The average rate of change between two points is the gradient of the chord joining them. The rate of change at one point is the gradient of the tangent there.
Average rate: draw the chord and find its gradient.
Rate at a point: draw the tangent to the curve at that point.
Find the tangent's gradient using two points far apart on it.
Watch out At a single instant you need a tangent, not a chord.
See it worked
Worked exampleCalculator allowed·2 marks
The graph shows the volume of water, V litres, in a tank t seconds after a tap is opened. A tangent has been drawn to the curve at t=6. It passes through (2,7) and (10,31).
02468101210203040t (seconds)V (litres)
Use the tangent to estimate the rate at which the volume is increasing at t=6.
10−231−7
The instantaneous rate of change is estimated by the gradient of the tangent.
Use its two stated points: (31 − 7)/(10 − 2) = 24/8 = 3 litres per second.
Answer3 litres/second
Your turn
Question 1Calculator allowed·4 marks
The temperature, T∘C, of a cup of tea t minutes after it is made is given by T=80×0.9t+20.
(a) Work out the average rate at which the tea cools during the first 5 minutes, in ∘C per minute. Give your answer to 2 decimal places.
(b) Is the tea cooling faster at t = 0 or at t = 5? Give a reason.
Question 2Non-calculator·3 marks
Volume V litres is modelled by V = 4t2 + 3, where t is time in seconds. The tangent at t = 3.75 has gradient 30 litres per second. By what percentage is this instantaneous rate greater than the average rate from t = 1 to t = 5? You must show your working.
Question 3Non-calculator·5 marks
A model has V=3t2+7, with volume V in litres and time t in seconds. At t = 4 a tangent has gradient 24 litres per second.
(a) Find the average rate of increase from t = 1 to t = 5.
(b) By what percentage is the tangent rate greater than this average?
In past papers
29 official questions in 29 papers since 2017. The latest: