Question

Figure 4 shows a container in the shape of an inverted right circular cone which contains some water.


The cone has an internal radius of 2.5 m and a vertical height of 4 m.


At time t seconds, the height of the water is h m, the volume of the water is V m3 and water is modelled as leaking from a hole in the bottom of the container at a rate of h m3 s –1.


see on picture!


Given that the container was initially full of water,


(b) find an equation, in terms of h and t, to model this situation.

(3)


It takes close to 43 minutes for the container to empty.


(c) Use this information to comment on the suitability of this model.

Answer 1

Answer: a) See below; b) 2/5($h^{\frac{5}{2} }$) = - $\frac{1}{200}$.t + 0.215

Step-by-step explanation:

a) To determine the radius (x) of the volume of water remaining, use proportionality of similar triangles:

$\frac{x}{2.5} = \frac{h}{4}$

x = 0.625h

Volume is

V = 1/3.π.(0.625h)².h

V = 1/3.π.0.391.h³

Using the chain rule to determine the rate of change of height in function of t:

$\frac{dV}{dt} = \frac{dV}{dh} . \frac{dh}{dt}$

$\frac{dh}{dt} = \frac{dV}{dt}/\frac{dV}{dh}$

$\frac{dh}{dt} =$ (π/512).$\sqrt{h}$ / 0.391.π.h²

$\frac{dh}{dt} =$ (π . $\sqrt{h}$) / 512.0.391.π.h²

$\frac{dh}{dt} =$ $\frac{h^{\frac{-3}{2} } }{200}$

$h^{\frac{3}{2} }\frac{dh}{dt}=\frac{-1}{200}$

b) To find an equation:

$h^{\frac{3}{2} }$ dh = -(1/200) dt

∫ $h^{\frac{3}{2} }$ dh = -1/200 ∫dt

$\frac{2.h^{\frac{5}{2} } }{5}$ = $\frac{-1.t}{200}$ + c

It is known that it takes 43 minutes to empty the cone, so:

0 =  $\frac{-1.43}{200}$ + c

c = 0.215

The equation is:

$\frac{2.\sqrt[5]{h^{2}} }{5}$ = $\frac{-t}{200}$ + 0.215

c) The equation shows how height and time are inversely proportional to each other: as time goes up, height goes down. Mathematically, though, time cannot be negative. So, the model is suitable for modeling time and height but is not mathematically correct.