Example 2 - Volume of a cube is increasing at a rate of 9 cubic

Example 2 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 2 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 2 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 2 - Chapter 6 Class 12 Application of Derivatives - Part 5

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

Transcript

Example 2 The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 10 centimeters ?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‘ Also it is given that Volume of cube is increasing at rate of 9 cubic cm/sec. Therefore ๐’…๐‘ฝ/๐’…๐’• = 9 Putting V = ๐’™๐Ÿ‘ (ใ€–๐‘‘(๐‘ฅใ€—^3))/๐‘‘๐‘ก = 9 ใ€–๐‘‘๐‘ฅใ€—^3/๐‘‘๐‘ก . ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ = 9 ใ€–๐‘‘๐‘ฅใ€—^3/๐‘‘๐‘ฅ . ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 9 3๐’™๐Ÿ . ๐’…๐’™/๐’…๐’• = 9 ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 9/ใ€–3๐‘ฅใ€—^2 ๐’…๐’™/๐’…๐’• = ๐Ÿ‘/๐’™^๐Ÿ Now, We need to find fast is the surface area increasing when the length of an edge is 10 centimeters i.e. ๐’…๐‘บ/๐’…๐’• for x = 10 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’™๐Ÿ = ๐Ÿ‘๐Ÿ”/๐’™ (From (1): ๐’…๐’™/๐’…๐’• = ๐Ÿ‘/๐’™^๐Ÿ ) For ๐‘ฅ= 10 cm ๐‘‘๐‘†/๐‘‘๐‘ก = 36/10 ๐‘‘๐‘†/๐‘‘๐‘ก = 3.6 Since surface area is in cm2 & time is in seconds, ๐‘‘๐‘†/๐‘‘๐‘ก = 3.6 ๐‘๐‘š2/๐‘  ๐’…๐‘บ/๐’…๐’• = 3.6 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.