Test 2
Test 2, worth 15% and based on Chapter 3, will now take place Week 12, 30 April. There is a sample test [to give an indication of length and layout only] in the notes (marking scheme) and the test will be based on Chapter 3 only.
More Q. 1s (on the test) can be found on p.112; more Q. 2s on p. 117; more Q. 3s on p.125 and p.172, Q.1; more Q. 4s on p.136, and more Q. 5s on p. 143.
Chapter 3 Summary p. 144.
Please feel free to ask me questions via email or even better on this webpage.
Homework
Once you are prepared for Test 2 you can start looking at Chapter 4:
- Revision of Integration, p.161.
- p.167, Q. 1-5
- p.182
Week 11
We had some tutorial time from 18:00-19:00 for further differentiation (i.e. for Test 2, after Easter).
We completed our review of antidifferentiation before starting Chapter 4 proper.
We looked at Integration by Parts and centroids.
For those who could not make it here is some video and slides from what we did after the video died.
Week 12
We will have some tutorial time from 18:00-19:00 for further differentiation.
We will have Test 2 from 19:00-20:05.
Then we will look at completing the square, centres of gravity, and work.
Week 13
We will look the Winter 2018 paper at the back of your manual.
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.
2 comments
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April 11, 2019 at 7:16 am
Student
Hi J.P.,
Stuck on p. 102, Q 1 (f)
I have attached my attempt. I’ve been checking out khan academy and it said I needed the change of base formula but I got confused from there.
Thanks.
April 11, 2019 at 7:23 am
J.P. McCarthy
The confusion here, largely my fault perhaps, is that when you see
in the calculus context that it is actually the natural log. So that, in this context:
This is the confusion.
So what you are differentiating is
. It isn’t 
and so you have a chain rule where the ‘outside’ is 
and the ‘inside’ is 
.
The chain rule says that
that is “differentiate the outside — evaluate that at the inside — and the multiply by the derivative of the inside’.
The derivative of the outside,
is in the tables and is 
. Evaluating this at the inside gives 
but we must multiply this by the derivative of the inside, the derivative of 
, which is 
.
All in all we have:
You probably need to move beyond this revision and tackle the new material — it might be the case that you don’t have to calculate more difficult derivatives such as these to do the new material.
Regards,
J.P.