Question of the day 4.09

AA HL · Calculus · Related Rates

Water is evaporating from a cup in the shape of an inverted cone. The rate of evaporation is proportional to the area of the surface of the water.

Show that the depth of the water decreases at a constant rate that does not depend on the dimensions of the cup.

Hint

Let $h$ be the depth of the water and $r$ the radius of its surface. Use similar triangles to express $r$ in terms of $h$ .

The evaporation rate is proportional to the exposed surface area:

$\frac{dV}{dt}=-k\pi r^2.$

Solution

Let $h$ be the depth of the water and $r$ the radius of its surface. Since the water forms a cone similar to the cup,

$\frac{r}{h}=c,$

where $c$ is a constant. Hence

$r=ch.$

The volume of the water is

$V=\frac13\pi r^2h.$

Substituting $r=ch$ ,

$V=\frac13\pi c^2h^3.$

Differentiating with respect to time,

$\frac{dV}{dt} = \pi c^2h^2\frac{dh}{dt}.$

Since $r=ch$ ,

$c^2h^2=r^2,$

$\frac{dV}{dt} = \pi r^2\frac{dh}{dt}.$

The rate of evaporation is proportional to the area of the surface of the water, so for some constant $k>0$ ,

$\frac{dV}{dt}=-k\pi r^2.$

Therefore,

$\pi r^2\frac{dh}{dt} = -k\pi r^2.$

Hence

$\boxed{\frac{dh}{dt}=-k}.$

Thus the depth of the water decreases at a constant rate, independent of the dimensions of the cup.

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