# Ex 6.1,8 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

Ex 6.1, 8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when radius is 15 cm. Balloon is spherical Let r be the radius of spherical balloon . & v be the volume of spherical balloon. It is given that The balloon is inflated by pumping in in 900 cubic cm of gas per sec i.e. volume of balloon increasing at the rate of 900 cm3/sec i.e. ๐๐ฃ/๐๐ก = 900 cm3/sec We need to calculate the rate of change at which radius of Balloon increases when radius is 15 cm i.e. we need to calculate ๐๐/๐๐ก when r = 15 cm we know that Volume of sphere = 4/3 ฯr3 i.e. V = 4/3 ฯr3 Differentiate w.r.t time(๐ก) ๐๐ฃ/๐๐ก = ๐(4/3 ๐๐3)/๐๐ก ๐๐ฃ/๐๐ก = 4/3 ฯ (๐ (๐3))/๐๐ก 900 = 4/3 ฯ . (๐(๐3))/๐๐ก ร ๐๐/๐๐ 900 = 4/3 ฯ . (๐(๐3))/๐๐ก ร ๐๐/๐๐ก 900 = 4/3 ฯ . 3r2 . ๐๐/๐๐ก 900 = 4 ร ฯ ร r2 . ๐๐/๐๐ก 900/(4 ร ๐ ร๐2) = ๐๐/๐๐ก ๐๐/๐๐ก = (900 ร 7)/(2 ร22 ร๐2) We need to find ๐๐/๐๐ก at r = 15 cm (From (1)) ๐๐/๐๐ก = 900/(4 ร๐ ร(15)^2 ) ๐๐/๐๐ก = 1/๐ Since radius in cm & time in sec So, ๐๐/๐๐ก = ๐/๐ ๐๐/๐๐๐ Hence ,the radius of the balloon is increasing at the rate of 1/๐ cm/sec when r = 15 cm.

Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.