Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

Transcript

Example 35 A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm.Let r be the radius of circle . & A be the Area of circle. Given that Radius increases at the rate of 0.05 cm/s Thus, ๐๐/๐๐ = 0.05 cm /sec We need to find rate of change of area of circle w. r. t time when r = 3.2 cm i.e. we need to find ๐๐จ/๐๐ when r = 3.2 cm We know that Area of circle = ฯr2 A = ฯr2 Differentiating w.r.t time ๐๐จ/๐๐ = ๐(๐๐๐)/๐๐ ๐๐ด/๐๐ก = ฯ ๐(๐2)/๐๐ก ๐๐ด/๐๐ก = ฯ ๐(๐2)/๐๐ก ร ๐๐/๐๐ ๐๐ด/๐๐ก = ฯ ๐(๐๐)/๐๐ ร ๐๐/๐๐ก ๐๐ด/๐๐ก = ฯ. 2r . ๐๐/๐๐ก ๐๐ด/๐๐ก = 2ฯr . ๐๐/๐๐ ๐๐ด/๐๐ก = 2ฯr . 0.05 ๐๐ด/๐๐ก = 0.1 ร ฯr When ๐ = 3.2 cm โ ๐๐ด/๐๐กโค|_(๐ =10) = 0.1 ร ฯ ร 3.2 โ ๐๐ด/๐๐กโค|_(๐ =10) = 0.320ฯ Since area is in cm2 & time is in seconds ๐๐จ/๐๐ = 0.320ฯ cm2/s Hence, Area is increasing at the rate of 0.320ฯ cm2/s when r = 0.32 cm