Note we don’t need the $+C$ at this point… if you do use it you get:

$\int u\,dv=u(v+C)-\int (v+C)\,du=uv+uC-\int v\,du-\int C\,du$
$=uv+uC-\int v\,du-Cu=uv-\int v\,du$ anyway.

The $dx$ is only recording what variable we are anti-differentiating with respect to (it has a more concrete meaning when we write $\displaystyle\int_a^b f(x)\,dx$ — namely the width of the ‘strips’).

Note that you have $\int dv=v$ here and it didn’t cause you a problem…

Now there are a two ways of dealing with this.

Firstly note if $dv=dx$ then $dv=1\,dx$ and so

$\int dv=\int 1\,dx\Rightarrow v=x$ as the derivative of $x$ with respect to $x$ is one.

Secondly, just like $\int dv=v$ , adding up all the little bits of $x$ gives you $x$ :

$\int dx=x$ .

Hope this helps.

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MATH6037: Week 2

Missing Manual

Quiz 1 Results and Solutions

Quiz 2 Question Bank

Week 6

Maple Labs

Week 2

Week 3

Notes

Study

Continuous Assessment

Student Resources

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Categories

J.P. McCarthy Maths

6 comments

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MATH6037: Week 2

Missing Manual

Quiz 1 Results and Solutions

Quiz 2 Question Bank

Week 6

Maple Labs

Week 2

Week 3

Notes

Study

Continuous Assessment

Student Resources

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