MATH6037: Week 12

May 7, 2015 in MATH6037

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.

Continuous Assessment Summary

The results of the individual components of the CA are lower down — in particular Quiz 11.

Unless you won or came second in the league, you are identified by the last four digits of your student number. QPP is your QPP out of 20, MM are your Maple Marks out of 100 (Maple is worth 10% so say 80% gives you 8 Marks for your final grade). GPP is your Gross Percentage Points for CA (out of 30). PP is your Passing Percentage for the final paper — what you need on the final paper to pass. Finally FP is your First Percentage — what you need on the final paper to get a first: 70%.

S/N	QPP	MM	GPP	PP	FP
Kelliher	20.0	93.2	29.3	15.3	58.1
Kiely	19.8	94.3	29.2	15.4	58.2
3281	19.9	89.8	28.9	15.9	58.7
5527	20.0	88.6	28.9	15.9	58.8
8416	19.7	84.1	28.1	17.0	59.8
8403	18.1	89.8	27.1	18.5	61.3
4198	18.5	85.2	27.0	18.5	61.4
6548	18.5	79.5	26.4	19.4	62.3
1864	17.2	86.0	25.8	20.3	63.1
8478	17.2	83.5	25.5	20.7	63.6
8556	17.5	76.1	25.1	21.3	64.2
7878	16.6	77.0	24.3	22.4	65.3
5546	16.1	77.3	23.8	23.1	66.0
2567	15.2	84.1	23.6	23.4	66.3
8603	15.7	77.3	23.4	23.7	66.6
1852	14.4	88.6	23.3	23.9	66.8
8455	14.4	78.4	22.2	25.4	68.3
2859	14.0	78.4	21.8	25.9	68.8
7950	10.4	63.4	16.7	33.2	76.1
4775	9.2	75.0	16.7	33.3	76.1
9464	8.6	60.0	14.6	36.2	79.1
7209	5.5	45.0	10.0	42.8	85.7
5553	1.3	45.0	5.8	48.9	91.7

Week 12

In Week 12 we finished off all but the last example. Then we spoke about applications — especially to damped harmonic oscillators.

Week 13

In Week 13 we will go over the exam paper after p.152. We haven’t done Q.1(b) and 2 (c) in the same way so I will have alternative questions.

Academic Learning Centre

The word on the Academic Learning Centre is that although the evening session perhaps might have been made exclusive to evening students, the fact of the matter is that they are not.

My departmental head suggested that if a group of ye want to get an improvement in your ALC experience, that ye should email questions to catherine.palmer@cit.ie in advance of the session. Dr Palmer said that this will allow her to more easily help ye.

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 Details

You are identified by the last four digits of your student number unless you are winning the league. The individual quiz 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 — now this includes missed quizzes — before I was doing the best non-zero but now I am including the zeros if a zero is in your best eight), MPP is your Maple Percentage Points for the biweekly lab, MT your mark on the Maple Test and MM your Maple Marks (as a percentage). GPP is your Gross Percentage Points (for best eight quizzes and Maple). Most of the columns are rounded but columns 11 & 12, for quizzes 10 & 11, are correct — as is GPP.

S/N	Q1	Q2	Q3	Q4	Q5	Q6	Q7	Q8	Q9	Q10	Q11	R %	QPP	MPP	MT	MM	GPP
Kelliher	3	3	3	3	2	3	3	3	3	2.5	2.5	100	20.0	7.5	1.8	93.2	29.3
Kiely	2	3	2	3	3	3	2	3	2	2.4	2.5	99	19.8	7.5	1.9	94.3	29.2
3281	2	3	3	3	2	3	3	3	2	2.4	2.5	100	19.9	7.5	1.5	89.8	28.9
5527	2	3	3	3	3	3	3	3	3	2.4	2.5	100	20.0	7.5	1.4	88.6	28.9
8416	2	1	2	3	3	1	2	3	3	2.5	2.5	99	19.7	7.5	0.9	84.1	28.1
8403	2	1	2	3	1	2	2	3	2	1.8	2.1	91	18.1	7.5	1.5	89.8	27.1
4198	0	1	2	3	2	3	3	3	2	1.1	2.5	93	18.5	7.5	1.0	85.2	27.0
6548	2	1	2	3	2	2	2	3	3	2.4	1.9	92	18.5	7.5	0.5	79.5	26.4
1864	1	2	3	2	1	1	2	3	1	2	2	86	17.2	7.5	1.1	86.0	25.8
8478	2	2	2	2	0	3	2	2	2	1.8	2	86	17.2	7.5	0.9	83.5	25.5
8556	1	1	2	2	0	2	3	3	1	2.1	2.4	87	17.5	7.5	0.1	76.1	25.1
7878	2	2	2	2	2	2	2	2	2	0.6	1	83	16.6	7.5	0.2	77.0	24.3
5546	0	0	1	1	2	2	3	2	3	2.2	2.5	81	16.1	7.5	0.2	77.3	23.8
2567	2	2	2	1	2	2	2	2	2	1.1	1.3	76	15.2	7.5	0.9	84.1	23.6
8603	1	2	2	2	0	2	3	2	2	0.8	0.7	78	15.7	7.5	0.2	77.3	23.4
1852	1	1	2	2	2	2	1	1	2	2.3	1.3	72	14.4	7.5	1.4	88.6	23.3
8455	0	1	1	1	2	2	2	1	2	1.5	2.5	72	14.4	7.5	0.3	78.4	22.2
2859	2	1	0	1	2	2	2	3	0	0.5	2.4	70	14.0	7.5	0.3	78.4	21.8
7950	0	0	1	0	2	1	3	3	0	1.3	0.3	52	10.4	6	0.3	63.4	16.7
4775	1	0	1	0	0	1	2	2	1	0.4	0.9	46	9.2	7.5	0.0	75.0	16.7
9464	1	1	2	1	1	1	2	0	0	0	0	43	8.6	6	0.0	60.0	14.6
7209	2	1	2	1	0	0	0	0	0	0	0	28	5.5	4.5	0.0	45.0	10.0
5553	0	1	0	0	0	0	0	0	0	0	0	7	1.3	4.5	0.0	45.0	5.8

Student Resources

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

8 comments

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May 16, 2015 at 11:50 am

Student

Hi J.P.

Just a quick question.

In the question attached that we did in class; Q. 1 (a) (ii); when we differentiated what was in the brackets should we not have kept the $2y$ as it was plus 0 not multiplied by?

Cheers.

May 16, 2015 at 11:53 am

J.P. McCarthy

We have using the chain rule

$\displaystyle \frac{\partial z}{\partial y}=-\frac12(x^2+y^2)^{-3/2}(0+2y)$ .

Note this is the same as

$\displaystyle (-1)\times \left(\frac12\right)\times (x^2+y^2)^{-3/2}\times(2)\times y$ .

Now these are all numbers so the order of multiplication does not matter so I can write this as

$\displaystyle (-1)\times \left(\frac12\right)\times 2\times y\times (x^2+y^2)^{-3/2}=-y(x^2+y^2)^{-3/2}$ .

Does this answer your question?

Regards,
J.P.

May 16, 2015 at 11:57 am

Student

J.P.,

Can you send me the worked solutions for questions 2(c) and 3(a) from Summer 2013?

Regards,

May 16, 2015 at 12:08 pm

J.P. McCarthy

I think you are close to Q.2 (c) but got mixed up with the 3000 and lost the square when calculating $\displaystyle \frac{\partial g}{\partial \ell}$ — everything else looks good.

Your errors are correct as well as your calculation of $g$ .

$\displaystyle \frac{\partial g}{\partial t}=-13.68$ is correct.

You should have

$\displaystyle \frac{\partial g}{\partial \ell}=\left.\frac{2(1)}{t^2}\right|_{t=1.43}=0.9780$ .

This gives

$\Delta g\approx dg=|-13.68|(0.01)+|0.9780|(0.001)=0.1378\approx 0.1\text{ m s}^{-2}$ .

This gives a final answer of

$g=9.8\pm0.1$ m s $^{-2}$ .

You have stopped on Q.3 (a)…

You are correct at

$s^2X(s)-2s+2sX(s)+X(s)=0$ .

To solve for $X(s)$ you need to do

$s^2X(s)+2sX(s)+X(s)=2s$

$\Rightarrow X(s)[s^2+2s+1]=2s$

$\displaystyle \Rightarrow X(s)=\frac{2s}{s^2+2s+1}$ .

Now you need to do partial fractions on $X(s)$ [HINT: It is a Rule II].

For part two, you can see before or after solving part (i), that the door closer is well designed because it is critically damped.

Regards,
J.P.

May 17, 2015 at 6:28 pm

Student

Hi J.P.,

I was just wondering if you could help us with one of the past exam paper questions. It’s from Winter 2011: Q3 (a) (i).

It’s a damped harmonic oscillator question
$\displaystyle x''(t)+x'(t)+\frac{17}{4}x(t)=0\,,\qquad x'(0)=3\,,\qquad x(0)=0$
I get to the point
$\displaystyle X(s)=\frac{3}{s^2+s+17/4}.$

and am just wondering where to go from there?

Thanks.

May 17, 2015 at 6:37 pm

J.P. McCarthy

On the one hand this is a messy question the likes of which you won’t get but on the other hand you should still know how to approach it in general.

So we get to here and this yoke isn’t in the tables. Therefore we need to apply a partial fraction expansion. Therefore we need to factorise $s^2+s+\frac{17}{4}$ but we can’t seem to do this using the method I showed ye… is it factorisable at all?

$\displaystyle b^2-4ac=1^2-4(1)\left(\frac{17}{4}\right)=-16<0$

and so there are no real roots and so no real factors. Therefore looking at the tables again perhaps we are dealing with shifted sines/cosines… we must complete the square.

So we want

$\displaystyle s^2+s+\frac{17}{4}\overset{!}{=}(s+a)^2+k^2$

$=s^2+2as+a^2+k^2$

$2a\overset{!}{=}1\Rightarrow a=\frac{1}{2}$ .

Also we want

$\displaystyle a^2+k^2\overset{!}{=}\frac{17}{4}$

$\displaystyle \Rightarrow \frac14+k^2+\frac{17}{4}$

$\displaystyle \Rightarrow k^2=\frac{16}{4}=4$

$\Rightarrow k=\pm2$ … take $k=2$ .

So we have

$\displaystyle X(s)=\frac{3}{\left(s+\frac12\right)^2+2^2}$ .

Can you take it from here?

Regards,
J.P.

May 18, 2015 at 10:04 pm

Student

Hey J.P.,

I have attached a scan with a problem if you could take a look at it please.

Thanks.

May 18, 2015 at 10:08 pm

J.P. McCarthy

Yes you are not doing this correctly.

The derivative of $e^y$ with respect to $e^y$ but as it is more complicated you need a Chain Rule. It might help to write it as $e^{x\cdot y^3}=\exp(x\cdot y^3)$ to see this:

$\displaystyle \frac{\partial}{\partial y}\exp(x\cdot y^3)=\exp(x\cdot y^3)\cdot \frac{\partial}{\partial y}(x\cdot y^3)$

$=\exp(x\cdot y^3)\cdot x(3y^2)=3xy^2e^{xy^3}$ .

Regards,
J.P.

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MATH6037: Week 12

Continuous Assessment Summary

Week 12

Week 13

Academic Learning Centre

Study

Continuous Assessment Details

Student Resources

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J.P. McCarthy Maths

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MATH6037: Week 12

Continuous Assessment Summary

Week 12

Week 13

Academic Learning Centre

Study

Continuous Assessment Details

Student Resources

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J.P. McCarthy Maths

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