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Ex 6.1, 12 - Radius of an air bubble is increasing at 1/2 cm/s

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

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Ex 6.1, 12 The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?Since Air Bubble is spherical Let r be the radius of bubble & V be the volume of bubble Given that Radius of an air bubble is increasing at the rate of 1/2 cm/s i.e. 𝒅𝒓/𝒅𝒕 = 𝟏/𝟐 cm/sec We need to calculate the rate is the volume of the bubble increasing when the radius is 1 cm i.e. we need to calculate 𝒅𝑽/𝒅𝒕 when r = 1 cm We know that Volume of sphere = V = πŸ’/πŸ‘ Ο€r3 Now, 𝑑𝑉/𝑑𝑑 = 𝑑(4/3 πœ‹π‘Ÿ3)/𝑑𝑑 𝑑𝑉/𝑑𝑑 = 4/3 Ο€ (𝑑 (π‘Ÿ3))/𝑑𝑑 𝑑𝑉/𝑑𝑑 = 4/3 Ο€ (𝑑 (π‘Ÿ3))/𝑑𝑑 𝑑𝑉/𝑑𝑑 = 4/3 Ο€ . (𝑑(π‘Ÿ3))/𝑑𝑑 Γ— 𝒅𝒓/𝒅𝒓 𝑑𝑉/𝑑𝑑 = 4/3 Ο€ . (𝑑(π‘Ÿ3))/𝑑𝑑 Γ— π‘‘π‘Ÿ/𝑑𝑑 𝑑𝑉/𝑑𝑑 = 4/3 Ο€ .3r2 . π‘‘π‘Ÿ/𝑑𝑑 𝑑𝑉/𝑑𝑑 = 4/3 Ο€ . 3r2 Γ— 1/2 𝑑𝑉/𝑑𝑑 = 2πœ‹π‘Ÿ^2 We need to find 𝑑𝑉/𝑑𝑑 at r = 1 cm 𝑑𝑉/𝑑𝑑 = 2πœ‹γ€–(1)γ€—^2 ("From (1): " π‘‘π‘Ÿ/𝑑𝑑=1/2 cm/s) 𝒅𝑽/𝒅𝒕 = πŸπ… Since Volume is in cm3 & time is in sec ∴ 𝑑𝑉/𝑑𝑑 = πŸπ… cm3/sec Hence, Volume is increasing at rate of 2πœ‹ cm3/sec

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.