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Sketch y=V²(x) on the radius domain using the extrema and inflection.

Read the idea, work independently, then explain what changed.

TOPIC 01

2024 JM02

The vertical coordinate is volume squared, not volume.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Sketch y=V²(x) on the radius domain using the extrema and inflection.

V(x)=π3x21−x2,0≤x≤1V(x)=\frac\pi3x^2\sqrt{1-x^2},\quad0\le x\le1

Official paper · jm02-2024 · 2(a)(iv) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 9

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Mark (0,0) and (1,0).
Hint 2
The inflection occurs before the maximum.
Worked solution
  1. The graph starts flat at zero, rises to its maximum and falls to zero.

    0<2/5<2/3<10<\sqrt{2/5}<\sqrt{2/3}<1
  2. It is concave up before √(2/5) and concave down afterwards.

    max⁡y=4π2/243,y(0)=y(1)=0\max y=4\pi^2/243,\quad y(0)=y(1)=0
2024 JM02 2(a)(iv): y=V²(x), 0≤x≤100.20.40.60.8100.050.10.150.2(0,0)(1,0)max: 4π²/243inflection2024 JM02 2(a)(iv): y=V²(x), 0≤x≤1

The plot shows the full domain, maximum, and inflection.

Checks and common pitfalls: The vertical coordinate is volume squared, not volume.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The graph starts flat at zero, rises to its maximum and falls to zero.
    0<2/5<2/3<10<\sqrt{2/5}<\sqrt{2/3}<1
  • It is concave up before √(2/5) and concave down afterwards.
    max⁡y=4π2/243,y(0)=y(1)=0\max y=4\pi^2/243,\quad y(0)=y(1)=0

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Curriculum and source notes ↗