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Chapter 6 Class 12 Application of Derivatives (Term 1)
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Ex 6.1,9 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 19, 2021 by

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

Transcript

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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