Question #a685f

I assume by the "plane" surface area you are referring to the flat surface of the hemisphere. The flat surface will clearly be a circle whose area is given by:

#A = pir^2=pi[r(t)]^2#

In the above we have made it explicit that #r# is time dependent.

We have that #(dr)/(dt)=0.02# from the question.

Integrate this with respect to #t# to get the explicit function of #r#:

#int(dr)/dtdt=int0.02dt=0.02t +C#

Assuming (as no other initial condition is stated) that the radius of the sphere is #0# when #t=0# then #C=0#.

So #r=0.02t#.

Using this information we can calculate the time derivative of #A#:

#(dA)/dt = d/dt{pi[r(t)]^2}=pid/(dt)(0.02t)^2#

#=0.0004pid/(dt)t^2=0.0008pit#

So, to work out the radius at which the circle's area is increasing at #0.04m^2 s^-1# we can set #(dA)/dt=0.04# and solve for #t#.

#-> (dA)/dt = 0.0008pit=0.04#

#-> t = 0.04/(0.0008pi)=50/pis #

So the time at which the sphere is increasing at #0.04m^2s^-1#
is at #50/ pi# seconds.

We know from above that #r=0.02t# so:

#r=0.02(50/pi)=pi m#

So the radius is #pi# metres.