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

Last updated at April 16, 2024 by Teachoo

Transcript

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. 𝒅𝒓/𝒅𝒕 = 𝟏/𝟐 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 𝒅𝑽/𝒅𝒕 when r = 1 cm
We know that
Volume of sphere = V = 𝟒/𝟑 πr3
Now,
𝑑𝑉/𝑑𝑡 = 𝑑(4/3 𝜋𝑟3)/𝑑𝑡
𝑑𝑉/𝑑𝑡 = 4/3 π (𝑑 (𝑟3))/𝑑𝑡
𝑑𝑉/𝑑𝑡 = 4/3 π (𝑑 (𝑟3))/𝑑𝑡
𝑑𝑉/𝑑𝑡 = 4/3 π . (𝑑(𝑟3))/𝑑𝑡 × 𝒅𝒓/𝒅𝒓
𝑑𝑉/𝑑𝑡 = 4/3 π . (𝑑(𝑟3))/𝑑𝑡 × 𝑑𝑟/𝑑𝑡
𝑑𝑉/𝑑𝑡 = 4/3 π .3r2 . 𝑑𝑟/𝑑𝑡
𝑑𝑉/𝑑𝑡 = 4/3 π . 3r2 × 1/2
𝑑𝑉/𝑑𝑡 = 2𝜋𝑟^2
We need to find 𝑑𝑉/𝑑𝑡 at r = 1 cm
𝑑𝑉/𝑑𝑡 = 2𝜋〖(1)〗^2
("From (1): " 𝑑𝑟/𝑑𝑡=1/2 cm/s)
𝒅𝑽/𝒅𝒕 = 𝟐𝝅
Since Volume is in cm3 & time is in sec
∴ 𝑑𝑉/𝑑𝑡 = 𝟐𝝅 cm3/sec
Hence, Volume is increasing at rate of 2𝜋 cm3/sec

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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