Hint
Differentiate each quantity with respect to r, then multiply by dr/dt at the stated instant.
Worked solution
(a)
The constants remain as multipliers: \(dV/dr=(4\pi/3)(3r^2)=4\pi r^2\). This is a geometric derivative, not yet a time rate.
Answer: \(dV/dr=4\pi r^2\).
(b)
\(dV/dt=(dV/dr)(dr/dt)=4\pi(4^2)(0.3)=19.2\pi=96\pi/5\) cm³/s. The positive sign agrees with expansion.
Answer: \(96\pi/5\) cm³/s, equivalently 19.2π cm³/s.
(c)
\(dA/dr=8\pi r\), so \(dA/dt=8\pi(4)(0.3)=9.6\pi=48\pi/5\) cm²/s. It is a different quantity with different units from the volume rate.
Answer: \(48\pi/5\) cm²/s, equivalently 9.6π cm²/s.
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) · 2 marks
- 1 mark: Apply the cubic power rule with the constant factor.
- 1 mark: Give 4πr².
Alternative methods: A correct difference-quotient derivation is accepted.
Common error: Using 4πr² directly as dV/dt without the radius rate.
(b) · 3 marks
- 1 mark: Use the chain rule linking volume rate and radius rate.
- 1 mark: Evaluate 96π/5 or an equivalent exact value.
- 1 mark: Give volume-per-time units cm³/s and the positive rate.
Alternative methods: A correctly evaluated decimal multiple of π is accepted.
Common error: Giving cm/s or omitting the factor 0.3.
(c) · 2 marks
- 1 mark: Use dA/dt=8πr dr/dt.
- 1 mark: Give 48π/5 cm²/s or its equivalent.
Alternative methods: An equivalent exact evaluation is accepted.
Common error: Reusing the volume derivative or volume units for surface area.
Total: 7 marks.