 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that air is escaping from a spherical balloon at 2 cubic centimeters per minute. At what rate is the surface area shrinking when the radius is 1 centimeter? We are given a spherical balloon. Let us make a sphere that represents the spherical balloon. We are given that air is escaping from the spherical balloon at a rate of 2 cubic centimeters per minute. It means volume of sphere is decreasing at a rate of 2 cubic centimeters per minute. Let volume is decreasing with respect to time t. Let V be the volume of the spherical balloon. Then dV by dt is equal to minus 2. We put a negative sign in the rate of change of volume V because the volume is decreasing. Let R be the radius of the spherical balloon. Now we have to find at what rate is the surface area shrinking when the radius is 1 centimeter. Let S be the surface area of the spherical balloon. Then we have to find dS by dt when R is equal to 1 centimeter. Now we know that surface area of a sphere is equal to 4 pi R square. That implies S is equal to 4 pi R square. Differentiate both sides with respect to t. We get dS by dt is equal to d by dt of 4 pi R square. This implies dS by dt is equal to 4 pi into 2R dr by dt which implies dS by dt is equal to 8 pi R dr by dt. Name this equation as 1. Now we need to find the value of dr by dt. We have dV by dt is equal to minus 2. Also we know that volume of a sphere is equal to 4 by 3 pi R cube. Differentiating both sides with respect to t. We get dV by dt is equal to 4 by 3 pi into 3R square dr by dt. That is dV by dt is equal to 4 pi R square dr by dt. Now we put dV by dt is equal to minus 2 in this equation. So this equation becomes minus 2 is equal to 4 pi R square dr by dt. Which implies minus 2 upon 4 pi R square is equal to dr by dt. That is dr by dt is equal to minus 1 upon 2 pi R square. So now put this value in equation 1 which is dS by dt equal to 8 pi R dr by dt. This implies dS by dt will be equal to 8 pi R into minus 1 upon 2 pi R square. Which is equal to minus 4 by R. So dS by dt is equal to minus 4 by R. And we have to find dS by dt when R is equal to 1 centimeter. So when we put R equal to 1 centimeter we obtain dS by dt is equal to minus 4 upon 1 which is equal to minus 4. This implies surface area of the spherical balloon is decreasing at a rate of 4 square centimeters per minute. This is the required answer. This completes our session. Hope you enjoyed this session.