Question 1186841
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A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. 
Water is flowing into the tank at a rate of 5 cubic feet per minute. 
Find the rate of change of the depth of the water when the water is 6 feet deep.
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                 <U>S t e p   b y   s t e p</U>



(1)  The formula for the tank radius as the function of the depth is

         R = {{{(5/12)*H}}}.     (1)


     Indeed, it gives the radius  R = 5 ft, when H = 12 ft.



(2)   The formula for the volume of the tank

          V = {{{(1/3)*pi*R^2*H}}} = {{{(pi/3)*(5/12)^2*H(t)^3}}}    (2)

       after substituting (1).



(3)  Differentiate it 

          {{{(dV)/(dt)}}} = {{{pi*(5/12)^2*H(t)^2*((dH)/(dt))}}}.



(4)  Substitute H(t) = 6 feet and {{{pi}}} = 3.14159265

           {{{(dV)/(dt)}}} = {{{3.14159265*(5/12)^2*6^2*((dH)/(dt))}}} = {{{19.63495406*((dH)/(dt))}}}.



(5)  Substitute  {{{(dV)/(dt)}}} = 5 cubic feet per minute

           5 = {{{19.63495406*((dH)/(dt))}}}.



(6)  From this, find

            {{{(dH)/(dt)}}} = {{{5/19.63495406}}} = 0.254647909  feet per minute.



(7)  Round and get the <U>ANSWER</U>:  the rate of change of the depth of the water 
     is  0.25465 feet per minute when the water is 6 feet deep.
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Solved.