 Hi and welcome to our session. Let us discuss the following question. The question says the volume of a spherical balloon is increasing at the rate of 25 cm Q per second. Find the rate of change of its surface area at the instant when its radius is 5 cm. Let us now begin with the solution. Let R be radius of balloon surface area V be its volume at any instant T. Volume of the spherical balloon is given by 4 by 3 pi RQ. Now differentiating both sides with respect to T we get V by DT equals to 4 by 3 pi 3R square into D R by DT. This implies V by DT is equal to 4 pi R square into D R by DT. Now in the question we are given that volume of spherical balloon is increasing at the rate of 25 cm Q per second. That means V by DT is equal to 25 cm Q per second. Substituting 25 in place of V by DT we get 25 equals to 4 pi R square into D R by DT. This implies D R by DT is equal to 25 divided by 4 pi R square. We have to find the rate of change of its surface area at the instant when its radius is 5 cm. Now surface area of this spherical balloon which we have denoted by S is given by 4 pi R square. Now differentiating both sides of this equation with respect to T we get DT equals to 8 pi R into D R by DT. Now D R by DT is equal to 55 divided by 4 pi R square. So T S by DT is equal to 50 pi R. Now when R is equal to S by DT is equal to 50 pi pi R square which completes this equation by integer.