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Posts Tagged ‘tutorial’

## Differentiation From First Principles (with example)

When you differentiate a function or curve you are finding the gradient of the tangent to the curve. If the curve you are differentiating a curve or function that is in terms of x, eg y=x2, then the differential is also a curve or function in terms of x, eg 2x. This allows you to find the gradient of the curve for any point on the curve with a known x co-ordinate.

Differentiating from first principles involves finding the gradient of a tangent to a curve from the basic definition of gradient,
grad = (y1 – y2) / (x1-x2)
ie) not by following a rule.

When differentiating from first principles we consider chords (lines passing from one point on the curve to another) from the point on the curve with the x coordinate x to points with a slightly larger x coordinate x+d. The gradient of these chords is an approximation of the gradient of the tangent to the curve at the point with x co-ordinate x and the approximation becomes better as d becomes smaller. If we take the limit as d tends to 0 we find the actual gradient of the tangent

Note: the derivative of a curve y = f(x) is written as dy/dx

## Example: y=x2

Consider 2 points P and Q on the curve y=x2 a small distance apart where the difference in their x co-ordinate is d. Then
P(x,x2)
Q(x+d,(x+d)2)
Then the gradient of the chord PQ is given by

To find the derivative we must now take the limit of this as d tends to 0.
So we find

By David Woodford

Categories: calculus, maths

## 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 9x9k -1=9x9k – 9+ 8=9(9k-1)+8
but from Pk we know 9k-1 = 8a so
9k+1-1 = 9(8a)+8=8(9a+1)
which is of the form 8b so is divisible by 8
Hence if Pk “9k-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

Categories: maths

## 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.

Categories: maths Tags: , , ,

## Auxiliary Angle Method for Solving Trigonometry Equations

April 19, 2009 2 comments

This is a method of solving equations in the form asinx+bcosx = c where a and b are constants and c is another expression.

It involves rewriting letting asinx + bcosx = rsin(x+y) (or you could use cos(x+y)) where y is acute and then finding values for r and y, then with only one trig function to deal with the equation can be solved more easily.

For example

consider 2sinx + 3cosx = 3

Let 2sinx + 3cosx = rsin(x+y)

Now expand the sin(x+y) to get

2sinx + 3cosx = rsinx cosy + rcosx siny

Since y is constant and therefore cosy and sin y are constant we can compare the coefficients to get

2 = rcosy —–(1)
3 = rsiny ——(2)

We can solve these to find values for r and y.
To find y consider (2)/(1) to get

3/2 = tany
since sin/cos = tan and the r’s cancel
so y = 56.3 °

To find r consider (1)2+(2)2 to get
22+32 = r2
since sin2+cos2 = 1
so r =√13

So we can write

2sinx + 3 cosx = √13 cos(x+56.3) = 3

so x = cos<sup>-1</sup>(3/√13) -56.3

so x = cos-1(3/√13) -56.3

since cos-1(3/√13) = 33.7 for solutions between 0° and 90°

x = ±33.7 -56.3 + 180n where n is an integer

In General

asinx + bcosx = √(a2+b2) sin(x+tan-1(b/a))

If you have any questions, comments or corrections please leave them as a comment below

By David Woodford

Categories: maths, trigonometry

## Rational and Irrational Numbers

All real numbers are either rational or irrational.

A number x is rational if it can be written in the form a/b where a and b are integers and b≠0 and a number which cant be written in this form is irrational.

Most of the time we deal with rational numbers, for example all the integers are rational as they can be written in the form of themselves over 1, all fractions are obviously rational and also are terminating or repeating decimals. When an irrational number is written out it will appear as an infinite non repeating decimal. Examples of irrational numbers are π (pi) and √2, in fact many square roots are irrational numbers.

Numbers can usually be proved to be irrational by contradiction, by assuming that they are rational and writing them in the form p/q where p and q are co-prime. Then a contradiction is usually produced by showing that they both have common factor other than 1. For example when proving √2 is irrational you can show that p and q must both have a common factor of 2.

There infinitely many rational and irrational numbers, however whilst it is possible to use a method to count all the rational numbers, if you had an infinite amount of time, this cannot be done for the irrational numbers.

Showing a repeating decimal is rational

A repeating decimal can be shown to be rational using the following method:

Consider the repeating decimal x,

Multiply x by 10 n where n is the number of repeating digits
Now subtract x from x10 n to get (10n-1)x
This number will be a finite decimal as all the repeating terms after the decimal point should cancel.
divide through by (10n-1)x to get an expression for x in the form of a fraction.

eg let x = 1/7 = 0.142857142857142857………
then 1000000x = 142857.142857142857………
so 999999x = 142857.0
x=142857/999999
so 1/7 = 142857/999999 therefore 1/7 is rational

if you have any questions, comment or corrections please leave them as a comment below

By David Woodford

Categories: maths, surds

## How to use Surds – Add, Multiply and Rationalise

Surds are roots to some power (usually square roots) that are left in their “root” form as opposed to being calculated as a decimal. In a way they are the power equivalent of fractions and are used when decimal value of a root isn’t needed, such as in the middle of a calculation, so it is better to leave the result more precisely as a surd.

Surds usually take the form a+b√c where b and c are non-zero

eg) √2
4+2√3

Simplifying Surds

It is easier to deal with surds if you can get the value in the square root to its lowest possible value by taking any square factors outside the root.

For example √50 = √(25×2) = 5√2

To do this I looked for any factors of 50 which are square, eg 25=5×5, can took these outside the square root – when taking a number outside the square root you obviously have to take the square root of the number so when you take out 25 you get a 5 left outside.

So in general

√(a2b) =a√b

This has the benefit of when multiple surds are in a calculation al the surds will be in their lowest terms so they will be easier to group

Adding and Subtracting Surds

Surds can easily be added just by considering then as another variable, so just add up all the coefficients.

eg) 2√3 +6√3 – 3√3 = 5√3

Note– only roots of the same number can be added and subracted in this way, roots of different numbers must be left as they are. This makes it important to simplify all surds as it means you will be able to spot all the surds that can be added or subracted.

eg) 2√3 + 4√5 cannot by simplified

but

3√2 +2√50 = 3√2 +2×5√2=13√2

Multiplying and Dividing Surds

To multiply or divide surds just multiply or divide the values within the roots with each other and the coefficents of the roots with each other. When surds contain roots and non root parts multiply as if they were brackets

eg) 3√2 x 4√5 = 12√10

3√2 ÷ 4√5 = ¾ √(2/5)

(2+4√3) x (3+2√5) = 6 +4√5+12√3+8√15

Note – you may be able to further simplify the surd after you multiply

eg) 2√6 x √3 = 2√18=2×3√2 = 6√2

In general

(a+b√c) x (d + e√f) = ad + ae√f + db√c + be√fc

Rationalising and Conjugates of Surds

When simplifying fractions all surds are usually removed from the bottom. In order to do this both the top and the bottom of the fraction are multiplied by the conjugate of the botom. If there is only a root on the bottom you can simply multiply the top and bottom by the root on the bottom.

The conjugate of a surd is simply the surd with the sign of the coefficient of the root reversed so

a+b√c becomes a-b√c

So to rationalise the fraction

(2+√3)/(3+√4)

we multiply the top and bottom by 3-√4 to get

(3-√4)(2+√3)/(3+√4)(3-√4)

which once the brackets are expanded becomes

(6+3√3-2√4-√12)/5

as the √4 x √4 becomes √16 = 4 and the bottom is the difference of two squares.

By David Woodford

Categories: maths, surds