**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.**

## Test 1

The first 15% test will take place at **4 pm MONDAY 20 October in B263**** **(Week 6). You can find a sample in the course notes, after the section on rates of change I think. It is a test that could arguably take 42 minutes but I’ll give ye from 16.05 — 17.00. You will be given a copy of these tables. Don’t worry I’ll scribble out the “UCC”!

Note that the format will be the same of this.

- Differentiation from First Principles
- Tangent Lines
- Differentiate by Rule
- Differentiate by Rule
- Differentiate by Rule
- Rates of Change
- Rate of Change/ Geometry of Graph,

but this is dependent on our progress in lectures.

## Week 5

In Week 5 we spoke about the duality/interplay between geometry & algebra and learnt how to find the maxima/minima of functions.

## Week 6

In Week 6 we will have Test 1 on Monday. In the remaining lectures we will do more examples of finding local maxima/minima of functions and perhaps begin some applied examples.

## Tutorials

The tutorial split is

**ON EVEN WEEKS (e.g. Week 4 is even)**

*– Group A’s tutorial is*~~Tuesday~~ Monday at ~~13:00~~ 17:00 in ~~B145~~ B189

*– Group B’s tutorial is Friday at 09:00 in B185*

**ON ODD WEEKS (e.g. Week 3 is odd)**

*– Group A’s tutorial is Friday at 09:00 in B145*

*– Group B’s tutorial is*~~Tuesday~~ Monday at ~~13:00~~ 17:00 in ~~B145~~ B189

## Continuous Assessment

As can be seen here in the Module Descriptor, there will be two 15% tests: one in Week 6 and one in Week 10. I hope to give you two week’s notice of each and there are sample tests in the notes.

## Student Resources

Student Resources are now collated at the Student Resources tab on the top of this page.

## 2 comments

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October 20, 2014 at 1:09 pm

StudentHi,

I am doing some last minute revision at the moment and I am finding Q7 on P.60 difficult.

Would appreciate any help.

Thanks.

October 20, 2014 at 1:20 pm

J.P. McCarthyStudent,

O.K. we are given that the displacement, (position/distance/distance travelled are the same thing for us), is given by

.

Now the rate of change of displacement is velocity, (speed) and we learnt that the rate of change of a function is it’s derivative. So we have

.

The rate of change of velocity then is acceleration (acceleration is a change in speed. Velocity is a change in displacement) so that

Now we can find the velocity after four seconds:

.

The units of displacement is metres and of time is seconds. So the units of velocity are

,

where means the units of … so the answer isn’t 136 but 136 .

The second part is to find when the velocity is zero. So we solve

.

To look for factors we examine . The factors of 12 are , and however none of these can ‘make’ 6. Hence we need the formula with , and .

.

so presumably I didn't include this negative-time solution.

Regards,

J.P.