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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 –1 (1/3) Let π‘Ÿ , h & l be the radius, height & slant height of a cone respectively And Let V & S be the volume & surface area & ΞΈ be a semi vertical angle of a cone Given surface Area of a cone is constant Surface Area of a cone = Ο€π‘Ÿ^2+πœ‹π‘Ÿπ‘™ S = Ο€π‘Ÿ^2+πœ‹π‘Ÿπ‘™ S – Ο€π‘Ÿ^2=πœ‹π‘Ÿπ‘™ (𝑆 βˆ’ πœ‹π‘Ÿ^2)/(\ πœ‹π‘Ÿ)=𝑙 𝑙 = (𝑆 βˆ’ πœ‹π‘Ÿ^2)/(\ πœ‹π‘Ÿ) We need to find minimize volume of a cone & show that semi vertical angle is sin (βˆ’1)/3 i.e. ΞΈ =𝑠𝑖𝑛 (βˆ’1)/3 sin ΞΈ =1/3 We know that sin ΞΈ =π‘Ÿ/𝑙 Volume of a cone = 1/3 πœ‹π‘Ÿ^2 β„Ž V = 1/3 πœ‹π‘Ÿ^2 √(𝑙^2βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2)/πœ‹π‘Ÿ)^2βˆ’π‘Ÿ^2 ) We need to find minimize volume of a cone & show that semi vertical angle is sin (βˆ’1)/3 i.e. ΞΈ =𝑠𝑖𝑛 (βˆ’1)/3 sin ΞΈ =1/3 We know that sin ΞΈ =π‘Ÿ/𝑙 Volume of a cone = 1/3 πœ‹π‘Ÿ^2 β„Ž V = 1/3 πœ‹π‘Ÿ^2 √(𝑙^2βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2)/πœ‹π‘Ÿ)^2βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √((𝑠 βˆ’ πœ‹π‘Ÿ^2 )^2/(πœ‹^2 π‘Ÿ^2 )βˆ’π‘Ÿ^2 ) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2 )^2 βˆ’ πœ‹π‘Ÿ^2 (π‘Ÿ^2 ))/(πœ‹^2 π‘Ÿ^2 )) V = 1/3 πœ‹π‘Ÿ^2 √(((𝑠 βˆ’ πœ‹π‘Ÿ^2 )^2 βˆ’ πœ‹^2 π‘Ÿ^4)/(πœ‹^2 π‘Ÿ^2 π‘Ÿ)) V = (πœ‹π‘Ÿ^2)/3πœ‹π‘Ÿ √((π‘ βˆ’πœ‹π‘Ÿ^2 )^2βˆ’πœ‹^2 π‘Ÿ^4 ) V = ((π‘Ÿ))/3 √(γ€–(𝑠)^2+(πœ‹π‘Ÿ^2 )γ€—^2βˆ’2𝑆 γ€–πœ‹π‘Ÿγ€—^2βˆ’πœ‹^2 π‘Ÿ^4 ) V = π‘Ÿ/3 √(𝑠^2+πœ‹^2 π‘Ÿ^4βˆ’2π‘†πœ‹π‘Ÿ^2βˆ’πœ‹^2 π‘Ÿ^4 ) (From (1)) V = π‘Ÿ/3 √(𝑠^2βˆ’2 π‘†πœ‹π‘Ÿ^2 ) V = 1/3 √(π‘Ÿ^2 (𝑠^2βˆ’2 𝑠 πœ‹π‘Ÿ^2 ) ) V = 1/3 √(π‘Ÿ^2 𝑠^2βˆ’2 π‘ πœ‹π‘Ÿ^4 ) Since V has square root It will be difficult to differentiate So, we take Z = V2 Z = 1/3 (π‘Ÿ^2 𝑠^2βˆ’2 π‘ πœ‹π‘Ÿ^4 ) Since V is positive, Z is maximum if V2 is maximum So, we maximize Z = V2 Diff. Z w.r.t π‘Ÿ 𝑑Z/π‘‘π‘Ÿ=𝑑(1/3 (π‘Ÿ^2 𝑠^2 βˆ’ 2π‘ πœ‹π‘Ÿ^4 ))/π‘‘π‘Ÿ 𝑑Z/π‘‘π‘Ÿ=1/3 [𝑠^2 (2π‘Ÿ)βˆ’2π‘ πœ‹ (4π‘Ÿ^3 )] 𝑑Z/π‘‘π‘Ÿ=1/3 [2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3 ] Putting 𝒅𝒁/𝒅𝒓 = 0 1/3 [2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3 ]=0 2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3=0 2π‘Ÿπ‘ ^2=8π‘ πœ‹π‘Ÿ^3 (2𝑠^2)/(4π‘ πœ‹ )=π‘Ÿ^3/π‘Ÿ 𝑠/(4πœ‹ )=π‘Ÿ^2 𝑠=4πœ‹π‘Ÿ^2 Finding (𝒅^𝟐 𝐙)/(𝐝𝒓^𝟐 ) 𝑑Z/π‘‘π‘Ÿ=1/3 [2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3 ] Diff w.r.t π‘₯ (𝑑^2 Z)/(π‘‘π‘Ÿ^2 ) = 𝑑/π‘‘π‘Ÿ [1/3 [2π‘Ÿπ‘ ^2βˆ’8π‘ πœ‹π‘Ÿ^3 ] " " ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2𝑠^2βˆ’8π‘ πœ‹(3π‘Ÿ^2) ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2𝑠^2βˆ’24π‘ πœ‹π‘Ÿ^2 ] Putting 𝑠=4πœ‹π‘Ÿ^2 (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [2γ€–(4πœ‹π‘Ÿ^2)γ€—^2βˆ’24(4πœ‹π‘Ÿ^2)πœ‹π‘Ÿ^2 ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [32πœ‹^2 π‘Ÿ^4βˆ’96πœ‹^2 π‘Ÿ^4 ] (𝑑^2 Z)/(𝑑r^2 ) = 1/3 [βˆ’64πœ‹^2 π‘Ÿ^4 ] Since (𝑑^2 Z)/(π‘‘π‘Ÿ^2 ) < 0 for 𝑠=4πœ‹π‘Ÿ^2 Volume is maximum for 𝑠=4πœ‹π‘Ÿ^2 Now, Surface area of cone = πœ‹π‘Ÿ^2+πœ‹π‘Ÿπ‘™ 𝑆=πœ‹π‘Ÿ^2+πœ‹π‘Ÿπ‘™ Putting S = 4πœ‹π‘Ÿ^2 4πœ‹π‘Ÿ^2=πœ‹π‘Ÿ^2+πœ‹π‘Ÿπ‘™ πœ‹π‘Ÿ^2+πœ‹π‘Ÿπ‘™=4πœ‹π‘Ÿ^2 Dividing both sides by πœ‹π‘Ÿ (πœ‹π‘Ÿ^2+ πœ‹π‘Ÿπ‘™)/πœ‹π‘Ÿ=(4πœ‹π‘Ÿ^2)/πœ‹π‘Ÿ π‘Ÿ+𝑙=4π‘Ÿ 𝑙=4π‘Ÿβˆ’π‘Ÿ 𝑙=3π‘Ÿ 𝑙/π‘Ÿ=3 π‘Ÿ/𝑙=1/3 But we know that sin ΞΈ =π‘Ÿ/𝑙 Putting value of π‘Ÿ/𝑙 sin ΞΈ =1/3 ΞΈ =γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑ 𝟏/π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.