Question 1125312
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You are given  V(t) = V(r(t)) = {{{(4/3)*pi*r(t)^3}}},  where the radius  r  varies  such that  {{{(dr)/(dt)}}} = 1 millimeter per minute.


Then  {{{(dV)/(dt)}}} = {{{4*pi*r^2}}}.{{{(dr)/dt)}}} = {{{4*3.14*2^2}}}.{{{1}}} = 4*3.14*4 = {{{16pi}}} = 50.24 mm^3 per minute.


Thus <U>the instantaneous volume growing rate</U> is  50.24 mm^3/minute.



Interesting, that if you calculate the <U>averaged volume growing rate</U> between  t= 2 seconds  and  t= 3 seconds, you will get another value


{{{(V(3) - V(2))/(3-2)}}} = {{{(4/3)*pi*3^3}}} - {{{(4/3)*pi*2^3}}} = {{{(4/3)*pi*(27-8)}}} = {{{(76/3)*pi}}} = {{{(76/3)*3.14}}} = 79.55  mm^3/minute.


But it only demonstrates the difference between these two conceptions: instantaneous and averaged rates.
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<U>Answer</U>.  &nbsp;&nbsp;The <U>instantaneous volume growing rate</U> at this moment is  &nbsp;50.24 mm^3 per minute.