Misc 15 - Chapter 5 Class 12 Continuity and Differentiability

Transcript

Misc 15 If 𝑥 – 𝑎﷯﷮2﷯+ 𝑦 – 𝑏﷯﷮2﷯= 𝑐2, for some 𝑐 > 0, prove that 1 + 𝑑𝑦﷮𝑑𝑥﷯﷯﷮2﷯﷯﷮ 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯﷯﷮ 3﷮2﷯﷯is a constant independent of a and b. First we will calculate 𝑑𝑦﷮𝑑𝑥﷯ 𝑥 – 𝑎﷯﷮2﷯+ 𝑦 – 𝑏﷯﷮2﷯= 𝑐2 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑 𝑥 – 𝑎﷯﷮2﷯+ 𝑦 – 𝑏﷯﷮2﷯﷯﷮𝑑𝑥﷯ = 𝑑 𝑐﷮2﷯﷯﷮𝑑𝑥﷯ 𝑑 𝑥 – 𝑎﷯﷮2﷯﷯﷮𝑑𝑥﷯ + 𝑑 𝑦 – 𝑏﷯﷮2﷯﷯﷮𝑑𝑥﷯ = 0 2 𝑥 – 𝑎﷯﷮2 −1 ﷯ . 𝑑 𝑥 − 𝑎﷯﷮𝑑𝑥﷯ + 2 𝑦 – 𝑏﷯﷮2 −1﷯ . 𝑑 𝑦 − 𝑏﷯﷮𝑑𝑥﷯ = 0 2 𝑥 – 𝑎﷯ 1 −0﷯ + 2 𝑦 – 𝑏﷯ . 𝑑𝑦﷮𝑑𝑥﷯ −0﷯ = 0 2 𝑥 – 𝑎﷯ + 2 𝑦 – 𝑏﷯ . 𝑑𝑦﷮𝑑𝑥﷯﷯ = 0 2 𝑦 – 𝑏﷯ . 𝑑𝑦﷮𝑑𝑥﷯ = −2 𝑥 – 𝑎﷯ 𝑑𝑦﷮𝑑𝑥﷯ = −2 𝑥 – 𝑎﷯﷮2 𝑦 – 𝑏﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = − 𝒙 − 𝒂﷯﷮𝒚 − 𝒃﷯ Again Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑﷮𝑑𝑥﷯ − 𝑥 − 𝑎﷯﷮𝑦 − 𝑏﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = − 𝑑﷮𝑑𝑥﷯ 𝑥 − 𝑎﷮𝑦 − 𝑏﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 𝑥 – 𝑎﷯﷮𝑑𝑥﷯ 𝑦 – 𝑏﷯ − 𝑑 𝑦 – 𝑏﷯﷮𝑑𝑥﷯ . 𝑥 – 𝑎﷯﷮ 𝑦 − 𝑏﷯﷮2﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1 − 0﷯ 𝑦 – 𝑏﷯ − 𝑑𝑦﷮𝑑𝑥﷯ − 0﷯ 𝑥 – 𝑎﷯﷮ 𝑦 − 𝑏﷯﷮2﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑦 – 𝑏﷯ − 𝑑𝑦﷮𝑑𝑥﷯﷯ 𝑥 – 𝑎﷯﷮ 𝑦 − 𝑏﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 𝑦 – 𝑏﷯ − − 𝑥 – 𝑎﷯﷮ 𝑦 – 𝑏﷯﷯ 𝑥 – 𝑎﷯﷮ 𝑦 − 𝑏﷯﷮2﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯= 𝑦 – 𝑏﷯﷮2﷯ + 𝑥 – 𝑎﷯﷮2﷯﷮ 𝑦 − 𝑏﷯﷮2﷯ 𝑦 − 𝑏﷯﷯﷯ 𝒅﷮𝟐﷯𝒚﷮𝒅 𝒙﷮𝟐﷯﷯= 𝒄﷮𝟐﷯﷮ 𝒚 − 𝒃﷯﷮𝟑﷯﷯ (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 )= (−𝒄^𝟐)/(𝒚 − 𝒃)^𝟑 Now, 〖[1+ (𝑑𝑦/𝑑𝑥)^2 ]/((𝑑^2 𝑦)/〖𝑑𝑥〗^2 )〗^(3/2) Putting values = 〖[1+ ((−(𝑥 – 𝑎))/(𝑦 – 𝑏))^2 ]/((−𝑐^2)/(𝑦 − 𝑏)^3 )〗^(3/2) = − 〖[((𝑦 − 𝑏)^2 + (𝑥 – 𝑎)^2)/(𝑦 – 𝑏)^2 ]/(𝑐^2/(𝑦 − 𝑏)^3 )〗^(3/2) = − 〖[𝑐^2/(𝑦 – 𝑏)^2 ]/(𝑐^2/(𝑦 − 𝑏)^3 )〗^(3/2) = − [𝑐^2/(𝑦 – 𝑏)^2 ]^(3/2) × (𝑦 − 𝑏)^3/𝑐^2 = − (𝑐/(𝑦 – 𝑏))^(2 × 3/2) × (𝑦 − 𝑏)^3/𝑐^2 "= −" (𝑐/(𝑦 – 𝑏))^3 " × " (𝑦 − 𝑏)^3/𝑐^2 "= −" 𝑐^3/𝑐^2 × (𝑦 − 𝑏)^3/(𝑦 − 𝑏)^3 = − 𝒄 = k Which is constant independent of a & b . Hence proved .