Math-MateSchool maths, made clearer

Search the complete textbook

Find a Math-Mate lesson

Type at least two letters to search 240+ lessons.

Topic 57 · overview

Topic overview

Clear explanations and worked examples from the Math-Mate school maths textbook.

Free lessonNo sign-upPrint friendly

Rates of change are very important in mathematics.  Take for example the ‘speed’ of a car.  It is a measure of how far the car travels over a certain time, usually expressed in km/hr.  It literally means how many kilometres the car will travel per hour.  It could also be expressed in km/s or m/s or mm/day.  The important thing is that it is a measure of how much the distance changes for a certain change (increase) in time.

For instance, 100 km/hr says that for every change in time of one hour, the distance travelled changes by 100 km.

If you graph the distance travelled by a car on the y-axis, and the time taken on the x-axis, for some journey you might get a graph like this:

Mathematical diagram

Now, for this graph (and any graph of distance vs. time), the gradient at any point represents the speed of the car at that time.  For instance, between hours 0 and 4, the car is travelling at 25 km/hr.  Between hours 4 and 6, the gradient is 0, telling us the car is not moving.  Then, between hours 6 and 8 the gradient is 50 km/hr.  Finally, between hours 8 and 10 the gradient is again 25 km/hr.  In general, the slope of a graph can tell us the rate of change of the y-axis variable, with respect to the x-axis variable.

What about for graphs with curves?

You can still find the rate of change the same way, but you have to draw a tangent to the point where you want to find the gradient, like follows:

Mathematical diagram

This graph shows a petrol tank being filled up at a gas station.  Now we know that the gradient represents the rate of change of the amount of fuel with respect to the change in time.  A steeper gradient means the change in fuel per second is greater.  Since the line gets steeper and steeper the tank is being filled at a higher and higher rate as time progresses.

Say we want to find the filling rate after 6 seconds.  We need the slope at that point.  We construct our own straight line parallel to the curved line at that point – called a tangent.  We then find the slope of our straight line and use that as our filling rate.  For this example, at time = 6 seconds, the filling rate is about 12 L/s.