Printable · GCSE Higher · ages 14-16
Gradients and areas under graphs worksheet — GCSE Higher
Fifteen questions on "gradients and areas under graphs" — DfE statement A15. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Gradients and areas under graphs worksheet — GCSE Higher
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- 1.A car is decelerating. The tangent to its velocity-time graph at t = 12 seconds passes through the points (8, 22) and (16, 6), where velocity is in m/s and time is in seconds. Work out the gradient of this tangent, in m/s².
- 2.A cyclist's speed rises from rest to a maximum — quickly at first, then more and more slowly — so her speed-time graph is a curve that is concave down (its gradient decreases as time goes on). A trapezium estimate is made for the distance travelled during this phase, using two points on the curve. Is the trapezium estimate an overestimate or an underestimate of the true distance, and why?
- 3.A tram travels between two stops. Its velocity-time graph consists of straight line segments joining the points (0, 0), (5, 20), (12, 20), (16, 4) and (20, 4), where time is in seconds and velocity is in m/s. Work out the average speed of the tram over the whole 20 seconds. Give your answer to 1 decimal place.
- 4.The tangent to a curve at the point (6, 1) has gradient −3. Work out the y-coordinate of the point where this tangent crosses the y-axis.
- 5.A tap fills a tank at a varying rate. A graph shows the rate of flow, in litres per minute, against time, in minutes. What does the area under this graph represent?
- 6.A speed-time graph shows a constant speed of 15 m/s for 20 seconds. Work out the distance travelled, using the area under the graph.
- 7.A bath is filled with water; the graph of volume (litres) against time (minutes) is a curve, since the flow rate changes. The tangent to the curve at t = 10 minutes passes through the points (8, 130) and (12, 190), and the volume in the bath at t = 10 minutes is 160 litres. Use the gradient of this tangent to estimate the volume in the bath 5 minutes after t = 10 minutes.
- 8.A company's cost, in £, for producing x items is shown on a graph. The tangent to the curve at x = 50 passes through (30, 400) and (70, 800). Interpret the gradient of this tangent in the context of the company's costs.
- 9.The area under a speed-time graph between t = 0 and t = 6 seconds is estimated using three strips of equal width, using the speeds, in m/s, at t = 0, 2, 4 and 6: 0, 5, 9 and 12. Using the trapezium rule with these three trapezia, estimate the distance travelled.
- 10.The table shows the velocity, in m/s, of a train at various times, in seconds: 0 at t = 0, 20 at t = 4, 35 at t = 10, 27 at t = 14. Assuming the velocity changes at a constant rate between each pair of readings, estimate the distance the train travels between t = 0 and t = 12, using the trapezium rule.
- 11.A car's speed, in m/s, during a journey is: 0 at t = 0 s, 8 at t = 4 s, and 8 (constant) from t = 4 s to t = 10 s, increasing at a constant rate between t = 0 and t = 4. Estimate the total distance the car travels, using the areas of a triangle and a rectangle.
- 12.A tap's flow rate, in litres per minute, is plotted against time, in minutes, on a graph. What are the units of the area under this graph?
- 13.The curve y = x² − 3x. Use the chord between x = 1 and x = 4 to estimate the gradient of the curve at x = 2.5.y = x² − 3x
- 14.The table shows the speed, in m/s, of a cyclist at times, in seconds, 1 second apart: 0 at t = 0, 7 at t = 1, 12 at t = 2, 15 at t = 3, 16 at t = 4, 14 at t = 5, 9 at t = 6. Using the trapezium rule with strip width 1 second, estimate the distance the cyclist travels between t = 2 and t = 5 only.
- 15.The tangent to a curve at the point where x = 4 passes through the points (2, 5) and (6, 21). Use these two points to estimate the gradient of the curve at x = 4.
Answer key
- (a) −2 m/s² — The gradient of a line through two points is the change in the y-value divided by the change in the x-value: (6 − 22)/(16 − 8) = −16/8 = −2 m/s². The negative sign confirms the deceleration mentioned in the question. Writing 2 m/s² comes from subtracting the velocities in the wrong order, using (22 − 6) instead of (6 − 22), which loses the sign that shows the car is slowing down. Writing −0.5 m/s² comes from inverting the fraction, dividing the change in time by the change in velocity instead of the other way round. Writing −16 m/s² comes from finding the change in velocity, −16, but forgetting to divide by the change in time, 8 seconds.
- (a) Underestimate: the chord lies below the curve — For a curve that is concave down (bending downward), the straight line joining any two points on the curve lies BELOW the curve. The trapezium's top edge is this straight line, so the trapezium's area is smaller than the true area under the curve — the trapezium rule UNDERESTIMATES the distance in this case. This is because a curve whose gradient keeps decreasing bends away from any chord joining two of its points, dropping the chord below every point of the curve in between. The reverse, a chord lying above the curve, is the rule for a concave-up curve, not this one — check which way the curve bends before deciding. The trapezium rule does not always overestimate or underestimate, and it is not exact except when the graph really is a straight line.
- (b) 12.7 m/s — Find the total distance from the area under the graph, then divide by the total time. Break the graph into its four straight sections: a triangle from (0, 0) to (5, 20), where 5 × 20 = 100 gives an area of 50 m when halved; a rectangle from (5, 20) to (12, 20), area 7 × 20 = 140 m; a trapezium from (12, 20) to (16, 4), where (20 + 4) × 4 = 96 gives an area of 48 m when halved; and a rectangle from (16, 4) to (20, 4), area 4 × 4 = 16 m — a total distance of 50 + 140 + 48 + 16 = 254 m. Dividing by the 20 seconds gives an average speed of 254 ÷ 20 = 12.7 m/s. Writing 15.1 m/s comes from forgetting to halve the trapezium area for the third section, using 96 instead of 48: a total of 302 m, and 302 ÷ 20 = 15.1 m/s. Writing 11.9 m/s comes from leaving out the final section, from (16, 4) to (20, 4), entirely: a total of 238 m, and 238 ÷ 20 = 11.9 m/s. Writing 12.1 m/s comes from dividing the correct total distance by 21 instead of 20 — a fencepost slip, counting the whole seconds from t = 0 to t = 20 inclusive as 21 seconds of travel rather than reading the journey as a duration of 20 seconds: 254 ÷ 21 = 12.1 m/s (1 d.p.).
- (b) 19 — A line through (x₁, y₁) with gradient m has equation y − y₁ = m(x − x₁). Substituting (6, 1) and m = −3: y − 1 = −3(x − 6), so y = −3x + 18 + 1, which simplifies to y = −3x + 19; at x = 0 this gives 19. Making a sign error when distributing, writing −3(x − 6) as −3x − 18 instead of −3x + 18, gives the wrong line y = −3x − 17, so −17 at x = 0. Forgetting to add the y₁ = 1 at the end, using y = −3(x − 6) alone, gives 18 at x = 0. Substituting the y-coordinate into the gradient term instead of using x, working out 1 + (−3 × 1), gives −2.
- (b) The total volume of water, in litres, that has flowed in. — On a rate-time graph, the y-axis is in litres per minute and the x-axis is in minutes; multiplying a rate by a time gives litres per minute × minutes = litres, a total volume. So the area under the graph represents the total volume of water that has flowed in. Thinking the area itself represents the rate, rather than what the rate accumulates to, gives the wrong claim about the average rate of flow. Confusing the area with the gradient of the graph — which measures how the rate is changing — gives the wrong claim about litres per minute squared. Ignoring the flow-rate axis and focusing only on the time axis gives the wrong claim that the area is simply the total time.
- (d) 300 m — For a constant speed, the speed-time graph is a horizontal line, and the area underneath is a rectangle: distance = speed × time = 15 × 20 = 300 m. Adding the two numbers instead of multiplying gives 35 m; dividing instead of multiplying gives 1.33 m; halving the product, as you would for a triangle, gives 7.5 m — but this section of the graph is a rectangle, not a triangle, so there is no halving to do.
- (b) 235 litres — The flow rate at t = 10 is the gradient of the tangent: change in volume ÷ change in time = (190 − 130) ÷ (12 − 8) = 60 ÷ 4 = 15 litres per minute. Treating this rate as roughly constant for a short interval, the volume 5 minutes after t = 10 is estimated as 160 + 5 × 15 = 235 litres. Using the tangent's own point spacing — 2 minutes, from t = 10 to t = 12 — instead of the 5 minutes actually asked for gives 160 + 2 × 15 = 190, which is just the volume already given at one of the tangent's own points, not an answer to the question asked. Multiplying the gradient by the time WITHOUT adding the starting volume, 5 × 15 = 75, forgets that a rate estimates a CHANGE, which must be added to the starting volume, not given as the answer on its own. Subtracting instead of adding, 160 − 5 × 15 = 85, extrapolates backward in time rather than forward.
- (d) The cost increases by about £10 per extra item — Gradient = change in cost ÷ change in items = (800 − 400) ÷ (70 − 30) = 400 ÷ 40 = 10. The units of the gradient are £ per item, so the cost is increasing by about £10 for every extra item produced, near x = 50. Subtracting in the wrong order, (400 − 800) ÷ (70 − 30) = −10, gives the right size but the wrong sign — check which point comes first each time. Leaving out the division by 40 gives £400 per item; dividing the wrong way round, 40 ÷ 400 = 0.1, gives £0.10 per item.
- (d) 40 m — Each trapezium has width 2. Its area is width × the average of its two heights. Strip 1: average of 0 and 5 is 2.5, so area = 2 × 2.5 = 5. Strip 2: average of 5 and 9 is 7, so area = 2 × 7 = 14. Strip 3: average of 9 and 12 is 10.5, so area = 2 × 10.5 = 21. Total distance = 5 + 14 + 21 = 40 m. Leaving out the division by 2 in the averaging step doubles every strip, giving 80 m instead of 40 m. Using only the LEFT height of each strip as a rectangle, 0 × 2 + 5 × 2 + 9 × 2 = 28, or only the RIGHT height, 5 × 2 + 9 × 2 + 12 × 2 = 52, both ignore that the graph curves between the two ends of each strip — always average the two heights of a trapezium, never use just one of them.
- (a) 271 m — The reading at t = 12 isn't in the table, so it must be interpolated between t = 10 (35 m/s) and t = 14 (27 m/s): the velocity falls by 8 m/s over the 4-second gap, so over 2 seconds it falls by 4 m/s, giving 35 − 4 = 31 m/s. The trapezium rule then uses three strips of unequal width, 4, 6 and 2 seconds, with areas of 40, 165 and 66: 40 + 165 + 66 = 271 m. Writing 267 m comes from skipping the interpolation and using the table's t = 14 reading, 27, directly as the height at t = 12. Writing 282 m comes from treating all three strips as though they were 4 seconds wide, instead of using the true gaps of 4, 6 and 2 seconds between the readings. Writing 275 m comes from using the t = 10 reading, 35, at both ends of the final strip, instead of interpolating a new value at t = 12.
- (c) 64 m — The distance travelled is the area under the speed-time graph. From t = 0 to t = 4, the shape is a triangle with base 4 and height 8, area 1/2 × 4 × 8 = 16. From t = 4 to t = 10, the shape is a rectangle with base 6 and height 8, area 6 × 8 = 48. Total distance: 16 + 48 = 64 m. Treating the whole 10 seconds as a single trapezium with parallel sides 0 and 8 and width 10, instead of splitting it into the triangle and rectangle, gives 40 m. Ignoring the acceleration phase completely and assuming the car travels at a constant 8 m/s for all 10 seconds gives 8 × 10 = 80 m. Misreading the second interval as running from t = 4 to t = 8 instead of t = 4 to t = 10 gives a rectangle area of 4 × 8 = 32, plus the correct triangle of 16, totalling 48 m.
- (a) Litres — The area under a graph is found by multiplying a y-value by an x-value, so its units are the y-axis units multiplied by the x-axis units: litres per minute × minutes = litres, since the 'per minute' cancels with the 'minutes'. Answering 'litres per minute' keeps the y-axis units unchanged, as if multiplying by time did nothing to the units at all. Answering 'litres per minute squared' treats the x-axis as also being measured 'per minute', squaring a unit that should instead cancel. Answering 'minutes' keeps only the x-axis units and drops the rate altogether.
- (c) 2 — At x = 1, y = 1² − 3(1) = 1 − 3 = −2. At x = 4, y = 4² − 3(4) = 16 − 12 = 4. Gradient of the chord = change in y ÷ change in x = (4 − (−2)) ÷ (4 − 1) = 6 ÷ 3 = 2. Writing down the change in y, 6, and stopping there without dividing by the change in x gives 6. Losing the negative sign on y = −2 at x = 1 and treating it as +2 gives (4 − 2) ÷ (4 − 1) = 2 ÷ 3 = 2/3. Dividing the wrong way round, change in x ÷ change in y, gives (4 − 1) ÷ (4 − (−2)) = 3 ÷ 6 = 1/2.
- (d) 44 m — The trapezium rule between t = 2 and t = 5 uses only the speeds at t = 2, 3, 4 and 5 — 12, 15, 16 and 14 — with three strips of width 1: adding the first and last readings and twice the sum of the middle readings gives 12 + 14 + 2 × (15 + 16) = 88, and half of that is 44 m. Writing 68.5 m comes from applying the trapezium rule to the whole table, from t = 0 to t = 6, instead of restricting it to the interval t = 2 to t = 5 that the question asks for. Writing 28.5 m comes from forgetting to double the two middle readings, 15 and 16, in the trapezium rule formula. Writing 39 m comes from averaging only the first and last speeds in the interval, 12 and 14, and multiplying by the 3-second interval, ignoring the readings at t = 3 and t = 4 in between.
- (a) 4 — The gradient of a straight line through two points is the change in y divided by the change in x. Change in y = 21 − 5 = 16. Change in x = 6 − 2 = 4. Gradient = 16 ÷ 4 = 4. Dividing x by y instead of y by x gives 0.25; forgetting to divide by the change in x at all leaves 16; dividing by only one of the two x-coordinates, 16 ÷ 2 = 8, uses the wrong denominator.
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