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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 6.3, 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 & θ be the semi vertical angle of the cone. Now, Height of cone = h = 𝑙 cos θ Radius of cone = r = 𝑙 sin θ We need to maximize volume of cone V = 1/3 𝜋𝑟^2 ℎ V = 1/3 𝜋𝑙^2sin2𝜃 l cos 𝜃 V= 1/3 𝜋𝑙^3sin2𝜃 cos 𝜃 Differentiating 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑙^3 (2 sin 𝜃 cos 𝜃 . cos 𝜃 + sin2 𝜃 (sin 𝜃)) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑙^3 [2 sin⁡𝜃 cos^2⁡𝜃−sin^3⁡𝜃 ] 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑙^3 (2 sin 𝜃 cos2 𝜃 sin3 𝜃) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑙^3 sin 𝜃 (2 cos2 𝜃 − sin2 𝜃) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑙^3 sin 𝜃 (√2 cos⁡〖𝜃+sin⁡〖𝜃)(〗 〗 √2 cos 𝜃 − sin 𝜃) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑙^3 sin 𝜃 cos 𝜃 ((√2 cos⁡〖𝜃+sin⁡𝜃 〗))/cos⁡𝜃 × cos 𝜃 ((√2 cos⁡〖𝜃−sin⁡𝜃 〗))/cos⁡𝜃 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑙^3 sin 𝜃 cos2 𝜃 (√2⁡〖+〖tan 𝜃〗⁡〖 )(〗 〗 √2 − tan 𝜃) Putting 𝑑𝑣/𝑑𝜃 = 0 1/3 𝜋𝑙^3 sin 𝜃 cos2 𝜃 (√2⁡〖+〖tan 𝜃〗⁡〖 )(〗 〗 √2 − tan 𝜃) = 0 sin 𝜃 cos2 𝜃 (√2⁡〖+〖tan 𝜃〗⁡〖 )(〗 〗 √2 − tan 𝜃) = 0 sin 𝜃 cos2 𝜃 (√2⁡〖+〖tan 𝜃〗⁡〖 )(〗 〗 √2 − tan 𝜃) = 0 sin 𝜽 = 0 𝜃 = 0° 𝜃 cannot be 0° for cone. cos2 𝜽 = 0 𝜃 = 90° 𝜃 cannot be 90° for cone. √𝟐 + tan 𝜽 = 0 tan 𝜃 = −√2 For cone, 0° < 𝜃 < 90° tan 𝜃 is (−) ve in II & IV quadrant so tan 𝜃 = − √2 is not possible √𝟐 − tan 𝜽 = 0 tan 𝜃 = √2 𝜃 = tan−1 √2 ∴ tan 𝜃 = √2 is the possible value for cone. sin 𝜽 0° < 𝜃 < 90° Since 𝜃 is in 1st quadrant sin 𝜃 > 0 cos 𝜽 0° < 𝜃 < 90° Since 𝜃 is in 1st quadrant ∴ cos 𝜃 > 0 (√𝟐+𝒕𝒂𝒏 𝜽) 0° < θ < 90° Since θ is in 1st quadrant So, tan θ > 0 So (√2+𝑡𝑎𝑛 θ) is also (+)ve. Since 𝑑𝑣/𝑑θ = changes sign from (+) ve to (−) ve 𝜽 = tan− 1√𝟐 is the maxima. Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.