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

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.1,13 A balloon, which always remains spherical, has a variable diameter 3﷮2﷯ (2𝑥 +1). Find the rate of change of its volume with respect to 𝑥. Let d be the diameter of the balloon Given d = 3﷮2﷯ (2x + 1) Let r be the radius of the balloon r = 𝑑﷮2﷯ = 3﷮4﷯ (2x + 1) The balloon is a sphere volume of the balloon = volume of sphere = 4﷮3﷯𝜋 𝑟﷮3﷯ We need to find rate of change of volume with respect to x that is 𝑑𝑣﷮𝑑𝑥﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑﷮𝑑𝑥﷯ 4﷮3﷯𝜋 𝑟﷮3﷯﷯ = 4𝜋﷮3﷯ 𝑑 𝑟﷮3﷯﷮𝑑𝑥﷯ = 4𝜋﷮3﷯ 𝑑﷮𝑑𝑥﷯ 27﷮64﷯ 2𝑥+1﷯﷮3﷯﷯ = 9𝜋﷮16﷯ 𝑑﷮𝑑𝑥﷯ 2𝑥+1﷯﷮3﷯ = 9𝜋﷮16﷯ 3(2x + 1)2 (2) = 𝟐𝟕𝝅﷮𝟖﷯ 𝟐𝒙+𝟏﷯﷮𝟐﷯ Hence, 𝑑𝑣﷮𝑑𝑥﷯ = 27𝜋﷮8﷯ 2𝑥+1﷯﷮2﷯

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