Ex 6.1, 2 - Volume of a cube is increasing at 8 cm3/s. How fast

Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives - Part 5

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

Transcript

Ex 6.1, 2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?Let ๐’™ be length of side V be Volume t be time per second We know that Volume of cube = (Side)3 V = ๐’™๐Ÿ‘ Given that Volume of cube is increasing at rate of 8 cm3/sec. Therefore ๐’…๐‘ฝ/๐’…๐’• = 8 Putting V = ๐’™๐Ÿ‘ (ใ€–๐‘‘(๐‘ฅใ€—^3))/๐‘‘๐‘ก = 8 ใ€–๐‘‘๐‘ฅใ€—^3/๐‘‘๐‘ก . ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ = 8 ใ€–๐‘‘๐‘ฅใ€—^3/๐‘‘๐‘ฅ . ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 8 3๐’™๐Ÿ . ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 8 ๐’…๐’™/๐’…๐’• = ๐Ÿ–/ใ€–๐Ÿ‘๐’™ใ€—^๐Ÿ Now, We need to find fast is the surface area increasing when the length of an edge is 12 centimeters i.e. ๐’…๐‘บ/๐’…๐’• for x = 12 We know that Surface area of cube = 6 ร— Side2 S = 6๐‘ฅ2 Finding ๐’…๐‘บ/๐’…๐’• ๐‘‘๐‘†/๐‘‘๐‘ก = (๐‘‘(6๐‘ฅ^2))/๐‘‘๐‘ก = (๐‘‘(6๐‘ฅ2))/๐‘‘๐‘ก . ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ = 6. (๐‘‘(๐‘ฅ2))/๐‘‘๐‘ฅ . ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 6 . (2x) . ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 12๐‘ฅ . ๐’…๐’™/๐’…๐’• = 12๐‘ฅ . ๐Ÿ–/๐Ÿ‘๐’™๐Ÿ = ๐Ÿ‘๐Ÿ/๐’™ For ๐‘ฅ= 12 cm ๐‘‘๐‘†/๐‘‘๐‘ก = 32/12 (From (1): ๐’…๐’™/๐’…๐’• = ๐Ÿ–/(๐Ÿ‘๐’™^๐Ÿ )) ๐‘‘๐‘†/๐‘‘๐‘ก = 8/3 Since surface area is in cm2 & time is in seconds, ๐’…๐‘บ/๐’…๐’• = ๐Ÿ–/๐Ÿ‘ cm2 /s

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.