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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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222 GRADIENT
Road G rad ients The gradient of a road is defined in the same way as a mathematical gradient. The fraction for the gradient of a road is defined as rise/ran. The gradients of roads may be expressed as percentages. For example, if the gradient of a road is |, the gradient may be expressed as 50%.
Example. Find the average gradient of the viaduct through Otira Gorge from the following data (see figure d). The roadway is 452 meters long. At the start of the viaduct, point A it is 744 m above sea level, and at the end, point B is 797 m above sea level.
B
53 m C
(d)
Solution. The height of point A above sea level is 744 m and the height of B is 797 m, so the length of BC is 797 — 744 = 53 m. To obtain the gradient of AB the length of AC is required. Write
AB2 = BC2 + AC2 4522 = 532 + AC2 AC2 = 204,304 - 2809 AC = -y/201,495 AC = 449 m
Now
Gradient of the road = Using the formula for the gradient
AC
53 “ 449
The gradient of the Otira viaduct is 53/449, or 53
x 100 = 11.8% (to 1 dp)
449 1
Coordinate Geometry Suppose two points A and B have coordinates A(xi, yi) and B(x2, y2)- Then the gradient of the line through the points A and B is given by the formula
xi -x2
Pythagoras’ theorem in triangle ABC Substituting AB = 452 and BC = 53 Rearranging equation, and squaring 452 and 53 Subtracting, and then taking the square root To nearest whole number
GRADIENT-INTERCEPT FORM 223
Example. Find the gradient of the line through the two points A (2, — 1) and £(-5,3).
Solution. A diagram is not necessary, because there is no need to decide whether or not the gradient is positive or negative, since the formula will take care of it. By comparing A and B with the formula we can see that X\ = 2, yx = — 1 and x2 = —5, y2 =3. Write
yi — y2
m =---------------- The formula applies equally well with the order reversed,
X2 -Xi
-1-3
m = ——— Substituting values into the formula
m = — ^ Simplifying
The gradient of the line through AB is m = —
References: Average Speed, Canceling, Gradient-Intercept Form, Graphs, Parallel, Perpendicular Lines, Trigonometry.
GRADIENT-INTERCEPT FORM
This entry refers to a method of drawing straight-line graphs without using a table of values. The equations of straight-line graphs can be written in the form y = mx + c, and this is called the gradient-intercept form of the straight-line graph. The graph of a straight line is often called a linear graph.
In the equation y = mx + c, the gradient of the straight-line graph is m and its y intercept is c. When we draw the graph of a straight line we first need to make sure that its equation is written in the form y = mx + c in order to identify the values of m and of c. The graph can be drawn if the values of m and c are known. The whole process is explained in the following examples.
Example 1. Draw the graph of the straight line whose equation is y = 2x — 3.
Solution. Compare the equation y = 2x — 3 with the formula y = mx + c. We can see that the y intercept is c = — 3, and m = 2 (see figure a). The gradient should be expressed as a fraction m = 2/1 so that the gradient can be drawn. Always start with fixing the point c = — 3 on the y-axis, which tells us that the graph crosses this axis at the point y = — 3. From this point on the y-axis we draw a straight line with a gradient of m = 2/1. Then we continue the straight line with a series of gradients m = 2/1 until we have an accurate graph of the straight line y = 2x — 3. The line should be extended at each end to show that it continues.
224
GRADIENT-INTERCEPT FORM
(a)
Example 2. Draw the graph of the straight line x + 2y = 4.
Solution. The equation of the straight line must first be rearranged until it is in the form y = mx + c. Write
lx + 2y = 4 Writing x as la:
2y = — la: + 4 Subtracting la: from both sides of the equation
y = — \x + 2 Dividing both sides of equation by 2
m = — | and c = 2 Comparing with the formula y = mx + c.
Start by fixing the point y = 2 on the y-axis, and from this point draw a straight line with a gradient of m = — | (see figure b). A line with a negative gradient slopes to the left, \. Then continue the straight line with another gradient of m = — | until you have an accurate graph of the straight line x + 2y = 4.
(b)
GRADIENT OF A CURVE 225
Special case. A straight line which passes through the origin with a gradient of m is of the form y = mx, because c = 0.
References: Changing the Subject of a Formula, Gradient, Graphs, Intercepts.
GRADIENT OF A CURVE
The gradient of a curve changes along its length, so a curve does not have one gradient, but a different gradient at each point on the curve. To find the gradient at a point on a curve, we draw a tangent to the curve at that point and find the gradient of the tangent, which is defined to be the gradient of the curve at that point. This process is explained in the following example.
Example. Harry is growing a marrow and hoping to enter it in due course in the garden show. To monitor its growth he decided to weigh it every week, and the first weighing is 50 grams. The marrow’s progress is displayed in the following table of values. After 6 weeks the marrow weighed in at 302 grams. Find the rate at which the marrow is growing at the end of week 4.
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