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Example 49 - A circular disc of radius 3 cm is being heated - Finding rate of change

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  1. Chapter 6 Class 12 Application of Derivatives
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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=𝜋﷐𝑟﷮2﷯ Diff w.r.t 𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐𝑑﷐𝜋𝑥2﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = 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 = ﷐𝑑𝑦﷮𝑑𝑥﷯ ∆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 ﷐𝑑𝑟﷮𝑑𝑡﷯ = 0.05cm/s We need to find the rate at which its area is increasing i.e. ﷐𝑑𝐴﷮𝑑𝑡﷯ Now, ﷐𝑑𝐴﷮𝑑𝑡﷯ = ﷐𝑑﷐𝜋𝑟2﷯﷮𝑑𝑡﷯ = π . ﷐𝑑𝑟2﷮𝑑𝑡﷯ = π . ﷐𝑑𝑟2﷮𝑑𝑟﷯ . ﷐𝑑𝑟﷮𝑑𝑡﷯ = π ﷐2𝑟﷯ ﷐𝑑𝑟﷮𝑑𝑡﷯ We need to find rate at which area is increasing when radius = 3.2cm ﷐𝑑𝐴﷮𝑑𝑡﷯ = π . 2 × ﷐3.2﷯ ×0.05 = 0.320π Since area is in cm2 & time is in seconds ﷐𝒅𝑨﷮𝒅𝒕﷯ = 0.320π cm2/s

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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