Ex 6.1,3 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 19, 2021 by Teachoo

Transcript

Ex 6.1, 3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.Let r be the radius of circle .
& A be the Area of circle.
Given that
Radius of a circle is increasing at the rate of 3 cm/s
Thus, 𝒅𝒓/𝒅𝒕 = 3 cm /sec
We need to find rate of change of area of circle w. r. t time when r = 10 cm
i.e. we need to find 𝒅𝑨/𝒅𝒕 when r = 10 cm
We know that
Area of circle = πr2
A = πr2
Differentiating w.r.t time
𝒅𝑨/𝒅𝒕 = 𝒅(𝝅𝒓𝟐)/𝒅𝒕
𝑑𝐴/𝑑𝑡 = π 𝑑(𝑟2)/𝑑𝑡
𝑑𝐴/𝑑𝑡 = π 𝑑(𝑟2)/𝑑𝑡 × 𝒅𝒓/𝒅𝒓
𝑑𝐴/𝑑𝑡 = π 𝒅(𝒓𝟐)/𝒅𝒓 × 𝑑𝑟/𝑑𝑡
𝑑𝐴/𝑑𝑡 = π. 2r . 𝑑𝑟/𝑑𝑡
𝑑𝐴/𝑑𝑡 = 2πr . 𝒅𝒓/𝒅𝒕
𝑑𝐴/𝑑𝑡 = 2πr . 3
𝑑𝐴/𝑑𝑡 = 6πr
When 𝒓 = 10 cm
├ 𝑑𝐴/𝑑𝑡┤|_(𝑟 =10) = 6 × π × 10
├ 𝑑𝐴/𝑑𝑡┤|_(𝑟 =10) = 60 π
(From (1): 𝑑𝑟/𝑑𝑡 = 3)
Since area is in cm2 & time is in sec
𝑑𝐴/𝑑𝑡 = 60π cm2/sec
Hence, Area is increasing at the rate of 60π cm2/sec when r = 10 cm

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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