Question 1172158
Assuming that "1 in 2/min" is supposed to be {{{ 1 (in^2/min) }}}...

{{{ V }}} = {{{ (4/3)pi*r^3 }}}
{{{ A }}} = {{{  4*pi*r^2 }}}<br>

V in terms of A:
{{{ V }}} = {{{ matrix(3,3,"","","", "", sqrt(A^3)/(6sqrt(pi)), "","","","") }}}<br>

Taking the derivative of V wrt A:
{{{ dV/dA }}} = {{{ sqrt(A)/(4*sqrt(pi)) }}}<br>

Need an expression for dV/dt in terms of information given:
{{{ dV/dt }}} = {{{ (dV/dA) * (dA/dt) }}} <br>

At r=3, A={{{4*pi*3^2}}} = {{{36pi}}} --->  {{{sqrt(A)}}} = {{{ 6*sqrt(pi) }}}<br>
dA/dt is given as {{{1 (in^2/min) }}} <br>

So, finally,  {{{ dV/dt }}} = {{{ (6*sqrt(pi) / (4*sqrt(pi))) * 1 }}} = {{{highlight(3/2)}}}{{{in^3/min}}}

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Another approach would be to use 'r' instead of 'A':   dV/dt = (dV/dr)*(dr/dt)
It is about equally messy to go this route.  Because you are given dA/dt, you
 have the additional step dr/dt = (dA/dt)*(dr/dA),  so dV/dt = (dV/dr)*(dA/dt)*(dr/dA) 

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