Example 2 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 15, 2021 by Teachoo
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 = ๐๐ Also it is given that Volume of cube is increasing at rate of 9 cubic cm/sec. Therefore ๐ ๐ฝ/๐ ๐ = 9 Putting V = ๐๐ (ใ๐(๐ฅใ^3))/๐๐ก = 9 ใ๐๐ฅใ^3/๐๐ก . ๐๐ฅ/๐๐ฅ = 9 ใ๐๐ฅใ^3/๐๐ฅ . ๐๐ฅ/๐๐ก = 9 3๐๐ . ๐ ๐/๐ ๐ = 9 ๐๐ฅ/๐๐ก = 9/ใ3๐ฅใ^2 ๐ ๐/๐ ๐ = ๐/๐^๐ Now, We need to find fast is the surface area increasing when the length of an edge is 10 centimeters i.e. ๐ ๐บ/๐ ๐ for x = 10 We know that Surface area of cube = 6 ร Side2 S = 6๐ฅ2 Finding ๐ ๐บ/๐ ๐ ๐๐/๐๐ก = (๐(6๐ฅ^2))/๐๐ก = (๐(6๐ฅ2))/๐๐ก . ๐๐ฅ/๐๐ฅ = 6. (๐(๐ฅ2))/๐๐ฅ . ๐๐ฅ/๐๐ก = 6 . (2x) . ๐๐ฅ/๐๐ก = 12๐ฅ . ๐ ๐/๐ ๐ = 12๐ฅ . ๐/๐๐ = ๐๐/๐ (From (1): ๐ ๐/๐ ๐ = ๐/๐^๐ ) For ๐ฅ= 10 cm ๐๐/๐๐ก = 36/10 ๐๐/๐๐ก = 3.6 Since surface area is in cm2 & time is in seconds, ๐๐/๐๐ก = 3.6 ๐๐2/๐ ๐ ๐บ/๐ ๐ = 3.6 cm2 /s
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Chapter 6 Class 12 Application of Derivatives (Term 1)
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About the Author
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.