Hint
Differentiate V with respect to time through r. A decrease in radius must keep its negative sign.
Worked solution
(a)
\(dV/dt=24\pi r\,dr/dt\). Thus −24π=24π(3)dr/dt, giving dr/dt=−1/3 cm/s.
Answer: \(-1/3\) cm/s.
(b)
The diameter is D=2r, so dD/dt=2dr/dt=−2/3 cm/s.
Answer: \(-2/3\) cm/s.
(c)
\(\Delta r\approx(-1/3)(0.06)=-0.02\) cm, so r≈3−0.02=2.98 cm. This uses the initial instantaneous rate as a local linear approximation; it does not assert an exact constant radius rate over time.
Answer: Approximately 2.98 cm.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Form dV/dt=24πr dr/dt.
- 1 mark: Substitute the instantaneous radius and solve correctly.
- 1 mark: Give −1/3 cm/s with the decreasing sign.
Alternative methods: Using dr/dV as the reciprocal of the nonzero dV/dr is accepted.
Common error: Dividing by r² or discarding the negative sign.
(b) · 2 marks
- 1 mark: Relate the diameter rate to twice the radius rate.
- 1 mark: Give −2/3 cm/s.
Alternative methods: Differentiating D=2r directly with respect to t is accepted.
Common error: Using the same numerical rate for diameter and radius.
(c) · 2 marks
- 1 mark: Use Δr≈(dr/dt)Δt with the initial rate and Δt=0.06.
- 1 mark: Give the approximate updated radius 2.98 cm.
Alternative methods: A correctly stated linear model through the initial radius is accepted.
Common error: Giving 0.02 as the new radius or claiming the finite update is exact.
Total: 7 marks.