## Overestimates and Underestimates of Integrals

We can estimate an integral using the trapezium or the mid – ordinate rule. Sometimes however it is desirable to find underestimates and overestimates for an integral. The closer these are, the more accurately can the value for the integral be found and we can also quantify the error for the estimate. The estimate below shows the graph ofWe can find an underestimate for by finding the minimum value of y on each of the intervals and adding the areas of the rectangles on each interval with these heights.

Ifis the actual value of the integral, then

Similarly an overestimate for I can be found by finding the maximum value ofon each of the intervals and adding the areas of the triangles on each interval with these heights.

Hence

## Radius of a Circle Inscribed in an Isosceles Right Angled Triangle

The diagram below shows a circle inscribed in an isosceles right angled triangle.

The vertical distance from the base of the triangle at O to the point A where the circle is a tangent to the triangle in the diagram below is

By symmetry the vertical distance from O to A is half the height (or base) of the triangle,

Hence

## Calculating Equal Monthly Payments on a Loan

Loans are usually arranged to be repaid at a constant fixed monthly rate. This means that to start with a large fraction of the repayment made each month is used to pay interest. As time progresses the amount used to pay interest decreases and with the last payment the loan is paid off completely.

It is quite tricky to calculate the month repayment. The method is illustrated in this example.

A loan of £500 is taken out at 11% annual interest to be repaid at a fixed monthly rate. Interest is to be charged from the date of the loan and the first repayment is to be made a month after the loan is taken out. Find the monthly repayment.

Call the monthly repayment x. Since the annual rate of interest is 11%, the monthly rate of interest is the solution to

After a month the amount owing in £ isand after the first monthly payment the amount owing is

After two months the amount owing in £ isand after the second monthly payment the amount owing is

After three months the amount owing in £ isand after the second monthly payment the amount owing is

We will carry on in the same way, until after two years, twenty four monthly repayments, the whole loan is paid off. After the last monthly payment there is no money left owing so

The expression on the right is a geometric series with first termand common ratioso the right hand side equalsand

Hence

Substituting the value ofinto this expression gives a monthly repayment of £23.18 to the nearest penny. The total amount repaid is £556.41 to the nearest penny.

## Angles in the Quadrilateral Formed By Centres and Intersections of Two Circles

Two circles of raddi R and r respectively. Intersect in the point A and B as shown below.

From the Cosine Rule, for the triangle CAB,and for the triangle OAB,

Setting these equal gives

Now use the identity cos 2x = 1-2 sin^2 x rearranged asto give

Dividing by 4 and square rooting gives

## Savings Schemes

Many saving schemes require a regular deposit to be made in each set time period – typically week, month or year. A very good example is saving for a pension. Money is deducted from your wages and invested. It is hoped that over a working life enough will be saved to provide a good income in retirement.

If the rate of interest is fixed, we can reduce the problem to consideration of a geometric series. Suppose for example that £1,000 is invested at the start of each year for a period of ten years. Money is paid at the end of each year on the money in the account at the start of the year.

At the start of the 1st year £1,000 is invested, gathering interest for the entire ten years. At the end of the ten years this £1,000 will have become

At the start of the 2nd year £1,000 is invested, gathering interest for the last nine years. At the end of the ten years this £1,000 will have become

At the start of the 3rd year £1,000 is invested, gathering interest for the last eight years. At the end of the ten years this £1,000 will have become

At the start of the 4th year £1,000 is invested, gathering interest for the last seven years. At the end of the ten years this £1,000 will have become

At the start of the 5th year £1,000 is invested, gathering interest for the last six years. At the end of the ten years this £1,000 will have become

At the start of the 6th year £1,000 is invested, gathering interest for the last five years. At the end of the ten years this £1,000 will have become

At the start of the 7th year £1,000 is invested, gathering interest for the last four years. At the end of the ten years this £1,000 will have become

At the start of the 8th year £1,000 is invested, gathering interest for the last three years. At the end of the ten years this £1,000 will have become

At the start of the 9th year £1,000 is invested, gathering interest for the last two years. At the end of the ten years this £1,000 will have become

At the start of the 10th year £1,000 is invested, gathering interest for the last year. At the end of the ten years this £1,000 will have become

The total amount has grown over the period of ten years to the sum of all the above expressions. It is convenient to write the sum in reverse order:

Obviously the terms form a geometric sequence with first termand common ratioThere are ten terms so, using the formula for the sum of a geometric series,

we have

## Proof of the Remainder Theorem

The remainder theorem states that when a polynomialis divided by a linear expression the remainder is

Example: Whenis devider bysubstituteto obtain the remainder

We can prove it quite easily by performing long division ofbyto obtain the quotientand remainderWe can write

or equivalently

(1)

If the degree ofisthen the degree ofis and the degree ofmust be one less than the degree ofie the degree ofis 0 sois a constant. We can write

Now substitute

so thatas required.

## Growth and Sustainability

‘Geometric’ is perhaps the wrong term for a sequences with terms defined by the ruleEach term is being multiplied by a constant factor to obtain the next term.

A quantity is said to grow or decay exponentially if the quantity at the start of each time period is multiplied by a constant factor to obtain the quantity at the end of the period, or the start of the next period.

If the constant factor is less than 1, then the quantity is decaying exponentially.

If the quantity is greater than 1, the quantity is growing exponentially.

Exponential growth is what the World economy has been experiencing for the past several hundred years. Every year since 1700 the world economy has grown by about 2.1% on average. 2.1% does not sound like a large increase, but a growth rate of 2.1% means that the World economy has been doubling every 33 years.

Economies have always grown by finding new lands to conquer and resources to exploit. Now however, all the land on Earth essentially belongs to someone, so no – one can gain title by just sitting on it and shouting, ‘this is mine’. Essentially all the natural resources on Earth are being exploited or overexploited. The future we face can be illustrated by considering oil. Roughly half of all the oil that was in the ground has been exploited. Demand for oil is growing year on year, but there are estimated to be only about 40 years worth of recoverable oil reserves in the ground. With no feasible alternative to oil in sight, the World economy faces a crunch when available supplies of oil inevitably dry up.

Suppose then, that astronomers found a ‘twin Earth’ on the other side of the Sun, and scientists could work out how to exploit it for its oil. Would we be saved? No. Assuming oil consumption grows at 2.1% a year, and in the current year 1/40 of the Earth’s oil reserves are used up, equal to 1/120 of our augmented reserves, the 1.5 ‘Earth’s worth of oil’ now available to us would only last n years, where n is the solution to the equation

Rearrangement givesyears.

## Proof of the Cosine Rule

With a triangle labelled as below

the Cosine Rule states

to prove the rule, drop a perpendicular fromto the side

Pythagoras Theorem gives

Equating these expressions gives

hence

From the above diagram,so

## The Factor Theorem

The Factor Theorem

Ifis a factor ofthensois a root ofor equivalently, a solution of the equation

Example: Show thatis a factor of

henceis a factor of

Example:andare both roots of the quadratic expressionFind

Sinceandare both rootsandare both factors. A quadratic equation has at most two real factors/roots hence

More complicated questions may involve simultaneous equations:

andare both roots of Find a and b and factorise f(x)..

divided byremainder is

divided byremainder is

We now solve the simultaneous equations

(1)

(2)

3*(1)+(2) gives

Then from (1)

is a cubic and has the two given roots so must have a third linear factor

by considering the coefficient ofand by considering the constant term 1

## Bearings

Bearings involve using trigonometry, generally the cosine or sine rules:

Cosine Rule:

Sine Rule:

For the above diagram, find

a)The distance BC

b)The bearing of A from B and the bearing of A from C.

a)Label the triangle as above, with sides labelled by little letters opposite angles labelled by big letters.

so

b)

We draw a right angled triangle between A and B. Construct a right angled triangle at A and a horizontal line starting from A to the right.The internal angles of the right angled triangle are 70 and 20 degrees. Hence the bearing of A from B is 360-20=340 degrees.

To find the bearing of A from C, use the Sine Rule to find C.

Construct the right angled triangle as shown, with the angles at C then the bearing is 56+32.75=88.75 degrees.