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

Last updated at April 16, 2024 by Teachoo

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 πr3
Now,
𝑑𝑉/𝑑𝑟 = (𝑑 (4/3 𝜋𝑟3))/𝑑𝑟
𝑑𝑉/𝑑𝑟 = 4/3 π 𝑑(𝑟3)/𝑑𝑟
𝑑𝑉/𝑑𝑟 = 4/3 π 3𝑟^2
𝑑𝑉/𝑑𝑟 = 4𝜋𝑟^2
When r = 10
𝑑𝑉/𝑑𝑟 = 4 × π × (10)2
𝑑𝑉/𝑑𝑟 = 400π
Since volume is in cm3 & Radius is in cm
So, 𝑑𝑉/𝑑𝑟 = 400π cm3/cm
Hence, volume is increasing at the rate of 400 π cm3/cm when r = 10 cm

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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