Find the inflection point of V² within its radius domain.

01 / Standard#Your turn

V(x)=\frac\pi3x^2\sqrt{1-x^2},\quad0\le x\le1

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Differentiate V² twice.

Hint 2

Require a change of concavity on both sides.

Worked solution

Solve the interior second-derivative equation.
$(V^2)''=\frac{2\pi^2}3x^2(2-5x^2),\quad x=\sqrt{2/5}=\frac{\sqrt{10}}5$
Concavity changes from up to down; evaluate the ordinate.
$V^2(\sqrt{2/5})=\frac{\pi^2}9\left(\frac4{25}-\frac8{125}\right)=\frac{4\pi^2}{375}$

Inflection point (√10/5,4π²/375).

Checks and common pitfalls: x=0 is a boundary and is not an inflection point of this radius-domain curve.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Solve the interior second-derivative equation.
$(V^2)''=\frac{2\pi^2}3x^2(2-5x^2),\quad x=\sqrt{2/5}=\frac{\sqrt{10}}5$
Concavity changes from up to down; evaluate the ordinate.
$V^2(\sqrt{2/5})=\frac{\pi^2}9\left(\frac4{25}-\frac8{125}\right)=\frac{4\pi^2}{375}$

Think first. Reveal a hint when the class is ready.

2024 JM02