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The A to Z of mathematics a basic guide - Sidebotham T.H.

Sidebotham T.H. The A to Z of mathematics a basic guide - Wiley publishing , 2002. - 489 p.
ISBN 0-471-15045-2
Download (direct link): theatozofmath2002.pdf
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Square Golden
removed rectangle
Square
removed
-I—Golden rectangle
(e)
The ancient Greeks attributed special merits to the golden rectangle, because they believed it was a rectangle in perfect proportion. This made it visually very pleasing to the eye, and they used the golden rectangle in their architecture and designs. It appears in the architecture of the Parthenon, which was built about 400 BC on the Acropolis in Athens, and in many famous paintings, particularly those of Leonardi da Vinci.
The ratio of terms in the Fibonacci sequence is in the golden ratio. The terms of the sequence are 1,1,2, 3, 5, 8, 13, 21, 34, 55, 89,144,233, — Suppose we take
GRADIENT 219
one term and divide it by the following term:
144 = 233 = 0.618 (to 3 dp)
89= 144 = 0.618 (to 3 dp)
34 = 55 = 0.618, (to 3 dp)
And so on. To an accuracy of two decimal places the result is the golden ratio. The pattern breaks down for the early terms of the sequence.
References: Enlargement, Fibonacci Sequence, Quadratic Formula, Ratio, Rectangle.
GOLDEN RECTANGLE
Reference: Golden Ratio.
GOODNESS OF FIT
Reference: Line of Best Fit.
GRADIENT
Another name for gradient is slope. The gradient of a straight line is expressed as a fraction, and is a measure of how steep the line is. The gradient may be positive, negative, zero, or undefined. The gradient of a line may be less than 1 or greater than
1. The gradient of a straight line is best described using squares on a grid, or x-y axes.
The symbol for the gradient of a straight line is m. The convention is that m is positive for lines that slope in the direction / and m is negative for lines that slope in the direction \. In other words, if a line is increasing from left to right it has a positive gradient, and if it is decreasing from left to right it has a negative gradient.
B
(a)
To find the gradient of the straight line AS in figure a, we first complete a right-angled triangle ACS. The vertical length SC is called the rise and the horizontal length
220 GRADIENT
AC is called the run. The gradient of the line AB is equal to the fraction
rise
m =-------
run
This particular line AB has a positive gradient because it slopes toward the right.
In the examples that follow the gradients of a variety of straight lines will be calculated using the formula m = rise/run. Some of the gradients will be positive and some will be negative according to the direction of the line. The fraction for the gradient is usually canceled down to its simplest form; it may be written as a mixed number or as an improper fraction, but the latter is preferable for ease in drawing the gradient.
is undefined.
(b)
The examples in figure b show that vertical lines have an undefined gradient, which may be referred to as a gradient of infinity, the symbol of which is oo. Also, horizontal lines have a gradient of zero. It can be seen that a line with a gradient of | is steeper than a line with gradient |, because | is greater than |.
To find the gradient of a longer line segment, we can choose any two points on the line and find the gradient of that line segment, which will be the same as the gradient of the longer line.
Parallel Lines Two lines are parallel if they both have the same gradient. If one line has a gradient of mi = | and another line has a gradient of m2 = f then the two lines are parallel, because mi = m2.
Perpendicular Lines Two lines are perpendicular if the product of their gradients is — 1. If one line has a gradient of mi = | and another line has a gradient of m2 = — |, then the two lines are perpendicular, because mi x m2 = — 1.
GRADIENT 221
In the examples so far we have counted squares to obtain the gradient of a line, but when different scales are involved, as in graph questions, it is important to measure units rather than count squares.
Example. Harry takes his family out in the car for a picnic. They leave home and travel 80 km to Boulder Bay at a constant speed, and arrive there after l| h. Their picnic lasts 2 h. Afterward they travel home in 1 h. On the graph in figure c the journey out to Boulder Bay is marked by A, the picnic by B, and the journey home by C. Find: (a) the speed out, (b) the speed back, (c) the average speed for the whole trip.
Distance-Time Graph of Picnic
80
60
Distance from home in km 40
20
0 1 2 3 4 ‘ J
Time in 2
(c)
Solution, (a) The speed for part A is the gradient of the distance-time graph. Write
m
Substituting rise = 80 and run in the formula for gradient
1;
53.3 (to 1 dp)
For the journey out the speed is 53.3 km/h.
(b) Similarly, for part C, the gradient is —80/1 = —80. For the journey back the speed is 80 km/h. This gradient is of course negative, but since speed cannot be negative (unlike velocity), we ignore the negative sign.
(c) Write
total distance covered
Average speed =--------------------------
total time taken
— A.^.9 _^rn Total time = l^+2+lh
4{h
= 35.6 km/h (to 1 dp)
Average speed = 35.6 km/h, including the time for the picnic.
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