Question 1127031
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The volume of a cone   V = {{{(1/3)*pi*r^2*h}}}.


Since we are given that  h = 2r (where 2r is the diameter), 


    V = {{{(1/3)*pi*(h/2)^2*h}}} = {{{(1/3)*pi*(1/4)*h^3}}} = {{{(pi/12)*h^3}}}.    (1)


The rate of the volume change in time is given;  it is 15 {{{(ft)^3/(min)}}};  therefore


15 = {{{(dV)/(dt)}}} = {{{((dV)/(dh))*((dh)/(dt))}}} = (calculate {{{(dV)/(dh)}}} from the formula (1)) = {{{(pi/4)*h^2}}}.{{{(dh)/(dt)}}} = (substitute here h= 5 ft) = {{{(pi/4)*25}}}.{{{(dh)/(dt)}}}.


It implies  {{{(dh)/(dt)}}} = {{{15/((pi/4)*25)}}} = {{{(3*4)/(5*pi)}}} = {{{12/(5*pi)}}} = 0.76 ft/min.  (rounded)


<U>Answer</U>.  At this moment, the height of the cone is increasing at the rate of  0.76 ft/min. (approximately)
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Solved.