Question 1188936
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Water flows into a tank having the form of frustum of a right circular cone. The tank is 4 m tall with upper radius of 1.5 m 
and the lower radius of 1 m. When the water in the tank is 1.2 m deep, the surface rises at the rate of 0.012 m/s. 
Calculate the discharge of water flowing into the tank in m3/s.
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        I will give a solution here different from that in the post by @CPhill.

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Here's how to calculate the discharge of water flowing into the conical frustum tank step by step.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>The geometry connections</U>


<pre>
We have the great cone with the base radius of 1.5 m and the height H m  (H is measured from the vertex).

We have the small cone with the base radius of 1 m   and the height (H-4) m.  This cone is cut out.

Therefore, for the small cone we have this proportion from similarity

     {{{1.5/H}}} = {{{1/(H-4)}}},  which gives  1.5H- 1.5*4 = H,  1.5H - H = 6,  0.5H = 6,  H = 6/0.5 = 12.

So, the great cone has the height H = 12 m;  the small cone has the height 12-4 = 8 m,  
and for every current h from vertex and r we have 

    {{{r/h}}} = {{{1.5/12}}} = {{{0.5/4}}} = 0.125.


Therefore, for any current h and r,  r = 0.125h, where h is measured from the vertex.

In particular, when r = 1 m (the lower radius),  h = 12-4 = 8 meters from the cone vertex.

When the height of the water in the tank is 1.2 meters, h = 8 + 1.2 = 9.2 meters and  r = 0.125*(8+1.2) = 1.15 m.
</pre>


    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>The volume of the water in the tank</U>


<pre>
When the level of the water is h meters from the vertex, the volume of the water is

    V = {{{(pi/3)*r^2*h}}} - {{{(pi/3)*1^2*8^2}}} m^3.


Substituting here  r = 0.125h, we have this formula for the volume of the water in the tank

    V = {{{(pi/3)*0.125^2*h^3}}} - {{{(pi/3)*64}}} m^3.


We can consider here V and h as functions of time

    V(t) = {{{(pi/3)*0.125^2*(h(t))^3}}} - {{{(pi/3)*64}}} m^3.    (1)
</pre>


    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Differentiate the volume of the water in the tank</U>


<pre>
Write the derivative of the volume of the water in the tank in respect to the time.
The constant term in formula (1) does not matter, so forget about it.

    {{{(dV)/dt)}}} = {{{(pi/3)*0.125^2*3h^2*((dh)/(dt))}}},

Apply it for h = 8+1.2 = 9.2 meters.  It will give

    {{{(dV)/(dt)}}} = {{{(3.14159/3)*0.125^2*3*9.2^2*((dh)/(dt))}}} = {{{4.154752775*((dh)/(dt))}}}


Now substitute the given value of the rate of the surface rise  {{{(dh)/(dt)}}} = 0.012 m/s.  You will get this equation

    {{{((dV)/(dt))}}} = {{{4.154752775*0.012}}} = 0.049857  (rounded).


<U>ANSWER</U>.  The inflow rate into the tank is about 0.049857 m^3/s.
</pre>

Solved.