Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12



  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


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 ๐‘ฅ 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, ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 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 ๐‘‘๐‘‰/๐‘‘๐‘ก when ๐‘ฅ = 10 cm We know that Volume of cube = (Edge)3 V = ๐‘ฅ3 Differentiate w.r.t time ๐‘‘๐‘ฃ/๐‘‘๐‘ก = (๐‘‘(๐‘ฅ3))/๐‘‘๐‘ก ๐‘‘๐‘ฃ/๐‘‘๐‘ก = (๐‘‘(๐‘ฅ3))/๐‘‘๐‘ก ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ก = (๐‘‘(๐‘ฅ3))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ก ๐‘‘๐‘ฃ/๐‘‘๐‘ก = 3๐‘ฅ2 . ๐’…๐’™/๐’…๐’• ๐‘‘๐‘ฃ/๐‘‘๐‘ก = 3๐‘ฅ2 ร— 3 (From (1)) ๐‘‘๐‘ฃ/๐‘‘๐‘ก = 9๐‘ฅ2 When ๐‘ฅ = 10 โ”œ ๐‘‘๐‘ฃ/๐‘‘๐‘กโ”ค|_(๐‘ฅ =10) = 9(10)2 โ”œ ๐‘‘๐‘ฃ/๐‘‘๐‘กโ”ค|_(๐‘ฅ =10) = 900 Since value is in cm3 & time is in sec ๐‘‘๐‘ฃ/๐‘‘๐‘ก = 900 (๐‘๐‘š^3)/๐‘ ๐‘’๐‘ 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

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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.