Example 35 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Examples
Example 1
You are here
You are here
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 ๐จ/๐ ๐ when r = 3.2 cm We know that Area of circle = ฯr2 A = ฯr2 Differentiating w.r.t time ๐ ๐จ/๐ ๐ = ๐ (๐ ๐๐)/๐ ๐ ๐๐ด/๐๐ก = ฯ ๐(๐2)/๐๐ก ๐๐ด/๐๐ก = ฯ ๐(๐2)/๐๐ก ร ๐ ๐/๐ ๐ ๐๐ด/๐๐ก = ฯ ๐ (๐๐)/๐ ๐ ร ๐๐/๐๐ก ๐๐ด/๐๐ก = ฯ. 2r . ๐๐/๐๐ก ๐๐ด/๐๐ก = 2ฯr . ๐ ๐/๐ ๐ ๐๐ด/๐๐ก = 2ฯr . 0.05 ๐๐ด/๐๐ก = 0.1 ร ฯr When ๐ = 3.2 cm โ ๐๐ด/๐๐กโค|_(๐ =10) = 0.1 ร ฯ ร 3.2 โ ๐๐ด/๐๐กโค|_(๐ =10) = 0.320ฯ Since area is in cm2 & time is in seconds ๐ ๐จ/๐ ๐ = 0.320ฯ cm2/s Hence, Area is increasing at the rate of 0.320ฯ cm2/s when r = 0.32 cm
