Hint

Square the radius before integrating. The derivative with respect to the cut-off coordinate k measures cross-sectional area, not elapsed time.

Worked solution

(a)

The radius is nonnegative on [0,4]. Its square is \(\displaystyle x^{2}(4-x)^{2}=16x^{2}-8x^{3}+x^{4}\). Hence \(\displaystyle V(k)= \pi \int_{0}^{k} (16x^{2}- 8x^{3}+ x^{4})dx= \pi (\frac{16k^{3}}{3}- 2k^{4}+ \frac{k^{5}}{5})\). The lower limit fixes \(\displaystyle V(0)=0\).

Answer: \(\displaystyle V(k)= \pi (\frac{16k^{3}}{3}- 2k^{4}+ \frac{k^{5}}{5})\).

(b)

\(\displaystyle V(4)= \pi (\frac{1024}{3}- 512+ \frac{1024}{5})= \frac{512\pi }{15}\mathrm{cm}^{3}\).

Answer: \(\displaystyle \frac{512\pi }{15}\) \(\displaystyle \mathrm{cm}^{3}\).

(c)

\(\displaystyle V'(k)=\pi k^{2}(4-k)^{2}\) is the disk area. Its derivative is 4πk(k−2)(k−4), positive for 0<k<2 and negative for 2<k<4. With endpoint values zero, the global maximum disk area is \(\displaystyle V'(2)=16\pi \mathrm{cm}^{2}\). However V′ is positive throughout (0,4), so V is strictly increasing and its greatest value occurs at \(\displaystyle k=4\). This is growth per unit length, not a time rate.

Answer: Greatest \(V^{\prime}=16\pi\) \(\displaystyle \mathrm{cm}^{2}\) at \(\displaystyle k=2 \mathrm{cm}\); V itself is greatest at \(\displaystyle k=4 \mathrm{cm}\).

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: Expand the squared radius as \(\displaystyle 16x^{2}-8x^{3}+x^{4}\).
• 1 mark: Integrate every term correctly.
• 1 mark: Evaluate the variable limits and retain π without a free constant.

Alternative methods: An equivalent exact antiderivative expression evaluated from 0 to k is accepted.

Common error: Using the unsquared profile or introducing an arbitrary volume constant.

(b) · 2 marks

• 1 mark: Substitute the upper cut-off 4 into the complete expression.
• 1 mark: Give \(\displaystyle \frac{512\pi }{15}\mathrm{cm}^{3}\) exactly.

Alternative methods: A separate definite integral of the squared radius over [0,4] is accepted.

Common error: Rounding intermediate fractions or omitting π.

(c) · 4 marks

• 1 mark: Identify V′(k) as \(\displaystyle \pi k^{2}(4-k)^{2}\).
• 1 mark: Find the maximum-growth candidate \(\displaystyle k=2\) with valid differentiation or an equivalent bound.
• 1 mark: Give the global greatest growth value \(\displaystyle 16\pi \mathrm{cm}^{2}\) with endpoint/sign reasoning.
• 1 mark: Explain that V remains increasing and therefore is greatest at \(\displaystyle k=4\).

Alternative methods: Bound the nonnegative radius by 4 using \(\displaystyle x(4-x)=4-(x-2)^{2}\), then distinguish the accumulated volume.

Common error: Claiming V is largest at \(\displaystyle k=2\) because its derivative is largest there.

Total: 9 marks.