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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Answer key: Gradients and areas under graphs worksheet — GCSE Higher
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- (c) The object was stationary (not moving) — On a distance-time graph, the gradient at any point represents the speed at that point. A gradient of 0 means distance is not changing over time, so the object is stationary. A straight, sloped line (not flat) shows constant nonzero speed; a curve bending one way shows acceleration or deceleration; a flat section is not a maximum speed — it is no speed at all. Read the shape of the graph, not just how steep it looks.
- (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) 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.
- (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.
- (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.
- (b) 1.6 m/s² — Acceleration is the change in velocity divided by the time taken: (11 − 3) ÷ 5 = 8 ÷ 5 = 1.6 m/s². Forgetting to subtract the initial velocity and dividing the final velocity by the time instead gives 11 ÷ 5 = 2.2 m/s². Inverting the fraction, dividing the time by the change in velocity, gives 5 ÷ 8 = 0.625 m/s². Finding the change in velocity, 8 m/s, but stopping without dividing by the time gives 8 m/s².
- (a) Overestimates — trapezium edges lie above the curve. — For a concave-up curve, each chord connecting two points on the curve lies ABOVE the curve between those points, because the curve bends away from the chord underneath it. The trapezium rule uses these chords as the top edges of each trapezium, so the trapeziums cover a larger region than the true area under the curve — the estimate is an overestimate. Saying it underestimates because the edges lie below the curve gets the direction backwards: for a concave-up curve the edges lie above, not below. Saying it overestimates 'because more strips always increase it' gives the right verdict for the wrong reason — using MORE, narrower strips makes a trapezium estimate more accurate and brings it closer to the true area, it does not simply increase the total. Saying the estimate is exact ignores that straight edges cannot follow a curved line perfectly; there is always a gap between the chord and the curve on a bend, which is exactly what makes the rule an estimate rather than an exact calculation.
- (c) Fastest at t = 2 min — steepest gradient. — The rate of cooling is given by the size (magnitude) of the gradient, ignoring its sign — the steeper the tangent, the faster the temperature is changing. Of −8, −3 and −0.5, the gradient −8 has the greatest magnitude, so the tea is cooling fastest at t = 2 minutes. 'Fastest at t = 20 min — largest gradient' confuses the signed value with the size of the rate: −0.5 is the largest NUMBER of the three, but it's the smallest in magnitude, meaning the tea is barely cooling at all by then. 'Cools at the same rate throughout' ignores that the three gradients are different sizes, not just all negative. 'Fastest at t = 10 min — the middle reading' isn't a mathematical reason at all — the gradients themselves have to be compared, not their position in the list.
- (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) 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) 168 m — Split the area under the graph into two parts. The rising section (0 to 8 s) is a triangle: area = 0.5 × 8 × 12 = 48. The constant section, from 8 s to 18 s (10 s), is a rectangle: area = 12 × 10 = 120. Total distance = 48 + 120 = 168 m. Leaving out the 0.5 doubles the triangle, giving 8 × 12 + 120 = 216 m; ignoring the triangle section altogether gives only 120 m; averaging the start and final speeds over the whole 18 seconds, (0 + 12) ÷ 2 = 6, then 6 × 18 = 108 m, wrongly treats the tram as accelerating the whole time, when the graph is flat for the last 10 seconds.
- (b) Student B's — narrower strips fit the curve more closely. — The trapezium rule replaces the curve with straight-line segments; the narrower each strip, the more closely its straight edge follows the curve, so Student B's estimate with 8 narrower strips is more likely to be closer to the true area. Reasoning from the number of arithmetic steps rather than from how well the straight lines fit the curve gives the wrong claim that Student A makes fewer rounding errors. Believing the trapezium rule is exact, rather than an estimate that improves with narrower strips, gives the wrong claim that the two are always the same. Believing wider strips smooth out the curve better, rather than following it less closely, gives the wrong claim in favour of Student A's wider strips.
- (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.
- (c) 81 m — Split the velocity-time graph into its three phases. The accelerating phase (0 to 4 s) is a triangle: 1/2 × 4 × 6 = 12. The constant phase (4 s at 6 m/s, for 10 s) is a rectangle: 10 × 6 = 60. The decelerating phase (3 s) is a triangle: 1/2 × 3 × 6 = 9. Total distance: 12 + 60 + 9 = 81 m. Forgetting the final decelerating phase entirely and adding only the first two areas gives 12 + 60 = 72 m. Treating all three phases as rectangles, forgetting to halve either triangular phase, gives 4 × 6 + 10 × 6 + 3 × 6 = 102 m. Halving only the deceleration triangle correctly but treating the acceleration phase as a rectangle too gives 4 × 6 + 60 + 9 = 93 m.
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