Unit 7
5 marks · EasyA sector of a circle has a radius of 5 cm and angle 72°. A water tank is connected to a pipe with outflow rate modelled by a reciprocal graph.
Calculate the length of the arc of the sector. Give your answer correct to 2 decimal places.
The volume of water in the tank at time t hours after midnight is given by the equation \[ v = \frac{600}{t} \] where v is in litres and t ≥ 1. Using t = 5, find the volume v of water left in the tank at 5 a.m.
Explain one feature of the graph of (such as its intercept or asymptote).
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) 6.28 cm
(b) 120 litres
(c) The graph has a vertical asymptote at t = 0 (the y-axis), and as t increases, v decreases but never reaches 0.
Common mistakes: Calculating arc length using the wrong angle or forgetting to convert to radians.; Substituting t = 5 incorrectly.
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Uses correct formula: arc length = (angle/360) × 2πr or equivalent.
A1 (a) · 1 mark · Obtains correct arc length: 6.28 cm (to 2 decimal places).
M2 (b) · 1 mark · Substitutes t = 5 correctly into v = 600 / t.
A2 (b) · 1 mark · Correct answer: v = 120 litres.
B1 (c) · 1 mark · States a correct feature of a reciprocal graph (e.g. asymptote or behaviour as t increases).
Worked solution
a) Arc length = (angle/360) × 2 × π × r = (72/360) × 2 × 3.14 × 5
b) v = 600 / t. Substitute t = 5: v = 600 / 5 = 120 litres.
c) The graph of v = 600/t has a vertical asymptote at t = 0. As t gets larger, v gets closer to 0 but never reaches it.