Question 1207271
.
water runs into a conical tank at the rate of 8 cubic meters per hour. 
if the height of the cone is 10 meters, and the diameter of its opening is 12 meters, 
how fast is the water level rising when the water is 3 meters deep?
~~~~~~~~~~~~~~~~~~~


<pre>
The part of the conical tank, occupied by water, is the cone with the ratio radius to the height
of  {{{r/h}}} = {{{6/10}}} = 0.6.


Hence, the volume of the water in the tank at every time moment t is

    V(t) = {{{(1/3)*pi*r^2*h}}} = {{{(1/3)*pi*(0.6*h)^2*h}}} = {{{0.12*pi*h^3}}}.    (1)


Differentiate it by the time

    {{{(dV)/(dt)}}} = {{{0.12*pi*3h^2*((dh)/(dt))}}}.    (2)


{{{(dV)/(dt)}}} is given: it is the inflow rate, or 8 m^3/hour.  
Hence, when h = 3 meters deep, we have from (2)


    8 = {{{0.12*pi*3*3^2*((dh)/(dt))}}} = {{{0.12*27*pi*((dh)/(dt))}}}.


Hence,  {{{(dh)/(dt)}}} = {{{8/(0.12*27*3.14159265)}}} = 0.78595 meters per hour.


Rounding to 3 decimals, you get the 


<U>ANSWER</U>.  When the water depth is 3 meters, the water rising rate is about 0.786 m/hour.
</pre>

Solved.