The rate a which the surface area, S, decreases is
[tex] \frac{dS}{dt} =-8 \pi \, cm^{2}/s[/tex]
The surface area is
S = 4ΟrΒ²
where r =Β the radius at timeΒ t.
Therefore
[tex] \frac{dS}{dt} = \frac{dS}{dr} \frac{dr}{dt} =8 \pi r \frac{dr}{dt} \\\\ -8 \pi = 8 \pi r \frac{dr}{dt} \\\\ \frac{dr}{dt} =- \frac{1}{r} [/tex]
When r = 3 cm, obtain
[tex] \frac{dr}{dt}]_{r=3} = - \frac{1}{3} \, cm/s [/tex]
Answer:Β -1/3Β cm/sΒ (or -0.333 cm/s)
