The storm cost us one lecture. Please watch this cantilever example and this summary of beams to catch up. Wednesday, we looked at Euler’s Method in the morning and had an extra tutorial in the evening.

In Week 7 we will look at the Three Term Taylor Method and begin Chapter 3 on Probability and Statistics.

Assessment 2 will be handed out on Friday but doesn’t have to be completed until the Monday of Week 11, giving you more than five full weeks to complete it.

Please feel free to ask me questions about the exercises via email or even better on this webpage — especially those of us who struggled in the test.

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

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The storm interrupted us, costing us two lectures. You must watch the following videos to catch up: Inclusion-Exclusion Principle and the properties of relations. We have/had our test this Friday. A sample has been handed out.

In Week 7 w will start the chapter on Functions.

Test 1 will be held at 09:00, Friday 20 October in Week 6. Everything up to but not including relations is examinable. I have emailed ye a sample, a hard copy of which I gave out in tutorials.

We are probably all going to have to put in some extra study before the test. Please try and find some extra time to try exercises, particularly with any new or harder material such as logs, cartesian products, and power sets. Please feel free to ask me questions about the exercises via email or even better on this webpage.

Anyone who is missing notes is to email me.

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc. There are some excellent notes on Blackboard for MATH6055.

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In Week 5 we looked more at sets and explored Cartesian Products We introduced relations. Then on Friday we looked harder at elements, subsets, power sets and the empty set.

In Week 6 we will look at the Inclusion-Exclusion Principle and the properties of relations. We should start the chapter on Functions and then have our Test 1 on Friday.

Test 1 will be held at 09:00, Friday 20 October in Week 6. Everything up to but not including relations is examinable. I have emailed ye a sample, a hard copy of which I will provide you with next week.

We are probably all going to have to put in some extra study before the test. Please try and find some extra time to try exercises, particularly with any new or harder material such as logs, cartesian products, and power sets. We all have only one hour of tutorials before the test and finding two hours to do exercises will make a big difference. Please feel free to ask me questions about the exercises via email or even better on this webpage.

Anyone who is missing notes is to email me.

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc. There are some excellent notes on Blackboard for MATH6055.

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Fundamental Principle of Solving ‘Easy’ Equations

Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with thesamesolution.

There are a number of subtleties here: basically sometimes you get extra ‘solutions’ (that are not solutions at all), and sometimes you can lose solutions.

Let us write the squaring function, e.g. , by and the square-rooting function by . It appears that are an inverse pair but not quite exactly. While

and

,

check out O.K. note that

,

does not bring us back to where we started.

This problem can be fixed by restricting the allowable inputs to to positive numbers only but for the moment it is better to just treat this as a subtlety, namely while , … in fact I recommend that we remember that with an there will generally be *two *solutions.

The other thing we look out for as much as possible is that *we cannot divide by zero*.

There are other issues around such as the fact that , so that the equation has no solutions (no, is not a solution! Check.). This equation has *no *solutions.

Often, in context, these subtleties are not problematic. For example, equations with no solutions rarely arise and quantities might be positive so that if we have , only need be considered (for example, might be a length).

Highlighting these small subtleties is not therefore to scare you off but to allow you full mastery. Such as in the next example.

*Solve*

*Solution: *Take out the common factor of :

.

Assume that is non-zero (you can’t divide by zero).

Let us see what it means if this is zero:

or

Now perhaps from the context of the equation, both and are strictly positive (bigger than zero), so the second possibility can’t occur. The first possibility can happen: if we have that the original equation reads:

Therefore you can assume that because if they are equal and the equation can’t be solved for .

So assuming you do have:

Now remember because of the square there are *two* solutions (i.e. — you need plus *and * minus). Therefore the correct implication is that:

.

However, if you assume that , then it cannot be the negative square root.

In fact, under these assumptions:

- ,

we can tidy this up a little further. Multiply the number inside the square root by :

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Have been emailed to you along with the marking scheme.

I want to make the following remarks. Below we have a plot of Final Grade vs Test 1 Mark for last year’s class.

Everyone under the horizontal line failed MATH6040. Note the students in the bottom right. They got ~70% and ~80% in the Vectors test and still failed. There were various students with ~55-100% in Vectors who barely passed. *On average*, the final grade was about 0.6 times the Test 1 Result + 8. This means that *on average *students with below 53% on Test 1 failed MATH6040. That is people to the left of the vertical line.

Therefore, note vectors is the easiest chapter in MATH6040 so don’t get too carried away with your mark. Conversely, if you have done poorly you need to take immediate action: by attending all your tutorials and possibly the Academic Learning Centre. The three students in the top left would have taken this advice. The seven students in the bottom left would not have.

The second thing to note about the results is the impact of attendance on grades. *On average*:

- those with satisfactory attendance got 74% — a distinction
- those with one attendance warnings got 63% — a merit 1
- those with two attendance warnings got 37% — a FAIL

The third thing to note is that perceived ability is not as important as attendance. Of those who attended the quick test, the correlation coefficient between Quick Test mark and Test 1 mark was only 0.27 while the correlation coefficient between Attendance Warnings and Test 1 mark was -0.59.

Roughly, this suggests that attendance is twice as important as ability.

We looked at Determinants and their use in Cramer’s Rule.

We will start Chapter 3 with a quick review of differentiation followed by looking at Parametric Differentiation.

On Chapter 3, not until Week 11.

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

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

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In Week 5 we finished looking at simply supported beams. We then looked at fixed end beams and cantilevered beams.

In Week 6 we will summarise what we learnt about beam and look at numerical approximations to solutions of differential equations that we cannot solve exactly.

Assessment 1 has been emailed to all of you. The hand in date is, tomorrow, 17:30 Thursday 12 October 2017. I will be in A283 from 10:00-12:00 and 13:00-17:30. Work handed in late shall be assigned a mark of zero.

Assessment 2 will be handed out on Friday but doesn’t have to be completed until the Monday of Week 11, giving you more than five full weeks to complete it.

Please feel free to ask me questions about the exercises via email or even better on this webpage — especially those of us who struggled in the test.

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

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*A car has to travel a distance on a straight road. The car has maximum acceleration and maximum deceleration . It starts and ends at rest.*

*Show that if there is no speed limit, the time is given by*

.

*Solution: *We are going to use two pieces of information:

- the area under the velocity-time graph is the distance travelled,
- the slope of the velocity-time graph is the acceleration.

First, a rough sketch of the velocity-time graph:

*The slope of the first line segment is , while the slope of the second line segment is . The area under the curve is given by , the total distance. The base is , the total time, and the height is the maximum speed.*

Let be the maximum speed reached. Using the area of a triangle, where is the total time:

.

Now using the fact that the slope of the line segments, rise/run, is equal to the accelerations, we derive:

, and

.

Thus

.

Now, using :

.

Taking square roots (and noting ) completes the proof.

*A particle, moving in a straight line, accelerates uniformly from rest to a speed . It continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance traveled while accelerating is 6 m. The total distance traveled is 30 m and the total time taken is 6 s. *

i. *Draw a velocity-time graph and hence, or otherwise, find .*

ii. *Calculate the distance travelled at constant velocity.*

*Solution: *

i. First the velocity-time graph:

There are number of things we need to use:

- the area under the velocity-time graph is the distance traveled
- the slope of the velocity-time graph is the acceleration

The question gives the following:

- the total distance is 30 m
- the total time is 6 s
- the distance traveled from to is 6 m.

In the first triangle:

.

In the second triangle:

,

and also . Define Let so that . Therefore, using the fact that the total time is 6 s:

.

Now, using the fact that the area under the first triangle is 6:

,

and the total area is 30:

m/s.

ii. Using we have so that the time spent at constant velocity is:

,

and therefore the distance traveled is:

.

Alternatively note that and so m.

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In Week 4 we finished looking at logarithms and started the second chapter on Sets and Relations. I tried and failed to tell you about the barber paradox.

In Week 5 we will look more at sets and set identities, and explore Cartesian Products and perhaps introduce relations.

Test 1 will be held at 09:00, Friday 20 October in Week 6. Note this is different to your assessment schedule provided by the head of department, Tim Horgan. Expect a sample this time next week. Only material from Weeks 1-5 will be examined.

We are probably all going to have to put in some extra study before the test. Please try and find some extra time to try exercises, particularly with any new or harder material such as logs. We all have two hours of tutorials before the test and finding two extra hours to do exercises will make a big difference. Please feel free to ask me questions about the exercises via email or even better on this webpage.

Anyone who is missing notes is to email me.

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc. There are some excellent notes on Blackboard for MATH6055.

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The 15% Test 1 will take place at 16:00 on 9 October, Week 5, in B263. There is a sample test in the notes.

We did some examples of matrix arithmetic and looked at Matrix Inverses — “dividing” for Matrices. This allowed us to solve matrix equations. Here find a note that answers the question: why do we multiply matrices like we do?

We will look at linear systems, and determinants.

You should certainly consider putting in some extra study for the Test: in particular BioEng2B who have only one more tutorial before the test. Please feel free to ask me questions about the exercises via email or even better on this webpage.

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In Week 4 we started looking at simply supported beams. We had a tutorial instead of a lecture Wednesday morning.

In Week 5 we might finish looking at simply supported beams and have a look at fixed end beams.

Assessment 1 has been emailed to all of you. The hand in date in Thursday 12 October. Work handed in late shall be assigned a mark of zero.

Please feel free to ask me questions about the exercises via email or even better on this webpage — especially those of us who struggled in the test.

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