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

Transcript

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

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.