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R15

Interpret the gradient at a point on a curve as the instantaneous rate of change; apply average and instantaneous rates of change (gradients of chords and tangents) (not calculus) [Higher only]

Higher only

Instantaneous rate of change

Worked answers, methods and verified real exam appearances for R15 on Edexcel GCSE Maths 1MA1.

Explanation

  • For a curve, the gradient changes. The gradient of the chord joining two curve points gives the average rate of change over that interval.
  • The gradient of a tangent at one point estimates the instantaneous rate of change there; GCSE questions use a drawn tangent, not calculus. Choose two well-separated, readable points on the chord or tangent and calculate ΔyΔx\dfrac{\Delta y}{\Delta x}.
  • The chosen tangent points need not lie on the original curve.
  • Include compound units and compare rates using their values.
  • Examiners allow a sensible range when answers depend on drawing and reading a tangent.
A chord estimates average rate over an interval; a tangent estimates instantaneous rate at one point.

Worked example

A tangent to a curve at x=7x=7 passes through (4,11)(4,11) and (10,32)(10,32). Estimate the instantaneous rate of change.

  1. 1.Use two clear points on the tangent.
  2. 2.Change in y=32−11=21y=32-11=21 and change in x=10−4=6x=10-4=6.
  3. 3.Tangent gradient =216=3.5=\dfrac{21}{6}=3.5.

Answer: 3.53.5 units of yy per unit of xx.

Common mistakes

  • Don't use two points on the curve instead of two points on the drawn tangent.
  • Don't call a chord gradient the instantaneous rate of change.
  • Don't read points too close together, magnifying graph-reading error.

Exam tip

Draw a large tangent triangle and show ΔyΔx\dfrac{\Delta y}{\Delta x}; a sensible estimate range is normally accepted.

Worked practice

Q1
Tier 1 · Easy

1

A curve passes through (2,5)(2,5) and (8,23)(8,23). Calculate the average rate of change of yy with respect to xx between these points.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkMark scheme
1
  • 33 units of yy per unit of xx
2The chord gradient is (23−5)/(8−2)=18/6=3(23-5)/(8-2)=18/6=3.
Q2
Tier 2 · Standard

2

A tangent to a curve at x=7x=7 passes through the grid points (4,11)(4,11) and (10,32)(10,32). Estimate the instantaneous rate of change at x=7x=7.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
QuestionAnswerMarkMark scheme
2
  • 3.53.5 units of yy per unit of xx
3Use two points on the tangent: (32−11)/(10−4)=21/6=3.5(32-11)/(10-4)=21/6=3.5. This tangent gradient estimates the instantaneous rate at x=7x=7.
Q3
Tier 3 · Hard

3

A curve shows water volume VV litres after tt minutes and passes through (2,46)(2,46) and (8,118)(8,118). The tangent at t=5t=5 passes through (4,70)(4,70) and (7,112)(7,112). Find the average rate from t=2t=2 to t=8t=8, estimate the instantaneous rate at t=5t=5, and compare them.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
QuestionAnswerMarkMark scheme
3
  • Average rate =12=12 litres per minute
  • Instantaneous rate =14=14 litres per minute
  • The instantaneous rate is 22 litres per minute greater.
5The chord gradient is (118−46)/(8−2)=72/6=12(118-46)/(8-2)=72/6=12 litres per minute. The tangent gradient is (112−70)/(7−4)=42/3=14(112-70)/(7-4)=42/3=14 litres per minute. Therefore the instantaneous rate is 14−12=214-12=2 litres per minute greater.
Q4
Tier 1 · Easy

4

Write down whether a chord or a tangent is used to estimate the instantaneous rate of change at a point on a curve.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
QuestionAnswerMarkMark scheme
4
  • A tangent
1The gradient of a tangent at the point estimates the instantaneous rate of change.
Q5
Tier 2 · Standard

5

A tangent to a temperature-time curve passes through (1,46)(1,46) and (7,19)(7,19), where time is in minutes and temperature is in degrees Celsius. Estimate the instantaneous rate of change of temperature.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
QuestionAnswerMarkMark scheme
5
  • −4.5 ∘C-4.5\,{}^\circ\text{C} per minute
3The tangent gradient is 19−467−1=−276=−4.5\dfrac{19-46}{7-1}=\dfrac{-27}{6}=-4.5. The instantaneous rate of change is −4.5 ∘C-4.5\,{}^\circ\text{C} per minute.
Q6
Tier 3 · Hard

6

A graph shows distance dd metres travelled after time tt seconds. The tangent at t=4t=4 passes through (2,15)(2,15) and (8,51)(8,51), while the tangent at t=10t=10 passes through (8,55)(8,55) and (14,73)(14,73). Compare the two instantaneous speeds.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
QuestionAnswerMarkMark scheme
6
  • Speed at t=4t=4 is 6 m/s6\,\text{m}/\text{s}
  • Speed at t=10t=10 is 3 m/s3\,\text{m}/\text{s}
  • The speed at t=10t=10 is half the speed at t=4t=4.
4At t=4t=4, the tangent gradient is 51−158−2=366=6 m/s\dfrac{51-15}{8-2}=\dfrac{36}{6}=6\,\text{m}/\text{s}. At t=10t=10, it is 73−5514−8=186=3 m/s\dfrac{73-55}{14-8}=\dfrac{18}{6}=3\,\text{m}/\text{s}. Therefore the later instantaneous speed is half the earlier speed.
Q7
Tier 2 · Standard

7

A curve shows the fuel remaining in a vehicle, in litres, after travelling dd kilometres. The tangent at d=170d=170 passes through (80,52)(80,52) and (260,43)(260,43). Estimate the instantaneous rate at which fuel is being used, in litres per 100 km100\,\text{km}.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
QuestionAnswerMarkMark scheme
7
  • 55 litres per 100 km100\,\text{km}
3The tangent gradient is 43−52260−80=−9180=−0.05\dfrac{43-52}{260-80}=\dfrac{-9}{180}=-0.05 litre per kilometre. The negative sign shows fuel is being used. Its rate of use is 0.05×100=50.05\times100=5 litres per 100 km100\,\text{km}.
Q8
Tier 3 · Hard

8

A distance-time curve has a tangent at t=45t=45 seconds passing through (20,140)(20,140) and (70,390)(70,390), where distance is in metres. Estimate the instantaneous speed at t=45t=45. If this speed is maintained, work out the time needed to travel a further 1.2 km1.2\,\text{km}.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
QuestionAnswerMarkMark scheme
8
  • 5 m/s5\,\text{m}/\text{s}
  • 240240 seconds (or 44 minutes)
4The tangent gradient is 390−14070−20=25050=5 m/s\dfrac{390-140}{70-20}=\dfrac{250}{50}=5\,\text{m}/\text{s}. A further 1.2 km1.2\,\text{km} is 1200 m1200\,\text{m}, so the time is 1200÷5=2401200\div5=240 seconds, which is 44 minutes.
Q9
Tier 3 · Hard

9

At a chosen point on a distance-time curve, the drawn tangent goes through (8,1.9)(8,1.9) and (23,2.5)(23,2.5), where time is in minutes and distance is in kilometres. Estimate the instantaneous speed there, in kilometres per hour.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • 2.4 km/h2.4\,\text{km}/\text{h}
4The tangent gradient is 2.5−1.923−8=0.615=0.04\dfrac{2.5-1.9}{23-8}=\dfrac{0.6}{15}=0.04 kilometres per minute. Multiplying by 6060 gives an instantaneous speed of 2.4 km/h2.4\,\text{km}/\text{h}.
Q10
Tier 3 · Hard

10

A graph shows the volume of liquid in a vertical tank against time. A tangent to the curve passes through (12,920)(12,920) and (27,560)(27,560), where time is in minutes and volume is in litres. The tank has constant horizontal cross-sectional area 0.8 m20.8\,\text{m}^2. Estimate the instantaneous rate at which the liquid level is falling, in centimetres per minute.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
QuestionAnswerMarkMark scheme
10
  • 3 cm3\,\text{cm} per minute
5The tangent gradient is 560−92027−12=−24\dfrac{560-920}{27-12}=-24 litres per minute, so volume is falling at 2424 litres per minute. This is 0.024 m30.024\,\text{m}^3 per minute. Using change in height=change in volumearea\text{change in height}=\dfrac{\text{change in volume}}{\text{area}} gives 0.024÷0.8=0.03 m0.024\div0.8=0.03\,\text{m} per minute, which is 3 cm3\,\text{cm} per minute.

Verified exam appearances

SeriesPaperQuestionMarksCalculatorTierLinks
2021-113HQ153AllowedHigherQPMS
2022-061HQ144Non-calculatorHigherQPMS
2022-112HQ214AllowedHigherQPMS
2019-113HQ193AllowedHigherQPMS
2024-062HQ146AllowedHigherQPMS
2019-062HQ147AllowedHigherQPMS

Other points in R Ratio, proportion and rates of change

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