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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Answer key: Instantaneous rate of change: gradients of curves worksheet — GCSE Higher
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- (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.
- (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.
- (b) No — rate fell by 3.7 thousand/month — Each tangent gradient is the instantaneous growth rate, in thousand subscribers per month. To compare them, subtract the later rate from the earlier one: 4.8 − 1.1 = 3.7. Since 1.1 is less than 4.8, the growth rate has fallen by 3.7 thousand subscribers per month, so the manager is wrong — the app is growing more slowly at 18 months, not faster. Subtracting the other way round and calling the result a rise, 'rate rose by 3.7 thousand/month', gets the direction backwards: the later gradient is the smaller of the two. Adding the two gradients, 4.8 + 1.1 = 5.9, and calling this a combined rate that shows speeding up, is the wrong operation for comparing two rates. Treating the difference 3.7 as a total number of subscribers lost, rather than a rate in thousands per month, confuses a rate with a count. Always subtract the two rates in a sensible order and keep the units in thousands per month.
- (c) 3 — Method: for a tangent written in the form y = mx + c, the coefficient m is the gradient of the line, and the gradient of the tangent at its point of contact equals the curve's instantaneous rate of change there. Working: y = 3x − 2 has gradient 3, so the instantaneous rate of change of y with respect to x at x = 4 is 3. Reading the constant term as the rate instead of the coefficient of x gives −2, but −2 is only where the tangent crosses the y-axis, not a rate. Reading the x-coordinate of the point of contact as the rate gives 4, but 4 only tells you where on the curve the tangent touches, not how fast y is changing there. Substituting x = 4 into the tangent equation, 3 × 4 − 2 = 10, gives the y-coordinate of the point of contact, not the rate; a candidate who works out the height of the point instead of the gradient gives 10. Whenever a tangent is given as an equation, the rate of change is always the coefficient of x — do not let the constant term, the x-value or a substituted y-value stand in for it.
- (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.
- (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.
- (d) £3.60 per component — The gradient of a cost-against-components graph has units of pounds per component, since cost is measured in pounds and the horizontal axis counts components. So 3.60 means it costs an extra £3.60 to produce one more component at that point. Calling it '£3.60 total cost' confuses the gradient, a rate, with the y-value on the graph, which is the total cost itself. Giving it as 3.60 components per pound swaps which axis is on top, giving the units of the reciprocal gradient, not the gradient itself. Turning 3.60 into a percentage invents a unit that has no basis in the graph's axes — a gradient here is a number of pounds, not a percentage. Always build the gradient's units from the two axes' own units, in the order y-axis over x-axis.
- (c) 1.2 cm/s — To estimate an instantaneous rate of change at a point from a table of readings, use the chord that spans the point symmetrically — equal steps either side — because the over-estimate on one side and the under-estimate on the other largely cancel. Here that is the chord from t = 10 to t = 30. The change in depth is 56 − 32 = 24 and the change in time is 30 − 10 = 20, so the estimate is 24 ÷ 20 = 1.2 cm/s. The one-sided chord from t = 20 to t = 30 gives (56 − 45) ÷ (30 − 20) = 11 ÷ 10 = 1.1 cm/s, which estimates the rate somewhere between t = 20 and t = 30 rather than at t = 20 itself. Dividing the 20-second change in depth by the 10-second gap between consecutive readings gives 24 ÷ 10 = 2.4, mixing the change from one interval with the time from another. Reporting the change in depth, 24, on its own is not a rate at all, because it has not been divided by a time. The best estimate of the instantaneous rate of change of depth at t = 20 is 1.2 cm/s.
- (c) Height rising at 2 m/s at t = 1.5 s — A tangent's gradient on a height-time graph is the instantaneous rate of change of height, in metres per second, so gradient 2 means the ball's height is increasing at 2 m/s at t = 1.5 s. Saying the height 'is 2 m' confuses the gradient, a rate, with the y-value on the graph, which is the ball's height itself. Saying the ball 'travelled 2 m from t = 1 to t = 2' treats the instantaneous gradient at one instant as if it were the total distance risen over a whole one-second interval, which is a different quantity found from two height readings, not from one tangent. Saying the speed 'is 2 m/s²' uses the wrong units — m/s² measures acceleration, the rate of change of speed, not speed itself. Always check that the units quoted match what a height-time graph's gradient can actually give you: metres per second.
- (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.
- (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) Faster at t = 8s — still accelerating — The gradient of a tangent on a distance-time graph is the instantaneous speed, in m/s. At t = 2 seconds the speed is 6 m/s; at t = 8 seconds it is 9.5 m/s, which is faster, so the sprinter is still accelerating between these two times. Saying the sprinter is slower at t = 8s reverses the comparison — 9.5 is greater than 6, not less. Writing 9.5 − 6 = 3.5 and calling this 'metres further covered' turns the difference of two speeds into a distance, which the units do not support: a difference of two speeds is itself a speed, not a distance. Taking 9.5 m/s, the larger of the two instantaneous speeds, as the average speed for the whole race confuses a speed at one instant with an average over the whole distance, which would need the total distance and total time, not two tangent gradients.
- (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) 12 — First find the y-values at the two ends of the chord. At x = 2, y = 2 × 2² + 5 = 2 × 4 + 5 = 8 + 5 = 13. At x = 4, y = 2 × 4² + 5 = 2 × 16 + 5 = 32 + 5 = 37. The gradient of the chord is the change in y divided by the change in x: 37 − 13 = 24, then 24 ÷ (4 − 2) = 24 ÷ 2 = 12. Stopping after finding the change in y and not dividing by the change in x gives 24, which is not a gradient at all. Inverting the fraction, change in x divided by change in y, gives 2 ÷ 24 ≈ 0.08. Averaging the two y-values instead of finding the change between them gives (13 + 37) ÷ 2 = 50 ÷ 2 = 25. A gradient is always change in y over change in x — never the other way round, and never a single y-value.
- (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.
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