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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
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RATE 377
Solution. The rate at which the bath is filling is the gradient of the first straight line:
^ rise
run
90 liters 3 minutes
= 30 liters/minute
The bath is filling at a rate of 30 liters per minute.
The rate at which the bath is emptying is the gradient of the third straight line, which is negative. The negative sign indicates that the bath is emptying rather than filling:
rise
run
90 liters 2 minutes
= —45 liters/minute
The bath is emptying at a rate of 45 liters per minute, (the negative sign is not written in the final answer, because it is stated that the bath is emptying).
For straight-line graphs the rates are constant throughout the length of the straight line. Rates can also be calculated for curved graphs. For curved graphs the rate is changing and has a different value at each point on the graph. Therefore, the rate can only be found at an instant. For this reason rates for curved graphs are called instantaneous rates. An example of instantaneous rates is included in the entries Gradient of a Curve and Rate of Change.
References: Average, Gradient of a Curve, Proportion, Ratio.
378 RATE OF CHANGE
RATE OF CHANGE
Suppose you are painting a wall of your house and observe how the area painted and the time spent on the job are related to each other. At different times of the day you are probably painting at a faster rate than at other times. A rate of change is the rate at which one quantity changes (say the area of the wall you have painted) with respect to the change in another quantity (say the time spent on painting that area of the wall). If a graph is drawn of the two quantities with area on the vertical axis and time on the horizontal axis, the rate of change of area with respect of time is the gradient of the graph. If the graph is a straight line, the rate of change is the gradient of the straight line, which is constant throughout its length. For a curved graph the gradient is changing throughout its length, so the gradient can only be calculated at a specific point on the graph and is called an instantaneous rate of change at that point.
Suppose a tap is turned on so that the water flows at a steady rate, and the water is caught in a container that is in the shape of a cylinder (see figure a). Let the depth of water in the container be d centimeters when the water has been flowing for t seconds. A graph is then drawn with time t seconds on the horizontal axis and depth d centimeters on the vertical axis. The resulting graph for this kind of container is a sloping straight line as shown in figure a.
(a)
The rate of change of the depth of water in the cylinder with respect to time is the gradient of the graph. Write
Rate of change
50
10
rise
run
The rate of change of depth with respect to time is 5 centimeters/second.
For containers of different shapes the graphs are different, and a graph has a negative gradient if the container is emptying. The containers and their graphs are shown side by side in figure b. In each case the water is flowing into the containers at a constant rate and the graphs show how the depth of water in the containers changes with time. When the flow of water stops the graph will be horizontal, which has a zero gradient.
RATE OF CHANGE 379
The rate increases more rapidly as the
The rate increases rapidly at first and then container gets narrower, more slowly as the container gets wider
The rate increasesly quite rapidly at first, then slows as the container widens, then increases again. Finally the rate becomes steady and the graph straight at the neck of the container.
(b)
If water is leaking fom the container at a steady rate the gradient of the graph will be negative.
We now look at the method of finding the rate, which is the gradient, when a graph is curved. Also see the entry Acceleration.
Example. Farmer John has a pig that he is fattening up for Christmas. He weighs the pig every week and enters the weight in a table:
Week 0 1 2 3 4 5
Weight in kg 30 32 38 48 62 80
Using the data in the table, draw a graph with the number of weeks (N) on the horizontal axis and weight (W) on the vertical axis (see figure c). From the graph estimate the rate of increase in the weight of the pig at 3 weeks.
(c)
380 RATIO
Solution. The graph is curved, so the rate is changing with time. To find the rate at 3 weeks we need to draw a tangent to the curve at the point on the curve when t = 3 weeks. A right-angled triangle is completed with the tangent as its hypotenuse. Try to arrange the size of the triangle so that its horizontal side is a whole number. This will make the next process easier. The sides of the triangle are 4 weeks and 48 kg. Write
rise