Finding rate of change
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Finding rate of change
Last updated at April 19, 2021 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. π π/π π = π/π 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 π π½/π π when r = 1 cm We know that Volume of sphere = V = π/π Οr3 Now, ππ/ππ‘ = π(4/3 ππ3)/ππ‘ ππ/ππ‘ = 4/3 Ο (π (π3))/ππ‘ ππ/ππ‘ = 4/3 Ο (π (π3))/ππ‘ ππ/ππ‘ = 4/3 Ο . (π(π3))/ππ‘ Γ π π/π π ππ/ππ‘ = 4/3 Ο . (π(π3))/ππ‘ Γ ππ/ππ‘ ππ/ππ‘ = 4/3 Ο .3r2 . ππ/ππ‘ ππ/ππ‘ = 4/3 Ο . 3r2 Γ 1/2 ππ/ππ‘ = 2ππ^2 We need to find ππ/ππ‘ at r = 1 cm ππ/ππ‘ = 2πγ(1)γ^2 ("From (1): " ππ/ππ‘=1/2 cm/s) π π½/π π = ππ Since Volume is in cm3 & time is in sec β΄ ππ/ππ‘ = ππ cm3/sec Hence, Volume is increasing at rate of 2π cm3/sec