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Chapter 6 Class 12 Application of Derivatives (Term 1)
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Ex 6.1,13 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 19, 2021 by

Ex 6.1, 13 - A balloon has a variable diameter 3/2 (2x + 1)

Ex 6.1,13 - Chapter 6 Class 12 Application of Derivatives - Part 2

Transcript

Ex 6.1, 13 A balloon, which always remains spherical, has a variable diameter 3/2 (2π‘₯ +1). Find the rate of change of its volume with respect to π‘₯.Let d be the diameter of the balloon Given that Diameter = d = 3/2 (2x + 1) Let r be the radius of the balloon r = 𝑑/2 = πŸ‘/πŸ’ (2x + 1) The balloon is a spherical Volume of the balloon = 4/3 πœ‹π‘Ÿ^3 We need to find rate of change of volume with respect to x i.e. 𝑑𝑉/𝑑π‘₯ Now, 𝑑𝑉/𝑑π‘₯ = 𝑑/𝑑π‘₯ (4/3 πœ‹π‘Ÿ^3 ) = 4πœ‹/3 Γ— (π‘‘π‘Ÿ^3)/𝑑π‘₯ = 4πœ‹/3 Γ— 𝑑/𝑑π‘₯ (27/64 (2π‘₯+1)^3 ) = 9πœ‹/16 Γ— (𝑑(2π‘₯ + 1)^3)/𝑑π‘₯ = 9πœ‹/16 Γ— 3(2x + 1)2 Γ— 2 = πŸπŸ•π…/πŸ– (πŸπ’™+𝟏)^𝟐

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.