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1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.1

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 3cm/s Thus, 𝑑𝑟﷮𝑑𝑡﷯ = 3cm /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 Differentiate 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 = 10cm 𝑑𝐴﷮𝑑𝑡﷯﷯﷮𝑟 =10﷯ = 6 × π × 10 𝑑𝐴﷮𝑑𝑡﷯﷯﷮𝑟 =10﷯ = 60 π Since area is in cm2 & time is in sec 𝑑𝐴﷮𝑑𝑡﷯ = 60 π 𝑐𝑚2﷮𝑠𝑒𝑐﷯ 𝑑𝐴﷮𝑑𝑡﷯ = 60 π cm2/sec Hence Area is increasing at the rate of 60 π cm2/sec when r = 10cm

Ex 6.1 