 Hello and welcome to the session. Let us understand the following question which says an edge of a variable cube is increasing at the rate of 3 cm per second. How fast is the volume of the cube increasing when the edge is 10 cm long? Now let us proceed on to the solution. Given to us is edge of the variable cube is increasing at the rate of 3 cm per second. Let x be the edge of the cube. Therefore it is given to us that dx by dt is equal to 3. Volume of a cube is given by v is equal to x cube. Now differentiating v with respect to t we get dv by dt is equal to d by dt of x cube. Now by chain rule this implies dv by dt is equal to 3x square dx by dt. dx by dt is given to us as 3. So this implies dv by dt is equal to 3x square multiplied by 3 which is equal to 9x square. Now we have to find the volume of the cube when edge is 10 cm long. Therefore dv by dt x is equal to 10 is equal to 9 multiplied by 10 multiplied by 10 which is equal to 900 cm cube per second. Hence volume of the cube is increasing at the rate of 900 cm cube per second when edge is 10 cm and this is the required answer. I hope you understood this question. Bye and have a nice day.