Miscellaneous

Chapter 6 Class 12 Application of Derivatives
Serial order wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Misc 9 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is γγ(πγ^(2/3) + π^(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 Letβs construct PL β₯ AB & PM β₯ BC β΄ PL = a & PM = b Let β  ACB = ΞΈ Thus, β  APL = ΞΈ We need to find maximum length of the hypotenuse Let l be the length of the hypotenuse , β΄ l = AP + PC In β π¨π·π³ cos ΞΈ = (ππππ ππππππππ‘ π‘π π )/(π»π¦πππ‘πππ’π π ) cos ΞΈ = ππΏ/π΄π cos ΞΈ = π/π΄π AP = π/cosβ‘π AP = π πππβ‘π½ In β π·π΄πͺ sin ΞΈ = (ππππ πππππ ππ‘π π‘π π)/(π»π¦πππ‘πππ’π π ) sin ΞΈ = ππ/ππΆ sin ΞΈ = (π )/ππΆ PC = π/sinβ‘π PC = π πππππ π½ Now, l = AP + PC l = a sec ΞΈ + b cosec ΞΈ Differentiating w.r.t ΞΈ ππ/ππ½=π(π secβ‘γπ + π πππ ππ πγ )/ππ ππ/ππ= a sec ΞΈ tan ΞΈ β b cosec ΞΈ cot ΞΈ Putting ππ/ππ½=π a sec ΞΈ tan ΞΈ β b cosec ΞΈ cot ΞΈ = 0 a 1/cosβ‘π . sinβ‘π/γ cosγβ‘π βπ . 1/sinβ‘γπ γ Γ cosβ‘π/sinβ‘π =0 (π sinβ‘π)/cos^2β‘π β(π cosβ‘π)/sin^2β‘π =0 (π sinβ‘π)/cos^2β‘π = (π cosβ‘π)/sin^2β‘π a sin ΞΈ Γ sin2 ΞΈ = b cos ΞΈ cos2 ΞΈ a sin3 ΞΈ = b cos3 ΞΈ sin^3β‘π/cos^3β‘π = π/π tan3 ΞΈ = π/π tan ΞΈ = (π/π)^(π/π) Finding (π^π π)/(π^π π½) ππ/ππ=π secβ‘γΞΈ tanβ‘πβπ πππ ππ π cotβ‘π γ Differentiating w.rt ΞΈ (π^2 π)/(π^2 π)= π(π secβ‘γπ tanβ‘γπ β π πππ πππ cotβ‘π γ γ )/ππ = a π(secβ‘γπ tanβ‘π γ )/ππβπ π(πππ ππ π cotβ‘π )/ππ = a((secβ‘π )^β² tanβ‘π+(tanβ‘π )^β² secβ‘π )βπ((πππ ππ π)^β² cotβ‘π+(cotβ‘π )^β² πππ ππ π) = a (secβ‘π.tanβ‘γπ.tanβ‘π+sec^2β‘γπ.secβ‘π γ γ )βπ((βπππ ππ π.cotβ‘π ) cotβ‘π+(βπππ ππ^2 π).πππ ππ π) = a(secβ‘γπ tan^2β‘γπ+sec^3β‘π γ γ )βπ(βπππ ππ π cot^2β‘γπβπππ ππ^3 πγ ) Using Product Rule as (π’π£)^β²=π’^β² π£+π£^β² π’ = a sec ΞΈ (γπππγ^πβ‘γπ½+γπππγ^πβ‘π½ γ )+π πππ ππ π(γγπππγ^π π½γβ‘γ+πππππ^π π½γ ) Here, square terms β (γπππγ^πβ‘γπ½+γπππγ^πβ‘π½ γ ) (γγπππγ^π π½γβ‘γ+πππππ^π π½γ ) are always positive And Since ΞΈ is acute, i.e. 0 < ΞΈ < π/2 β΄ ΞΈ lies in 1st quadrant So, sec ΞΈ & cosec ΞΈ will be positive Thus, (π^π π)/(π^π π½) > 0 at tan ΞΈ = (π/π)^(1/3) β΄ l is least when tan ΞΈ = (π/π)^(1/3) Now, tan ΞΈ = π^(π/π)/π^(π/π) tan ΞΈ = π»πππβπ‘/π΅ππ π Height = π^(π/π) & base π^(π/π) Using Pythagoras theorem π»π¦πππ‘πππ’π π^2=π»πππβπ‘^2+π΅ππ π^2 π»π¦πππ‘πππ’π π^2=(π^(1/3) )^2+(π^(1/3) )^2 π―πππππππππ = β(π^(π/π) + π^(π/π) ) Least value of l Least value of l = a sec ΞΈ + b cosec ΞΈ = a Γ (π―πππππππππ )/(π©πππ )+π "Γ " (π―πππππππππ )/(π―πππππ ) = a Γ β(π^(2/3) + π^(2/3) )/π^(1/3) +b Γβ(π^(2/3) + π^(2/3) )/π^(1/3) l = β(π^(2/3)+π^(2/3) ) (π^(1 β 1/3)+π^(1 β 1/3) ) l = β(π^(2/3)+π^(2/3) ) (π^(2/3)+π^(2/3) ) l = (π^(2/3)+π^(2/3) )^(1/2 + 1) l = (π^(π/π)+π^(π/π) )^(π/π) Hence l = (π^(π/π)+π^(π/π) )^(π/π) Hence proved..