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


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 γ€–γ€–(π‘Žγ€—^(2/3) + 𝑏^(2/3)) γ€—^(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 βŠ₯ AB & PM βŠ₯ BC ∴ PL = a & PM = b Let ∠ ACB = ΞΈ Thus, ∠ APL = ΞΈ (Corresponding angles) We need to find maximum length of the hypotenuse Let l be the length of the hypotenuse , ∴ l = AP + PC In βˆ† 𝑨𝑷𝑳 cos ΞΈ = (𝑆𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ πœƒ )/(π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ ) cos ΞΈ = 𝑃𝐿/𝐴𝑃 cos ΞΈ = π‘Ž/𝐴𝑃 AP = π‘Ž/cosβ‘πœƒ AP = π‘Ž secβ‘πœƒ In βˆ† 𝑷𝑴π‘ͺ sin ΞΈ = (𝑆𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ πœƒ)/(π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ ) sin ΞΈ = (𝑏 )/𝑃𝐢 PC = 𝑏/sinβ‘πœƒ PC = 𝑏 π‘π‘œπ‘ π‘’π‘ πœƒ Now, l = AP + PC l = a sec ΞΈ + b cosec ΞΈ Differentiating w.r.t ΞΈ 𝑑𝑙/π‘‘πœƒ=𝑑(π‘Ž secβ‘γ€–πœƒ + 𝑏 π‘π‘œπ‘ π‘’π‘ πœƒγ€— )/π‘‘πœƒ 𝑑𝑙/π‘‘πœƒ= a sec ΞΈ tan ΞΈ – b cosec ΞΈ cot ΞΈ Putting 𝑑𝑙/π‘‘πœƒ=0 a sec ΞΈ tan ΞΈ – b cosec ΞΈ cot ΞΈ = 0 a 1/cosβ‘πœƒ . sinβ‘πœƒ/γ€– cosγ€—β‘πœƒ βˆ’π‘ . 1/sinβ‘γ€–πœƒ γ€— Γ— cosβ‘πœƒ/sinβ‘πœƒ =0 (π‘Ž sinβ‘πœƒ)/cos^2β‘πœƒ βˆ’(𝑏 cosβ‘πœƒ)/sin^2β‘πœƒ =0 (π‘Ž sinβ‘πœƒ)/cos^2β‘πœƒ = (𝑏 cosβ‘πœƒ)/sin^2β‘πœƒ a sin ΞΈ Γ— sin2 ΞΈ = b cos ΞΈ cos2 ΞΈ a sin3 ΞΈ = b cos3 ΞΈ sin^3β‘πœƒ/cos^3β‘πœƒ = 𝑏/π‘Ž tan3 ΞΈ = 𝑏/π‘Ž tan ΞΈ = (𝑏/π‘Ž)^(1/3) Finding (𝑑^2 𝑙)/(𝑑^2 πœƒ) 𝑑𝑙/π‘‘πœƒ=π‘Ž secβ‘γ€–π‘‘π‘Žπ‘›πœƒβˆ’π‘ π‘π‘œπ‘ π‘’π‘ πœƒ cotβ‘πœƒ γ€— Differentiating w.rt ΞΈ (𝑑^2 𝑙)/(𝑑^2 πœƒ)= 𝑑(π‘Ž secβ‘γ€–πœƒ tanβ‘γ€–πœƒ βˆ’ 𝑏 π‘π‘œπ‘ π‘’π‘πœƒ cotβ‘πœƒ γ€— γ€— )/π‘‘πœƒ = a 𝑑(secβ‘γ€–πœƒ tanβ‘πœƒ γ€— )/π‘‘πœƒβˆ’π‘ 𝑑(π‘π‘œπ‘ π‘’π‘ πœƒ cotβ‘πœƒ )/π‘‘πœƒ = a((secβ‘πœƒ )^β€² tanβ‘πœƒ+(tanβ‘πœƒ )^β€² secβ‘πœƒ )βˆ’π‘((π‘π‘œπ‘ π‘’π‘ πœƒ)^β€² cotβ‘πœƒ+(cotβ‘πœƒ )^β€² π‘π‘œπ‘ π‘’π‘ πœƒ) = a (secβ‘πœƒ.tanβ‘γ€–πœƒ.tanβ‘πœƒ+sec^2β‘γ€–πœƒ.secβ‘πœƒ γ€— γ€— )βˆ’π‘((βˆ’π‘π‘œπ‘ π‘’π‘ πœƒ.cotβ‘πœƒ ) cotβ‘πœƒ+(βˆ’π‘π‘œπ‘ π‘’π‘^2 πœƒ).π‘π‘œπ‘ π‘’π‘ πœƒ) = a(secβ‘γ€–πœƒ tan^2β‘γ€–πœƒ+sec^2β‘πœƒ γ€— γ€— )βˆ’π‘(βˆ’π‘π‘œπ‘ π‘’π‘ πœƒ cot^2β‘γ€–πœƒβˆ’π‘π‘œπ‘ π‘’π‘^3 πœƒγ€— ) using product rule as (𝑒𝑣)^β€²=𝑒^β€² 𝑣+𝑣^β€² 𝑒 = a sec ΞΈ (tan^2β‘γ€–πœƒ+sec^2β‘πœƒ γ€— )+𝑏 π‘π‘œπ‘ π‘’π‘ πœƒ(γ€–cot^2 πœƒγ€—β‘γ€–+π‘π‘œπ‘ π‘’π‘^2 πœƒγ€— ) = a sec ΞΈ + b cosec ΞΈ Since ΞΈ is acute , i.e. 0 < ΞΈ < πœ‹/2 β‡’ ΞΈ lie in 1st quadrant So, sec ΞΈ & cosec ΞΈ will be positive β‡’ (𝑑^2 𝑙)/(𝑑^2 πœƒ) > 0 β‡’ (𝑑^2 𝑙)/(𝑑^2 πœƒ) > 0 at tan ΞΈ = (𝑏/π‘Ž)^(1/3) β‡’ l is least when tan ΞΈ = (𝑏/π‘Ž)^(1/3) "As tan2 ΞΈ + sec2 ΞΈ = 1" "& cot2 ΞΈ + cosec2 ΞΈ =" 1 Now, tan ΞΈ = 𝑏^(1/3)/π‘Ž^(1/3) tan ΞΈ = β„Žπ‘’π‘–π‘”β„Žπ‘‘/π‘π‘Žπ‘ π‘’ Height = 𝑏^(1/3) & base π‘Ž^(1/3) Using Pythagoras theorem π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’^2=π»π‘’π‘–π‘”β„Žπ‘‘^2+π΅π‘Žπ‘ π‘’^2 π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’^2=(𝑏^(1/3) )^2+(π‘Ž^(1/3) )^2 π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ = √(π‘Ž^(2/3) + 𝑏^(2/3) ) Least value of l Least value of l = a sec ΞΈ + b cosec ΞΈ = a Γ— (β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ )/(π‘π‘Žπ‘ π‘’ )+𝑏 "Γ— " (β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ )/(β„Žπ‘’π‘–π‘”β„Žπ‘‘ ) = a Γ— √(π‘Ž^(2/3) + 𝑏^(2/3) )/π‘Ž^(1/3) +b Γ—βˆš(π‘Ž^(2/3) + 𝑏^(2/3) )/𝑏^(1/3) l = √(π‘Ž^(2/3)+𝑏^(2/3) ) (π‘Ž^(1 βˆ’ 1/3)+𝑏^(1 βˆ’ 1/3) ) l = √(π‘Ž^(2/3)+𝑏^(2/3) ) (π‘Ž^(2/3)+𝑏^(2/3) ) l = (π‘Ž^(2/3)+𝑏^(2/3) )^(1/2 + 1) l = (π‘Ž^(2/3)+𝑏^(2/3) )^(3/2) Hence l = (𝒂^(𝟐/πŸ‘)+𝒃^(𝟐/πŸ‘) )^(πŸ‘/𝟐) 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.