Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Last updated at Jan. 7, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Transcript

Example 38 Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that cone. Let OC = r be the radius of cone & OA = h be height of cone & ∠ OAQ = α be the semi − vertical angle of cone Let OE = x be the radius of cylinder Height of cylinder = OO’ From (1) & (2) 𝑥/(ℎ − 𝑂^′ 𝑂)=𝑟/ℎ ℎ𝑥/𝑟=ℎ−𝑂^′ 𝑂 𝑂^′ 𝑂=ℎ−ℎ𝑥/𝑟 𝑂^′ 𝑂= (ℎ𝑟 − ℎ𝑥)/𝑟 In Δ AO’Q tan α = (𝑂^′ 𝑄)/𝐴𝑂′ tan α = 𝑂𝐸/(𝑂𝐴 − 𝑂^′ 𝑂) tan α = 𝑥/(ℎ − 𝑂^′ 𝑂) In ∆𝑨𝑶𝑪 tan α = 𝑂𝐶/𝑂𝐴 tan α = 𝑟/ℎ 𝑂^′ 𝑂= ℎ(𝑟 − 𝑥)/𝑟 Now, Curved Surface Area Of cylinder = 2𝜋 × Radius × Height of S = 2𝜋" × " 𝑥" × " 𝑂^′ 𝑂 S = 2𝜋𝑥 ℎ(𝑟 − 𝑥)/𝑟 S = 2𝜋ℎ/𝑟 (𝑟𝑥−𝑥^2 ) S = 𝑘(𝑟𝑥−𝑥^2 ) We need to minimize S, So, finding S’(x) S’ = 𝑑(𝑘(𝑟𝑥 − 𝑥^2 ))/𝑑𝑥 (Where k = 2𝜋ℎ/𝑟 is a constant) S’ = 𝑘 𝑑(𝑟𝑥 − 𝑥^2 )/𝑑𝑥 S’ = 𝑘(𝑟−2𝑥) Putting S’ = 0 0 = 𝑘(𝑟−2𝑥) 𝑟−2𝑥 = 0 𝑥 = 𝑟/2 Now, Finding S’’(x) at x = 𝑟/2 S’’ = 𝑑(𝑘(𝑟 − 2𝑥))/𝑑𝑥 S’’ = 𝑘 𝑑(𝑟 − 2𝑥)/𝑑𝑥 S’’ = 𝑘 (0−2) S’’ = −2𝑘 So, S’’ = −2𝑘 Hence, at 𝑥=𝑟/2 ├ 〖𝑆′′〗_(𝑥 = 𝑟/2) ┤|<0 ∴ 𝒙=𝒓/𝟐 is maxima of S. Hence, radius of cylinder with greatest curved surface area which can be inscribed in a given cone is half of that cone.

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Chapter 6 Class 12 Application of Derivatives

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.