Cartesian Equation of a Circle

Since a circle is made up of all the points a fixed distance (its radius) from a given point (its centre) then the equation of a circle simply needs to ensure this is true. This can be done using Pythagoras’s theorem. This is because we can draw a right angled triangle with the centre of the circle at one corner and the point on the circle at the opposite corner as shown below. The radius is then the hypotenuse, the vertical side is the difference between the y co-ordinate of the point and that of the centre and the horizontal side is the difference between the x co-ordinate of the point and centre. From Pythagoras we therefore know that a circle of radius r and centre (a,b) must have a Cartesian equation

r² = (x-a)² + (y-b)²

Circle on Cartesian axis

However, we can expand these brackets out to get

r² = x² – 2ax + a² + y² – 2by + b²

but since a²+b²+r², -a and -b are all constant we can let

c = a²+b²-r²,
f = -a
g = -b

to get

x² + y² + 2gx + 2fy + c = 0
where the circle has a centre (-g,-f) and radius √(a²+b²-c²)

Sine Graph

The sine function is a periodic function meaning that it repeats itself every so many (in the case of sine 2pi radians or 360^o). It has a range of -1 to 1 and has a domain for -∞ to ∞. Starting at the origin it increase to 1 at 90^o or pi/2 radians and then decrease to -1 at 270⁰ or 3pi/2 radians and then returns to 0 and 360^o or 2pi radians.

On the graph below the angle, in radians, is along the x axis and the value of the sine function for that angle is on the y axis.

Graph of y=sin(x)

Proof by Mathematical Induction

Proof by induction involves proving that if a statement is true in one case then it must also be true for the next case. Then by showing that it is true for a base case, eg 0, you can conclude that it is true for all positive integral cases, since if it is true for case 0 it must be true for case 1 and if it is true for case 1 it must also be true for case 2 etc

To do this it is normal to let Pn be the statement you are trying to prove in a general case n. Let us use the example Pn: 9ⁿ – 1 is divisible by 8 so we may right 9ⁿ-1 = 8a where a is an integer

Then we must prove that if Pk is true Pk+1 is also true. This will require using the expression Pk otherwise it isn’t induction since you are not showing that the truth of each statement can be found from the previous one.

so consider 9^(k+1) – 1
This is equal to 9×9k -1=9×9k – 9+ 8=9(9k-1)+8
but from Pk we know 9k-1 = 8a so
9^k+1-1 = 9(8a)+8=8(9a+1)
which is of the form 8b so is divisible by 8
Hence if Pk “9^k-1 is divisible by 8″ is true then Pk+1 is also true

Finally we must prove that P1 is true and we then know that Pn is true for all k >=1 where k is an integer.

P1 is the statement 9¹ – 1 is divisible by eight
9¹-1=9-1=8 so is true

therefore Pn is true for all positive integral value of n

Why anything to the power 0 is eqaul to 1

It is a common identity that x⁰ = 1 for all x.
Here is one way in which to see why this is so.

We know

x/x = 1

but x= x¹ and 1/x = x^-1

x/x = x¹x^-1

We also know that x^ax^b = x^a+b which gives

x/x = x^1-1=x⁰

We initally stated that x/x = 1 and have shown x/x = x⁰ so

x⁰ =x/x = 1

so

x⁰ = 1

Implicit Differentiation

Implicit differentiation involves differentiating an equation that hasn’t been arranged such that all of one variable, eg y, is on one side and all of the other variable, eg x, is on the other side. For example differentiating the equation with respect to x (ie find dy/dx):

3x² +2y³ + 6 = 3x²y

TO do this you need to remember that the derivative of y with respect to x is dy/dx, hence when you differentiate a function such as

y=3x

you get

dy/dx = 3

Where the right has gone to the derivative of 3x and the left has gone to the derivative of y, dy/dx.

If you want to differentiate more complex terms involving x you can use the chain rule, since y can be written as a function of x.
so if f(x) = g(y) then
f’(x) = dy/dx g’(y)

(or a simple way of doing it is treat any y’s like x’s and stick a dy/dx on the end).

So for example y² differentiated becomes 2ydy/dx

All the other rules like the product rule and quotient rule still apply.
So to finish lets consider our original equation

3x² +2y³ + 6 = 3x²y

This becomes

6x + 6y²dy/dx = 6xy +3x²dy/dx

Which we can re-arrange to get

dy/dx = (6xy-6x)/(6y²-3x²)=(2xy-2x)/(2y²-1x²)

By David Woodford

Confindence Intervals

Confidence intervals are a range of values within which you can say an unknown value is expected to lie with a specified degree of certainty or probability. For example, from a sample of 10 journeys you might say that the you are 95% certain that the average time it takes me to get to school (from all journeys I have made to school not just the sample of 10) lies within the range 10-12.5 minutes.

When taking a random sample it is much better to use a confidence interval for your results rather than just giving the mean, because it gives an idea of how reliable your mean is. This is because you can calculate an average value from a set of completely random results but this doesn’t mean that you can have any certainty that the next result will be similar to the mean (since we have stated that the results are completely random they are no more likely to be close to the mean than any other value).

Calculating a confidence interval (for a normal distribution)

When calculating a confidence interval you must first decide on the percentage certainty that you are going to use for the interval (a common value to use is a 95% confidence interval)

You then use the reverse tables for the normal distribution to work the value of the standardised normally distributed variable to use.
Note: You must use the value half way between the certainty level and 100%, ie if you want a 95% confidence interval use 97.5% since you only want 2.5% on either side of the distribution.

You can now calculate the interval. To do this you need the standard deviation and mean of the sample. However to correct the standard deviation for the entire sample of possible tests divide by the square root of the number of items in your sample

ie) if you sample of n items is Y and the entire sample of possible results is X
sd(X) = sd(Y)/√n

Now the confidence interval is

X – z sd(X), X + z sd(X)

Where z is the value obtained from the inverse tables.

Equation of a Line: y = mx + c

The equation of all straightlines can be written in the form

y=mx+c

where m is the gradient, c is the intercept on the y-axis and you are plotting xagainst y. m and c ae constants meaning their value is fixed.

graph of y=mx+c

The y intercept, c, is how far up, or down if its negative, he line crosses the veritcal yaxis.
The gradient, m, is how step the line is. A gradient equal to 1 means that the graph is at 45 degress to the axis, a gradient greater than 1 is steeper and a gradient less than 1 is shallower.

To find where the graph cuts the x-axis simply let y=0 and find the value of x since at the x-axis the grap has zero height so y=0.

The easiest way to sketch a graph of the form y=mx + c is to find the x and y intercepts and then draw a straight line through these points.

Find the gradient of a line

The gradient, m, of  a straight line can be found using the equation

m=change in y/change in x = (y₁-y₂)/(x₁-x₂)

Where x1 and y1 are the co-ordinates of one point on the line and x2 and y2 are the coordinate of another point on the line.

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If you have any comments, questions or improvements please them as a comment below.

By David Woodford

First Order Differential Equations

First order differential equations are equations which include a first derivative ie) dy/dx if the equation is in terms of x and y.

Solving these equations involves finding a expression for one of the variables in terms of the other without including a derivative.

To do this usually a general solution is found, one which includes a constant from integration, and then using a given set of conditions (usually initial conditions) the particular solution can be found.

This is done by “splitting the variables” and then integrating the resulting equation. To do this you treat the derivative like a fraction and then multiply through by the bottom of that one derivative is on one side (eg dx) and the derivative is on the other side (eg dy), you then want all the x’s to be multiplied by the dx and all the y’s to multiplied by the dy such that there are no terms not being multiplied by either. You can then integrate both sides, the one by dx and the other by dy – remembering to add a constant to one side.

Then simply re arranging will give the general solution. If you need the particular solution substitute the given values into the equation to find out the value of the constant and then substitute this value for the constant into the general solution.

If all of that was a bit hard to follow here is s a worked example.

Example

Find the general solution to
dy/dx = 3x + 4
and the particular solution in the case y = 4 when x = 0 .
So we can rearrange to get

dy = (3x+4)dx

And then by integrating both sides we get

∫dy = ∫3x+4 dx

y = (3x+4)²/6 + c

Which is the general solution.
We can use boundary conditions, ie) y = 4 when x = 0 to find a particular solution
4 = (0+4)²/6 +c
c = 4-16/6 = 8/6 = 4/3

so the particular solution is
y = (3x+4)²/6 +4/3

Area of a Triangle

Here is a general formula to calculate the area, A, of a triangle with width w and height h as shown in the diagram.

A = ½ wh

triangle of width w and height h

Proof

In order to work out the area of a triangle we can draw it within a rectangle that touches all three corners and has a base of equal width as shown below.

triangle inside rectangle

From this diagram we can now consider the are of the triangle on the left and the one on the right seperatley and then add them together to get the area of the whole triangle. We can see that each of these triangles cuts the rectangle they are in in half so they half the area of that rectangle,

ie) the left hand triangle has the area ha/2 and the right hand one has the area hb/2 so the total are of the triangle is

A = ha/2 + hb/2

Now we can take out the h/2 as a common factor to get

A =(a+b) h/2

but a+b = w since that was how the triangle was constructed hence we get the formula for the area

A = wh/2

Proof by Contradiction

The aim of proof by contradiction is to prove a statement is true by assuming it is false and showing that this leads to a contradiction so that the statement must be true. This is often easier than proving the statement directly.

For example consider the proof √2 is irrational that follows

Assume root 2 is rational, ie that it can be written as r/s where s≠0 and r and s are both integers. We can choose r and s such that they have no common factors, since any common factors can be cancelled out.

then 2=r2/s2
so r2 = 2s2 —(1)
hence 2 is a factor of r2 and since 2 is prime 2 must also be a factor of r so we can write r = 2k where k is an integer.

From (1) we can now write
4k2=2s2
so s = 2k2
so 2 is also a factor of s
But we assumed r and s had no common factors, Contradiction therefore root 2 must be irrational.

So to conclude, here we have taken the statement root 2 is irrational, assumed it to be false, shown this leads to a contradiction and therefore concluded that root 2 must be irrational.