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









Last updated at April 15, 2021 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Ex 6.5, 26 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is tan โ1 (1/3) Let ๐ , 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 = ฯ๐^2+๐๐๐ S = ฯ๐^2+๐๐๐ S โ ฯ๐^2=๐๐๐ (๐ โ ๐๐^2)/(\ ๐๐)=๐ ๐ = (๐ โ ๐๐^2)/(\ ๐๐) We need to find minimize volume of a cone & show that semi vertical angle is sin (โ1)/3 i.e. ฮธ =๐ ๐๐ (โ1)/3 sin ฮธ =1/3 We know that sin ฮธ =๐/๐ Volume of a cone = 1/3 ๐๐^2 โ V = 1/3 ๐๐^2 โ(๐^2โ๐^2 ) V = 1/3 ๐๐^2 โ(((๐ โ ๐๐^2)/๐๐)^2โ๐^2 ) We need to find minimize volume of a cone & show that semi vertical angle is sin (โ1)/3 i.e. ฮธ =๐ ๐๐ (โ1)/3 sin ฮธ =1/3 We know that sin ฮธ =๐/๐ Volume of a cone = 1/3 ๐๐^2 โ V = 1/3 ๐๐^2 โ(๐^2โ๐^2 ) V = 1/3 ๐๐^2 โ(((๐ โ ๐๐^2)/๐๐)^2โ๐^2 ) We need to find minimize volume of a cone & show that semi vertical angle is sin (โ1)/3 i.e. ฮธ =๐ ๐๐ (โ1)/3 sin ฮธ =1/3 We know that sin ฮธ =๐/๐ Volume of a cone = 1/3 ๐๐^2 โ V = 1/3 ๐๐^2 โ(๐^2โ๐^2 ) V = 1/3 ๐๐^2 โ(((๐ โ ๐๐^2)/๐๐)^2โ๐^2 ) V = 1/3 ๐๐^2 โ((๐ โ ๐๐^2 )^2/(๐^2 ๐^2 )โ๐^2 ) V = 1/3 ๐๐^2 โ(((๐ โ ๐๐^2 )^2 โ ๐๐^2 (๐^2 ))/(๐^2 ๐^2 )) V = 1/3 ๐๐^2 โ(((๐ โ ๐๐^2 )^2 โ ๐^2 ๐^4)/(๐^2 ๐^2 ๐)) V = (๐๐^2)/3๐๐ โ((๐ โ๐๐^2 )^2โ๐^2 ๐^4 ) V = ((๐))/3 โ(ใ(๐ )^2+(๐๐^2 )ใ^2โ2๐ ใ๐๐ใ^2โ๐^2 ๐^4 ) V = ๐/3 โ(๐ ^2+๐^2 ๐^4โ2๐๐๐^2โ๐^2 ๐^4 ) V = ๐/3 โ(๐ ^2โ2 ๐๐๐^2 ) V = 1/3 โ(๐^2 (๐ ^2โ2 ๐ ๐๐^2 ) ) V = 1/3 โ(๐^2 ๐ ^2โ2 ๐ ๐๐^4 ) Since V has square root It will be difficult to differentiate So, we take Z = V2 Z = 1/3 (๐^2 ๐ ^2โ2 ๐ ๐๐^4 ) Since V is positive, Z is maximum if V2 is maximum So, we maximize Z = V2 Diff. Z w.r.t ๐ ๐Z/๐๐=๐(1/3 (๐^2 ๐ ^2 โ 2๐ ๐๐^4 ))/๐๐ ๐Z/๐๐=1/3 [๐ ^2 (2๐)โ2๐ ๐ (4๐^3 )] ๐Z/๐๐=1/3 [2๐๐ ^2โ8๐ ๐๐^3 ] Putting ๐ ๐/๐ ๐ = 0 1/3 [2๐๐ ^2โ8๐ ๐๐^3 ]=0 2๐๐ ^2โ8๐ ๐๐^3=0 2๐๐ ^2=8๐ ๐๐^3 (2๐ ^2)/(4๐ ๐ )=๐^3/๐ ๐ /(4๐ )=๐^2 ๐ =4๐๐^2 Finding (๐ ^๐ ๐)/(๐๐^๐ ) ๐Z/๐๐=1/3 [2๐๐ ^2โ8๐ ๐๐^3 ] Diff w.r.t ๐ฅ (๐^2 Z)/(๐๐^2 ) = ๐/๐๐ [1/3 [2๐๐ ^2โ8๐ ๐๐^3 ] " " ] (๐^2 Z)/(๐r^2 ) = 1/3 [2๐ ^2โ8๐ ๐(3๐^2) ] (๐^2 Z)/(๐r^2 ) = 1/3 [2๐ ^2โ24๐ ๐๐^2 ] Putting ๐ =4๐๐^2 (๐^2 Z)/(๐r^2 ) = 1/3 [2ใ(4๐๐^2)ใ^2โ24(4๐๐^2)๐๐^2 ] (๐^2 Z)/(๐r^2 ) = 1/3 [32๐^2 ๐^4โ96๐^2 ๐^4 ] (๐^2 Z)/(๐r^2 ) = 1/3 [โ64๐^2 ๐^4 ] Since (๐^2 Z)/(๐๐^2 ) < 0 for ๐ =4๐๐^2 Volume is maximum for ๐ =4๐๐^2 Now, Surface area of cone = ๐๐^2+๐๐๐ ๐=๐๐^2+๐๐๐ Putting S = 4๐๐^2 4๐๐^2=๐๐^2+๐๐๐ ๐๐^2+๐๐๐=4๐๐^2 Dividing both sides by ๐๐ (๐๐^2+ ๐๐๐)/๐๐=(4๐๐^2)/๐๐ ๐+๐=4๐ ๐=4๐โ๐ ๐=3๐ ๐/๐=3 ๐/๐=1/3 But we know that sin ฮธ =๐/๐ Putting value of ๐/๐ sin ฮธ =1/3 ฮธ =ใ๐๐๐ใ^(โ๐)โก ๐/๐ Hence proved
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