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Example 49 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at March 16, 2023 by Teachoo
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Chapter 6 Class 12 Application of Derivatives
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Transcript
Checking sign for numerator cos π₯ β sin π₯ > 0 cos π₯ > sin π₯ 1 > sinβ‘π₯/cosβ‘π₯ 1 > tan π₯ πππ§β‘π<π This is possible only if π<π<π /π Thus, fβ(π) = ((+))/((+) ) > 0 in x β (0 , π/4) Hence, f is strictly increasing function in (π , π /π) 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, π π/π π = 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 π π¨/π π when r = 3.2 cm We know that Area of circle = Οr2 A = Οr2 Differentiating w.r.t time π π¨/π π = π (π ππ)/π π ππ΄/ππ‘ = Ο π(π2)/ππ‘ ππ΄/ππ‘ = Ο π(π2)/ππ‘ Γ π π/π π ππ΄/ππ‘ = Ο π (ππ)/π π Γ ππ/ππ‘ ππ΄/ππ‘ = Ο. 2r . ππ/ππ‘ ππ΄/ππ‘ = 2Οr . π π/π π ππ΄/ππ‘ = 2Οr . 0.05 ππ΄/ππ‘ = 0.1 Γ Οr When π = 3.2 cm β ππ΄/ππ‘β€|_(π =10) = 0.1 Γ Ο Γ 3.2 β ππ΄/ππ‘β€|_(π =10) = 0.320Ο Since area is in cm2 & time is in seconds π π¨/π π = 0.320Ο cm2/s Hence, Area is increasing at the rate of 0.320Ο cm2/s when r = 0.32 cm
