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MATHEMATHICS BLOG 4KN 2013

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Gradient

Gradient of a Straight Line

The gradient of a straight line is the rate at which the line rises (or falls) vertically for every unit across to the right. That is:

Gradient = Rise / Run = Change in y / Change in x = (y2 - y1) / (x2 - x1)		A linear graph on the Cartesian plane shows the line PQ defined by the point P(x1, y1) and Q(x2, y2). The rise and run between the two points is shown.
Note:
The gradient of a straight line is denoted by m where:
m = (y2 - y1) / (x2 - x1)


Example 1

Find the gradient of the straight line joining the points P(– 4, 5) and Q(4, 17).

Solution:
Let (x1, y1) = (-4, 5) and (x2, y2) = (4, 17).m = (y2 - y1) / (x2 - x1) = (17 - 5) / (4 - (-4)) = 12 / 8 = 1.5
So, the gradient of the line PQ is 1.5.
The points P(-4, 5) and Q(4, 17) form a straight line on the Cartesian plane.
Note: If the gradient of a line is positive, then the line slopes upward as the value of x increases.


Example 2

Find the gradient of the straight line joining the points A(6, 0) and B(0, 3).

Solution:
Let (x1, y1) = (6, 0) and (x2, y2) = (0, 3).
m = (y2 - y1) / (x2 - x1) = (3 - 0) / (0 - 6) = 3 / -6 = - 1/2So, the gradient of the line AB is - 1/2.

The points A(6,0) and B(0,3) form a straight line on the Cartesian plane.

Note: If the gradient of a line is negative, then the line slopes downward as the value of x increases.


Applications of Gradients

Gradients are an important part of life. The roof of a house is built with a gradient to enable rain water to run down the roof. An aeroplane ascends at a particular gradient after take off, flies at a different gradient and descends at another gradient to safely land. Tennis courts, roads, football and cricket grounds are made with a gradient to assist drainage.

Example 3

A horse gallops for 20 minutes and covers a distance of 15 km, as shown in the diagram.
Find the gradient of the line and describe its meaning.
The graph of distance, d, in km against time, t, in minutes depicts the horse's journey from the point (0,0) to the point (20,15).
Solution:
Let (t1, d1) = (0,0) and (t2, d2) = (20,15).
Now, m = (d2 - d1) / (t2 - t1) = (15 - 0) / (20 - 0) = 15 / 20 = 3 / 4
So, the gradient of the line is 3 / 4 km/min.In the above example, we notice that the gradient of the distance-time graph gives the speed (in kilometres per minute); and the distance covered by the horse can be represented by the equation:
d = 3/4t {Because distance = speed * time}


Example 4

The cost of transporting documents by courier is given by the line segment drawn in the diagram. Find the gradient of the line segment; and describe its meaning.
The graph of cost($), c, against distance, d, in km shows the points A(0,5) and B(6,23) forming a straight line.
Solution:
Let (d1,c1) = (0,5) and (d2,c2) = (6,23).
m = (c2 - c1) / (d2 - d1) = (23 - 5) / (6 - 0) = 18 / 6 = 3                 So, the gradient of the line is 3. This means that the cost of transporting documents is $3 per km plus a fixed charge of $5, i.e. it costs $5 for the courier to arrive and $3 for every kilometre travelled to deliver the documents.

An alternative definition of the gradient is: The gradient is the rate of change of one variable with respect to another.
-Melodie-
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