The volume of the balloon is a function of the radius
[tex]V= \frac{4}{3} \pi r^{3} [/tex]
So since the volume is changing in time, both volume and radius are functions of time
[tex]V(t)= \frac{4}{3} \pi r(t)^{3} [/tex]
Now take the derivative of both sides with respect to t, and not forgetting the chain rule
[tex] \frac{dV}{dt}=4 \pi r^{2} \frac{dr}{dt} [/tex]
and now we use dV/dt=1.5 and solve for dr/dt
[tex] \frac{dr}{dt} = \frac{1.5}{4 \pi r^{2} } [/tex]
since we want the rate of surface area change when r=0.04m, we can put that value in for r
[tex] \frac{dr}{dt} = \frac{1.5}{4 \pi (0.04)^{2} } = \frac{1.5}{0.0064 \pi }[/tex]
Ok now look at the surface area formula:
[tex]A=4 \pi r^{2} [/tex]
Now since we want its rate of change, we take the derivative again remembering to use the chain rule
[tex] \frac{dA}{dt}=8 \pi r \frac{dr}{dt} [/tex]
Now we use the value of dr/dt we found earlier and put in 0.04 for r[tex]\frac{dA}{dt}=8 \pi 0.04\frac{1.5}{0.0064 \pi } =75 \frac{ m^{2} }{min}[/tex]
