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

Transcript

Ex 6.5, 25 Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan –1 √2 Let 𝑙 be the slant height & ΞΈ be the semi vertical angle of the cone. Now, Height of cone = h = 𝑙 cos ΞΈ Radius of cone = r = 𝑙 sin ΞΈ We need to maximize volume of cone V = 1/3 πœ‹π‘Ÿ^2 β„Ž V = 1/3 πœ‹π‘™^2sin2πœƒ l cos πœƒ V= 1/3 πœ‹π‘™^3sin2πœƒ cos πœƒ Differentiating 𝑑𝑣/π‘‘πœƒ = 1/3 πœ‹π‘™^3 (2 sin πœƒ cos πœƒ . cos πœƒ + sin2 πœƒ (sin πœƒ)) 𝑑𝑣/π‘‘πœƒ = 1/3 πœ‹π‘™^3 [2 sinβ‘πœƒ cos^2β‘πœƒβˆ’sin^3β‘πœƒ ] 𝑑𝑣/π‘‘πœƒ = 1/3 πœ‹π‘™^3 (2 sin πœƒ cos2 πœƒ sin3 πœƒ) 𝑑𝑣/π‘‘πœƒ = 1/3 πœ‹π‘™^3 sin πœƒ (2 cos2 πœƒ βˆ’ sin2 πœƒ) 𝑑𝑣/π‘‘πœƒ = 1/3 πœ‹π‘™^3 sin πœƒ (√2 cosβ‘γ€–πœƒ+sinβ‘γ€–πœƒ)(γ€— γ€— √2 cos πœƒ βˆ’ sin πœƒ) 𝑑𝑣/π‘‘πœƒ = 1/3 πœ‹π‘™^3 sin πœƒ cos πœƒ ((√2 cosβ‘γ€–πœƒ+sinβ‘πœƒ γ€—))/cosβ‘πœƒ Γ— cos πœƒ ((√2 cosβ‘γ€–πœƒβˆ’sinβ‘πœƒ γ€—))/cosβ‘πœƒ 𝑑𝑣/π‘‘πœƒ = 1/3 πœ‹π‘™^3 sin πœƒ cos2 πœƒ (√2⁑〖+γ€–tan πœƒγ€—β‘γ€– )(γ€— γ€— √2 βˆ’ tan πœƒ) Putting 𝑑𝑣/π‘‘πœƒ = 0 1/3 πœ‹π‘™^3 sin πœƒ cos2 πœƒ (√2⁑〖+γ€–tan πœƒγ€—β‘γ€– )(γ€— γ€— √2 βˆ’ tan πœƒ) = 0 sin πœƒ cos2 πœƒ (√2⁑〖+γ€–tan πœƒγ€—β‘γ€– )(γ€— γ€— √2 βˆ’ tan πœƒ) = 0 sin πœƒ cos2 πœƒ (√2⁑〖+γ€–tan πœƒγ€—β‘γ€– )(γ€— γ€— √2 βˆ’ tan πœƒ) = 0 sin 𝜽 = 0 πœƒ = 0Β° πœƒ cannot be 0Β° for cone. cos2 𝜽 = 0 πœƒ = 90Β° πœƒ cannot be 90Β° for cone. √𝟐 + tan 𝜽 = 0 tan πœƒ = βˆ’βˆš2 For cone, 0Β° < πœƒ < 90Β° tan πœƒ is (βˆ’) ve in II & IV quadrant so tan πœƒ = βˆ’ √2 is not possible √𝟐 βˆ’ tan 𝜽 = 0 tan πœƒ = √2 πœƒ = tanβˆ’1 √2 ∴ tan πœƒ = √2 is the possible value for cone. The value is either maxima or minima. Finding maxima or minima by first derivative test 𝑑𝑣/π‘‘πœƒ = 1/3 πœ‹π‘™^3 sin πœƒ cos2 πœƒ (√2 + tanπœƒ) (√2 βˆ’ tan πœƒ) sin 𝜽 0Β° < πœƒ < 90Β° Since πœƒ is in 1st quadrant sin πœƒ > 0 cos 𝜽 0Β° < πœƒ < 90Β° Since πœƒ is in 1st quadrant ∴ cos πœƒ > 0 (√𝟐+𝒕𝒂𝒏 𝜽) 0Β° < ΞΈ < 90Β° Since ΞΈ is in 1st quadrant So, tan ΞΈ > 0 So (√2+π‘‘π‘Žπ‘› ΞΈ) is also (+)ve. Since 𝑑𝑣/𝑑θ = changes sign from (+) ve to (βˆ’) ve 𝜽 = tanβˆ’ 1√𝟐 is the maxima. Hence proved

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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.