A capsule changes inside a fixed cone · source reference

Printed question

A capsule made of two similar hemispheres of radius $r$ cm, one at each end of a cylinder, is inscribed into a fixed cone of height 12 cm and base radius 9 cm. The two figures have a common axis of symmetry and the cross-section is shown below.

(i) Show that the height of the cylinder of the capsule, $x$ cm, is related to the radius of the hemisphere through the equation $x=12-\frac83r$. [3]

(ii) Given that, when $r=2$, $x$ is decreasing at a rate of $0.1\,\mathrm{cm\,s^{-1}}$, find the rate of change at this instant of the volume of the capsule. [4]

[Given: Volume of sphere $=\frac43\pi r^3$.]

[2008/AJC/Prelim/I/7]

Printed working

(i) Pythagoras gives $AD=\sqrt{9^2+12^2}=15$. Similar triangles $ABC$ and $ADE$ give $AB/AD=BC/DE$:

$$\frac{12-x-r}{15}=\frac r9\Rightarrow9(12-x-r)=15r\Rightarrow x=12-r-\frac53r=12-\frac83r$$

(shown).

Alternative solution: let $\angle EAD=\theta$. Using triangles $ABC$ and $ADE$, $\sin\theta=r/AB=9/15\Rightarrow AB=5r/3$. Thus $x=12-r-5r/3=12-8r/3$.

(ii)

$$V=\frac43\pi r^3+\pi r^2x=\frac43\pi r^3+\pi r^2(12-8r/3)=12\pi r^2-\frac43\pi r^3,$$ $$\frac{dV}{dr}=24\pi r-4\pi r^2,$$ $$\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}=\frac{dV}{dr}\frac{dr}{dx}\frac{dx}{dt}.$$

Since $x=12-8r/3$, $dx/dr=-8/3$, so

$$\frac{dV}{dt}=(24\pi r-4\pi r^2)(-3/8)\frac{dx}{dt}.$$

At $r=2$, $dx/dt=-0.1$:

$$\frac{dV}{dt}=[24\pi(2)-4\pi(2)^2](-3/8)(-0.1)=3.77$$

(3 s.f.). The capsule volume is increasing at $3.77\,\mathrm{cm^3\,s^{-1}}$ when $r=2$.

Teacher notes

Express $V$ in terms of only one variable.

Apply Chain Rule.

$x$ decreases ⇒ $dx/dt<0$, therefore $dx/dt=-0.1$ when $r=2$.

Answer key

$3.77\,\mathrm{cm^3\,s^{-1}}$.

Native diagrams