Monday, May 26, 2008
Paper 2 Question 11 - Linear Programming
Here are the notes on Linear Programming:
Introduction - here and here
Looking at part a) half-planes
Worked example of a part b).
NB: when answering part b) (i) which is usually "write down two inequalities in x and y and illustrate these on graph paper" you must
a) use graph paper and
b) shade in the region which represents all the inequalities (including x≥0 and y≥0) or at least indicate the half planes with arrows. If you just draw the lines you could lose 3 marks per inequality ... or 1%!
Introduction - here and here
Looking at part a) half-planes
Worked example of a part b).
NB: when answering part b) (i) which is usually "write down two inequalities in x and y and illustrate these on graph paper" you must
a) use graph paper and
b) shade in the region which represents all the inequalities (including x≥0 and y≥0) or at least indicate the half planes with arrows. If you just draw the lines you could lose 3 marks per inequality ... or 1%!
Paper 2 Question 7 - Statistics
Paper 2 Question 6 - Probability
Here are the links to notes on Probability:
A Tricky probability question done using rules of probability and sample space technique.
A Tricky probability question done using rules of probability and sample space technique.
Paper 2 Question 5 - Trigonometry
Paper 2 Question 2 and 3 - Co-ordinate Geometry of the Line and Circle
Paper 2 Question 1 - Perimeter Area and Volume
Paper 1 Question 6, 7 and 8 - Functions and Differentiation
Paper 1 Question 4 - Complex Numbers
This is the only blog entry on complex numbers. It gives the 1-page overview.
Paper 1 Question 2 and 3 - Algebra
Here are all the links to algebra entries on this blog.
There is an overview of algebra here.
An example of simultaneous equations one linear one quadratic and awkward algebraic fractions can be found here. (sorry about the formatting in the second part, if anyone can tell me why " " doesn't work, I'm all ears).
There are some notes on numberlines and manipulating indices here.
There is an overview of algebra here.
An example of simultaneous equations one linear one quadratic and awkward algebraic fractions can be found here. (sorry about the formatting in the second part, if anyone can tell me why " " doesn't work, I'm all ears).
There are some notes on numberlines and manipulating indices here.
Paper 1 Question 1 Arithmetic
Here are the links to arithmetic to help you revise for this question:
There is an overview of the arithmetic question and in particular compound interest here.
There is an overview of the arithmetic question and in particular compound interest here.
Tuesday, May 20, 2008
Linear Programming Class - Thursday morning
There will be a special Linear Programming class on Thursday (22nd May) morning for anyone who is interested. The class will be in Room 1 at 8:40 (normal maths class time).
If you would like a class focussing on another topic, or just want to have a meeting with me, contact me at the school asap.
If you would like a class focussing on another topic, or just want to have a meeting with me, contact me at the school asap.
Wednesday, May 14, 2008
Checklist for exam
Make sure you have your exam strategy ready long before the exam. Be familiar with which questions you are planning to answer. |
Have your pens, pencil, calculator (check battery), geometry set ready well before the exam time. |
2½ hours for exam = 25 mins per question. Exam starts at 9:30, you should be well into your 2nd question by 10am. If you have choices, do not spend too long deciding which questions you are going to do. Circle the questions you are going to do and draw a pencil line through the questions you are definitely not going to attempt. |
Write clearly. Number all questions and parts of questions. Don't waste time rewriting the question. Highlight your answer clearly. |
Keep all parts of each question together. It is a good idea to keep questions in order. If you have to come back to re-attempt a question later in the exam, it will be easier to find it. |
Read questions carefully, do your work and then read question again to make sure you have done what was asked and to make sure you format your answer properly (correct to 2 decimal places, as a fraction, in surd form, in km/h etc.). |
Attempt all questions. Where there is a part (i), (ii), (iii) etc., don't assume that if you cannot finish part (i) that you cannot attempt part (ii), (iii). If you are running out of time remember how to get attempt marks - “any correct substitution”. |
Only attempt a 7th question after you have checked the first 6. |
Some students find it reassuring to write all the formulae you have memorised down as soon as you are told that you may start. |
Remember, for every blunder (-3 marks) you lose ½%, so be careful! |
How NOT to lose marks in your LC exam
Question | How I lose marks ... | How I get them back. | ||
Arithmetic | ▼ | Not giving answer in correct format. | ▲ | Recheck what was asked. |
▼ | Not handling percentages, percentage error and taxes properly. | ▲ | Practice all type of percentage questions | |
Algebra | ▼ | Failing to square an expression properly (x+y)² = x² + y² + 2xy | ▲ | Practice |
▼ | Getting sloppy towards the end of a simplification, mixing up signs, adding instead of multiplying.... | ▲ | Practice and be careful. Then, substitute test values to verify results. | |
▼ | Failing to find corresponding y-values after solving for x, esp when one simultaneous equation is quadratic. | ▲ | Pay attention to steps of solution – always include the checking answers step. | |
▼ | Not handling laws of indices properly | ▲ | Learn these – check them on your calculator! Make sure you can enter complicated indices on your calculator correctly. | |
Complex numbers | ▼ | Failing to set i² = -1 | ▲ | Practice. |
▼ | Not recognising symbols and their meanings: complex conjugate z-bar, modulus |2+3i| | ▲ | Go over chapter – it is quite self-contained. | |
▼ | Not handling substitution properly w=2+3i, what is w² | ▲ | Look at past exam questions. | |
▼ | Not knowing quadratic formula properly | ▲ | Write it down and check it every day. | |
Functions | ▼ | Not showing marks on graph or understanding how to use graph, | ▲ | Use a pencil and ruler to show how you interpret graph. |
▼ | Uneven scale on x- or y-axis | ▲ | Use graph paper, watch that all intervals are even, including 0 to 1, 1 to 2 etc. | |
Differentiation | ▼ | Not taking care applying rules for differentiation, or not knowing chain rule (it is not in tables) | ▲ | Learn it and practice. |
▼ | Leaving out LHS or Limit idea in first principles. | ▲ | Practice. | |
Perimeter, area & volume | ▼ | Not knowing simpson's rule inside out. | ▲ | Practice, especially with unknown heights etc. |
▼ | Not handling Π as requested. | ▲ | Re-check what you are asked. | |
▼ | General errors applying formulae. | ▲ | Practice using tables for formulae, and sanity check your answers. | |
Co-ordinate geometry of line | ▼ | Not knowing formulae and how to apply them. | ▲ | As for quadratic formula, write them out each day – make sure you understand logic of slope, midpoint and distance formulae. |
▼ | Uneven scale on x- and y-axis (must be both to same scale) | ▲ | Use graph paper and check this. | |
Co-ordinate geometry of circle | ▼ | Not taking minus signs into account in formula for equation of circle (e.g. (x-2)² + (y+3)² = 8 means centre of circle is (2,-3) and radius = √8 | ▲ | Practice. |
▼ | Failing to understand geometry of circle (tangent at a point perpendicular to radius, tangents at opp end of diameter are parallel) | ▲ | Go over JC geometry if necessary. | |
Trigonometry | ▼ | Not knowing or correctly using Sin= 0pp/Hyp etc | ▲ | Learn mnemonic. |
▼ | Failing to “fill out” the sine rule/ cosine rule correctly. | ▲ | Practice. | |
▼ | Being caught out by inverse sine of an angle that is >90º | ▲ | Be alert. | |
▼ | Having calculator set to radians or not knowing how to use degrees/minutes function. | ▲ | Check it and practice. | |
Probability | ▼ | Not showing your work | ▲ | Can be difficult, but write something in case your answer is wrong. |
▼ | Failing to see that there are two ways a combined event can happen and adding probabilities of each. | ▲ | Use a sample space to check answers | |
Statistics | ▼ | Forgetting to divide by sum of frequencies | ▲ | Sanity check answers |
▼ | Not showing marks on graph (eg median from ogive) | ▲ | Indicate how you use graph clearly | |
Linear programming | ▼ | Not indicating half-plane when graphing inequalities | ▲ | Include it in your graph. |
▼ | Leaving out the “obvious” inequalities, x≥0 | ▲ | Make sure you give the number of inequalities you were asked for. | |
▼ | Not using simultaneous equations to calculate point of intersection. | ▲ | You must not use your graph for this part. | |
▼ | Not writing english sentence at the end | ▲ | Explain what x=10, y=7 means, 10 what, 7 what and why (“to maximise profit” or whatever). |
Sunday, May 11, 2008
Probability Question
While the probability question can be very straightforward, you need to pay attention to the detail of what you are being asked. I gave this question (1999) in the mini-mock exam and nobody got it right.
In your exam, you can answer questions like this using the sum or product of probabilities as appropriate or using a sample space. While the sample space might take some time to set up, it is a foolproof method. To illustrate how the probabilities work, I have created the matching sample space for each part of the question. (you will need to click on the image to be able to read the text within the sample space diagram)
Note that there are 66 possible combinations, as you cant pick the same exact sample twice and selecting e.g. A1 and B2 is the same as selecting B2 and A1.
Looking at part (i):From the sample space you can count the 10 cases where this can happen. So the probability is 10/66 = 5/33
Using probabilities only, you work it out like this:
Looking at part (ii):
Counting from the sample space you get 12/66 = 2/11.
Working this out using probabilities is trickier.
Here you have to remember that there are two ways this could happen. The first way is that you get a B first, then an O, OR the second is the other way around. In both cases you have to remember that you are choosing from 12 the first time you pick and from 11 the second time.
The probabilities work out like this:
P(B then O OR O then B) :
Work out P(B then O) first:
P(1st is type B) = 4/12
P(2nd is type O) = 3/11
So P(B AND O) = 4/12 x 3/11 = 12/132
Then work out P(O then B)
P(1st is type O) = 3/12
P(2nd is type B) = 4/11
So P(O AND B) = 3/12 x 4/11 = 12/132
(Note you could have assumed that you "B followed by O" is equally likely as "O followed by B" so that probabilities would be the same).
Now work out P(B then O OR O then B)
= P(B then O) + P(O then B)
= 12/132 + 12/132
= 24/132
= 12/66 (same as sample space above)
Simplify to 2/11.
Looking at part (iii):From the sample space you can see that this can occur 19 ways out of 66 so P(both the same type) = 19/66.
Using probabilities, you need to consider what is a successful outcome. If you write out what "of the same blood type" means, you will see that a successful outcome is "both A OR both B OR both O", so you need to add P(both A) + P(both B) + P(both O).
You have already found the probabilty of both samples being type A in part (i).
The answer was 10/66 or 5/33.
Now you need to find the probability of both being B and the probability of both being O.
P(Both B) = 4/12 x 3/11 = 12/132 = 6/66
P(Both O) = 3/12 x 2/11 = 6/132 = 3/66
(I am keeping all fractions with 66 as common denominator on purpose).
Now add
P(Both A OR both B OR both O) = 10/66 + 6/66 + 3/33 = 19/66
In your exam, you can answer questions like this using the sum or product of probabilities as appropriate or using a sample space. While the sample space might take some time to set up, it is a foolproof method. To illustrate how the probabilities work, I have created the matching sample space for each part of the question. (you will need to click on the image to be able to read the text within the sample space diagram)
Note that there are 66 possible combinations, as you cant pick the same exact sample twice and selecting e.g. A1 and B2 is the same as selecting B2 and A1.
Looking at part (i):From the sample space you can count the 10 cases where this can happen. So the probability is 10/66 = 5/33
Using probabilities only, you work it out like this:
P(1st is type A) = 5/12
P(2nd is type A) = 4/11 [If the first one was an A, then there will only be 4 As left. Also, there will only be 11 samples left, as the first sample isn't replaced.]
P(1st is type A AND 2nd is type A) = 5/12 x 4/11 = 20/132 = 5/33
Looking at part (ii):
Counting from the sample space you get 12/66 = 2/11.
Working this out using probabilities is trickier.
Here you have to remember that there are two ways this could happen. The first way is that you get a B first, then an O, OR the second is the other way around. In both cases you have to remember that you are choosing from 12 the first time you pick and from 11 the second time.
The probabilities work out like this:
P(B then O OR O then B) :
Work out P(B then O) first:
P(1st is type B) = 4/12
P(2nd is type O) = 3/11
So P(B AND O) = 4/12 x 3/11 = 12/132
Then work out P(O then B)
P(1st is type O) = 3/12
P(2nd is type B) = 4/11
So P(O AND B) = 3/12 x 4/11 = 12/132
(Note you could have assumed that you "B followed by O" is equally likely as "O followed by B" so that probabilities would be the same).
Now work out P(B then O OR O then B)
= P(B then O) + P(O then B)
= 12/132 + 12/132
= 24/132
= 12/66 (same as sample space above)
Simplify to 2/11.
Looking at part (iii):From the sample space you can see that this can occur 19 ways out of 66 so P(both the same type) = 19/66.
Using probabilities, you need to consider what is a successful outcome. If you write out what "of the same blood type" means, you will see that a successful outcome is "both A OR both B OR both O", so you need to add P(both A) + P(both B) + P(both O).
You have already found the probabilty of both samples being type A in part (i).
The answer was 10/66 or 5/33.
Now you need to find the probability of both being B and the probability of both being O.
P(Both B) = 4/12 x 3/11 = 12/132 = 6/66
P(Both O) = 3/12 x 2/11 = 6/132 = 3/66
(I am keeping all fractions with 66 as common denominator on purpose).
Now add
P(Both A OR both B OR both O) = 10/66 + 6/66 + 3/33 = 19/66
Mini-mock Paper 2 Feedback
For the Trigonometry question, you just had to know your Trig ratios sin=opp/hyp etc and apply them to the triangle. It wasn't necessary to work out the exact answer. You lost marks if you didn't give the answer as a fraction.
The question on Co-ordinate Geometry of the circle was well answered. You had to use the midpoint formula to find the centre of the circle.
The Linear programming question was tricky, because it was easy to overlook the "number of drivers available" as a limitation. Most people who got this second inequality right got the correct solution.
NB: Don't forget that you must use simultaneous equations to find the vertex at the intersection of the two lines.
By the way, it looks like we will have only one class left before Friday, so have your questions ready.
The question on Co-ordinate Geometry of the circle was well answered. You had to use the midpoint formula to find the centre of the circle.
Centre of S is midpoint of diameter. (1,-2)
Radius = distance fom either endpoint of diameter (e.g. (-1,-1)) to the centre (1,-2). (or half the diameter)
= √(1+1)² + (-2+1)²
= √5
Now use radius √5 and centre (1,-2) to form equation of circle:
(x – 1)² + (y + 2)² = 5
The simpson's rule was straightforward and was well answered in general.
The Linear programming question was tricky, because it was easy to overlook the "number of drivers available" as a limitation. Most people who got this second inequality right got the correct solution.
NB: Don't forget that you must use simultaneous equations to find the vertex at the intersection of the two lines.
By the way, it looks like we will have only one class left before Friday, so have your questions ready.
Thursday, May 8, 2008
Paper 1 feedback
Following the mini-mock exam yesterday, there were a few noteworthy common errors.
Firstly, lots of students lost marks by not following instructions exactly.
In the differentiation question there are still issues.
Right throughout the first principles question there are opportunities to lose marks. The main one that we've covered before is not having a left-hand-side, and not mentioning the concept of a limit.
Also lots of students cancelled
3h² + 6xh + h
--------------------
h
to get
3h + 6x
forgetting that h/h = 1
The complex numbers question was poorly answered. This needs to be revised. The previous entry on complex numbers is here.
Also, the performance on the laws of indices question was weak.
Almost everyone got caught on the second question where each part of the equation was already in the form 2 to the power of something. This was a part (c) question so it is unlikely to be so simple. You must work out the right-hand-side
(64 - 32 = 32 = 2^5) and then equate the powers.
This is also a good place to demonstrate the value of checking your answer.
If you do the question wrong:
3p-7 = 6-5
3p = 8
p = 8/3
Try substituting this back in to the original and you will see that it doesn't balance.
The correct way is to recognise that the RHS ends up as 2^5 and then equating the powers correctly, you get:
3p -7 = 5
3p = 12
p = 4
Also, don't forget ... always make an attempt. Many attempts are work 3 marks - which is ½%
In tomorrow's double, we will take a general look at Paper 2 and then we'll do a Paper 2 mini-mock in our second period.
Firstly, lots of students lost marks by not following instructions exactly.
- The solution set of an inequality isn't n<4. You have to list the elements as {0,1,2,3} or graph this on a number line.
- The cooking instructions question asked for the time in hours and minutes - if you gave it in minutes only, you lost marks.
- On the functions question, you were asked for the co-ordinates of the local max and min - if you only gave the values of x for which the max and min occur, you lose marks.
In the differentiation question there are still issues.
Right throughout the first principles question there are opportunities to lose marks. The main one that we've covered before is not having a left-hand-side, and not mentioning the concept of a limit.
Also lots of students cancelled
3h² + 6xh + h
--------------------
h
to get
3h + 6x
forgetting that h/h = 1
The complex numbers question was poorly answered. This needs to be revised. The previous entry on complex numbers is here.
Also, the performance on the laws of indices question was weak.
Almost everyone got caught on the second question where each part of the equation was already in the form 2 to the power of something. This was a part (c) question so it is unlikely to be so simple. You must work out the right-hand-side
(64 - 32 = 32 = 2^5) and then equate the powers.
This is also a good place to demonstrate the value of checking your answer.
If you do the question wrong:
3p-7 = 6-5
3p = 8
p = 8/3
Try substituting this back in to the original and you will see that it doesn't balance.
The correct way is to recognise that the RHS ends up as 2^5 and then equating the powers correctly, you get:
3p -7 = 5
3p = 12
p = 4
Also, don't forget ... always make an attempt. Many attempts are work 3 marks - which is ½%
In tomorrow's double, we will take a general look at Paper 2 and then we'll do a Paper 2 mini-mock in our second period.
Tuesday, May 6, 2008
Last bits of revision
In tomorrow's class we will have a Paper 1 mini-mock exam.
On Thursday we will do a proper revision of co-ordinate geometry of the circle as this topic seems to be still causing problems for many.
On Friday we will have a Paper 2 mini-mock and then resume revision, including probability.
We'll also cover general exam approach and prep over the next few days.
On Thursday we will do a proper revision of co-ordinate geometry of the circle as this topic seems to be still causing problems for many.
On Friday we will have a Paper 2 mini-mock and then resume revision, including probability.
We'll also cover general exam approach and prep over the next few days.
Thursday, May 1, 2008
Perimeter Area and Volume
For my one-page overview you can look at the previous entry.
In class on Thursday, we looked at the different types of questions you can get asked for perimeter, area and volume.
Finding area and perimeter of rectangular shapes and triangles
Irregular rectangular shapes: break them into smaller rectangles.
Triangles: half the base by the perpendicular height.
[Area of a triangle of where base = 8cm and perpendicular height = 3cm multiply as follows
1/2 x 8 x 3 = 12cm²]
Volume of a prism (any rectangular solid with a uniform cross-section)
Find the area of the uniform cross-section and multiply this by the length of the prism.
Circular shapes
The first question is how to handle π, pi.
The most important thing is to follow instructions, using 22/7, 3.14 or giving your answer in terms of π.
Know where to find the formulae you need.
Trickier questions
Changing shapes
If a solid in one shape is being melted down and recast into another shape you need to set up an equation. If both shapes are circular (eg a sphere being melted down into a cylinder, or liquid in a cylindrical jug being poured into smaller cylindrical tumblers) then the πs on both sides of the equation will cancel. In questions like this you will not be told how to handle π.
Liquid in a pipe
Water flows in a pipe at the rate if 50cm/sec. The pipe has a radius of 2cm. How long will it take to fill a tank measuring 30cm x 30cm x 100cm.
To answer a question like this, consider a single molecule of H2O in the pipe. It takes 1 second to travel 50cm - that means that in the space of 1 second all the molecules are replaced. If you find the volume of 50cm of pipe you will have 1 second's worth of water.
[This is π x 4² x 50]
Then find the volume of the tank
[This is 30 x 30 x 50]
Finally divide the first answer into the second answer to find the time in seconds.
Then convert to minutes or hours as required.
In class on Thursday, we looked at the different types of questions you can get asked for perimeter, area and volume.
Finding area and perimeter of rectangular shapes and triangles
Irregular rectangular shapes: break them into smaller rectangles.
Triangles: half the base by the perpendicular height.
[Area of a triangle of where base = 8cm and perpendicular height = 3cm multiply as follows
1/2 x 8 x 3 = 12cm²]
Volume of a prism (any rectangular solid with a uniform cross-section)
Find the area of the uniform cross-section and multiply this by the length of the prism.
Circular shapes
The first question is how to handle π, pi.
The most important thing is to follow instructions, using 22/7, 3.14 or giving your answer in terms of π.
Know where to find the formulae you need.
Trickier questions
Changing shapes
If a solid in one shape is being melted down and recast into another shape you need to set up an equation. If both shapes are circular (eg a sphere being melted down into a cylinder, or liquid in a cylindrical jug being poured into smaller cylindrical tumblers) then the πs on both sides of the equation will cancel. In questions like this you will not be told how to handle π.
Liquid in a pipe
Water flows in a pipe at the rate if 50cm/sec. The pipe has a radius of 2cm. How long will it take to fill a tank measuring 30cm x 30cm x 100cm.
To answer a question like this, consider a single molecule of H2O in the pipe. It takes 1 second to travel 50cm - that means that in the space of 1 second all the molecules are replaced. If you find the volume of 50cm of pipe you will have 1 second's worth of water.
[This is π x 4² x 50]
Then find the volume of the tank
[This is 30 x 30 x 50]
Finally divide the first answer into the second answer to find the time in seconds.
Then convert to minutes or hours as required.
Wednesday, April 30, 2008
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.
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.
Tuesday, April 29, 2008
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.
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.
Monday, April 28, 2008
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.
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.
Sunday, April 27, 2008
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
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:
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:
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).
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 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.
Thursday, April 24, 2008
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.
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.
Wednesday, April 23, 2008
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.
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.
Monday, April 21, 2008
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.
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.
Friday, April 18, 2008
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.
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.
Wednesday, April 16, 2008
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
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.
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.
Tuesday, April 15, 2008
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.
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.
Monday, April 14, 2008
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.
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.
Tuesday, April 8, 2008
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.
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.
Monday, April 7, 2008
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.
Thursday, April 3, 2008
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:
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.
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
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.
Tuesday, April 1, 2008
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)
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)
Sunday, March 30, 2008
One last day on co-ordinate geometry
In the last class before the break, we did a co-ordinate geometry test - with a Q2 and Q3 type question.
Most people did well on Q2 (co-ordinate geometry of the line) but not well on Q3 (co-ordinate geometry of the circle).
It is worth putting more work into this topic, especially the co-ordinate geometry of the circle as all you need to remember is
Don't forget that you are supposed to be attempting Leaving Cert 2003 over the Easter break.
Most people did well on Q2 (co-ordinate geometry of the line) but not well on Q3 (co-ordinate geometry of the circle).
It is worth putting more work into this topic, especially the co-ordinate geometry of the circle as all you need to remember is
- the formula of a circle (x-h)² + (y-k)² = r²
- all the co-ordinate geometry of the line you know so well anyway for Q2
- the relationship between radius, diameter, tangents etc that you learned in Junior Cert
- the intersection of a line and circle which you need to know anyway as it can come up in the algebra section of Paper 1.
Don't forget that you are supposed to be attempting Leaving Cert 2003 over the Easter break.
Tuesday, March 11, 2008
Co-ordinate geometry
This topic is examined on Paper 2 question 2 and 3.
Question 2 is co-ordinate geometry of the line only - this is mainly the same as Junior Cert Higher Level co-ordinate geometry, with the addition of finding the area of a triangle.
Question 3 is co-ordinate geometry of the circle.
Once you remember that the equation of a circle with centre (h,k) is
Here is a 1-page overview.
Question 2 is co-ordinate geometry of the line only - this is mainly the same as Junior Cert Higher Level co-ordinate geometry, with the addition of finding the area of a triangle.
Question 3 is co-ordinate geometry of the circle.
Once you remember that the equation of a circle with centre (h,k) is
(x-h)² + (y-k)² = r²
and some basic rules about tangents to circles there is not a lot of extra work required for this question.Here is a 1-page overview.
Sunday, March 2, 2008
Schedule of revision
This is what I propose for the next 8 weeks:
I haven't counted all 6 classes each week as there will inevitably be interruptions to classes due to oral exams etc so we can use those extra classes as contingency. I also haven't counted 2 days for reviewing the mock exams, since I don't know when they will be returned.
If you feel strongly about re-ordering some of these topics, let me know.
Week starting | Topics to be covered | |
Week 1 | 03/Mar | 3 classes complex numbers / 2 classes probability / (1 class geometry on fri) |
Week 2 | 10/Mar | 1 class probability / 4 classes co-ord geometry / 1 class arithmetic |
17/Mar | ||
24/Mar | ||
Week 3 | 31/Mar | 1 class arithmetic / 3 classes trigonometry / 1 class algebra |
Week 4 | 07/Apr | 3 classes algebra / 2 classes functions + graphs |
Week 5 | 14/Apr | 1 class functions + graphs / 4 classes differentiation |
Week 6 | 21/Apr | 3 classes statistics / 2 classes linear programming |
Week 7 | 28/Apr | 1 class linear programming / 3 classes perim, area + vol |
Week 8 | 05/May | General revision |
12/May | Graduation week |
I haven't counted all 6 classes each week as there will inevitably be interruptions to classes due to oral exams etc so we can use those extra classes as contingency. I also haven't counted 2 days for reviewing the mock exams, since I don't know when they will be returned.
If you feel strongly about re-ordering some of these topics, let me know.
Complex Numbers
Geometry overview
We discussed the approach to geometry in Friday's class.
Study the theorems one by one at home.
In class, we covered the basics you need for Section A, Q4: essentially JC higher level geometry and the concept of enlargement.
On Friday 7th during the 2nd maths period, we'll take a look at the Section B, Q8.
Study the theorems one by one at home.
In class, we covered the basics you need for Section A, Q4: essentially JC higher level geometry and the concept of enlargement.
On Friday 7th during the 2nd maths period, we'll take a look at the Section B, Q8.
Wednesday, February 27, 2008
Geometry revisited
Since quite a few students are interested in geometry, I've looked into whether we can cover this section of the course in the time remaining.
Basically, if you know your theorems from higher level junior cert and remember the basics of geometry, you should have few problems with LC ordinary level geometry.
We won't have time to go through each theorem individually, but they should be easy to study on your own. The geometry questions (4 and 8) usually have a part b which involves proving a theorem.
We will do the constructions and enlargements in class.
If your preference was to stick with the questions and options we have already covered, don't worry. We will spend just a few classes on geometry and you can work on other maths material during those classes.
Basically, if you know your theorems from higher level junior cert and remember the basics of geometry, you should have few problems with LC ordinary level geometry.
We won't have time to go through each theorem individually, but they should be easy to study on your own. The geometry questions (4 and 8) usually have a part b which involves proving a theorem.
We will do the constructions and enlargements in class.
If your preference was to stick with the questions and options we have already covered, don't worry. We will spend just a few classes on geometry and you can work on other maths material during those classes.
Tuesday, February 26, 2008
After the mocks ...
With the mock exams completed, we have to settle on a schedule of revision.
Not including 2 weeks at Easter and the week starting May 12th, we will have 8 weeks of normal class time, give or take a day here or there.
We are currently prepared for the following.
Paper 1:
Q 1,2,3,4,6,7,8 (must do 6 of these 7 questions)
We left out Q5 Sequences and Series
Paper 2:
Section A Q 1,2,3,5,6,7 (must do 5 of these 6 questions)
We left out Q4 Theorems and Enlargements.
Section B Q 11 (must do 1, there are 3 other options)
We left out Q8 Further Geometry, Q9 Vectors and Q10 Further Sequences and Series.
Do we want to prepare for any further questions?
Not including 2 weeks at Easter and the week starting May 12th, we will have 8 weeks of normal class time, give or take a day here or there.
We are currently prepared for the following.
Paper 1:
Q 1,2,3,4,6,7,8 (must do 6 of these 7 questions)
We left out Q5 Sequences and Series
Paper 2:
Section A Q 1,2,3,5,6,7 (must do 5 of these 6 questions)
We left out Q4 Theorems and Enlargements.
Section B Q 11 (must do 1, there are 3 other options)
We left out Q8 Further Geometry, Q9 Vectors and Q10 Further Sequences and Series.
Do we want to prepare for any further questions?
Thursday, February 7, 2008
Linear Programming
The other topic we will cover in Friday's class is Linear Programming.
You need to know all the inequalities, simultaneous equations and co-ordinate geometry stuff for other parts of the paper anyway, so Linear Programming gives you an extra question with very little extra work.
It is the only option in Section B which we will be doing for now.
This is the 1-pager.
Perimeter, Area + Volume
Here is my 1-page overview of what you need to know.
Practice using the formulae for surface area, volume etc as they are presented in the mathematical tables.
Simpson's rule is in the tables also, but not in the form we are used to. h/3(1st+last+TOFE) is a good way to remember it, but you have to practice.
Alegbra class
Last night's homework question 2 had a simultaneous equation - one linear one quadratic - type question. Notethat it was xy and not x², but it has to be handled the same way.
x + 3 = 2y
xy – 7y + 8 = 0 <-this is the quadratic (can be x² or xy)
x = 2y -3 [rearrange linear equation]
(2y-3)(y) – 7y + 8 = 0 [sub in new value for x]
2y² – 3y -7y +8 = 0
2y² – 10y + 8 = 0 [always simplify equation if you can]
y² -5y + 4 = 0 [factor -4 x -1 = 4 and -4 -1 = -5]
(y-1)(y-4) = 0
So y = 1,4
If y= 1 then
x = 2(1) – 3, so x = -1 values: x = -1, y = 1
if y= 4 then
x = 2(4) – 3, so x = 5 values: x = 5, y = 4
And always check:
(-1)(1) – 7(1) + 8 = 0
(5)(4) – 7(4) + 8 = 0
Part c of this question works out as follows:
(i) √x + 1 [Turn them into same type of fraction ... LCM of 1 and √x is √x]
1 √x
= √x√x + 1 [multiply above and below leftmost fraction by √x]
√x √x
= x + 1 [don't forget √x√x = x]
√x
ii) ( 2√x ) • (x + 1 ) [cancel (x + 1) with (1+x) and cancel √x with 2√x ]
(1+x ) ( √x )
= 2
iii) 2 = x -3
x = 5
Tomorrow's double class will be Perimeter area + volume (1st period) and Linear Programming (2nd period) - Homework for tomorrow is 2006 Paper 2 - q1 and 11.
x + 3 = 2y
xy – 7y + 8 = 0 <-this is the quadratic (can be x² or xy)
x = 2y -3 [rearrange linear equation]
(2y-3)(y) – 7y + 8 = 0 [sub in new value for x]
2y² – 3y -7y +8 = 0
2y² – 10y + 8 = 0 [always simplify equation if you can]
y² -5y + 4 = 0 [factor -4 x -1 = 4 and -4 -1 = -5]
(y-1)(y-4) = 0
So y = 1,4
If y= 1 then
x = 2(1) – 3, so x = -1 values: x = -1, y = 1
if y= 4 then
x = 2(4) – 3, so x = 5 values: x = 5, y = 4
And always check:
(-1)(1) – 7(1) + 8 = 0
(5)(4) – 7(4) + 8 = 0
Part c of this question works out as follows:
(i) √x + 1 [Turn them into same type of fraction ... LCM of 1 and √x is √x]
1 √x
= √x√x + 1 [multiply above and below leftmost fraction by √x]
√x √x
= x + 1 [don't forget √x√x = x]
√x
ii) ( 2√x ) • (x + 1 ) [cancel (x + 1) with (1+x) and cancel √x with 2√x ]
(1+x ) ( √x )
= 2
iii) 2 = x -3
x = 5
Tomorrow's double class will be Perimeter area + volume (1st period) and Linear Programming (2nd period) - Homework for tomorrow is 2006 Paper 2 - q1 and 11.
Wednesday, February 6, 2008
Leaving Cert Algebra
There are two algebra questions on Paper 1 and algebra permeates almost every question on the exam to a certain extent. It is worthwhile investing time and effort in practising algebra.
This is my 1-page overview of algebra - an outline of the basic concepts you need to master.
We will review this in Thursday's class.
Homework for Thursday is 2005 Paper 1 questions 2 and 3.
Statistics wrap-up
For the most part the statistics homework was well done.
It was question 7 on this paper.
The two difficulties seemed to be
Thursday's class will be all algebra.
It was question 7 on this paper.
The two difficulties seemed to be
- creating the cumulative frequency table from the original data given. Make sure that the last number in the cumulative frequency table is the sum of all the frequencies - which was 80 (already used this value in part ii)
- figuring out how many cars were priced between the mean and the median. This was an unusual question. You had to take the mean you calculated in part ii (26.25 or 26,250), find this value on the X-axis, trace a vertical line to the curve and then find the corresponding position on the Y-axis. This should be 50. That tells you that the "50th car" is priced at the mean price. The number between the mean and median is therefore 50-40= 10 cars.
Thursday's class will be all algebra.
Tuesday, February 5, 2008
Statistics again
Today's class didn't go to plan - too many students didn't do the weekend work and we didn't get through the corrections.
We will have to do a full revision of Statistics in tomorrow's class and set the homework on Algebra.
Homework for the evening of 6th Feb will be two algebra questions: the 2005 paper I, q2 and q3.
We will have to do a full revision of Statistics in tomorrow's class and set the homework on Algebra.
Homework for the evening of 6th Feb will be two algebra questions: the 2005 paper I, q2 and q3.
Monday, February 4, 2008
Leaving Cert Statistics
There are lies, damned lies and statistics - attributed to Mark Twain or Disraeli depending on how you google it.
Statistics is the art of interpreting data - summarising complex data in numbers (by calculating a mean or a standard deviation) or presenting it graphically in graphs.
If you understand statistics, you'll know when someone is using their statistical knowledge to warp and twist the truth for their own advantage ... if you master statistics, you'll be able to warp and twist the truth to your own advantage.
We'll revise Statistics in class on Feb 5th.
Homework will be LC 2006 Paper 2 Q 7
Here is my 1 pager
Statistics is the art of interpreting data - summarising complex data in numbers (by calculating a mean or a standard deviation) or presenting it graphically in graphs.
If you understand statistics, you'll know when someone is using their statistical knowledge to warp and twist the truth for their own advantage ... if you master statistics, you'll be able to warp and twist the truth to your own advantage.
We'll revise Statistics in class on Feb 5th.
Homework will be LC 2006 Paper 2 Q 7
Here is my 1 pager
Introduction
This blog is aimed at my Leaving Cert students who are taking Ordinary Level Maths gearing up for their Mock exams and then for the Leaving Cert exams in June.
In real-world class, we will be preparing for the Mock exams which take place during the week starting Feb 18th.
Students can keep up with the schedule of revision using this blog. Any other students who may find this blog useful are welcome to join in.
The questions we are working toward are as follows:
Paper I
Q1 Arithmetic (completed mid Jan)
Q2 Algebra (Thurs 7th Feb)
Q3 Algebra (..)
Q4 Complex Numbers (Thurs 31 Jan)
Q6 Functions (Fri 1 and Tue 5 Feb)
Q7 Calculus (Fri 1 and Tue 5 Feb)
Q8 Calculus (Fri 1 and Tue 5 Feb)
Paper II
Q1 Perimeter Area + Volume (Fri 8 Feb)
Q2 Co-ord geom of Line (Fri 25 Jan)
Q3 Co-ord geom of Circle (Mon 28 Jan)
Q5 Trigonometry (Tue 29 Jan)
Q6 Probability (Wed 30 Jan)
Q7 Statistics (Wed 6 Feb)
Q11 Linear Programming (Fri 8 Feb)
All your comments and constructive feedback are welcome!
In real-world class, we will be preparing for the Mock exams which take place during the week starting Feb 18th.
Students can keep up with the schedule of revision using this blog. Any other students who may find this blog useful are welcome to join in.
The questions we are working toward are as follows:
Paper I
Q1 Arithmetic (completed mid Jan)
Q2 Algebra (Thurs 7th Feb)
Q3 Algebra (..)
Q4 Complex Numbers (Thurs 31 Jan)
Q6 Functions (Fri 1 and Tue 5 Feb)
Q7 Calculus (Fri 1 and Tue 5 Feb)
Q8 Calculus (Fri 1 and Tue 5 Feb)
Paper II
Q1 Perimeter Area + Volume (Fri 8 Feb)
Q2 Co-ord geom of Line (Fri 25 Jan)
Q3 Co-ord geom of Circle (Mon 28 Jan)
Q5 Trigonometry (Tue 29 Jan)
Q6 Probability (Wed 30 Jan)
Q7 Statistics (Wed 6 Feb)
Q11 Linear Programming (Fri 8 Feb)
All your comments and constructive feedback are welcome!
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