O-LevelAdditional MathematicsTrigonometry4037/22 · May/June 2018 · Q128 Marks

In this question all lengths are in centimetres. The volume of a cone of height h and base radius r is given by V = ⅓πr²h. It is known that sin (π/12) = (√6 - √2)/4, cos (π/12) = (√6 + √2)/4, tan (π/12) = 2 - √3. A water cup is in the shape of a cone with its axis vertical. The diagrams show the cup and its cross-section. The vertical angle of the cone is π/6 radians. The depth of water in the cup is h. The surface of the water is a circle of radius r. [Figure 12.1]

(i)Find an expression for r in terms of h and show that the volume of water in the cup is given by V = π(7−4√3)h³/3.[4]
(ii)Water is poured into the cup at a rate of 30 cm³s⁻¹. Find, correct to 2 decimal places, the rate at which the depth of water is increasing when h = 5.[4]

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Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

[M1] tan(π/12) = r/h oe
[A1] r = h(2 - √3) or r = h tan(π/12) oe
[M1] Correctly uses their expression for r in terms of h in formula for volume of a cone dependent on finding an expression connecting r and h
[A1] V = π/3 (2-√3)² h² × h oe
[A1] V = π(4 - 4√3 + 3)h³ / 3 oe, correctly leading to V = π(7 - 4√3)h³ / 3 AG
[B1] Correct derivative of V e.g. dV/dh = 3π(7 - 4√3)h² / 3 oe isw OR dV/dh = π(7 - 4√3)h²
[B1] dh/dt = dh/dV × dV/dt soi
[M1] 1 / (their dV/dh) × 30 (if correct implies B1 B1; if incorrect, a correct FT statement implies the second B1)
[A1] 5.32

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