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Shodor Graph Sketcherr

This enables you to plot most functions in Cartesian form. It requires you to enter the graph as a function of x and then clicking “graph” will plot it for you. Using the tools in the top right hand corner you can zoom in/out and more around the graph, you can also adjust the axis.

A screen shot of the graph y=x^2 using the Shodor Graph Sketcher

Polar Co-ordinates Graph Sketcher

This applet will allow you to sketch graphs using polar co-ordinates. You can enter a range and your graph as a function of t (the angle). It also allows you to zoom in and out of the graph and includes a list of the functions it supports and how to enter them.

Screenshot of the graph r=sin(t) as drawn by the polar co-ordinates graph sketcher

Also see

• Equations using Polar Co-ordinates (0)
• Exponential Functions (1)
• Using Polar Co-ordinates and Converting to and from Cartesian (0)
• Tan Graph – y=tan(x) (0)

It works by taking an “initial line” (shown in red — the equivalent of the positive x axis in Cartesian co-ordinates) with the origin at one end. Then any point can be found by drawing a line from the point to the origin and quoting the length of this line (the radius) and the angle the line makes with the origin. These numbers are usually written in brackets in the same way as Cartesian co-ordinates with the radius first followed by the angle. For example look at the point (3,π/3) below.

The point (3,pi/3) in Polar Co-ordinates

Note// The angles are usually measured in radians

Cartesian Equivalent of Polar Co-ordinates

Since both Cartesian and polar co-ordinates are a way of describing a point position in a 2D plane it is possible to convert between then. When doing this the initial line is taken as the x-axis.

To find the Cartesian co-ordinates we must use trigonometry by drawing a vertical line down from the point to the x-axis to form a right angled triangle. The length of the vertical line then gives the y- coordinate and its distance from the origin gives the x-coordinate.

To find the x and y values for the point (r,Θ) we must therefore use the equations

x = rcosΘ

and

y = rsinΘ

If we want to go in reverse to find the polar co-ordinates of the point (x,y) in a Cartesian system we must solve these equations simultaneously.

We can eliminate Θ by squaring both of the equations to obtain

x² = r²cos²Θ
and
y² = r²sin²Θ

and then adding these equations to get

x² + y² = r²(cos²Θ + sin²Θ)

by substituting the trigonometric identity sin²+cos² = 1 to get

x² + y² = r²

we have removed Θ and can therefore calculate r using
r =√(x² + y²)

To find Θ we can divide the two equations given at the start such that the r’s cancel to get
y/x = sin Θ/cos Θ
which using the identity tan Θ=sin Θ/cos Θ gives

Θ = tan^-1(y/x)

Also see

• Online Graph Sketchers (0)
• Equations using Polar Co-ordinates (0)

1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,
46368,75025,121393,196418,317811,514229,832040,1346269….

which quickly become very large.

The Fibonacci numbers, f_r can be defined by
f₁ = f₂ = 1
and
f_k = f_k-1 + f_k-2 for k≥3

PHI the golden ratio

PHI or Φ is said to be the golden ratio since so many things in nature seem to naturally arrange themselves in this ratio. It is approximately equal to 1.618033988749895.

Phi is also the positive solution to the equation

x² = 1 + x

which has the irrational solution

phi = (1+√5)/2

Interestingly an irrational number is one which cant be written as the ratio of two integers so the golden ratio is not in fact a ratio meaning ratios in nature can only become very close to it but cant actually equal it.

PHI and Division of Fibonacci numbers

It is interesting to calculate the ratio of consecutive numbers in the Fibonacci sequence. These ratios quickly approach the golden ratio know as PHI or Φ.

The start of list of value obtained from these divisions are:
1
2
1.5
1.6666666666666667
1.6
1.625
1.6153846153846154
1.619047619047619
1.6176470588235294
1.6181818181818182
1.6179775280898876
1.6180555555555556
1.6180257510729614
1.6180371352785146

Calculation of PHI from limit of Fibonacci divisions

If we suppose that the ratio of consecutive terms in the Fibonacci sequence do approach a limit we can use this to find the value of phi.

IF we denote the nth Fibonacci number by f_n and the nth ratio as r_nthen

r_n = f_n+1 / f_n

But using the definition of a Fibonacci number:

f_n+1 = f_n + f_n-1

then the ratio is

r_n = (f_n + f_n-1 )
f_n

Which can be simplified to be

r_n = 1 + f_n-1/f_n

but f_n-1/ f = 1/r_n-1

so

r_n = 1 + 1/r_n-1

Now supposing that the ratios tend to a limit p as n tends to infinity then p is the solution of the equation

p = 1 + 1/p

which can be re-arranged to give

p² – p – 1 = 0

and can be solved using the quadratic equation to give

p = (1 + √5)/2
and
p = (1 – √5)/2

The first of these solutions happens to be the golden ratio PHI or Φ

If anyone has a proof that these ratios do infact approach a limit please include it in the comments of this post.

Also see

• Arithmetic Series and Progression (2)

A series of trapeziums can be used to approximate the area under a curve

When using the trapezium method the widths of the trapeziums used can be different, however it is often easier to calculate the total area (using the formula explained later) if all of the trapeziums are of equal width.

To improve the accuracy of the approximation you can use more trapeziums (the process of integration is simply allowing there to be an infinite number of trapeziums)

The area A of a trapezium with height (width when vertical in the approximation) h, and parallel sides a and b is given by

A=h(A+B)/2

To calculate the approximation we can let the x-values where each of the sides of the trapeziums touch the x-axis be x₀,x₁,x₂… and the y values where they cut the curve be y₀,y₁…

This means the the width of the first trapezium is x₁-x₀ which we will let equal d which is the same for all the trapeziums if we let their widths be equal. The parallel sides are of length y₀ and y₁ so the area A₀ is given by

A₀ = d(y₀+y₁)/2

The total area A under the curve is found by adding the areas of all of the trapeziums.

A = d(y₀+y₁)/2 + d(y₁+y₂)/2 + d(y₂+y₃)/2 + d(y₃+y₄)/2 +…….

However since d is a common factor it can brought outside in a bracket. Also all the y co-ordinates occur twice (in two consecutive trapeziums) apart from y₀ and the last y co-ordinate but all the terms are also halved this means that all but the first and last y values should be counted once and the first and last halved so we get the trapezium rule as follows

A = d(y₁+y₂+y₃ + … + y_n-1 +(y₀+y_n)/2)

where there are n trapeziums.

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1. All angles are the same
2. All corresponding sides between the two shapes are of the same ratio.

For example the two shapes on the left are similar but the ones on the right are not.

These tri angles are similar because all sides are of the ratio 1:2

These triangles are dis-similar because sides are of differen lengths

The Sides and angles of one triangle can be worked out from another if they are known to be similar. The angles are all the same. To find the sides you must have one side which is known in both triangles. From this you can work out the ratio of the lengths of the sides of the two triangles and then calculate any unknown sides of one triangle if they correspond to known sides of the first by multiplying by the ratio.

The areas of similar triangles are also related. However because the area is two dimensional instead we must first square the ratio of the lengths of the sides before we can calculate the two areas.

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Pyramids with different base shapes

The volume, V, of any of these pyramids is given by a third of base area, A, multiplied by the height, h or can be written as

V=Ah/3

The most usual way to prove this is to split the pyramid up into layers which can be approximated as cuboids, then by increasing the number of layers and therefore reducing the thickness of each layer the approximation becomes better untill when the the width of each layer tends to zero you get the exact volume of the pyramid.

To see an actual proof see

http://www.trans4mind.com/personal_development/mathematics/geometry/pyramids.htm

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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)

Also see

• Area And Circumference of a Circle : pi (38)
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• Integration by Parts (0)
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• Area of a Triangle (3)

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

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• Proof by Contradiction (0)
• Rational and Irrational Numbers (5)
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• Trigonometry Identities (11)
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P(h +a cos t ,k+ b sint), Where P is a point of the ellipse

or

where h and k are the amount the center of the ellipse is offset from the origin in the x and y directions and a and b the points on the x and y axis where the ellipse cuts.

An ellipse has to focal points (+/- ae) and can be defined as the loci of all the points P such that the sum of the distances from P to each focal point is constant.

The eccentricity of an ellipse e defines how sharp the ends of it are. e is always < 1 and can be calculated from the equation b² = a²(1-e²).

The directrices of an ellipse are x = ±a/e. These are lines such that the ratio of the distance from every point P on the ellipse from the directic and foci is the constant e, the eccentricity of the ellipse.

For example at the point (a,0) on the ellipse the distance to the directirc is
a/e – a = a(1/e – 1) = a(1-e)/e
and the distance from the ellipse to the foci is
a – ae= a(1-e)
the ration of these is
a(1-e)
a(1-e)/e
which equals e

and for any point p,
the distance to the directric is
a/e – acost = a(1/e – cost) = a(1-ecost)/e
when squared becomes
a²cost(e²cost-2e)+a²
e²

the distance to the foci is

√{(acost-ae)² + b²sin²t}
which with substitution for b given above becomes
a²cost(e²cost-2e)+a²

hence the ratio of the square of these 2 distances e² so the ratio of the distances themselves is e.

Also see

• Area of a Triangle (3)
• Proof by Contradiction (0)
• Rational and Irrational Numbers (5)
• Derive Quadratic Formula (16)
• Trigonometry Identities (11)
• Proof of Pythagorases Theroem (0)

Square used to prove Pythagoras theorem

This can be proved quite easily by drawing a square into which fit 4 of the same right angled triangle as shown below

As you can see the area of the whole square is equal to the the sum of the 2 shorter sides squared or (a+b)². The area of the green square left is the square of the longest side c². We also know that the area of each of the triangles is 1/2 x base x height = ab/2

From these 3 areas we can prove the theorem. The know that the total area of the square is equal to the area of the green square plus 4 of the triangles ie)

(a+b)²=c²+ 4ab/2
a² + b² + 2ab = c²+ 2ab

The 2ab ‘s cancel and we are left with Pythagoras theorem
a² + b² = c²

Also see

• Trigonometry Identities (11)
• Proof of Cosine Rule (4)
• Area of a Triangle (3)
• Proof by Contradiction (0)
• Rational and Irrational Numbers (5)
• Derive Quadratic Formula (16)