# Ex 6.1,4 - 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, 4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of cube increasing when the edge is 10 cm long? Let ๐ฅ be the edge of cube. & V be the volume of cube. Given that edge of cube is increasing at the rate of 3 cm/ sec Thus, ๐๐ฅ/๐๐ก = 3 cm/sec We need to calculate How fast volume of cube increasing w. r. t time when edge is 10 cm i.e. we need to calculate ๐๐/๐๐ก when ๐ฅ = 10 cm We know that Volume of cube = (Edge)3 V = ๐ฅ3 Differentiate w.r.t time ๐๐ฃ/๐๐ก = (๐(๐ฅ3))/๐๐ก ๐๐ฃ/๐๐ก = (๐(๐ฅ3))/๐๐ก ร ๐๐ฅ/๐๐ฅ ๐๐ฃ/๐๐ก = (๐(๐ฅ3))/๐๐ฅ ร ๐๐ฅ/๐๐ก ๐๐ฃ/๐๐ก = 3๐ฅ2 . ๐ ๐/๐ ๐ ๐๐ฃ/๐๐ก = 3๐ฅ2 ร 3 (From (1)) ๐๐ฃ/๐๐ก = 9๐ฅ2 When ๐ฅ = 10 โ ๐๐ฃ/๐๐กโค|_(๐ฅ =10) = 9(10)2 โ ๐๐ฃ/๐๐กโค|_(๐ฅ =10) = 900 Since value is in cm3 & time is in sec ๐๐ฃ/๐๐ก = 900 (๐๐^3)/๐ ๐๐ ๐ ๐/๐ ๐ = 900 cm3/sec Hence, volume of a cube is increasing at the rate of 900 cm3/sec when edge is 10cm

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.