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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Misc 15 If (๐‘ฅ โ€“ ๐‘Ž)^2+ (๐‘ฆ โ€“ ๐‘)^2= ๐‘2, for some ๐‘ > 0, prove that ใ€–[1 + (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^2 ]/((๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 )ใ€—^(3/2)is a constant independent of a and b.First we will calculate ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (๐‘ฅ โ€“ ๐‘Ž)^2+ (๐‘ฆ โ€“ ๐‘)^2= ๐‘2 Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘((๐‘ฅ โ€“ ๐‘Ž)^2+ (๐‘ฆ โ€“ ๐‘)^2 )/๐‘‘๐‘ฅ = ๐‘‘(๐‘^2 )/๐‘‘๐‘ฅ ๐‘‘((๐‘ฅ โ€“ ๐‘Ž)^2 )/๐‘‘๐‘ฅ +" " ๐‘‘((๐‘ฆ โ€“ ๐‘)^2 )/๐‘‘๐‘ฅ = 0 2(๐‘ฅ โ€“ ๐‘Ž). ๐‘‘(๐‘ฅ โˆ’ ๐‘Ž)/๐‘‘๐‘ฅ + 2 (๐‘ฆ โ€“ ๐‘). ๐‘‘(๐‘ฆ โˆ’ ๐‘)/๐‘‘๐‘ฅ = 0 2 (๐‘ฅ โ€“ ๐‘Ž) (1 โˆ’0) + 2(๐‘ฆ โ€“ ๐‘) . (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ’0) = 0 2 (๐‘ฅ โ€“ ๐‘Ž) + 2(๐‘ฆ โ€“ ๐‘) . (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = 0 2(๐‘ฆ โ€“ ๐‘) . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’2 (๐‘ฅ โ€“ ๐‘Ž) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’2 (๐‘ฅ โ€“ ๐‘Ž))/2(๐‘ฆ โ€“ ๐‘) ๐’…๐’š/๐’…๐’™ = (โˆ’(๐’™ โˆ’ ๐’‚))/(๐’š โˆ’ ๐’ƒ) Again Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = ๐‘‘/๐‘‘๐‘ฅ ((โˆ’(๐‘ฅ โˆ’ ๐‘Ž))/(๐‘ฆ โˆ’ ๐‘)) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’ ๐‘‘/๐‘‘๐‘ฅ ((๐‘ฅ โˆ’ ๐‘Ž)/(๐‘ฆ โˆ’ ๐‘)) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 )= โˆ’ ((๐‘‘(๐‘ฅ โ€“ ๐‘Ž)/๐‘‘๐‘ฅ (๐‘ฆ โ€“ ๐‘) โˆ’ ๐‘‘(๐‘ฆ โ€“ ๐‘)/๐‘‘๐‘ฅ . (๐‘ฅ โ€“ ๐‘Ž))/(๐‘ฆ โˆ’ ๐‘)^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’ (((1 โˆ’ 0) (๐‘ฆ โ€“ ๐‘) โˆ’ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ’ 0)(๐‘ฅ โ€“ ๐‘Ž))/(๐‘ฆ โˆ’ ๐‘)^2 ) Using Quotient rule As (๐‘ข/๐‘ฃ)โ€ฒ = (๐‘ข^โ€ฒ ๐‘ฃ โˆ’ ๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 where u = x โˆ’ ๐‘Ž & v = y โˆ’ b (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’ (((๐‘ฆ โ€“ ๐‘) โˆ’ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)(๐‘ฅ โ€“ ๐‘Ž))/(๐‘ฆ โˆ’ ๐‘)^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’ (((๐‘ฆ โ€“ ๐‘) โˆ’ (โˆ’ (๐‘ฅ โ€“ ๐‘Ž))/((๐‘ฆ โ€“ ๐‘) ) (๐‘ฅ โ€“ ๐‘Ž))/(๐‘ฆ โˆ’ ๐‘)^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 )= โˆ’ (((๐‘ฆ โ€“ ๐‘)^2 + (๐‘ฅ โ€“ ๐‘Ž)^2)/((๐‘ฆ โˆ’ ๐‘)^2 (๐‘ฆ โˆ’ ๐‘) )) (๐’…^๐Ÿ ๐’š)/(๐’…๐’™^๐Ÿ )= (โˆ’๐’„^๐Ÿ)/(๐’š โˆ’ ๐’ƒ)^๐Ÿ‘ Now, finding value of ใ€–[๐Ÿ+ (๐’…๐’š/๐’…๐’™)^๐Ÿ ]/((๐’…^๐Ÿ ๐’š)/ใ€–๐’…๐’™ใ€—^๐Ÿ )ใ€—^(๐Ÿ‘/๐Ÿ) (Given (๐‘ฅ โ€“ ๐‘Ž)^2+ (๐‘ฆ โ€“ ๐‘)^2= ๐‘2) ใ€–[๐Ÿ+ (๐’…๐’š/๐’…๐’™)^๐Ÿ ]/((๐’…^๐Ÿ ๐’š)/ใ€–๐’…๐’™ใ€—^๐Ÿ )ใ€—^(๐Ÿ‘/๐Ÿ) Putting values = ใ€–[1+ ((โˆ’(๐‘ฅ โ€“ ๐‘Ž))/(๐‘ฆ โ€“ ๐‘))^2 ]/((โˆ’๐‘^2)/(๐‘ฆ โˆ’ ๐‘)^3 )ใ€—^(3/2) = โˆ’ ใ€–[((๐‘ฆ โˆ’ ๐‘)^2 + (๐‘ฅ โ€“ ๐‘Ž)^2)/(๐‘ฆ โ€“ ๐‘)^2 ]/(๐‘^2/(๐‘ฆ โˆ’ ๐‘)^3 )ใ€—^(3/2) = โˆ’ ใ€–[๐‘^2/(๐‘ฆ โ€“ ๐‘)^2 ]/(๐‘^2/(๐‘ฆ โˆ’ ๐‘)^3 )ใ€—^(3/2) = โˆ’ [๐‘^2/(๐‘ฆ โ€“ ๐‘)^2 ]^(3/2) ร— (๐‘ฆ โˆ’ ๐‘)^3/๐‘^2 = โˆ’ (๐‘/(๐‘ฆ โ€“ ๐‘))^(2 ร— 3/2) ร— (๐‘ฆ โˆ’ ๐‘)^3/๐‘^2 "= โˆ’" (๐‘/(๐‘ฆ โ€“ ๐‘))^3 " ร— " (๐‘ฆ โˆ’ ๐‘)^3/๐‘^2 "= โˆ’" ๐‘^3/๐‘^2 ร— (๐‘ฆ โˆ’ ๐‘)^3/(๐‘ฆ โˆ’ ๐‘)^3 = โˆ’๐’„ = Constant Which is constant independent of a & b 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 10 years. He provides courses for Maths and Science at Teachoo.