# Example 49

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 49 (Method 1) A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm. Let r be the radius of a circular disc & y be the area of a circular disc. We know that Area of a circle = πr2 𝑦=𝜋𝑟2 Diff w.r.t 𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝜋𝑥2𝑑𝑥 𝑑𝑦𝑑𝑥 = π 2𝑥 Given that Due to expansion its radius increases at the rate of 0.05 cm/s i.e. ∆𝑥 = 0.05 when 𝑥= 3.2 cm We have to find approximate expansion in area when 𝑥 = 3.2 cm ∆y = 𝑑𝑦𝑑𝑥 ∆𝑥 Putting values ∆y = 2π𝑥 . ∆𝑥 ∆y = 2π 3.2 × 0.05 ∆y = 0.320 π Since area in cm2 & time is in seconds Thus the approximate change in volume is 0.320π cm2/s. Example 49 (Method 2) A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm. Let A be the area & radius of circular disc A = πr2 Given radius increases at 0.05 cm/s 𝑑𝑟𝑑𝑡 = 0.05cm/s We need to find the rate at which its area is increasing i.e. 𝑑𝐴𝑑𝑡 Now, 𝑑𝐴𝑑𝑡 = 𝑑𝜋𝑟2𝑑𝑡 = π . 𝑑𝑟2𝑑𝑡 = π . 𝑑𝑟2𝑑𝑟 . 𝑑𝑟𝑑𝑡 = π 2𝑟 𝑑𝑟𝑑𝑡 We need to find rate at which area is increasing when radius = 3.2cm 𝑑𝐴𝑑𝑡 = π . 2 × 3.2 ×0.05 = 0.320π Since area is in cm2 & time is in seconds 𝒅𝑨𝒅𝒕 = 0.320π cm2/s

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Example 49 You are here

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Chapter 6 Class 12 Application of Derivatives

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About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .