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Find equation of tangent to a curve

November 19, 2009 Leave a comment

The tangent to a curve is a line which touches the curve at a point without intersecting it at that point so the gradient of the curve at that point and the gradient of the tangent are the same. So we can work out the point the tangent passes though and the gradient of the tangent from the equation of the curve, which will give us enough information to find the equation of the tangent.

Example y=x2
Find the equation of the tangent to the curve y=x^2 when x=4?

To do this we first need to find the gradient of the curve which we can do by differentiating it.
\frac{d}{dx}(x^2) = 2x
so at the point x=t the gradient is 2t.

From this we can get a general equation for the tangent using the equation for the gradient of a straight line
grad = $latex \frac{y – y_1}{x – x_1}
to get the general equation for the tangent at the point x=t by substituting x1=t, y1=t^sup>2 and m=2t
2t = \frac{y - t^2}{x - t} \rightarrow 2xt - 2t^2 + t^2 = y \rightarrow y=2xt-t^2

Then we can substitute in t=4 to find the equation of the tangent when x=4 to get
y=8x-16
which is our final answer.

Complex roots of unity

September 22, 2009 Leave a comment

Without complex numbers taking the square root of any positive integer, such as 1, will give you two answers, in this case +/- 1, but taking the cube roots will only five you one answer, 1. However when we consider the complex roots you will find that the nth root of any number will give you n roots.

This can be shown using De Moivres theorem. Firstly consider the complex number z = cosΘ + isinΘ and let

zn = 1
(cosΘ+isinΘ)n = 1

which using De Moivres theomrm gives

cos nΘ + isin nΘ = 1

We can now compare the real and imaginary parts to find the values of n

cos nΘ = 1 and sin nΘ = 0

Therefore nΘ = 2pπ for integral values of p
So \theta = \frac{2p\pi}{n}

which gives

\theta = \frac{2\pi}{n} , \frac{4\pi}{n} , \frac{6\pi}{n} , \frac{8\pi}{n} , ... , 2\pi

so the roots of unity are
z_1 = cos \frac{2\pi}{n} + i sin \frac{2\pi}{n}
z_2 = cos \frac{4\pi}{n} + i sin \frac{4\pi}{n}
z_3 = cos \frac{6\pi}{n} + i sin \frac{6\pi}{n}
etc

However what is interesting about these is when you plot them on an argand diagram they are evenly spaced around a circle of radius 1 and centred at the origin. An example for n=8 is shown below

The roots of unity are space evenly around a circle of radius 1 and center at the origin

The roots of unity are space evenly around a circle of radius 1 and centre at the origin

This happens becuase the increase in the angle for each successive root is equal since we divided 2pi by n.

Using Polar Co-ordinates and Converting to and from Cartesian

September 19, 2009 Leave a comment

Polar co-ordinates are a different co-ordinate scheme to the standard Cartesian co-ordinates.  Again any point in a 2D plane can be located using only two numbers.

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

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)

Trapezium Method for Approximating the Area Under a Curve

The trapezium method allows you to approximate the area under a curve by breaking the curve up into and number of trapeziums whose areas can be easily calculated and then adding these areas up. This can be seen below.

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

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 + … + yn-1 +(y0+yn)/2)

where there are n trapeziums.

Cartesian Equation of a Circle

June 28, 2009 2 comments

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

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,
g = -a
f = -b

to get

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