 Hi and welcome to our session. Let us discuss the following question. The question says surface area of spherical bubble is increasing at a rate of 2 cm square per second. Find the rate at which volume of the bubble is increasing at the instant when its radius is 6 cm. Now begin with the solution. Let radius of the bubble be r cm. The surface area of the bubble let us denoted by s is equal to 4 pi r square as the bubble is spherical in shape. Now s equals to 4 pi r square implies ds by dt is equal to 8 pi r into dr by dt. We are given that surface area of the spherical bubble is increasing at the rate of 2 cm square per second. That means ds by dt is equal to 2 cm square per second. And thus ds by dt equals to 8 pi r into dr by dt becomes 2 equals to 8 pi r into dr by dt. dr by dt is equal to 1 by 4 pi r. Let's name this equation as equation number 1. We have to find the rate at which the volume of the bubble is increasing at the instant when its radius is 6 cm. Now volume of the bubble that is denoted by v is equal to 4 by 3 pi r cube. Now v equals to 4 by 3 pi r cube implies dv by dt is equal to 4 by 3 into 3 pi r square into dr by dt. Now this is equal to 4 pi r square into dr by dt. Let's name this equation as equation number 2. From 1 to dv by dt equals to 4 pi r square into 1 by 4 pi r and this is equal to r. We have to find the rate at which the volume of the bubble is increasing at the instant when its radius is 6 cm. So that means we have to find dv by dt when radius is 6 cm. Now dv by dt r equals to 6 cm is equal to 6 cm cube per second. The required answer is 20 m cube per second. This completes the session. Bye and take care.