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Ex 6.1, 9 - A balloon has a variable radius. Find rate - Ex 6.1

Ex 6.1,9 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,9 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Ex 6.1, 9 A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.Since Balloon is spherical Let r be the radius of balloon . & V be the volume of balloon. We need to find rate at which balloon volume is increasing when radius is 10cm i.e. We need to find change of volume w.r.t radius when r = 10 i.e. we need to find 𝒅𝑽/𝒅𝒓 when r = 10 cm We know that Volume of sphere = V = 4/3 Ο€r3 Now, 𝑑𝑉/π‘‘π‘Ÿ = (𝑑 (4/3 πœ‹π‘Ÿ3))/π‘‘π‘Ÿ 𝑑𝑉/π‘‘π‘Ÿ = 4/3 Ο€ 𝑑(π‘Ÿ3)/π‘‘π‘Ÿ 𝑑𝑉/π‘‘π‘Ÿ = 4/3 Ο€ 3π‘Ÿ^2 𝑑𝑉/π‘‘π‘Ÿ = 4πœ‹π‘Ÿ^2 When r = 10 𝑑𝑉/π‘‘π‘Ÿ = 4 Γ— Ο€ Γ— (10)2 𝑑𝑉/π‘‘π‘Ÿ = 400Ο€ Since volume is in cm3 & Radius is in cm So, 𝑑𝑉/π‘‘π‘Ÿ = 400Ο€ cm3/cm Hence, volume is increasing at the rate of 400 Ο€ cm3/cm when r = 10 cm

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.