Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives

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 base Let 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(𝑟𝑎𝑑𝑖𝑢𝑠)^2 (ℎ𝑒𝑖𝑔ℎ𝑡) V = 1/3 𝜋(𝑟)^2 ℎ 3𝑉/𝜋=𝑟^2 ℎ h = (3𝑉/𝜋) 1/𝑟^2 ℎ= 𝑘/𝑟^2 We need to show Curved surface Area is least & Altitude = √2 times of radius of base i.e. ℎ=√2 𝑟 Now, Curved Surface Area of Cone = π 𝑟𝑙 Putting 𝑙 = √(ℎ^2+𝑟^2 ) S = π 𝑟 √(ℎ^2+𝑟^2 ) S = π 𝑟 √((𝑘/𝑟^2 )^2+𝑟^2 ) S = π 𝑟 √(𝑘^2/𝑟^4 +𝑟^2 ) S = π 𝑟 √((𝑘^2 + 𝑟^6)/𝑟^4 ) S = π 𝑟/𝑟^2 √(𝑘^2+𝑟^6 ) S = 𝜋/𝑟 √(𝑘^2+𝑟^6 ) S = π [(√(𝑘^2 + 𝑟^6 ) " " )/𝑟] Diff w.r.t r 𝑑𝑠/𝑑𝑟=𝜋 𝑑/𝑑𝑟 [(√(𝑘^2 + 𝑟^6 ) " " )/𝑟] =𝜋[((√(𝑘^2 + 𝑟^6 ))^′ (𝑟) − (𝑟)^′ (√(𝑘^2 + 𝑟^6 )))/𝑟^2 ] =𝜋[(𝑟/(2√(𝑘^2 + 𝑟^6 )) × 𝑑(𝑘^2 + 𝑟^6 )/( 𝑑𝑟) . 𝑟 −1 .√(𝑘^2 + 𝑟^6 ))/𝑟^2 ] =𝜋[(𝑟/(2√(𝑘^2 + 𝑟^6 )) . (0 + 6𝑟^5 ) − √(𝑘^2 + 𝑟^6 ))/𝑟^2 ] =𝜋[((6𝑟^6)/(2√(𝑘^2 + 𝑟^6 )) − √(𝑘^2 + 𝑟^6 ))/𝑟^2 ] =𝜋[(6𝑟^6 − 2(√(𝑘^2 + 𝑟^6 ))^2)/(2√(𝑘^2 + 𝑟^6 ) × 𝑟^2 )] =𝜋[(6𝑟^6− 2(𝑘^2 + 𝑟^6 ))/(2𝑟^2 √(𝑘^2 + 𝑟^6 ))] =𝜋[(6𝑟^6− 2𝑘^2− 2𝑟^6)/(2𝑟^2 √(𝑘^2+〖 𝑟〗^6 ))] =𝜋[(4𝑟^6− 2𝑘^2)/(2𝑟^2 √(𝑘^2 + 𝑟^6 ))] =𝜋[2(2𝑟^6−𝑘^2 )/(2𝑟^2 √(𝑘^2+ 𝑟^2 ) )] =𝜋[((2𝑟^6−𝑘^2 ))/(𝑟^2 √(𝑘^2+ 𝑟^2 ) )] Putting 𝑑𝑠/𝑑𝑟=0 𝜋[(2𝑟^6− 𝑘^2)/(𝑟^2 √(𝑘^2+ 𝑟^2 ))]=0 2𝑟^6−𝑘^2=0 ×𝜋 ×𝑟^2 √(𝑘^2+𝑟^2 ) 2𝑟^6−𝑘^2=0 𝑟^6=𝑘^2/2 Now, Finding (𝑑^2 𝑠)/(𝑑𝑟^2 ) 𝑑𝑠/𝑑𝑟=𝜋[((2𝑟^6− 𝑘^2 ))/(𝑟^2 √(𝑘^2+ 𝑟^6 ))] 𝑑𝑠/𝑑𝑟=𝜋[(2𝑟^6− 𝑘^2)/√(𝑘^2 𝑟^4 + 𝑟^10 )] Diff Again w.r.t 𝑟 (𝑑^2 𝑠)/(𝑑𝑟^2 )=𝜋[((𝑑(2𝑟^6− 𝑘^2 )/𝑑𝑟 .√(𝑘^2 𝑟^4 + 𝑟^10 )) − ((𝑑√(𝑘^2 𝑟^4 + 𝑟^10 ))/( 𝑑𝑟) . (2𝑟^6− 𝑘^2 )) )/(√(𝑘^2 𝑟^4 + 𝑟^10 ))^2 ] =𝜋[((12𝑟^5+ 0) √(𝑘^2 𝑟^4 + 𝑟^10 ) − 1/(2√(𝑘^2 𝑟^4+𝑟^10 )) 𝑑(𝑘^2 𝑟^4+𝑟^10 )/𝑑𝑟 (2𝑟^6 − 𝑘^2 ))/(√(𝑘^2 𝑟^4 + 𝑟^10 ))^2 ] =𝜋[(12𝑟^5 √(𝑘^2 𝑟^4 + 𝑟^10 ) − 1/(2√(𝑘^2 𝑟^4 + 𝑟^10 )) (4𝑘^2 𝑟^3 + 10𝑟^9 )(2𝑟^6 − 𝑘^2 ))/((𝑘^2 𝑟^4 + 𝑟^10 ) )] =𝜋[(12𝑟^5 √(𝑘^2 𝑟^4 + 𝑟^10 ) × √(𝑘^2 𝑟^4 + 𝑟^10 )−(4𝑘^2 𝑟^3 + 10𝑟^9 )(2𝑟^6 − 𝑘^2 ))/(2√(𝑘^2 𝑟^4 + 𝑟^10 ) (𝑘^2 𝑟^4 + 𝑟^10 ) )] =𝜋[(12𝑟^5 (𝑘^2 𝑟^4 + 𝑟^10 )−(4𝑘^2 𝑟^3 + 10𝑟^9 )(2𝑟^6 − 𝑘^2 ))/(2√(𝑘^2 𝑟^4 + 𝑟^10 ) (𝑘^2 𝑟^4 + 𝑟^10 ) )] Putting 𝑘^2=2𝑟^6 =𝜋[(12𝑟^5 ((2𝑟^6 ) 𝑟^4 + 𝑟^10 )−(4(2𝑟^6 ) 𝑟^3 + 10𝑟^9 )(2𝑟^6 − (2𝑟^6 )))/(2√((2𝑟^6)𝑟^4 + 𝑟^10 ) ((2𝑟^6)𝑟^4 + 𝑟^10 ) )] =𝜋[(12𝑟^5 ((2𝑟^6 ) 𝑟^4 + 𝑟^10 )−(4(2𝑟^6 ) 𝑟^3 + 10𝑟^9 )(0))/(2√((2𝑟^6)𝑟^4 + 𝑟^10 ) ((2𝑟^6)𝑟^4 + 𝑟^10 ) )] =𝜋[(12𝑟^5 (2𝑟^10+ 𝑟^10 ) − 0)/(2√(2𝑟^10+ 𝑟^10 ) (2𝑟^10+ 𝑟^10 ) )] =𝜋[(12𝑟^5 (2𝑟^10+ 𝑟^10 ) )/(2√(2𝑟^10+ 𝑟^10 ) (2𝑟^10+ 𝑟^10 ) )] = 𝜋[(12𝑟^5)/(2√(3𝑟^10 )) ] = 𝜋[(12𝑟^5)/(2𝑟^5 √3) ] = 𝜋[6/√3 ] > 0 Therefore, (𝑑^2 𝑠)/(𝑑𝑠^2 )>0 when 𝑘^2=2𝑟^6 S is minimum when 〖 𝑘〗^2=2𝑟^6 Now, ℎ=𝑘/𝑟^2 ℎ𝑟^2=𝑘 (ℎ𝑟^2 )^2=𝑘^2 ℎ^2 𝑟^4=𝑘^2 Putting value of 𝑘^2=2𝑟^6 ℎ^2 𝑟^4=2𝑟^6 ℎ^2=(2𝑟^6)/𝑟^4 ℎ^2=2𝑟^2 ℎ=√(2𝑟^2 ) 𝒉=𝒓√𝟐 Hence altitude equal to √2 times the radius of the base Hence proved