Water rises through a conical container

Conical water and similar trianglesGiven inverted-cone cross-section: full height24cm and base radius9cm, water depthh. Water radius relation is not supplied.h cm24 cm9 cm
Question diagram 1
Conical water and similar trianglesGiven inverted-cone cross-section: full height24cm and base radius9cm, water depthh. Water radius relation is not supplied.h cm24 cm9 cm

Given inverted cone and water cross-section, labelled9cm radius,24cm height andh depth.

The diagram shows a vertical cross-section of a container in the form of an inverted cone of height 24 cm and base radius 9 cm. The container is initially empty. Water is poured into it at a constant rate of 30 cm3 s−130\,\mathrm{cm^3\,s^{-1}} and the height of water level in the container at time tt is hh cm.

(i) Show that the volume of water in the container is given by V=364πh3V=\frac3{64}\pi h^3. marks

(ii) Find the value of hh at the instant when the rate of increase in the depth of water is 5.65 cm s−15.65\,\mathrm{cm\,s^{-1}}, leaving your answer in 2 decimal places. marks

[2009/SRJC/Prelim/II/1]

Total marks: 5

Solution

(i)

Similar triangles

An inverted cone has height 24 cm and base radius 9 cm. Show that the volume of water of depth hh is V=364πh3V=\frac3{64}\pi h^3.

Hint

  1. The water forms a smaller cone similar to the whole cone.

  2. Its radius is r=924hr=\frac{9}{24}h.

Answer · (i)

V=3πh3/64V=3\pi h^3/64.

(ii)

Related rates

Water is poured at 30 cm3 s−130\ \mathrm{cm^3\,s^{-1}}. Find hh when the depth is increasing at 5.65 cm s−15.65\ \mathrm{cm\,s^{-1}}.

Explore this idea

Hint

  1. dVdt=dVdhdhdt\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}.

  2. Substitute dVdt=30\frac{dV}{dt}=30 and dhdt=5.65\frac{dh}{dt}=5.65 and solve for hh.

Answer · (ii)

h=3.47h=3.47 cm (2 d.p.).

Explore the idea

How fast does the level rise at each depth?

dh/dt = 1920/(9πh²) falls steeply as the water gets deeper. Slide the line to find the depth for a given rate.

Height of the horizontal linek = 5.65
Try an example
Number of distinct depths
1

Each rate corresponds to one depth. At 5.65 cm/s the depth is h≈3.47h\approx3.47 cm; the deeper the water, the slower the rise.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
3.466815.65
  • Dh/dt against the depth h
  • Chosen rate
  • Depth with that rate
  • Intersection selected in the coordinate table
Key idea

Constant inflow in a widening container means a slowing rise.

Ideas to try
At what depth is the level rising at 1 cm/s?