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

Transcript

Misc 12 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is 〖〖(𝑎〗^(2/3) + 𝑏^(2/3)) 〗^(3/2) Let P be the point on the hypotenuse Ac Given point P is at distance a & b from sides of triangle Construct PL ⊥ AB & PM ⊥ BC ∴ PL = a & PM = b Let ∠ ACB = θ Thus, ∠ APL = θ (Corresponding angles) We need to find maximum length of the hypotenuse Let l be the length of the hypotenuse , ∴ l = AP + PC In ∆ 𝑨𝑷𝑳 cos θ = (𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝜃 )/(𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ) cos θ = 𝑃𝐿/𝐴𝑃 cos θ = 𝑎/𝐴𝑃 AP = 𝑎/cos⁡𝜃 AP = 𝑎 sec⁡𝜃 In ∆ 𝑷𝑴𝑪 sin θ = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝜃)/(𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ) sin θ = (𝑏 )/𝑃𝐶 PC = 𝑏/sin⁡𝜃 PC = 𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃 Now, l = AP + PC l = a sec θ + b cosec θ Differentiating w.r.t θ 𝑑𝑙/𝑑𝜃=𝑑(𝑎 sec⁡〖𝜃 + 𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃〗 )/𝑑𝜃 𝑑𝑙/𝑑𝜃= a sec θ tan θ – b cosec θ cot θ Putting 𝑑𝑙/𝑑𝜃=0 a sec θ tan θ – b cosec θ cot θ = 0 a 1/cos⁡𝜃 . sin⁡𝜃/〖 cos〗⁡𝜃 −𝑏 . 1/sin⁡〖𝜃 〗 × cos⁡𝜃/sin⁡𝜃 =0 (𝑎 sin⁡𝜃)/cos^2⁡𝜃 −(𝑏 cos⁡𝜃)/sin^2⁡𝜃 =0 (𝑎 sin⁡𝜃)/cos^2⁡𝜃 = (𝑏 cos⁡𝜃)/sin^2⁡𝜃 a sin θ × sin2 θ = b cos θ cos2 θ a sin3 θ = b cos3 θ sin^3⁡𝜃/cos^3⁡𝜃 = 𝑏/𝑎 tan3 θ = 𝑏/𝑎 tan θ = (𝑏/𝑎)^(1/3) Finding (𝑑^2 𝑙)/(𝑑^2 𝜃) 𝑑𝑙/𝑑𝜃=𝑎 sec⁡〖𝑡𝑎𝑛𝜃−𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃 cot⁡𝜃 〗 Differentiating w.rt θ (𝑑^2 𝑙)/(𝑑^2 𝜃)= 𝑑(𝑎 sec⁡〖𝜃 tan⁡〖𝜃 − 𝑏 𝑐𝑜𝑠𝑒𝑐𝜃 cot⁡𝜃 〗 〗 )/𝑑𝜃 = a 𝑑(sec⁡〖𝜃 tan⁡𝜃 〗 )/𝑑𝜃−𝑏 𝑑(𝑐𝑜𝑠𝑒𝑐 𝜃 cot⁡𝜃 )/𝑑𝜃 = a((sec⁡𝜃 )^′ tan⁡𝜃+(tan⁡𝜃 )^′ sec⁡𝜃 )−𝑏((𝑐𝑜𝑠𝑒𝑐 𝜃)^′ cot⁡𝜃+(cot⁡𝜃 )^′ 𝑐𝑜𝑠𝑒𝑐 𝜃) = a (sec⁡𝜃.tan⁡〖𝜃.tan⁡𝜃+sec^2⁡〖𝜃.sec⁡𝜃 〗 〗 )−𝑏((−𝑐𝑜𝑠𝑒𝑐 𝜃.cot⁡𝜃 ) cot⁡𝜃+(−𝑐𝑜𝑠𝑒𝑐^2 𝜃).𝑐𝑜𝑠𝑒𝑐 𝜃) = a(sec⁡〖𝜃 tan^2⁡〖𝜃+sec^2⁡𝜃 〗 〗 )−𝑏(−𝑐𝑜𝑠𝑒𝑐 𝜃 cot^2⁡〖𝜃−𝑐𝑜𝑠𝑒𝑐^3 𝜃〗 ) using product rule as (𝑢𝑣)^′=𝑢^′ 𝑣+𝑣^′ 𝑢 = a sec θ (tan^2⁡〖𝜃+sec^2⁡𝜃 〗 )+𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃(〖cot^2 𝜃〗⁡〖+𝑐𝑜𝑠𝑒𝑐^2 𝜃〗 ) = a sec θ + b cosec θ Since θ is acute , i.e. 0 < θ < 𝜋/2 ⇒ θ lie in 1st quadrant So, sec θ & cosec θ will be positive ⇒ (𝑑^2 𝑙)/(𝑑^2 𝜃) > 0 ⇒ (𝑑^2 𝑙)/(𝑑^2 𝜃) > 0 at tan θ = (𝑏/𝑎)^(1/3) ⇒ l is least when tan θ = (𝑏/𝑎)^(1/3) "As tan2 θ + sec2 θ = 1" "& cot2 θ + cosec2 θ =" 1 Now, tan θ = 𝑏^(1/3)/𝑎^(1/3) tan θ = ℎ𝑒𝑖𝑔ℎ𝑡/𝑏𝑎𝑠𝑒 Height = 𝑏^(1/3) & base 𝑎^(1/3) Using Pythagoras theorem 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒^2=𝐻𝑒𝑖𝑔ℎ𝑡^2+𝐵𝑎𝑠𝑒^2 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒^2=(𝑏^(1/3) )^2+(𝑎^(1/3) )^2 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = √(𝑎^(2/3) + 𝑏^(2/3) ) Least value of l Least value of l = a sec θ + b cosec θ = a × (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 )/(𝑏𝑎𝑠𝑒 )+𝑏 "× " (ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 )/(ℎ𝑒𝑖𝑔ℎ𝑡 ) = a × √(𝑎^(2/3) + 𝑏^(2/3) )/𝑎^(1/3) +b ×√(𝑎^(2/3) + 𝑏^(2/3) )/𝑏^(1/3) l = √(𝑎^(2/3)+𝑏^(2/3) ) (𝑎^(1 − 1/3)+𝑏^(1 − 1/3) ) l = √(𝑎^(2/3)+𝑏^(2/3) ) (𝑎^(2/3)+𝑏^(2/3) ) l = (𝑎^(2/3)+𝑏^(2/3) )^(1/2 + 1) l = (𝑎^(2/3)+𝑏^(2/3) )^(3/2) Hence l = (𝒂^(𝟐/𝟑)+𝒃^(𝟐/𝟑) )^(𝟑/𝟐) Hence proved..

Miscellaneous 