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Ex 6.5, 25 - Show that semi-vertical angle of cone of max volume

Ex 6.5,25 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,25 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,25 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.5,25 - Chapter 6 Class 12 Application of Derivatives - Part 5

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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 √2Let 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 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.