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

Transcript

Ex 6.5, 26 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is tan โ€“1 (1/3) Let ๐‘Ÿ , h & l be the radius, height & slant height of a cone respectively And Let V & S be the volume & surface area & ฮธ be a semi vertical angle of a cone Given surface Area of a cone is constant Surface Area of a cone = ฯ€๐‘Ÿ^2+๐œ‹๐‘Ÿ๐‘™ S = ฯ€๐‘Ÿ^2+๐œ‹๐‘Ÿ๐‘™ S โ€“ ฯ€๐‘Ÿ^2=๐œ‹๐‘Ÿ๐‘™ (๐‘† โˆ’ ๐œ‹๐‘Ÿ^2)/(\ ๐œ‹๐‘Ÿ)=๐‘™ ๐‘™ = (๐‘† โˆ’ ๐œ‹๐‘Ÿ^2)/(\ ๐œ‹๐‘Ÿ) We need to find minimize volume of a cone & show that semi vertical angle is sin (โˆ’1)/3 i.e. ฮธ =๐‘ ๐‘–๐‘› (โˆ’1)/3 sin ฮธ =1/3 We know that sin ฮธ =๐‘Ÿ/๐‘™ Volume of a cone = 1/3 ๐œ‹๐‘Ÿ^2 โ„Ž V = 1/3 ๐œ‹๐‘Ÿ^2 โˆš(๐‘™^2โˆ’๐‘Ÿ^2 ) V = 1/3 ๐œ‹๐‘Ÿ^2 โˆš(((๐‘  โˆ’ ๐œ‹๐‘Ÿ^2)/๐œ‹๐‘Ÿ)^2โˆ’๐‘Ÿ^2 ) We need to find minimize volume of a cone & show that semi vertical angle is sin (โˆ’1)/3 i.e. ฮธ =๐‘ ๐‘–๐‘› (โˆ’1)/3 sin ฮธ =1/3 We know that sin ฮธ =๐‘Ÿ/๐‘™ Volume of a cone = 1/3 ๐œ‹๐‘Ÿ^2 โ„Ž V = 1/3 ๐œ‹๐‘Ÿ^2 โˆš(๐‘™^2โˆ’๐‘Ÿ^2 ) V = 1/3 ๐œ‹๐‘Ÿ^2 โˆš(((๐‘  โˆ’ ๐œ‹๐‘Ÿ^2)/๐œ‹๐‘Ÿ)^2โˆ’๐‘Ÿ^2 ) V = 1/3 ๐œ‹๐‘Ÿ^2 โˆš((๐‘  โˆ’ ๐œ‹๐‘Ÿ^2 )^2/(๐œ‹^2 ๐‘Ÿ^2 )โˆ’๐‘Ÿ^2 ) V = 1/3 ๐œ‹๐‘Ÿ^2 โˆš(((๐‘  โˆ’ ๐œ‹๐‘Ÿ^2 )^2 โˆ’ ๐œ‹๐‘Ÿ^2 (๐‘Ÿ^2 ))/(๐œ‹^2 ๐‘Ÿ^2 )) V = 1/3 ๐œ‹๐‘Ÿ^2 โˆš(((๐‘  โˆ’ ๐œ‹๐‘Ÿ^2 )^2 โˆ’ ๐œ‹^2 ๐‘Ÿ^4)/(๐œ‹^2 ๐‘Ÿ^2 ๐‘Ÿ)) V = (๐œ‹๐‘Ÿ^2)/3๐œ‹๐‘Ÿ โˆš((๐‘ โˆ’๐œ‹๐‘Ÿ^2 )^2โˆ’๐œ‹^2 ๐‘Ÿ^4 ) V = ((๐‘Ÿ))/3 โˆš(ใ€–(๐‘ )^2+(๐œ‹๐‘Ÿ^2 )ใ€—^2โˆ’2๐‘† ใ€–๐œ‹๐‘Ÿใ€—^2โˆ’๐œ‹^2 ๐‘Ÿ^4 ) V = ๐‘Ÿ/3 โˆš(๐‘ ^2+๐œ‹^2 ๐‘Ÿ^4โˆ’2๐‘†๐œ‹๐‘Ÿ^2โˆ’๐œ‹^2 ๐‘Ÿ^4 ) (From (1)) V = ๐‘Ÿ/3 โˆš(๐‘ ^2โˆ’2 ๐‘†๐œ‹๐‘Ÿ^2 ) V = 1/3 โˆš(๐‘Ÿ^2 (๐‘ ^2โˆ’2 ๐‘  ๐œ‹๐‘Ÿ^2 ) ) V = 1/3 โˆš(๐‘Ÿ^2 ๐‘ ^2โˆ’2 ๐‘ ๐œ‹๐‘Ÿ^4 ) Since V has square root It will be difficult to differentiate So, we take Z = V2 Z = 1/3 (๐‘Ÿ^2 ๐‘ ^2โˆ’2 ๐‘ ๐œ‹๐‘Ÿ^4 ) Since V is positive, Z is maximum if V2 is maximum So, we maximize Z = V2 Diff. Z w.r.t ๐‘Ÿ ๐‘‘Z/๐‘‘๐‘Ÿ=๐‘‘(1/3 (๐‘Ÿ^2 ๐‘ ^2 โˆ’ 2๐‘ ๐œ‹๐‘Ÿ^4 ))/๐‘‘๐‘Ÿ ๐‘‘Z/๐‘‘๐‘Ÿ=1/3 [๐‘ ^2 (2๐‘Ÿ)โˆ’2๐‘ ๐œ‹ (4๐‘Ÿ^3 )] ๐‘‘Z/๐‘‘๐‘Ÿ=1/3 [2๐‘Ÿ๐‘ ^2โˆ’8๐‘ ๐œ‹๐‘Ÿ^3 ] Putting ๐’…๐’/๐’…๐’“ = 0 1/3 [2๐‘Ÿ๐‘ ^2โˆ’8๐‘ ๐œ‹๐‘Ÿ^3 ]=0 2๐‘Ÿ๐‘ ^2โˆ’8๐‘ ๐œ‹๐‘Ÿ^3=0 2๐‘Ÿ๐‘ ^2=8๐‘ ๐œ‹๐‘Ÿ^3 (2๐‘ ^2)/(4๐‘ ๐œ‹ )=๐‘Ÿ^3/๐‘Ÿ ๐‘ /(4๐œ‹ )=๐‘Ÿ^2 ๐‘ =4๐œ‹๐‘Ÿ^2 Finding (๐’…^๐Ÿ ๐™)/(๐๐’“^๐Ÿ ) ๐‘‘Z/๐‘‘๐‘Ÿ=1/3 [2๐‘Ÿ๐‘ ^2โˆ’8๐‘ ๐œ‹๐‘Ÿ^3 ] Diff w.r.t ๐‘ฅ (๐‘‘^2 Z)/(๐‘‘๐‘Ÿ^2 ) = ๐‘‘/๐‘‘๐‘Ÿ [1/3 [2๐‘Ÿ๐‘ ^2โˆ’8๐‘ ๐œ‹๐‘Ÿ^3 ] " " ] (๐‘‘^2 Z)/(๐‘‘r^2 ) = 1/3 [2๐‘ ^2โˆ’8๐‘ ๐œ‹(3๐‘Ÿ^2) ] (๐‘‘^2 Z)/(๐‘‘r^2 ) = 1/3 [2๐‘ ^2โˆ’24๐‘ ๐œ‹๐‘Ÿ^2 ] Putting ๐‘ =4๐œ‹๐‘Ÿ^2 (๐‘‘^2 Z)/(๐‘‘r^2 ) = 1/3 [2ใ€–(4๐œ‹๐‘Ÿ^2)ใ€—^2โˆ’24(4๐œ‹๐‘Ÿ^2)๐œ‹๐‘Ÿ^2 ] (๐‘‘^2 Z)/(๐‘‘r^2 ) = 1/3 [32๐œ‹^2 ๐‘Ÿ^4โˆ’96๐œ‹^2 ๐‘Ÿ^4 ] (๐‘‘^2 Z)/(๐‘‘r^2 ) = 1/3 [โˆ’64๐œ‹^2 ๐‘Ÿ^4 ] Since (๐‘‘^2 Z)/(๐‘‘๐‘Ÿ^2 ) < 0 for ๐‘ =4๐œ‹๐‘Ÿ^2 Volume is maximum for ๐‘ =4๐œ‹๐‘Ÿ^2 Now, Surface area of cone = ๐œ‹๐‘Ÿ^2+๐œ‹๐‘Ÿ๐‘™ ๐‘†=๐œ‹๐‘Ÿ^2+๐œ‹๐‘Ÿ๐‘™ Putting S = 4๐œ‹๐‘Ÿ^2 4๐œ‹๐‘Ÿ^2=๐œ‹๐‘Ÿ^2+๐œ‹๐‘Ÿ๐‘™ ๐œ‹๐‘Ÿ^2+๐œ‹๐‘Ÿ๐‘™=4๐œ‹๐‘Ÿ^2 Dividing both sides by ๐œ‹๐‘Ÿ (๐œ‹๐‘Ÿ^2+ ๐œ‹๐‘Ÿ๐‘™)/๐œ‹๐‘Ÿ=(4๐œ‹๐‘Ÿ^2)/๐œ‹๐‘Ÿ ๐‘Ÿ+๐‘™=4๐‘Ÿ ๐‘™=4๐‘Ÿโˆ’๐‘Ÿ ๐‘™=3๐‘Ÿ ๐‘™/๐‘Ÿ=3 ๐‘Ÿ/๐‘™=1/3 But we know that sin ฮธ =๐‘Ÿ/๐‘™ Putting value of ๐‘Ÿ/๐‘™ sin ฮธ =1/3 ฮธ =ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โก ๐Ÿ/๐Ÿ‘ 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.