Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Slide1.JPG

Slide2.JPG
Slide3.JPG Slide4.JPG Slide5.JPG Slide6.JPG Slide7.JPG Slide8.JPG Slide9.JPG Slide10.JPG Slide11.JPG Slide12.JPG

 

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

Transcript

Misc 8 Find the maximum area of an isosceles triangle inscribed in the ellipse ๐‘ฅ^2/๐‘Ž^2 + ๐‘ฆ^2/๐‘^2 = 1 with its vertex at one end of the major axis.Given equation of ellipse is ๐‘ฅ^2/๐‘Ž^2 +๐‘ฆ^2/๐‘^2 =1 where Major axis of ellipse is along x-axis Here, Coordinate of A = (a, 0) Coordinate of Aโ€™ = (โˆ’a, 0) An isosceles triangle inscribed in the ellipse its vertex at one end of major axis Let โˆ†๐‘ท๐‘จ๐‘ท^โ€ฒ be isosceles triangle Let coordinates of point P be (๐’‰, ๐’Œ) Since, ellipse is symmetric So, coordinates of point Pโ€™ is (๐’‰ , โˆ’๐’Œ) Since point M lies on the ๐’™-axis So, coordinates of point M = (๐’‰ , ๐ŸŽ) We need to Maximize Area of triangle Let A be the area of isosceles โˆ†๐‘ƒ๐ด๐‘ƒโ€ฒ A = 1/2 ร— ๐ต๐‘Ž๐‘ ๐‘’ ร— ๐ป๐‘’๐‘–๐‘”โ„Ž๐‘ก A = ๐Ÿ/๐Ÿ ร— PPโ€™ ร— AM By Distance Formula, AM = a โˆ’ h PPโ€™ = 2k Since h and k are two variables, we write h in terms of k Now, Point (๐’‰, ๐’Œ) lies on the ellipse โˆด (โ„Ž , ๐‘˜) will satisfy the equation of ellipse Putting ๐‘ฅ = โ„Ž & y = k in equation ๐‘ฅ^2/๐‘Ž^2 +๐‘ฆ^2/๐‘^2 =1 (โ„Ž)^2/๐‘Ž^2 +(๐‘˜)^2/๐‘^2 =1 (๐‘^2 โ„Ž^2 + ๐‘Ž^2 ๐‘˜^2)/(๐‘Ž^2 ๐‘^2 )=1 ๐‘2โ„Ž2+๐‘Ž2๐‘˜2 = ๐‘Ž2๐‘2 ๐’‚๐Ÿ๐’Œ๐Ÿ = ๐‘Ž2๐‘2 โ€“ ๐‘2โ„Ž2 ๐‘˜2 = (๐‘Ž2๐‘2" โ€“ " ๐‘2โ„Ž2" " )/๐‘Ž2 ๐‘˜ = โˆš((๐‘Ž2๐‘2" โ€“ " ๐‘2โ„Ž2" " )/๐‘Ž2) ๐’Œ = ๐Ÿ/๐’‚ โˆš((๐’‚๐’ƒ)^๐Ÿโˆ’(๐’ƒ๐’‰)^๐Ÿ ) Thus, Area of Triangle A = 1/2 ร— AM ร— PPโ€™ A = 1/2 ร— (๐‘Žโˆ’โ„Ž) ร—2๐‘˜ A = (๐’‚โˆ’๐’‰) ร—๐Ÿ/๐’‚ โˆš((๐’‚๐’ƒ)^๐Ÿโˆ’๐’ƒ^๐Ÿ ๐’‰^๐Ÿ ) We need to maximise A, but A has a square root Which will be difficult to differentiate Let Z = A2 Z = ((๐‘Žโˆ’โ„Ž) ร— 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘ โ„Ž)^2 ))^2 Z = (๐‘Žโˆ’โ„Ž)^2 ร— 1/๐‘Ž^2 ((๐‘Ž๐‘)^2โˆ’(๐‘^2 โ„Ž^2 )) Z = (๐‘Žโˆ’โ„Ž)^2 ร— 1/๐‘Ž^2 ร— ๐‘^2 (๐‘Ž^2โˆ’โ„Ž^2 ) Z = (๐’‚โˆ’๐’‰)^๐Ÿ ร— (๐’ƒ/๐’‚)^๐Ÿ (๐’‚^๐Ÿโˆ’๐’‰^๐Ÿ ) Z = (๐‘/๐‘Ž)^2 (๐‘Žโˆ’โ„Ž)^2 (๐‘Ž^2โˆ’โ„Ž^2 ) Since A is positive, A is maximum if A2 is maximum So, we maximise Z = A2 Differentiating w.r.t h ๐‘‘๐‘/๐‘‘โ„Ž= ๐‘‘((๐‘/๐‘Ž)^2 (๐‘Ž โˆ’ โ„Ž)^2 (๐‘Ž^2 โˆ’ โ„Ž^2 ))/๐‘‘โ„Ž ๐‘‘๐‘/๐‘‘โ„Ž=(๐‘/๐‘Ž)^2. [(๐‘‘(๐‘Žโˆ’โ„Ž)^2)/๐‘‘โ„Ž.(๐‘Ž^2โˆ’โ„Ž^2 )+๐‘‘(๐‘Ž^2โˆ’ โ„Ž^2 )/๐‘‘โ„Ž . (๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž=(๐‘/๐‘Ž)^2 [2(๐‘Žโˆ’โ„Ž). ๐‘‘(โˆ’โ„Ž)/๐‘‘โ„Ž (๐‘Ž^2โˆ’โ„Ž^2 )+(0โˆ’2โ„Ž) (๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= (๐‘/๐‘Ž)^2 [โˆ’2(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2 )โˆ’2โ„Ž(๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= (๐‘/๐‘Ž)^2 (โˆ’2)[(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2 )+โ„Ž(๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 [(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2+โ„Ž(๐‘Žโˆ’โ„Ž))] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 (๐‘Žโˆ’โ„Ž)[๐‘Ž^2โˆ’โ„Ž^2+โ„Ž๐‘Žโˆ’โ„Ž^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 (๐‘Žโˆ’โ„Ž)[๐‘Ž^2+โ„Ž๐‘Žโˆ’2โ„Ž^2 ] ๐‘‘๐‘/๐‘‘โ„Ž=(๐‘/๐‘Ž)^2. [(๐‘‘(๐‘Žโˆ’โ„Ž)^2)/๐‘‘โ„Ž.(๐‘Ž^2โˆ’โ„Ž^2 )+๐‘‘(๐‘Ž^2โˆ’ โ„Ž^2 )/๐‘‘โ„Ž . (๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž=(๐‘/๐‘Ž)^2 [2(๐‘Žโˆ’โ„Ž). ๐‘‘(โˆ’โ„Ž)/๐‘‘โ„Ž (๐‘Ž^2โˆ’โ„Ž^2 )+(0โˆ’2โ„Ž) (๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= (๐‘/๐‘Ž)^2 [โˆ’2(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2 )โˆ’2โ„Ž(๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= (๐‘/๐‘Ž)^2 (โˆ’2)[(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2 )+โ„Ž(๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 [(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2+โ„Ž(๐‘Žโˆ’โ„Ž))] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 (๐‘Žโˆ’โ„Ž)[๐‘Ž^2โˆ’โ„Ž^2+โ„Ž๐‘Žโˆ’โ„Ž^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 (๐‘Žโˆ’โ„Ž)[๐‘Ž^2+โ„Ž๐‘Žโˆ’2โ„Ž^2 ] a โ€“ h = 0 h = a a2 + ah โ€“ 2h2 = 0 a2 + 2ah โ€“ ah โ€“ 2h2 = 0 a(๐‘Ž+2โ„Ž)โˆ’โ„Ž(๐‘Ž+2โ„Ž)=0 (๐‘Žโˆ’โ„Ž)(๐‘Ž+2โ„Ž)=0 So, h = a &h = (โˆ’๐’‚)/๐Ÿ If h = a then k = 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘โ„Ž)^2 )= 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘๐‘Ž)^2 )= 0 Which is not possible Hence, h = (โˆ’๐’‚)/๐Ÿ only Finding (๐’…^๐Ÿ ๐’)/(๐’…๐’‰^๐Ÿ ) ๐‘‘๐‘/๐‘‘โ„Ž=โˆ’2(๐‘/๐‘Ž)^2 ((๐‘Žโˆ’โ„Ž)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 )) Differentiating w.r.t โ„Ž (๐‘‘^2 ๐‘)/(๐‘‘โ„Ž^2 ) = โˆ’2(๐‘/๐‘Ž)^2. ๐‘‘((๐‘Žโˆ’โ„Ž)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 ))/๐‘‘โ„Ž = โˆ’2(๐‘/๐‘Ž)^2 [๐‘Ž^2+2๐‘Ž^2+6/4 ๐‘Ž^2 ] = โ€“2 (๐‘/๐‘Ž)^2 [3๐‘Ž^2+3/2 ๐‘Ž^2 ] = โ€“2((๐‘^2 )/๐‘Ž^2 )((9๐‘Ž^2)/2) = โˆ’9b^2 < 0 Since ๐™^โ€ฒโ€ฒ < 0 for h = (โˆ’๐‘Ž)/2 โˆด Z is maximum when h = (โˆ’๐‘Ž)/2 Thus, A is maximum at h = (โˆ’๐’‚)/๐Ÿ Maximum value of Area A = (๐’‚โˆ’๐’‰) ร—๐Ÿ/๐’‚ โˆš((๐’‚๐’ƒ)^๐Ÿโˆ’๐’ƒ^๐Ÿ ๐’‰^๐Ÿ ) = (๐‘Žโˆ’((โˆ’๐‘Ž)/2)) ร—1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’๐‘^2 ((โˆ’๐‘Ž)/2)^2 ) = 3a/2 ร—1/๐‘Ž โˆš(๐‘Ž^2 ๐‘^2โˆ’(๐‘Ž^2 ๐‘^2)/4) = 3a/2 ร—1/๐‘Ž โˆš((3๐‘Ž^2 ๐‘^2)/4) = 3a/2 ร—1/๐‘Ž ร— (โˆš3 ๐‘Ž๐‘)/2 = (3โˆš3 ๐‘Ž๐‘)/4 Hence, maximum area is (๐Ÿ‘โˆš๐Ÿ‘ ๐’‚๐’ƒ)/๐Ÿ’ square units

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.