Printable · GCSE Higher · ages 14-16
Instantaneous rate of change: gradients of curves worksheet — GCSE Higher
Fifteen questions on "instantaneous rate of change: gradients of curves" — DfE statement R15. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Instantaneous rate of change: gradients of curves worksheet — GCSE Higher
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- 1.The population of a town, P thousand, is plotted against time, t years, since 2020. A tangent to the graph at t = 6 has gradient 2.5. What does this gradient represent?
- 2.A tangent to the graph of population against time at the point where t = 4 passes through (2, 180) and (6, 260). Work out the instantaneous rate of change of the population at t = 4, in people per year.
- 3.A cyclist's distance travelled, in metres, is recorded against time, in seconds. From t = 3 to t = 8 the distance increases from 12 m to 32 m. A tangent to the distance–time graph at t = 6 has gradient 3. Work out the average speed of the cyclist over the interval from t = 3 to t = 8.
- 4.A curve has equation y = x² + 2x. Work out the average rate of change of y with respect to x over the interval from x = 1 to x = 4.y = x² + 2x
- 5.An ice-cream van's daily takings, in £, are modelled by a curve plotted against the average temperature that day, in °C. At a temperature of 22°C, the gradient of the tangent to this curve is 14. What does this gradient tell you about the takings at 22°C?
- 6.A tangent to a curve at the point where x = 3 passes through the points (1, 4) and (5, 20). Work out the gradient of the tangent at x = 3, and hence the instantaneous rate of change of y with respect to x at that point.
- 7.The volume of water in a paddling pool, in litres, is plotted against the time since the tap was turned on, in minutes. What are the units of the gradient of a tangent to this graph?
- 8.The cost, in pounds, of hiring a minibus is modelled against the number of passengers booked. A tangent to the cost graph at 20 passengers passes through the points (16, 184) and (24, 216). Work out the instantaneous rate at which the cost increases with each extra passenger, at 20 passengers.
- 9.A firework rocket's height above the ground, in metres, t seconds after launch, is modelled by h = 30t − 5t². Use a chord between t = 1 second and t = 3 seconds to estimate the instantaneous rate of change of the height at t = 2 seconds, in m/s.
- 10.A cyclist's speed, in metres per second, is plotted against time, in seconds, since she sets off. Which statement correctly pairs the gradient of a tangent to this graph, and the area between the graph and the time-axis, with what each one represents?
- 11.An online shop's total number of orders received is recorded at the end of each day for five consecutive days: Day 1, 1,284 orders; Day 2, 1,509 orders; Day 3, 1,830 orders; Day 4, 2,296 orders; Day 5, 2,510 orders. Work out between which two consecutive days the average rate of increase in orders was greatest.
- 12.A cup of tea is cooling. Its temperature, in °C, is plotted against time, in minutes, since it was poured. At t = 4 minutes, the gradient of the tangent to the graph is −3.2. What does this tell you about the tea at t = 4 minutes?
- 13.The height of water in a tank, h metres, t minutes after a tap is opened is modelled by h = 0.02t² + 0.5. Estimate the instantaneous rate of change of the height at t = 10, using the gradient of the chord joining t = 9 and t = 11.
- 14.A shop's daily profit, in pounds, is plotted against the price charged for an item, in pounds. A tangent to the profit graph at a price of £15 passes through the points (12, 540) and (18, 480). Work out the instantaneous rate of change of profit with respect to price at £15, and state whether profit is increasing or decreasing there.
- 15.A tangent to a curve at the point where x = 3 passes through the points (1, 2) and (5, 14). A candidate says: "The gradient of the tangent at x = 3 is 3, because 14 − 2 = 12 and 12 ÷ 4 = 3." Which statement is correct?
Answer key
- (d) The population is growing at 2500 people per year. — The gradient of a tangent to a graph gives the instantaneous rate of change of the quantity on the vertical axis with respect to the quantity on the horizontal axis, at that exact point — not the total change and not an average. Here the vertical axis is population in thousands and the horizontal axis is time in years, so the gradient is measured in thousands of people per year. A gradient of 2.5 means the population is growing at an instantaneous rate of 2.5 thousand people per year, and since P is measured in thousands, 2.5 × 1000 = 2500 people per year. This describes the rate of change at that instant, not the total increase over the 6 years and not an average population.
- (a) 20 people/year — The gradient of a tangent to a graph at a point equals the instantaneous rate of change of the quantity there. A straight line's gradient is the change in the vertical value divided by the change in the horizontal value between two points on it. Here the tangent passes through (2, 180) and (6, 260), so the change in population is 260 − 180 = 80 and the change in time is 6 − 2 = 4. The gradient is 80 ÷ 4 = 20. Reporting the change in population, 80, on its own is not a rate, because that growth happened over 4 years and has not been divided by them. Adding the two changes instead of dividing gives 80 + 4 = 84, which is not a rate. Subtracting the coordinates in the wrong order, (180 − 260) ÷ (6 − 2), gives −80 ÷ 4 = −20, the wrong sign. The instantaneous rate of change of the population at t = 4 is 20 people per year.
- (c) 4 m/s — The average rate of change of distance with respect to time over an interval is the change in distance divided by the change in time — the gradient of the chord joining the two endpoints, not the gradient of any tangent inside the interval. From t = 3 to t = 8 the change in time is 8 − 3 = 5 and the change in distance is 32 − 12 = 20, so the average speed is 20 ÷ 5 = 4 m/s. Reporting the change in distance on its own, as 20 m/s, is not a speed: those 20 metres were covered over the whole 5 seconds, not in one second, so the 20 still has to be divided by the 5. The tangent's gradient of 3 m/s is the instantaneous speed at the single moment t = 6, not the average over the whole 5-second interval, so it must not be used here. Adding the change in distance and the change in time instead of dividing gives 20 + 5 = 25, which is not a speed. The average speed of the cyclist over the interval is 4 m/s.
- (a) 7 — The average rate of change of y with respect to x over an interval is the change in y divided by the change in x between its two endpoints — the gradient of the chord joining them, not a rate at a single point. At x = 1: 2 × 1 = 2, so y = 1 + 2 = 3. At x = 4: 2 × 4 = 8, so y = 16 + 8 = 24. The change in y is 24 − 3 = 21 and the change in x is 4 − 1 = 3, so the average rate of change is 21 ÷ 3 = 7. Reporting the change in y, 21, on its own is not a rate of change, because it has not been divided by the 3 units of x over which it happened. Reporting 3 is not a rate either — 3 is the width of the interval, and also the value of y at x = 1, and neither of those measures how fast y is changing. Subtracting in the wrong order gives −21 ÷ 3 = −7, the wrong sign. The average rate of change of y with respect to x over the interval is 7.
- (b) Takings rise about £14 per 1°C rise — The gradient here is positive, so as temperature rises, takings rise too: near 22°C, takings increase by about £14 for every 1°C rise in temperature. Reversing this to say takings rise for every 1°C FALL gets the direction of the independent variable backwards — a positive gradient means both quantities move the same way. Saying 'takings are £14 at 22°C' confuses the gradient, a rate of change, with the y-value on the graph, which is the takings itself. Saying takings 'rose £14 in total' from 0°C to 22°C treats the gradient at a single point as if it applied over the whole range from 0°C to 22°C, when it only describes the instant at 22°C. Always keep a rate, a total change and a single reading separate.
- (c) 4 — The gradient of a tangent at a point equals the instantaneous rate of change of y with respect to x at that point. A straight line's gradient is found from the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 4) and (5, 20), so the change in y is 20 − 4 = 16 and the change in x is 5 − 1 = 4. The gradient is 16 ÷ 4 = 4. Stopping after finding the change in y, 16, without dividing it by the change in x, is not a gradient at all — that rise happened across 4 units of x, not 1. Adding the two changes instead of dividing one by the other gives 16 + 4 = 20, which is not how a gradient is found. Subtracting in the wrong order, (4 − 20) ÷ (5 − 1), gives −16 ÷ 4 = −4, the wrong sign. The instantaneous rate of change of y with respect to x at x = 3 is 4.
- (b) litres per minute — The gradient of a tangent is the change in the quantity on the vertical axis divided by the change in the quantity on the horizontal axis, so its units come from both axes: litres on the vertical axis and minutes on the horizontal axis give litres per minute. Giving the units as minutes for each litre inverts the fraction, giving the units of the RECIPROCAL of the gradient, not the gradient itself. Writing just litres uses only the vertical axis's units and ignores that a gradient is a rate, not an amount. Writing just minutes uses only the horizontal axis's units. A gradient always combines both axes' units as one divided by the other.
- (d) £4 — The gradient of the tangent gives the instantaneous rate of change of cost with respect to the number of passengers, in pounds per passenger. The tangent passes through (16, 184) and (24, 216), so the change in cost is 216 − 184 = 32 and the change in passengers is 24 − 16 = 8. The gradient is 32 ÷ 8 = 4. Stopping after finding the change in cost, without dividing by the change in passengers, leaves 32, not a rate. Adding the two changes instead of dividing gives 32 + 8 = 40, which is not a rate either. Reading off only the change in passengers, 8, is not a rate at all — a rate needs the change in cost as well. The instantaneous rate is £4 per extra passenger.
- (d) 10 — First find the height at each end of the chord. At t = 1, h = 30 × 1 − 5 × 1² = 30 − 5 = 25. At t = 3, h = 30 × 3 − 5 × 3² = 90 − 45 = 45. The gradient of the chord estimates the instantaneous rate at the midpoint t = 2: 45 − 25 = 20, then 20 ÷ (3 − 1) = 20 ÷ 2 = 10 m/s. Finding the change in height but forgetting to divide by the change in time gives 20, which is a distance, not a rate. Averaging the two heights instead of finding the difference gives (25 + 45) ÷ 2 = 70 ÷ 2 = 35. Subtracting in the wrong order, 25 − 45 = −20, then −20 ÷ 2 = −10, gives the correct size with the sign flipped — the rocket is rising, not falling, at t = 2 seconds, so a negative rate cannot be right here.
- (d) Gradient = acceleration; area = distance travelled. — Method: on a speed–time graph, the gradient of the graph at an instant is the rate of change of speed with time, which is acceleration; the area between the graph and the time-axis over an interval is the total distance covered in that interval, because it accumulates speed × time. Working: gradient = acceleration and area = distance travelled is the correct pairing. Swapping the two quantities completely, gradient = distance travelled and area = acceleration, is the reverse of what each actually measures. Keeping gradient = acceleration correct but then also claiming area = acceleration too is wrong because the area is a different physical quantity, distance, not a second way of finding the same rate. Claiming the gradient itself gives the speed confuses the RATE OF CHANGE of the plotted quantity with the plotted quantity itself — the gradient is how fast the speed is changing, not the speed. On any rate graph, the gradient of the graph is always the RATE at that instant, and the area under the graph is always the TOTAL AMOUNT accumulated — keep straight which of the two questions each one answers.
- (b) Day 3 to Day 4 — Method: the average rate of increase between two consecutive days is the difference in the number of orders divided by the number of days between them, which here is just the difference itself, since each gap is one day; comparing all four differences finds which is greatest. Working: the differences are 1,509 − 1,284 = 225 (Day 1 to Day 2), 1,830 − 1,509 = 321 (Day 2 to Day 3), 2,296 − 1,830 = 466 (Day 3 to Day 4), and 2,510 − 2,296 = 214 (Day 4 to Day 5); 466 is the greatest of the four, so the rate of increase was greatest from Day 3 to Day 4. Choosing Day 4 to Day 5 comes from picking the interval that ends on the highest total number of orders, 2,510, confusing the SIZE of the total with the RATE at which it grew. Choosing Day 1 to Day 2 comes from assuming the rate must be greatest at the very start, without working out any of the four differences. Choosing Day 2 to Day 3 comes from comparing only the first two differences, 225 and 321, and stopping there without checking Day 3 to Day 4 or Day 4 to Day 5. Finding the greatest rate of change from a table always means computing every difference between consecutive values and comparing them all — the day with the highest total, or the first pair you check, is not a shortcut.
- (c) Falling at 3.2°C per minute — The gradient of a tangent gives the instantaneous rate of change, in °C per minute here, not a temperature and not a total change. The negative sign means the temperature is falling, not rising, so the tea is cooling at a rate of 3.2°C per minute at the instant t = 4. Reading the sign the wrong way round gives 'rising at 3.2°C per minute', which would mean the tea is heating up. Treating −3.2 as a total drop since the tea was poured confuses a rate with an accumulated change, which would need the temperatures at two different times, not the gradient at one instant. Treating −3.2 as the temperature reading itself confuses the gradient, a rate of change, with the y-value on the graph. Always check whether a number is a rate, a total, or a single reading before you use it.
- (d) 0.40 m/min — To estimate an instantaneous rate of change at a point without a diagram, use the gradient of a chord joining two points close to it, one on each side. At t = 9: 0.02 × 81 = 1.62, so h = 1.62 + 0.5 = 2.12. At t = 11: 0.02 × 121 = 2.42, so h = 2.42 + 0.5 = 2.92. The change in height is 2.92 − 2.12 = 0.80 and the change in time is 11 − 9 = 2, so the gradient of the chord is 0.80 ÷ 2 = 0.40. Reporting the change in height, 0.80, on its own is not a rate, because it has not been divided by the 2 minutes over which it happened. Using the chord from t = 0 (where h = 0.5) to t = 11 instead gives 2.92 − 0.5 = 2.42, and 2.42 ÷ 11 = 0.22, which is the average gradient over the whole 11 minutes, not the instantaneous rate at t = 10. Substituting t = 10 into the formula gives 0.02 × 100 + 0.5 = 2.50, which is the height of the water at that moment, not the rate at which the height is rising. The estimated instantaneous rate of change at t = 10 is 0.40 m/min.
- (d) Profit is decreasing by £10 per £1 rise in price. — The gradient of a tangent gives the instantaneous rate of change of profit with respect to price, found from the change in profit divided by the change in price between two points on the tangent. Here the tangent passes through (12, 540) and (18, 480), so the change in profit is 480 − 540 = −60 and the change in price is 18 − 12 = 6. The gradient is −60 ÷ 6 = −10. A negative gradient means profit is decreasing as price increases, so profit is decreasing at an instantaneous rate of £10 for every £1 rise in price. Subtracting the profits in the wrong order, 540 − 480 = 60, and dividing by the same change in price, 60 ÷ 6 = 10, gives a positive value and the wrong direction — profit is not increasing at £15. Stopping after finding only the change in profit, 480 − 540 = −60, without dividing by the change in price, is not a rate at all. Reading off the change in price, 6, and calling it the rate gives £6 per £1 rise in price, but 6 is only the width of the price interval — it is not a change in profit at all, and profit falls across that interval, so the direction is wrong too. The instantaneous rate of change of profit with respect to price at £15 is a decrease of £10 per £1 rise in price.
- (a) The gradient is 3; the candidate's method is right. — The gradient of a tangent, like any straight line, is the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 2) and (5, 14), so the change in y is 14 − 2 = 12 and the change in x is 5 − 1 = 4. The gradient is 12 ÷ 4 = 3, so the candidate's calculation is correct. Subtracting in the wrong order, (2 − 14) ÷ (5 − 1), gives −12 ÷ 4 = −3, the wrong sign. Adding the two changes instead of dividing them, 12 + 4 = 16, does not find a gradient at all. Dividing the change in x by the change in y instead of the other way round, 4 ÷ 12 = 1/3, inverts the calculation completely. Before accepting or rejecting a claimed gradient, always redo the calculation yourself in the same order — change in y over change in x — rather than trusting the arithmetic as given.
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