


Finding rate of change
Finding rate of change
Last updated at Dec. 16, 2024 by Teachoo
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?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 = 𝟒/𝟑 πr3 Now, 𝑑𝑉/𝑑𝑡 = 𝑑(4/3 𝜋𝑟3)/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π (𝑑 (𝑟3))/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π (𝑑 (𝑟3))/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π . (𝑑(𝑟3))/𝑑𝑡 × 𝒅𝒓/𝒅𝒓 𝑑𝑉/𝑑𝑡 = 4/3 π . (𝑑(𝑟3))/𝑑𝑡 × 𝑑𝑟/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π .3r2 . 𝑑𝑟/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π . 3r2 × 1/2 𝑑𝑉/𝑑𝑡 = 2𝜋𝑟^2 We need to find 𝑑𝑉/𝑑𝑡 at r = 1 cm 𝑑𝑉/𝑑𝑡 = 2𝜋〖(1)〗^2 ("From (1): " 𝑑𝑟/𝑑𝑡=1/2 cm/s) 𝒅𝑽/𝒅𝒕 = 𝟐𝝅 Since Volume is in cm3 & time is in sec ∴ 𝑑𝑉/𝑑𝑡 = 𝟐𝝅 cm3/sec Hence, Volume is increasing at rate of 2𝜋 cm3/sec