Miscellaneous

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

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 ) = 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..

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.