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Worked solution

Find the falling water level in an inverted cone

A vessel containing water has the shape of an inverted right circular cone with base radius 2 m2\text{ m} and height 5 m5\text{ m}. Water flows from its vertex at 0.2 m3/s0.2\text{ m}^3/\text{s}. Find the exact rate at which the water level is dropping when the water depth is 4 m4\text{ m}.

CJC · Lecture · Example 13

Use the labelled similar triangles

Let rr be the water-surface radius and hh its depth, both in metres. The source diagram labels the full cone radius 2 and height 5, and the changing radius r and depth h. Similar triangles give r/h=2/5r/h=2/5, so r=2h/5r=2h/5. Both copies advise drawing a diagram to visualise the problem; the Teacher supplies the remaining working.

How did this part go?

Express volume in one changing variable

V=13πr2h=13π(2h/5)2h=475πh3V=\frac13\pi r^2h=\frac13\pi(2h/5)^2h=\frac4{75}\pi h^3, so dV/dh=425πh2dV/dh=\frac4{25}\pi h^2. Since water leaves the vessel, dV/dt=−0.2=−1/5 m3/sdV/dt=-0.2=-1/5\text{ m}^3/\text{s}. The chain rule gives dh/dt=(dV/dt)/(dV/dh)=−5/(4πh2)dh/dt=(dV/dt)/(dV/dh)=-5/(4\pi h^2).

How did this part go?

Separate signed change from the dropping rate

At h=4h=4, dh/dt=−5/(64π) m/sdh/dt=-5/(64\pi)\text{ m/s}. Thus the water level is dropping at 5/(64π) m/s5/(64\pi)\text{ m/s}, the positive magnitude printed in the Teacher answer. The negative derivative records the direction of change.

How did this part go?

Compare your sketch

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Inverted cone with full radius 2 m and height 5 m, water-surface radius r and water depth h measured from the common bottom vertex. Similar triangles give r/h=2/5.
Inverted cone with full radius 2 m and height 5 m, water-surface radius r and water depth h measured from the common bottom vertex. Similar triangles give r/h=2/5.
Inverted cone with full radius 2 m and height 5 m, water-surface radius r and water depth h measured from the common bottom vertex. Similar triangles give r/h=2/5.

Common mistakes

The water radius is changing too: use similar triangles before differentiating. Outflow makes dV/dt negative.

Solution notes

Both original copies visually compared. Printed working and shared guidance retained; additional domain and verification explanations are independently authored. No marks or past-paper attribution printed.

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