Apologies for the lack of recent posts. I've actually done more reading and writing lately than I have maths.
Nonetheless, I have booked to do Pure 1 and Stats 1 on the 9th January so expect a trickle of revision related posts over the next two months!
Wednesday, 29 October 2008
Sunday, 14 September 2008
Geometry of the Line (4): Circles
Time to go back to straight line geometry for a bit.
With knowledge of how to calculate the gradient, the perpendicular gradient, the mid-point and the length of the line, you've also done quite a lot of co-ordinate geometry of the circle. This is an interesting concept. A circle on a set of cartesian co-ordinates.
You start from the premise of a circle with centre (0,0). The equation of this circle is x2 + y2 = r2. If you study a diagram of it, or even draw one yourself, the reason for this becomes clear. Like a lot of co-ordinate geometry, it's based on Pythagoras. In this case, drawing a triangle inside the circle to a point somewhere on the circumference. Say that this circle with centre (0,0) passes through the point (3,4).
The line from the centre to the point functions as the hypoteneuse of the right angled triangle, while line x measures the distance travelled horizontally to the point, and the line y measures the vertical distance to the point.
The equation of this particular circle then tells us the radius.
32 + 42 = r2
or 9 + 16 = r2
= 25
r = 5.
The radius of the circle is 5.
Things get a little more complicated if the circle's centre is not (0,0). Say we have a circle with centre (2,3). We still devise the equation of the circle based on its distance from (0,0).
This circle's centre has moved 2 units x-wards and 3 y-wards (across and up) from the origin. Any calculation of the radius needs to take into account the fact that we are already some way from the origin.
So if the circle passes through the point (5,8), the distance from the centre to the point will be 5-2 and 8-3 (otherwise we're talking about including space outside the circle).
So the equation will be: (x-2)2 + (x-3)2 = r2.
In this case: 9 + 25 = 34
So r = √34
It is perfectly acceptable to leave the answer as a surd at C1, indeed often you have to.
Usually you are given the equation, say (x-4)2 + (y-7)2 = 4
or something.
The centre of the circle is ALWAYS THE OPPOSITE OF THE NUMBERS IN BRACKETS, in this case (4,7). Worth remembering. If it is (x-a)2 + (y-b)2 = r2 the centre is always (a,b).
With knowledge of how to calculate the gradient, the perpendicular gradient, the mid-point and the length of the line, you've also done quite a lot of co-ordinate geometry of the circle. This is an interesting concept. A circle on a set of cartesian co-ordinates.
You start from the premise of a circle with centre (0,0). The equation of this circle is x2 + y2 = r2. If you study a diagram of it, or even draw one yourself, the reason for this becomes clear. Like a lot of co-ordinate geometry, it's based on Pythagoras. In this case, drawing a triangle inside the circle to a point somewhere on the circumference. Say that this circle with centre (0,0) passes through the point (3,4).
The line from the centre to the point functions as the hypoteneuse of the right angled triangle, while line x measures the distance travelled horizontally to the point, and the line y measures the vertical distance to the point.
The equation of this particular circle then tells us the radius.
32 + 42 = r2
or 9 + 16 = r2
= 25
r = 5.
The radius of the circle is 5.
Things get a little more complicated if the circle's centre is not (0,0). Say we have a circle with centre (2,3). We still devise the equation of the circle based on its distance from (0,0).
This circle's centre has moved 2 units x-wards and 3 y-wards (across and up) from the origin. Any calculation of the radius needs to take into account the fact that we are already some way from the origin.
So if the circle passes through the point (5,8), the distance from the centre to the point will be 5-2 and 8-3 (otherwise we're talking about including space outside the circle).
So the equation will be: (x-2)2 + (x-3)2 = r2.
In this case: 9 + 25 = 34
So r = √34
It is perfectly acceptable to leave the answer as a surd at C1, indeed often you have to.
Usually you are given the equation, say (x-4)2 + (y-7)2 = 4
or something.
The centre of the circle is ALWAYS THE OPPOSITE OF THE NUMBERS IN BRACKETS, in this case (4,7). Worth remembering. If it is (x-a)2 + (y-b)2 = r2 the centre is always (a,b).
Saturday, 13 September 2008
A Note on Revision
Since I've finished C1 and S1 I am sort of both doing C2 and revising C1.
If I were to give one piece of advice, it would be to treat revision as a dynamic process. Do it as soon as you finish a topic or a module. Do it randomly, answer questions out of the blue, keep coming back to it.
Revision is only boring and hard when you have piles of stuff you barely remember to do in a short space of time. If you do it regularly, treating it like another aspect of your learning, then it becomes easier and things go in better. Writing this blog has been a valuable source of revision for me as well - I hope - as some of my readers.
Keep on keeping on revising. That's the sum total of my advice.
If I were to give one piece of advice, it would be to treat revision as a dynamic process. Do it as soon as you finish a topic or a module. Do it randomly, answer questions out of the blue, keep coming back to it.
Revision is only boring and hard when you have piles of stuff you barely remember to do in a short space of time. If you do it regularly, treating it like another aspect of your learning, then it becomes easier and things go in better. Writing this blog has been a valuable source of revision for me as well - I hope - as some of my readers.
Keep on keeping on revising. That's the sum total of my advice.
Blog Purpose Update
Well I know that I don't have all that many readers here but I don't mind too much. To judge from recent Sitemeter reports some people are actually reading what I have written, even if they are not so interested in commenting.
A quick reminder. I am no expert, in anything. Just a vaguely mature student of A Level maths who is doing it in his spare time when he isn't down the pub or surfing the blogosphere with increasing frustration and disappointment. So this blog was intended to be a wholly positive space in which I could devote time and effort to something of thought, however superficial. I do have a couple of problems though, especially with graphics and symbols. I still haven't worked out how to show these in blogposts properly. I have heard of something called TeX or LaTeX but i don't know how to use it.
If you have found anything on here useful, then I'm glad. It will be the first geniune contribution a blog of mine has made to the world.
A quick reminder. I am no expert, in anything. Just a vaguely mature student of A Level maths who is doing it in his spare time when he isn't down the pub or surfing the blogosphere with increasing frustration and disappointment. So this blog was intended to be a wholly positive space in which I could devote time and effort to something of thought, however superficial. I do have a couple of problems though, especially with graphics and symbols. I still haven't worked out how to show these in blogposts properly. I have heard of something called TeX or LaTeX but i don't know how to use it.
If you have found anything on here useful, then I'm glad. It will be the first geniune contribution a blog of mine has made to the world.
Thursday, 11 September 2008
Dividing Polynomials (ii)
Alright then, let's have a crack at this properly.
We are armed with some fraction revision and a little bit of knowledge of polynomials, including factorising using the system of comparing coefficients.
We only need to divide a polynomial by a linear factor for C1.
Division has this nasty habit of not being exact, but of leaving remainders. Unfortunately, 5 does not go easily into 24 but leaves a remainder. Polynomials are also subject to this problem. A linear expression might not be a factor of P(x) as such, but might go into something else, leaving factors and a remainder when asked to be slotted into P(x).
Remember that QUADRATIC EQUATIONS sometimes factorise into TWO LINEAR expressions,
while CUBICS factorise into LINEAR + QUADRATIC (and sometimes three linear).
So if you divide a quadratic by a linear, this might happen:
x2 + 2x + 4 divided by x-2.
If we are to use the concept of comparing coefficients, which is very efficient, we first rewrite the equation into the form it would take expressed as factors and remainders:
x2 + 2x + 4 = (x+p)(x+q) + r.
Got it?
We need to divide it by x-2 though, so we already have one desired factor (THIS DOES NOT MEAN THAT x-2 WILL BE A FACTOR - just that we are trying to express the quadratic in terms of how much x-2 goes into it.
So:
x2 + 2x + 4 = (x-2)(x+q) + r.
Then we collect the different terms in this form. We are not completely, expanding the brackets, we are re-arranging them in terms of their own factors in order to look for coefficients which will tell us the other factors and remainders.
x2 + 2x + 4 = x2 + (-2 + q)x + (-2q+r)
F(x) is in this case equal to x2 and (-2+q) lots of x, and then the two constants, with no x variable, (-2q and r, the possible remainder).
Therefore 2= (-2+q) so q= 4.
4 = (-2x4 + r)
4=8+r
r = -4
x2 + 2x + 4 = (x-2)(x+4) -4.
This only expresses the quadratic as factors and a remainder. The division isn't over yet.
Now let's actually divide the quadratic by (x-2).
x2 + 2x + 4
-------------
x-2
=
(x+4) - 4
----
(x-2)
I'll explain it later. Proper work beckons.
We are armed with some fraction revision and a little bit of knowledge of polynomials, including factorising using the system of comparing coefficients.
We only need to divide a polynomial by a linear factor for C1.
Division has this nasty habit of not being exact, but of leaving remainders. Unfortunately, 5 does not go easily into 24 but leaves a remainder. Polynomials are also subject to this problem. A linear expression might not be a factor of P(x) as such, but might go into something else, leaving factors and a remainder when asked to be slotted into P(x).
Remember that QUADRATIC EQUATIONS sometimes factorise into TWO LINEAR expressions,
while CUBICS factorise into LINEAR + QUADRATIC (and sometimes three linear).
So if you divide a quadratic by a linear, this might happen:
x2 + 2x + 4 divided by x-2.
If we are to use the concept of comparing coefficients, which is very efficient, we first rewrite the equation into the form it would take expressed as factors and remainders:
x2 + 2x + 4 = (x+p)(x+q) + r.
Got it?
We need to divide it by x-2 though, so we already have one desired factor (THIS DOES NOT MEAN THAT x-2 WILL BE A FACTOR - just that we are trying to express the quadratic in terms of how much x-2 goes into it.
So:
x2 + 2x + 4 = (x-2)(x+q) + r.
Then we collect the different terms in this form. We are not completely, expanding the brackets, we are re-arranging them in terms of their own factors in order to look for coefficients which will tell us the other factors and remainders.
x2 + 2x + 4 = x2 + (-2 + q)x + (-2q+r)
F(x) is in this case equal to x2 and (-2+q) lots of x, and then the two constants, with no x variable, (-2q and r, the possible remainder).
Therefore 2= (-2+q) so q= 4.
4 = (-2x4 + r)
4=8+r
r = -4
x2 + 2x + 4 = (x-2)(x+4) -4.
This only expresses the quadratic as factors and a remainder. The division isn't over yet.
Now let's actually divide the quadratic by (x-2).
x2 + 2x + 4
-------------
x-2
=
(x+4) - 4
----
(x-2)
I'll explain it later. Proper work beckons.
Dividing Polynomials
As promised....
One of the key elements of this is the ability to split off fractions effectively.
Think of a normal, but improper fraction, such as 4/3. This could be expressed as 3/3 + 2/3 (or any other combination but this is clean and tidy and useful for us now). 3/3 is of course 1. We are left with 1 + 1/3.
The same thing can be done algebraically, so as to solve basic division questions.
Take (x + 6)/x. As above, we can think of this as x/x + 6/x. x/x = 1, so the answer is 1 + 6/x.
We use this property, making part of the numerator match the denominator, in dividing polynomials.
One of the key elements of this is the ability to split off fractions effectively.
Think of a normal, but improper fraction, such as 4/3. This could be expressed as 3/3 + 2/3 (or any other combination but this is clean and tidy and useful for us now). 3/3 is of course 1. We are left with 1 + 1/3.
The same thing can be done algebraically, so as to solve basic division questions.
Take (x + 6)/x. As above, we can think of this as x/x + 6/x. x/x = 1, so the answer is 1 + 6/x.
We use this property, making part of the numerator match the denominator, in dividing polynomials.
Sunday, 7 September 2008
Rounding Up Factorisation
...so after all that you now have the technology (all right, the skills) to factorise a polynomial completely. You first do trial and error to find the first linear factor, then you multiply it by ax2 + bx + c and compare coefficients to find the quadratic. Then you check if the quadratic factorises further.
Onto dividing polynomials proper....
Onto dividing polynomials proper....
Subscribe to:
Posts (Atom)
