Last updated at April 19, 2021 by Teachoo

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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 = ๐๐ Given that Volume of cube is increasing at rate of 8 cm3/sec. Therefore ๐ ๐ฝ/๐ ๐ = 8 Putting V = ๐๐ (ใ๐(๐ฅใ^3))/๐๐ก = 8 ใ๐๐ฅใ^3/๐๐ก . ๐๐ฅ/๐๐ฅ = 8 ใ๐๐ฅใ^3/๐๐ฅ . ๐๐ฅ/๐๐ก = 8 3๐๐ . ๐๐ฅ/๐๐ก = 8 ๐ ๐/๐ ๐ = ๐/ใ๐๐ใ^๐ Now, We need to find fast is the surface area increasing when the length of an edge is 12 centimeters i.e. ๐ ๐บ/๐ ๐ for x = 12 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 . ๐/๐๐๐ = ๐๐/๐ For ๐ฅ= 12 cm ๐๐/๐๐ก = 32/12 (From (1): ๐ ๐/๐ ๐ = ๐/(๐๐^๐ )) ๐๐/๐๐ก = 8/3 Since surface area is in cm2 & time is in seconds, ๐ ๐บ/๐ ๐ = ๐/๐ cm2 /s

Ex 6.1

Ex 6.1, 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.