Check sibling questions

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

Example 49 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 49 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 49 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Checking sign for numerator cos π‘₯ – sin π‘₯ > 0 cos π‘₯ > sin π‘₯ 1 > sin⁑π‘₯/cos⁑π‘₯ 1 > tan π‘₯ π­πšπ§β‘π’™<𝟏 This is possible only if 𝟎<𝒙<𝝅/πŸ’ Thus, f’(𝒙) = ((+))/((+) ) > 0 in x ∈ (0 , πœ‹/4) Hence, f is strictly increasing function in (𝟎 , 𝝅/πŸ’) Let r be the radius of circle . & A be the Area of circle. Given that Radius increases at the rate of 0.05 cm/s Thus, 𝒅𝒓/𝒅𝒕 = 0.05 cm /sec We need to find rate of change of area of circle w. r. t time when r = 3.2 cm i.e. we need to find 𝒅𝑨/𝒅𝒕 when r = 3.2 cm We know that Area of circle = Ο€r2 A = Ο€r2 Differentiating w.r.t time 𝒅𝑨/𝒅𝒕 = 𝒅(π…π’“πŸ)/𝒅𝒕 𝑑𝐴/𝑑𝑑 = Ο€ 𝑑(π‘Ÿ2)/𝑑𝑑 𝑑𝐴/𝑑𝑑 = Ο€ 𝑑(π‘Ÿ2)/𝑑𝑑 Γ— 𝒅𝒓/𝒅𝒓 𝑑𝐴/𝑑𝑑 = Ο€ 𝒅(π’“πŸ)/𝒅𝒓 Γ— π‘‘π‘Ÿ/𝑑𝑑 𝑑𝐴/𝑑𝑑 = Ο€. 2r . π‘‘π‘Ÿ/𝑑𝑑 𝑑𝐴/𝑑𝑑 = 2Ο€r . 𝒅𝒓/𝒅𝒕 𝑑𝐴/𝑑𝑑 = 2Ο€r . 0.05 𝑑𝐴/𝑑𝑑 = 0.1 Γ— Ο€r When 𝒓 = 3.2 cm β”œ 𝑑𝐴/𝑑𝑑─|_(π‘Ÿ =10) = 0.1 Γ— Ο€ Γ— 3.2 β”œ 𝑑𝐴/𝑑𝑑─|_(π‘Ÿ =10) = 0.320Ο€ Since area is in cm2 & time is in seconds 𝒅𝑨/𝒅𝒕 = 0.320Ο€ cm2/s Hence, Area is increasing at the rate of 0.320Ο€ cm2/s when r = 0.32 cm

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Davneet Singh

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.