The volume of a sphere is changing at a constant rate of #pi/3 \ cm^3s^-1#. How fats is the surface area changing when the volume is #(9pi)/2#?

Let us set up the following variables:

# {(r, "Radius of sphere at time t","(cm)"), (A, "Surface area of sphere at time t", "(cm"^2")"), (V, "Volume of sphere at time t", "(cm"^3")"), (t, "time", "(sec)") :} #

Our aim is to find #(dA)/dt# when #V=(9pi)/2# and #(dV)/dt=pi/3#.

The standard formula for Area & Volume of a sphere are:

# V=4/3pir^3 \ \ \ \ .... [1] #
#A=4pir^2 \ \ \ \ \ \ .... [2] #

When # V=(9pi)/2 => 4/3pir^3 =(9pi)/2 #

# :. r^3 =9/2*3/4 #
# :. r =3/2 #

Differentiating [1] and [2] wrt #r# we get;

# (dV)/(dr)=4pir^2 # and # (dA)/(dr) = 8pir #

And from the chain rule we get:

# (dA)/dt =(dA)/(dr) * (dr)/(dV)* (dV)/(dt) #
# \ \ \ \ \ \ \=8pir * 1/(4pir^2) * (dV)/(dt) #
# \ \ \ \ \ \ \=2/r * (dV)/(dt) #

So when #V=(9pi)/2#, #(dV)/dt=pi/3# and #r =3/2#, then:

# (dA)/dt =2/(3/2) * pi/3 #
# \ \ \ \ \ \ \=(4pi)/9 #

#V = 4/3 pi*r^3#, where V =volume of sphere and r =radius.

given that,
#(dV)/(dt) =pi/3# and #V =9/2 pi#

#4/3 pi *r^3 = 9/2 pi#

#r^3 = 9/2 pi * 3/(4 pi)#

#r^3 = 27/8#, #r =3/2#

#V=4/3 pi*r^3,# then # (dV)/(dr) = 4 pi*r^2#

when #r =3/2#, # (dV)/(dr) = 4 pi*(3/2)^2=9 pi#

#(dr)/(dt) = (dV)/(dt)/((dV)/(dr))#

# = (pi/3)/(9 pi)=1/27#

Area pf sphere, #A =4 pi r^2#

#(dA)/(dr) = 8 pi r#
when #r =3/2#, #(dA)/(dr) = 8 pi r =8 pi (3/2)=12 pi#

therefore,
#(dA)/(dt) = (dA)/(dr)*(dr)/(dt)#

#=12 pi*1/27 = 4/9 pi# #(cm)^2/s#