# Misc 12 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 23 + 𝑏 23) 32 Let P be the point of the hypotenuse AC of ∆ABC 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 We need to find maximum length of the hypotenuse Let l be the length of the hypotenuse , ⇒ l = AP + PC Now l = AP + PC l = a sec θ + b cosec θ Differentiating w.r.t θ 𝑑𝑙𝑑𝜃= 𝑑 𝑎 sec𝜃 + 𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃𝑑𝜃 𝑑𝑙𝑑𝜃= a sec θ tan θ – b cosec θ cot θ Putting 𝑑𝑙𝑑𝜃=0 a sec θ tan θ – b cosec θ cot θ = 0 a 1 cos𝜃 . sin𝜃 cos𝜃−𝑏 . 1 sin𝜃 × cos𝜃 sin𝜃=0 𝑎 sin𝜃 cos2𝜃− 𝑏 cos𝜃 sin2𝜃=0 𝑎 sin𝜃 cos2𝜃 = 𝑏 cos𝜃 sin2𝜃 a sin θ × sin2 θ = b cos θ cos2 θ a sin3 θ = b cos3 θ sin3𝜃 cos3𝜃= 𝑏𝑎 tan3 θ = 𝑏𝑎 tan θ = 𝑏𝑎 13 Finding 𝑑2𝑙 𝑑2𝜃 𝑑𝑙𝑑𝜃=𝑎 sec𝑡𝑎𝑛𝜃−𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃 cot𝜃 Differentiating w.rt θ 𝑑2𝑙 𝑑2𝜃= 𝑑 𝑎 sec𝜃 tan𝜃 − 𝑏 𝑐𝑜𝑠𝑒𝑐𝜃 cot𝜃𝑑𝜃 = a 𝑑 sec𝜃 tan𝜃𝑑𝜃−𝑏 𝑑 𝑐𝑜𝑠𝑒𝑐 𝜃 cot𝜃𝑑𝜃 = a sec𝜃′ tan𝜃+ tan𝜃′ sec𝜃−𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃′ cot𝜃+𝑐𝑜𝑠𝑒𝑐 𝜃 = a sec𝜃. tan𝜃. tan𝜃+ sec2𝜃. sec𝜃−𝑏 −𝑐𝑜𝑠𝑒𝑐 𝜃. cot𝜃+ −𝑐𝑜𝑠𝑒 𝑐2𝜃.𝑐𝑜𝑠𝑒𝑐 𝜃 = a sec𝜃 tan2𝜃+ sec2𝜃−𝑏 −𝑐𝑜𝑠𝑒𝑐 𝜃 cot2𝜃−𝑐𝑜𝑠𝑒 𝑐3𝜃 = a sec θ tan2𝜃+ sec2𝜃+𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃 cot2𝜃+𝑐𝑜𝑠𝑒 𝑐2𝜃 = a sec θ + b cosec θ Since θ is acute , i.e. 0 < θ < 𝜋2 ⇒ θ lie in 1st quadrant So, sec θ & cosec θ will be positive ⇒ 𝑑2𝑙 𝑑2𝜃 > 0 ⇒ 𝑑2𝑙 𝑑2𝜃 > 0 at tan θ = 𝑏𝑎 13 ⇒ l is least when tan θ = 𝑏𝑎 13 Now, tan θ = 𝑏 13 𝑎 13 tan θ = ℎ𝑒𝑖𝑔ℎ𝑡𝑏𝑎𝑠𝑒 Height = 𝑏 13 & base 𝑎 13 Using Pythagoras theorem 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠 𝑒2=𝐻𝑒𝑖𝑔ℎ 𝑡2+𝐵𝑎𝑠 𝑒2 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠 𝑒2= 𝑏 132+ 𝑎 132 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑎 23 + 𝑏 23 Least value of l = a sec θ + b cosec θ = a × ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑏𝑎𝑠𝑒 +𝑏 × ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 = a × 𝑎 23 + 𝑏 23 𝑎 13+b × 𝑎 23 + 𝑏 23 𝑏 13 l = 𝑎 23+ 𝑏 23 𝑎1 − 13+ 𝑏1 − 13 l = 𝑎 23+ 𝑏 23 𝑎 23+ 𝑏 23 l = 𝑎 23+ 𝑏 23 12 + 1 l = 𝑎 23+ 𝑏 23 32 Hence l = 𝒂 𝟐𝟑+ 𝒃 𝟐𝟑 𝟑𝟐 Hence proved..

Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.