Question

A vessel is in the shape of a right circular cone. The

Answer 1

The conic water vessel of height 40 cm and radius 8 cm filled at 20 cm^3/s gives;

• The rate at which the water level is rising when the water level is 12 cm from the vertex is approximately 1.1 cm/s.

• When the vessel is one-quarter filled the rate at which the water level is rising is approximately 0.21 cm/s.

How can the water level rate be found?

The volume of the vessel, v = π•r^2•h/3

$\frac{h}{r} = \frac{40}{8} = 5$

h = 5•r

Therefore;

v = π•r^2•(h)/3 = π•(h/5)^2•h/3

v = π•h^3/75

By chain rule of differentiation, we have;

$\frac{dv}{dh} = \frac{dv}{dt} \times \frac{dt}{dh}$

Which gives;

$π• \frac{ {h}^{2} }{25} = 20 × \frac{dt}{dh}$

$\frac{dh}{dt} = \frac{20}{\pi \frac{ {h}^{2} }{25} }$

When the height is 12 cm from the vertex, we have;

$\frac{dh}{dt} = \frac{20}{\pi \frac{ {12}^{2} }{25} } = 1.1$

• The rate at which the water level is rising when the water level is 12 cm from the vertex is approximately 1.1 cm/s.

When the vessel is one-quarter filled, we have;

v = π•h^3/75

π•(40)^3/(75×4) = π•h^3/75

10•(40)^2 = 16,000 = h^3

h = (16,000)^(1/3) = 25 (approx)

Which gives;

$\frac{dh}{dt} = \frac{20}{\pi \frac{ {25}^{2} }{25} } = 0.21$

When the vessel is one-quarter filled the rate at which the water level is rising is approximately 0.21 cm/s.

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