Ex 6.1,12 - Chapter 6 Class 12 Application of Derivatives

Transcript

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? Let r be the radius of air bubble Given 𝑑𝑟/𝑑𝑡 = 1/2 cm/s The Bubble is a sphere Volume of bubble = Volume of sphere = 4/3 𝜋𝑟^3 We need to find the rate at which volume of the bubble is increasing when radius is 1 cm …(1) i.e. 𝑑𝑣/𝑑𝑡 at r = 1 cm 𝑑𝑣/𝑑𝑡 = 𝑑(4/3 𝜋𝑟^3 )/𝑑𝑡 = 4/3 𝜋 (𝑑𝑟^3)/𝑑𝑡 = 4/3 𝜋 (𝑑𝑟^3)/𝑑𝑟 𝑑𝑟/𝑑𝑡 = 4𝜋/3 〖3𝑟〗^2 𝑑𝑟/𝑑𝑡 = 4𝜋𝑟^2 1/2 = 2𝜋𝑟^2 (From (1): 𝑑𝑟/𝑑𝑡=1/2) When radius is 1cm 𝑑𝑣/𝑑𝑡 = 2𝜋(1)^2 𝑑𝑣/𝑑𝑡 = 2𝜋 Since volume is in cm3 & time is in sec So 𝑑𝑣/𝑑𝑡 = 𝟐𝝅 cm3/sec Hence, volume is increasing at rate of 2𝜋 cm3/sec when radius = 1cm