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Ex 6.5, 24 - Show that cone of least curved surface, given volume

Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives - Part 5
Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives - Part 6
Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives - Part 7
Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives - Part 8


Transcript

Ex 6.5, 24 Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the baseLet r & h be the radius & height of a cone respectively And V & S be the volume & curved surface area of cone respectively Given volume of cone is constant Volume of cone = 1/3 πœ‹(π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘ )^2 (β„Žπ‘’π‘–π‘”β„Žπ‘‘) V = 1/3 πœ‹(π‘Ÿ)^2 β„Ž 3𝑉/πœ‹=π‘Ÿ^2 β„Ž h = (3𝑉/πœ‹) 1/π‘Ÿ^2 β„Ž= π‘˜/π‘Ÿ^2 We need to show Curved surface Area is least & Altitude = √2 times of radius of base i.e. β„Ž=√2 π‘Ÿ Now, Curved Surface Area of Cone = Ο€ π‘Ÿπ‘™ Putting 𝑙 = √(β„Ž^2+π‘Ÿ^2 ) S = Ο€ π‘Ÿ √(β„Ž^2+π‘Ÿ^2 ) S = Ο€ π‘Ÿ √((π‘˜/π‘Ÿ^2 )^2+π‘Ÿ^2 ) Since V is constant β‡’ (3𝑉/πœ‹) is also constant Let k = 3𝑉/πœ‹ S = Ο€ π‘Ÿ √(π‘˜^2/π‘Ÿ^4 +π‘Ÿ^2 ) S = Ο€ π‘Ÿ √((π‘˜^2 + π‘Ÿ^6)/π‘Ÿ^4 ) S = Ο€ π‘Ÿ/π‘Ÿ^2 √(π‘˜^2+π‘Ÿ^6 ) S = πœ‹/π‘Ÿ √(π‘˜^2+π‘Ÿ^6 ) S = Ο€ [(√(π‘˜^2 + π‘Ÿ^6 ) " " )/π‘Ÿ] Since S has square root It will be difficult to differentiate So, we take Z = S2 Z = πœ‹^2 [(π‘˜^2 + π‘Ÿ^6 " " )/π‘Ÿ^2 ] Z = πœ‹^2 [(π‘˜^2 " " )/π‘Ÿ^2 +π‘Ÿ^4 ] Z = πœ‹^2 [π‘˜^2 π‘Ÿ^(βˆ’2)+π‘Ÿ^4 ] Since S is positive, S is minimum if A2 is minium So, we minimize Z = A2 Diff. Z w.r.t π‘Ÿ 𝑑Z/π‘‘π‘Ÿ=𝑑(πœ‹^2 [π‘˜^2 π‘Ÿ^(βˆ’2)+π‘Ÿ^4 ])/π‘‘π‘Ÿ 𝑑Z/π‘‘π‘Ÿ=πœ‹^2 [π‘˜^2 (βˆ’2π‘Ÿ^(βˆ’2βˆ’1))+4π‘Ÿ^3 ] 𝑑Z/π‘‘π‘Ÿ=πœ‹^2 [βˆ’2π‘˜^2 π‘Ÿ^(βˆ’3)+4π‘Ÿ^3 ] Putting 𝒅𝒁/𝒅𝒓 = 0 πœ‹^2 [βˆ’2π‘˜^2 π‘Ÿ^(βˆ’3)+4π‘Ÿ^3 ] = 0 βˆ’2π‘˜^2 π‘Ÿ^(βˆ’3)+4π‘Ÿ^3= 0 4π‘Ÿ^3 = 2π‘˜^2 π‘Ÿ^(βˆ’3) 4π‘Ÿ^3 = (2π‘˜^2)/π‘Ÿ^3 4π‘Ÿ^3 Γ— π‘Ÿ^3 = 2π‘˜^2 4π‘Ÿ^6 = 2π‘˜^2 π‘Ÿ^6 = π‘˜^2/2 Finding (𝒅^𝟐 𝐙)/(𝐝𝒙^𝟐 ) 𝑑Z/π‘‘π‘Ÿ=πœ‹^2 [βˆ’2π‘˜^2 π‘Ÿ^(βˆ’3)+4π‘Ÿ^3 ] Diff w.r.t π‘₯ (𝑑^2 Z)/(π‘‘π‘Ÿ^2 ) = 𝑑/π‘‘π‘Ÿ [πœ‹^2 [βˆ’2π‘˜^2 π‘Ÿ^(βˆ’3)+4π‘Ÿ^3 ]] (𝑑^2 Z)/(𝑑π‘₯^2 ) = πœ‹^2 [βˆ’2π‘˜^2 (βˆ’3π‘Ÿ^(βˆ’4))+4(3π‘Ÿ^2)] (𝑑^2 Z)/(𝑑π‘₯^2 ) = πœ‹^2 [6π‘˜^2 π‘Ÿ^(βˆ’4)+12π‘Ÿ^2 ] Since everything is positive, (𝑑^2 Z)/(𝑑π‘₯^2 ) > 0 for π‘Ÿ^6 = π‘˜^2/2 Thus, Surface area is minimum when π‘Ÿ^6 = π‘˜^2/2 Now, β„Ž=π‘˜/π‘Ÿ^2 β„Žπ‘Ÿ^2=π‘˜ (β„Žπ‘Ÿ^2 )^2=π‘˜^2 β„Ž^2 π‘Ÿ^4=π‘˜^2 Putting value of π‘˜^2=2π‘Ÿ^6 β„Ž^2 π‘Ÿ^4=2π‘Ÿ^6 β„Ž^2=(2π‘Ÿ^6)/π‘Ÿ^4 β„Ž^2=2π‘Ÿ^2 β„Ž=√(2π‘Ÿ^2 ) 𝒉=π’“βˆšπŸ Hence altitude equal to √2 times the radius of the base 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 12 years. He provides courses for Maths and Science at Teachoo.