 Hello and welcome to the session. Let us understand the following question which says, A stone is dropped into a quiet lake and waves move in a circle at the speed of 5 cm per second. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing? Now let us proceed on to the solution. Let radius of the circle be denoted by R and area B A. It is given to us is that stone is moving in a circle at the speed of 5 cm per second. That is, dR by dt is equal to 5 and we have to find how fast is the enclosed area increasing? That is dA by dt. We know area of a circle is given by A is equal to pi R square. Now differentiating it with respect to T we get dA by dt is equal to 2 pi R dR by dt. dR by dt is given to us as 5 so substituting its value we get, this implies dA by dt is equal to 2 pi R multiplied by 5 which is equal to 10 pi R. Here we have to find the value of dA by dt when radius is equal to 8 cm. Therefore dA by dt at R is equal to 8 is equal to 10 multiplied by pi multiplied by 8 is equal to 80 cm square per second. Hence the enclosed area is increasing at the rate of 80 cm square per second when radius is equal to 8 cm. I hope you understood this question by and have a nice day.