Simpson's Rule

Today we looked at the application of Simpson's rule for working out the area of an irregular shape.

The shape is divided into an even number of strips which are of equal width.

You are given a shape like this (taken from LC paper 2005).
It is divided into an even number of strips (in this case 6) and each strip is bounded by a strip length. As there are 6 strips there are 7 strip lengths.

Note that the last strip length is zero.


First step is to label each strip length as odd or even.

<- Like this.






Now apply Simpson's rule.
The rule is that the area of this shape
= w/3(First + last + 2(sum of odds) + 4(sum of evens))
Filling this in:
area = 18/3 (10 + 0 + 2(30 + 36) + 4 (25 + 38 + 22))
= 6(10 + 132 + 340)
= 6(482)
= 2892m²

[As it is easy to make an error in the adding and multiplying, it is a good habit to do a rough check - to make sure that your answer is in the right ballpark. You can check this by approximating the average length of strip and multiplying by the total width. Width = 18 x 6 = 108 Average length = 26 (rough estimate) Ballpark for area = 108 x 26 = 2808m%sup2;]


In this diagram, there are 4 (an even number) strips, and there are two sets of measurements for each strip.

Label the strip lengths as odd or even and add the upper length to the lower one.

... giving you this
The total lengths are in green. You then sub these into the Simpson's Rule formula:








Area = 5/3(0 + 0 + 2(11) + 4(8 + 6))
= 5/3(22+ 56)
= 5/3(78)
= 130

Again ballpark estimate would be: 20x 7 = 140.



Simpson's rule with Missing Values.
There are two possibilities - either the width of each strip is missing or one of the strip lengths is missing. In either case, you will be given the area and you use it to build an equation around your unknown.

In this example the strip width is not given.
Let's say that you are told that the
area = 1432m²










You set up an equation like this:
1432 = x/3(0 + 0 + 2(37) + 4(30+41))
1432 = x/3(358)
4296 = 358x
x = 12



The missing value could be in the strip length.
In fact there could be two missing values, both expressed in terms of x. For example.

In this example, the area = 270m²

















Set up an equation:

270 = 3/3(12 + x + 2(2x+18) + 4(18+22+8))
270 = 1(12 + x + 4x + 36 + 192)
270 = 5x + 240
300 = 5x
x = 6









Note that there are a few variants of Simpson's rule.
You could be asked to work out the area between the graph of a function and the x-axis (see 2002 question).
You could be asked to copy a diagram, measure it and work out the area.

Linear Programming - part (b)

We looked at a few linear programming problems today.
There are 3 parts to tackling these questions.
STEP 1
Read through the text and extract 2 inequalities.
If you look back over the years, you will see that this question mostly comes up in a formulaic manner. In 2005 you were asked
"A manufacturer of garden furniture produces plastic chairs and tables. Each chair requires 2 kg of raw material and each table requires 5 kg of raw material. In any working period the raw material used cannot exceed 800 kg.
Each chair requires 4 minutes of machine time and each table requires 4 minutes of machine time. The total machine time available in any working period is 1000 minutes."
Identify the resctrictions: the manufacturer is restricted by amount of raw material they can use and by the minutes of machine time they have available.
x is going to be the number of chairs they manufacture, so as each chair reguires 2kg of raw material, they will require 2x kg for all the chairs they make (so if it turned out that they made 30 chairs, then they would need 2(30) = 60kg.)
Note that the words "cannot exceed" translate to "≤" in maths.
You build your two inequalities like this:
Raw materials: 2x + 5y ≤ 800
Time: 4x + 4y ≤ 1000

However in 2007, it was worded this way
"A developer is planning a holiday complex of cottages and apartments.
Each cottage will accommodate 3 adults and 5 children and each apartment will accommodate 2 adults and 2 children.
The other facilities in the complex are designed for a maximum of 60 adults and a maximum of 80 children."
Here the inequalities are not so obvious. A trick for working these out is to look at the two limiting factors - you are told that the max number of adults is 60 and the max number of children is 80.
So you know that your inequalities will be of the form
Adults: ...x + ....y ≤ 60
and
Children: ...x + ...y ≤ 80
(in other words the first inequality concerns restrictions of number of adults and the second inequality concerns restrictions of number of children)
Once you know that all the children info goes into the first inequality, you can find it more easily.

STEP 2
Graph the inequalities.
To do this you need to write the inequalities as equations. Then plot them the way you would in coordinate geometry - the easiest way is to use the intercept method (if you've forgotten this, it is described below in the section on Question 11 part(a).
Use the test-point to work out which half-plane you are interested in for each line. The resulting shaded in area is almost always a quadrilateral bounded by the x and y axes.

STEP 3
Evaluate an expression. After you've been asked to graph all the information you'll be told another piece of information about x and y, this time with no restriction.
In 2005 this was "The manufacturer sells each chair for €20 and each table for €40."
In 2007 it was "If the rental income per night will be €65 for a cottage and €40 for an apartment ...". In each case you are asked to maximise profit or revenue. You use this new information to create an expression in x and y.
In 2005 it would be 20x + 40y
In 2007 it would be 65x + 40y
The optimum values occur at the vertices (corners) of the shaded region. The vertex formed by the intersection of the two lines you just graphed must be calculated using simultaneous equations. The others can be read from the graph.
Then you take the (x,y) co-ordinates of each vertex and sub the x and y values into the expression.
Highlight the optimal value and very important, write the answer as an english sentence.
"The developer should build 10 cottages and 15 apartments".

There is usually a final part to this question - always something different, requiring you to evaluate another expression, find a percentage, deduct some amount from the overall profit etc.

Planning for the last few classes

I've made some updates to yesterday's post on half-planes. It is pretty comprehensive now.

We should be able to cover the main part of the linear programming question should in Tuesday's class.

If you take a look at the schedule of revision that we agreed back in March you'll see that the only topic left to cover is perimeter area and volume - paper 2 question 1.
I would like to get this started on Wednesday.

Homework for Tuesday is the 2005 linear programming question. If you have trouble with the part (a), take a look at the end of the previous entry in this blog.

Half-planes in linear programming

A half-plane is a region on a co-ordinated plane for which an inequality in x, y or both is true.

The idea of graphing a half-plane (such as 2x + 3y ≤ 6) is similar to graphing a line.
First change the inequality to an equation - graph 2x+3y = 6.
Remember how to do this?
Find two points on the line

• set x=0, see what y is - giving you (0,2)
• set y=0, see what x is - giving you (3,0).

Now you can graph the line.























But how do you find the half-plane 2x + 3y ≤ 6
In other words, on which side of this line will every single point obey the rule 2x + 3y ≤ 6 ?
To find out, you use a test point.
The easiest test point to use is (0,0) - remember you can't use this if the line goes through the origin and you will know if goes through the origin if there is no constant (e.g. the line 2x + 3y = 0 goes through the origin).
Substitute the test point (0,0) into 2x + 3y ≤ 6 and we get
2(0) + 3(0) ≤ 6
0≤6
This last line is true, (0 is less than 6) so if 2x + 3y ≤ 6 is true for (0,0), it is also true for all other points on the same side of the line as (0,0).
So, draw arrows pointing towards the (0,0) side like this.


If your test point had produced result that wasn't true such as 8 ≤ 6, then the arrows would point towards the side of the line that did not contain the test point.














A typical exam part (a) type question would be:
Shade in the region of a plane which simultaneously satisfies the following inequalities:

• 2x + 3y ≤ 6
• x ≥ 1
• y ≥ 0

The first one is already done.
The second one works like this:
Turn it into an equation x = 1 and graph this. Any point where x = 1 is on the line ((1,0), (1,1), (1,2) etc) so this line is the line that cuts the x-axis at 1 and is parallel to the y-axis.
Work out which way the arrow goes ... take a test point like (0,0) and see does it fit the inequality x ≥ 1.
Substituting into the inequality gives 0 ≥ 1 which is not true, so the half-plane we are looking for is on the other side, arrows pointing to the right.

Use the same logic to work out that y=0 is the equation for the x-axis. You can't use (0,0) as a test-point as it is on the line, so use another convenient point instead - e.g. (0,1).
Substituting into the inequality gives 1 ≥ 0 which is true, so this means that the test point is in the half-plane and that the arrows point upwards.
Last part of the question is to draw shade in the region which is in all 3 half-planes.



















Note that there are other ways this question could be asked. You could be given the diagram and be asked for the 3 inequalities that define a region.

This was the example we started looking at in class on Monday.

The vertical and horizontal lines should be easy now.
"Above the x-azis" means the inequality is y ≥ 0
"To the right of the y-axis" means the inequality is x ≥ 0

The diagonal line is more difficult.
You know that two points on the line are (4,0) and (0,3) so how can you get the equation of the line?

You have to resurrect your co-ordinate geometry.
First find the slope of the line
m = 3-0/0-4 = -3/4

Now put one of your points into the formula y-y1 = m(x-x1)
I'll choose (4,0)
This gives you:
y - 0 = -3/4(x - 4)
4y = -3x + 12
Rearranging this gives us the equation
3x + 4y = 12 ( which is the rule for being a point on the diagonal line).
But you want rule for being on the half-plane which is below and to the left of this line.
So you have to use a test-point. (0,0) is a good candidate here.

3x + 4y = 12
Left-hand-side works out as 3(0) + 4(0) which is 0.
Right hand side is 12.

Now you have to work out the direction of the inequality:
0 ? 12
Answer is that 0 < 12 so in this case the inequality that defines the half-plane in question is
3x + 4y ≤ 12

A few closing points:

• If you are doing this question in your exam, make sure that you write out all three inequalities clearly.
• On our course the half-planes are always defined using ≤ or ≥, not < or >.
• Also, note that part (b) of this question will be concerned with positive x and y values (since they refer to real-world problems) , but in part (a) you could have lines that intercept the x and/or y-axes at negative numbers.

Linear Programming

For most of the class, the linear programming question will be the only question we prepare for in section B of paper 2. You re-use your co-ordinate geometry and algebra skills (particularly linear inequalities and simultaneous equations) which you need anyway for other parts of the exam. The linear programming problem is quite formulaic - if you practice a few you will see the pattern and be able to get a good mark on this part of the exam. Don't forget to write the answer as an english sentence at the end of your answer.

The linear programming question is almost always in two parts.
a) worth 15 marks asks about half-planes
b) is the linear programming problem.

Homework this evening involves solving the question from 2000 (paper 2 Question 11).

Here is a 1-page overview of this question.

Finishing statistics

Today we looked at the 2006 question in statistics.
The main points points that we went over were:
calculating standard deviation
finding the mean using mid-interval values
producing histograms
For calculating the standard deviation you should format your work into columns, working out the differences from the mean, then the differences squared (then this value x the frequencies if applicable) etc. If doesn't matter if your work isn't in columns, as long as you follow the correct steps and show your work. As a rule of thumb, almost all your data should be within 2 std deviations above and below the mean. This means that the value of the std deviation should not be more than 1/2 the mean.
Whenever you are finding the mean value from a set of data you should check your answer to see if it is reasonable. The most common error when finding the mean of a frequency distribution is to divide by the number of categories instead of the sum of the frequencies. If you do this, your mean will be much too large and it should be obvious that you've made a mistake.
The purpose of histograms is to distribute data across uneven intervals. It is not a straightforward bar chart. You should write the frequency in a circle in each block. You can use the alternative format of histogram which has a key showing e.g. one square = 5 cars (in the 2006 question).

Homework for tonight is the 2005 statistics question. This one includes weighted mean and std deviation of a frequency distribution. If you can do the statistics questions from 2005 to 2007 you will have covered the topic pretty comprehensively.

On Friday in the double class 2nd period we will have a test (statistics + maybe lin programming also).
Tomorrow, once we have corrected the homework, we will go on to Linear Programming.

Statistics

We will revise statistics for the next few classes.
The main skills you need for this question are:
- being able to work with frequency distributions in their various forms
- being able to calculate standard deviations
- being able to graph data using histograms and ogives
- being able to interpret graphs ( remember to show the markings that indicate how you work out e.g. median or mean from your graphs)

We will go over the topics that came up in the 2007 paper in class tomorrow.

Here is the 1-page overview for statistics.

Finishing up Differentiation

To round off this topic, here is the 1-page overview.

Differentiation test

Everyone did well on the main question in today's test - a really strong performance on this question.

The differentiating from first principles was a little dodgy though.
I have annotated a typical marking scheme for this, so you can see where you would pick up blunders B(-3), slips S(-1) or misreadings M(-1). Nobody made the dreaded "worthless" W(0) grade.

The main issue is that many students leave out the left hand side (abbreviated as LHS). The process of differentiation only makes sense if you include the left hand side. In today's test, several students wrote things like 4x - x^2 = 4 - 2x which doesn't make any sense.

Read through the this marking scheme (taken from the 2005 exam). We will go through it on Monday and I will get everyone to mark their own first principles answer according to this outline.

Differentiating from first principles

This question comes up frequently (not every year) in the Leaving Cert. When it comes up, you usually have to differentiate a quadratic, though you should practice differentiating linear functions (e.g. f(x) = 4x -1) also. It is usually worth 20 marks.

The way I approach it is to set up the derivative function as if you were finding the slope using the slope formula m = (y2-y1)/(x2-x1).
Today's example which we did in class was

3x²-4x+2 (I think)

x1 = x
y1 = 3x²-4x+2
x2 = x + h
y2 = 3(x+h)²-4(x+h)+2

If you plug these into the slope formula and multiply out the y2 and watch your signs with y1, you'll get

m =
3x²+3h² +6xh -4x-4h+2 - 3x²+4x-2
------------------------------------------------
x + h - x

= 3h² +6xh -4h
---------------
h

= 3h +6x -4

Now apply the Limit as h tends towards zero (you'll lose marks if you don't do this part properly)

Lim [f(x+h) - f(x)]/h = 6x -4
h->0

Also, write down somewhere either f(x) = 3x²-4x+2 and f'(x) = 6x -4

or
y = 3x²-4x+2

dy
-- = 6x -4
dx

It is worth spending time practicing and understanding this process. You only have to do it if you are specifically asked to differentiate "from first principles", but it is central to figuring out other topics in functions and differentiation, and it gives you an application of coordinate geometry and practice at some tricky algebra.

I have a worksheet that I didn't have time to distribute today that gives a worked example and practice of differentiating from first principles.

We'll look at the rules for differentiating (product, quotient and chain rules) tomorrow.
Test on Friday.

Finishing Functions, Graphs and Differentiation.

We need to finish functions, graphs and our calculus.
In tomorrow's class we will look at differentiating from first principles and at how to apply the different rules (product, quotient and chain rules).
Everyone should do the 2007 Q6,7,8 and 1999 Q7,8 at least.

We'll have a class test on Thursday.

Functions and Graphs

We are working on functions and graphs at the moment.

In today's class we looked at how you can use a function to solve a related function.
Using graphs of functions to find the roots of a polynomial is a key skill.
When you graph a function, you find the output values for a selection of discreet input values.
When you join the dots with a smooth curve you are extrapolating the output values of the input values in between. That is why we graph functions.
Wherever your line crosses the x-axis, the output value is 0 - don't forget equation of the x-axis is y=0. So you can use these points to work out the roots of a cubic equation.
We looked at the 1999 q8 question on functions. This included working out a related function. Make sure you do this question and are able to follow it.
When you split the related cubic function into the original cubic and a related linear function, you can work out the roots of the related function by reading the points of intersection from your graph.

Other points on functions which you need to know are mentioned in the 1 page overview.

Algebra

We are spending a few days revising algebra. This is a really important topic as it comes up all over the paper 1 and 2 as well as in 2 algebra-only questions (Paper 1 Q2 and Q3).

We started by looking at the 2007 questions.
Q2 involved solving inequalities. Key points:
- know how to graph on the numberline or work out the solution set = { , , } if x is an element of N, Z and R
- know how to handle cases like -x <>3 ]

It also involved manipulating indices. It is too difficult to format the rules here... look them up in your book and make sure you know them. With equations, get the same number raised to different powers on each side of the =.
Eg 2^(x+1) = 4^x
rewrite 4^x as (2^2)^x or 2^2x
Then equate the powers:
x+1 = 2x
x = 1

We will look at the rest of Q2 and Q3 tomorrow. Here is a slightly revised version of the algebra 1-page overview.

More trigonometry

The unit circle, compound angles etc took longer than planned on Friday. As I said in class, these topics come up infrequently, so if you are aiming to pass this question, make sure that you are able to use sin=opp/hyp etc, sine rule, cosine rule and area of triangle.

I forgot to post the 1-page overview of trigonometry last time.



Also, I will mention this in tomorrow's class, but there is one thing you should be aware of when using the Sine rule:
If you use your calculator to work out the inverse sine of 0.9897, your calculator will tell you it is 81.7 degrees ... which it is. But, it is also the sine of 97.3 degrees. You know from the discussion of unit circle that there are two possible angles that have the same sine, cosine and tan. Our calculators always return the angle that is between 0 and 90 degrees only, not the obtuse alternative.
Look at 2003 P2 Q5 for an example where this can cause problems.
If you solve (c) (ii) by putting
Sin60/7 = SinA/8
you will get the wrong answer.
The problem here is that the shorter side of the triangle (3) is not used in this equation and so the triangle where we have a known angle (60) and two other sides (7 and 8) is not a well defined triangle. In effect, it is ASS (angle, side, side) rather than SAS (side, angle, side)! (remember the 4 conditions for congruence of triangles that you learned: SSS,SAS,ASA,RHS)
You have to use the acute angle to work it out.
Sin60/7 = SinB/3 ... which gives you 22 degrees for that angle and leaves 98 degrees for the obtuse angle (60 + 22 + 98 = 180).

How do you know which angle to go for? A triangle can have at most 1 obtuse angle (> 90 degrees) and that will be opposite the longest side.
If you have a choice, solve the acute angle, opposite the shortest side, first.

Trigonometry

There are two parts to the trigonometry course.

The first part relates to finding missing sides and angles or the area of a triangle:

• application of the trigonometric ratios for sin, cos and tan (sin=opp/hyp etc.) for right-angled triangles
• application of the sine rule for cases where you have one complete and one incomplete "angle and opposite side" pair
• application of cosine rule where you have 2 sides and the angle between them, or just 3 sides of a triangle
• application of area of triangle rule 1/2 ab SinC

This is relatively straightforward and covers about 90% of the questions that come up on the exam.

The second part relates to compound angles, angles > 90 degrees, compass directions, unit circle, handling special angles (30, 45 and 60 degrees) using surds. This is the second half of the chapter.

In today's class we went through the 2007 trigonometry question which involved material from the first list above. You have to go back quite a few years to find a question that involves compound angles, angles which are > 90 degrees etc.

Homework for tonight was the 2006 Trig question.

In tomorrow's class we will tackle the second part of the trigonometry course.
In the second period, we'll start revising Algebra.

Arithmetic

If you work through all the part a,b and c type questions you got in class today and practice from previous exam papers, you'll be well prepared for Paper 1 Question 1.

We went over 2 types of percentages problems in class:

a) An item is reduced by 10% and the new price is 1125. What was the original price?

and

b) An amount is invested for 1 year at 6% interest rate and amounts to 5618 at the end of the year. What was the original principle invested?

To answer both questions you need to look at the number you are given and work out what percentage this represents.

In a) the amount 1125 represents 90%, so you can write down
90% = 1125 <- This is the most important step. Now I know 90%, I can work out what 10% is by dividing both sides by 9 10% = 125 100% = 1250 <- 100% is the original amount before the reduction. In b) the amount at the end of the year is the original principle (100%) plus 6% so you can write 106% = 5618 1% = 53 [dividing both sides by 106 to find what 1% is] 100% = 5300 <- 100% is the original principle that was invested.

In both cases the first step is key .. look at the number you are given and work out what percentage it represents. Then use the equation technique of multiplying or dividing on both sides of the equals by the same amount to find the percentage you are asked for.

Here is the 1-page overview of arithmetic.


Note that the formula for compound interest can be expressed in two ways.
You can say

1 + (Rate/100)

or

(100 + Rate)/100

The key point is that if the rate is, say, 19%, then you will end up with
Amount = Principle(1.19)^Years

(Sorry about formatting , ^ means raised to a power)