


Finding rate of change
Ex 6.1, 1 Deleted for CBSE Board 2022 Exams
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Misc 3 Important
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Example 44 Important Deleted for CBSE Board 2022 Exams
Finding rate of change
Last updated at April 19, 2021 by Teachoo
Ex 6.1, 8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when radius is 15 cm.Since Balloon is spherical Let r be the radius of spherical balloon & V be the volume of spherical balloon. Given that The balloon is inflated by pumping in in 900 cubic cm of gas per sec i.e. Volume of balloon increasing at the rate of 900 cm3/sec i.e. π π½/π π = 900 cm3/sec We need to calculate the rate of change at which radius of Balloon increases when radius is 15 cm i.e. we need to calculate π π/π π when r = 15 cm We know that Volume of sphere = V = 4/3 Οr3 Now, ππ/ππ‘ = π(4/3 ππ3)/ππ‘ ππ/ππ‘ = 4/3 Ο (π (π3))/ππ‘ 900 = 4/3 Ο (π (π3))/ππ‘ 900 = 4/3 Ο . (π(π3))/ππ‘ Γ π π/π π 900 = 4/3 Ο . (π(π3))/ππ Γ ππ/ππ‘ 900 = 4/3 Ο . 3r2 . ππ/ππ‘ 900 = 4 Γ Ο Γ r2 . ππ/ππ‘ 900/(4 Γ π Γ π2) = ππ/ππ‘ π π/π π = (πππ Γ π)/(π Γ ππ Γ ππ) We need to find ππ/ππ‘ at r = 15 cm ππ/ππ‘ = 900/(4 Γ π Γ (15)^2 ) π π/π π = π/π Since radius in cm & time in sec ππ/ππ‘ = π/π cm/s Hence, the radius of the balloon is increasing at the rate of 1/π cm/sec when r = 15 cm.