Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.1,2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm? Let be the edge of a cube. Let V be the volume of cube. & S be the surface are of cube. Given that Volume of a cube is increasing at the rate of 8 cm3/ sec i.e. = 8 cm3/ sec We need to calculate how fast the surface Area is increasing when length of an edge is 12 cm i.e. We need to calculate when = 12 cm. Now, Volume of cube = (edge)3 ( 3) = 8 ( 3) = 8 ( 3) = 8 3x2 . = 8 = 8 3 2 We know that Surface Area of cube = 6 (edge)2 S =6 2 Different w.r.t = 6 2 = 6 2 = 6 . 2 = 6 . 2 = 6 . 2 . = 12 . = 12 . = 32 When = 12 =12 = 32 12 =12 = 8 3 Since surface area is in cm2 & time is in sec = 8 3 2 = cm2/sec Hence, surface area is increasing at the rate of 8 3 cm2/sec when edge is 12 cm