A capsule changes inside a fixed cone

Capsule inscribed in coneGiven unlabelled axial cross-section: two equal hemispheres at the ends of a cylinder inscribed in a cone, with common axis of symmetry. Shape is illustrative; no numerical capsule dimensions or solution construction is supplied.
Question diagram 1
Capsule inscribed in coneGiven unlabelled axial cross-section: two equal hemispheres at the ends of a cylinder inscribed in a cone, with common axis of symmetry. Shape is illustrative; no numerical capsule dimensions or solution construction is supplied.

Given unlabelled capsule/cone cross-section, without derived construction or numerical example.

A capsule made of two similar hemispheres of radius rr 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, xx cm, is related to the radius of the hemisphere through the equation x=12−83rx=12-\frac83r. marks

(ii) Given that, when r=2r=2, xx is decreasing at a rate of 0.1 cm s−10.1\,\mathrm{cm\,s^{-1}}, find the rate of change at this instant of the volume of the capsule. marks

[Given: Volume of sphere =43πr3=\frac43\pi r^3.]

[2008/AJC/Prelim/I/7]

Total marks: 7

Solution

(i)

Similar triangles

A capsule (two hemispheres of radius rr joined by a cylinder of height xx) is inscribed in a cone of height 12 cm and base radius 9 cm. Show that x=12−83rx=12-\frac83r.

Hint

  1. The slant height of the cone is 92+122=15\sqrt{9^2+12^2}=15.

  2. Use similar right triangles: the apex, the point where the top hemisphere touches the axis, and the point of tangency.

Answer · (i)

x=12−8r/3x=12-8r/3.

For a genuine cylinder 0<r<9/2.

(ii)

Related rates

When r=2r=2, xx is decreasing at 0.1 cm s−10.1\ \mathrm{cm\,s^{-1}}. Find the rate of change of the volume of the capsule at this instant.

Explore this idea

Hint

  1. The volume is two hemispheres (one sphere) plus a cylinder.

  2. Eliminate xx to get V(r)V(r).

  3. dVdt=dVdr⋅drdx⋅dxdt\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dx}\cdot\frac{dx}{dt}.

Answer · (ii)

dV/dt=6π/5≈3.77dV/dt=6\pi/5\approx3.77 cm³/s, increasing.

Explore the idea

Capsule volume against hemisphere radius r

V increases with r for r < 6; the steepness at r = 2 is dV/dr = 32π.

Position along the curver = 2
Try an example
f(r)
117.28613
Gradient f′(r)
100.53096
positive

For 0<r≤4.5 the derivative 4πr(6−r) is positive. At r=2 it equals 32π≈100.531.

Selected values are within the original drawing windows.

  • V(r)
  • Tangent
  • DV/dr
  • Join P to Q; the dashed legs show Δx and Δy
Key idea

dVdt\frac{dV}{dt} is the gradient of V(r)V(r) times how fast rr is changing: 32π×0.0375=6π532\pi\times0.0375=\frac{6\pi}5.

Ideas to try
Slide to r=2r=2 and read the gradient. Then multiply by drdt=0.0375\frac{dr}{dt}=0.0375.