The volume V m³ of a large circular mound of iron ore of radius r m is modelled by the equation V = (9/2)(r - 1/2)³ – 1 for r ≥ 2. Iron ore is added to the mound at a constant rate of 1.5 m³ per second.
- (a)Find the rate at which the radius of the mound is increasing at the instant when the radius is 5.5 m.[3]
- (b)Find the volume of the mound at the instant when the radius is increasing at 0.1 m per second.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [B1] dV/dr = 9/2 * 3(r - 1/2)² OE. Accept unsimplified.
- [M1] Correct use of chain rule with 1.5, their differentiated expression for dV/dr and using r = 5.5 .
- [A1] 0.0133 or 3/225 or 1/75 [metres per second]
- [B1 FT] Correct statement involving dV/dr or their dV/dr, 1.5 and 0.1.
- [B1] (r - 1/2)² = 15 => r = 0.5 + √(15)
- [B1] [Volume =] 8.13 AWRT
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