A liquid is filling a container that has the form of a cone of angle α around a vertical axis (Fig. 1.1). The liquid enters the cone from the apex through a hole of diameter d at a velocity v (t) = kt where k is a constant. When the surface of the liquid is at height h (t), the volume is V (t)

Question

A liquid is filling a container that has the form of a cone of angle α around a vertical axis (Fig. 1.1). The liquid enters the cone from the apex through a hole of diameter d at a velocity ν (t) = kt where k is a constant. When the surface of the liquid is at height h (t), the volume is [latex]V (t) = \frac{1}{3} π\ tan^2 α h^3 (t)[/latex]. Initially, at time t = 0, the height h (0) = 0. Find an expression for the rate of change of volume [latex]\dot{V} (t)[/latex] and determine h (t).

Accepted Answer

The volume rate of change of the liquid is obtained by taking the time derivative of the volume [latex]V (t) = \frac{1}{3} π\ tan^2 α h^3 (t)[/latex],

[latex]\dot{V}(t) = π\ tan^2 α h^2 (t) \dot{h}(t)[/latex].

where the angle α is a constant. The volume rate of change of the liquid inflow is expressed as,

[latex]\dot{V} (t)=\pi \Bigl(\frac{d}{2}\Bigr)^2 u (t) =\pi \frac{kd^2}{4} t[/latex].

Equating the two previous equations yields,

[latex] an^2 α h^2 (t) \dot{h}(t) = \frac{kd^2}{4} t[/latex].

which can be recast as,

[latex]h^2 (t) d h (t) = \frac{kd^2}{4 an^2 α } t dt[/latex].

When integrating, we obtain,

[latex]\int_{h(0)}^{h(t)}h^{\prime 2} (t^{\prime} ) d h^{\prime} (t^{\prime} ) = \frac{kd^2}{4 an ^2 \alpha } \int_{0}^{4}{t^{\prime} d t^{\prime} }[/latex].

The result of this integration is,

[latex]\frac{1}{3} h^3(t) = \frac{kd^2}{4 an ^2 \alpha } \frac{1}{2} t^2[/latex].

Thus, the height of liquid inside the cone is,

[latex]h(t) = \biggl(\frac{3kd^2}{8 an ^2 \alpha }\biggr)^{\frac{1}{3} } t^{\frac{2}{3} }[/latex].

Question 1.4: A liquid is filling a container that has the form of a cone ...

Related Answered Questions