The radius of the sphere decreases at the rate of 6 m/ sec
That is, [tex] \frac{dr}{dt}=-6 [/tex]
Surface Area of the sphere of radius r is given by [tex] S=4\pi r^2 [/tex]
Differentiating with respect to t,
[tex] \frac{dS}{dt}=4\pi (2r)\frac{dr}{dt}=8\pi r\frac{dr}{dt} [/tex]
[tex] \frac{dS}{dt}=8\pi (20)^2(-6)=-960\pi \: m^2/sec [/tex]
