 Hello and welcome to the session. In this session we will discuss rate of change of quantities. Here we will learn how the derivative can be used to determine rate of change of quantities. Consider the derivative ds upon dt. By this derivative we mean that the rate of change of the distance s with respect to the time t. So we have if a quantity y varies with another quantity x satisfying some rule y equal to fx then we have dy by dx or this could also be written as f dash x represents the rate of change of y with respect to and also dy by dx at x equal to x naught is also written as f dash x naught this represents the rate of change of y with respect to x at x equal to x naught. Consider that the side of square be given by a it's given that the side of the square is increasing at the rate of that is we have dA by dt is given as 0.2 centimeter per second and let p be the perimeter of the square then we need to find dp by dt that is the rate of increase of the perimeter of the square. We know that the perimeter of the square p is given by 4 into the side of the square that is a so p is equal to 4a from here we have dp by dt that is rate of increase of the perimeter of the square is given by 4 into dA by dt and we have dA by dt is equal to 0.2 centimeter per second so this becomes equal to 4 into 0.2 that comes out to be equal to 0.8 centimeter per second so we have dp by dt is equal to 0.8 centimeter per second that is we have got the rate of increase of the perimeter of the square and a very important point to discuss is that dy upon dx would be positive if y increases as x increases dy by dx would be negative if y decreases as x increases. Next we have the chain rule if two variables x and y are varying with respect to another variable t that is if we are given x is equal to f t and y is equal to gt that is two variables x and y are varying with respect to the another variable t then by chain rule we have dy by dx is equal to dy by dt upon dx upon dt if dx upon dt is not equal to 0. Consider Sbd surface area of a spherical bubble we are given the rate of increase of the surface area of the bubble that is vs upon dt as 2 centimeter square per second let v be the volume of the spherical bubble we need to find the rate of increase of the volume of the spherical bubble that is we need to find dv by dt. Consider R to be the radius of the spherical bubble we know that the surface area s is given by 4 pi r square so we have ds by dt would be equal to ds by dr into dr by dt. Now from here we have ds by dr is equal to 8 pi r so on substituting the values for ds upon dt which is given to us and ds upon dr we get dr upon dt is equal to ds upon dt that is 2 centimeter square per second upon ds upon dr that is 8 pi r so we get dr upon dt equal to 1 upon 4 pi r centimeter per second the volume v of the spherical bubble is 4 upon 3 pi r cube from here we have dv by dt is equal to dv by dr into dr by dt from this v equal to 4 upon 3 pi r cube we have dv by dr is equal to 4 pi r square and we already have dr upon dt equal to 1 upon 4 pi r so we get from here dv by dt equal to 4 pi r square into 1 upon 4 pi r so we get this equal to r that is rate of increase of the volume of the spherical bubble that is dv upon dt is equal to r now when we are given that the radius r of this spherical bubble is 6 centimeters then dv by dt at r equal to 6 is equal to 6 centimeter cube per second this completes the session hope you have understood the rate of change of quantities and the chain rule