I am emailing a link of this to everyone on the class list every Friday afternoon. 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.ie and I will add you to the mailing list.

## Notes

As of 12 October: MATH6000 Lecture Notes (with gaps).

Also an E-Book: Engineering Mathematics by John Bird.

Also I don’t know why I didn’t advise this earlier — you would be well worth investing in a ring binder (to be kept at home or whatever) for your notes. You can see already the amount of sheets. You should be organised with these and only bring the ones we are working on to class.

## Assessment 2

Your second assessment is on Wednesday 24 October in Week 6.

Trigonometry, Approximation and Statistics will be examined.

The sample is currently being drafted.

The policy on missed assessments is very strict in CIT as you can see here.

## Exercises

The following exercise sheets are available on Blackboard for Assessment 2:

You have the last four and I will give you the trigonometry exercise sheet on Monday.

Common Entry Science: We will work through Trigonometry first and then finish off the other sheets.

Biosciences: We should work through the exercise sheets in order.

Computing: We should work through the exercise sheets in order.

## Frequency Distributions

The third statistics exercise sheet is on frequency distributions. I hope to cover this in class but just in case you miss this here are the notes and formulas.

## Error Problems

Firstly what we learnt in lectures applie. When we approximate a quantity $Q$ by $A$ we might denote this by $Q\approx A$. We define the absolute error (of the approximation) as

$\displaystyle \Delta Q=|Q-A|$,

where $|\cdot|$ is the absolute value. Then we can take the relative or percentage error as

$\displaystyle \frac{\text{absolute error}}{\text{true value}}=\frac{\Delta Q}{Q}$.

For Q. 9 – 12 I said that we’d hold back until I found a better method of doing these questions as the one I was using was far from elegant. Just in case you didn’t see the mess I made let’s take a look at question 9 using the terrible method I used.

### Question 9

The side of a square is measured 5% too long. Find the approximate percentage error in the area of the square.

Car Crash Solution: Let $s_1$ be the approximate length of the side. Hence the approximate area is given by $A_1=s_1^2$. Now the approximation to the side length, $s_1$, was 5% too big so represented 105% of the true side length. Hence we have

$\displaystyle s_1=(1.05)s_0\Rightarrow s_0=\frac{s_1}{1.05}$.

Hence the true area is given by:

$\displaystyle A_0=\left(\frac{s_1}{1.05}\right)^2=\frac{s_1^2}{1.1025}$.

Hence the error is given by ($|A_0-A_1|=A_1-A_0$ as the approximation is an over-approximation. Hence $A_0-A_1$ is negative and the absolute value of a negative quantity $N$ is $-N$: $|N|=-N$ if $N$ is negative):

$\displaystyle |A_0-A_1|=A_1-A_0=s_1^2-\frac{s_1^2}{1.1025}$.

Now the percentage error is given by the absolute error over the true value:

$\displaystyle\frac{s_1^2-\frac{s_1^2}{1.1025}}{\frac{s_1^2}{1.1025}}$.

We want this to be a real number so try and eliminate the variable $s_1$. Divide above and below by $s_1^2$:

$\displaystyle\frac{1-\frac{1}{1.1025}}{\frac{1}{1.1025}}$.

We could put this into the calculator but I want to show how messy this solution truly is! I don’t like fractions in the numerator and denominator so I try and get the 1.1025s out of there. I can achieve this by multiplying above and below by 1.1025 AKA multiplying by one:

$\displaystyle\frac{1.1025-1}{1}=\frac{0.1025}{1}=0.1025$.

So the percentage error is 10.25%; or if you like, approximately 10%.

Slicker Solution: Suppose that the true side length is $s$. Then the approximate side length is 5% longer: $(1.05)s$. Hence the true area is $s^2$ but the approximate error is $(1.05s)^2=1.1025s^2$. Hence the approximation increased the area by 10.25% — the approximation had an error of 10.25%; or if you like, approximately 10%.