DERIVATIVES Thus, the rate of change of y with respect to x can be calculated using the rate of change of y and that of x both with respect to t . Let us consider some examples. Example Find the rate of change of the area of a circle per second with respect to its radius r when r = cm. Solution The area A of a circle with radius r is given by A = π r .
Therefore, the rate of change of the area A with respect to its radius r is given by A dr dr = π . When r = cm, A dr = π . Thus, the area of the circle is changing at the rate of π cm /s. Example The volume of a cube is increasing at a rate of cubic centimetres per second.
How fast is the surface area increasing when the length of an edge is centimetres ? Solution Let x be the length of a side, V be the volume and S be the surface area of the cube. Then, V = x and S = x , where x is a function of time t . Now V dt = 9cm /s (Given) Therefore = V dx dx ⋅ (By Chain Rule) dx ⋅ dx dt = ...
( ) Now dS dt = ( ( dx dx ⋅ (By Chain Rule) x ⋅ (Using ( )) Hence, when x = cm, . cm /s dS dt = Example A stone is dropped into a quiet lake and waves move in circles at a speed of 4cm per second. At the instant, when the radius of the circular wave is cm, how fast is the enclosed area increasing? Solution The area A of a circle with radius r is given by A = π r .
Therefore, the rate of change of area A with respect to time t is A dt = dr dr ⋅ = π r dr (By Chain Rule) It is given that dr dt = 4cm/s Therefore, when r = cm, A dt = π ( ) ( ) = π Thus, the enclosed area is increasing at the rate of π cm /s, when r = cm.