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Finding rate of change
Ex 6.1, 1
Ex 6.1,17 (MCQ)
Example 5
Ex 6.1,15 Important
Example 6
Ex 6.1,16
Ex 6.1, 18 (MCQ) Important
Example 2
Ex 6.1,2
Example 35
Example 3
Ex 6.1,5 Important
Ex 6.1,3
Ex 6.1,6
Ex 6.1,12 You are here
Ex 6.1,13 Important
Misc 16 (MCQ)
Ex 6.1,14
Example 31 Important
Ex 6.1,4 Important
Example 30 Important
Example 4 Important
Ex 6.1,7
Ex 6.1,8
Ex 6.1,9
Ex 6.1,11 Important
Misc 2 Important
Ex 6.1,10 Important
Example 32 Important
Finding rate of change
Last updated at May 29, 2023 by Teachoo
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