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Misc 15 - If (x - a)2 + (y - b)2 = c2, prove [1 + (dy/dx)2]3/2

Misc  15 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  15 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Misc  15 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Misc  15 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Misc  15 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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Misc 15 If (π‘₯ – π‘Ž)^2+ (𝑦 – 𝑏)^2= 𝑐2, for some 𝑐 > 0, prove that γ€–[1 + (𝑑𝑦/𝑑π‘₯)^2 ]/((𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 )γ€—^(3/2)is a constant independent of a and b.First we will calculate 𝑑𝑦/𝑑π‘₯ (π‘₯ – π‘Ž)^2+ (𝑦 – 𝑏)^2= 𝑐2 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑((π‘₯ – π‘Ž)^2+ (𝑦 – 𝑏)^2 )/𝑑π‘₯ = 𝑑(𝑐^2 )/𝑑π‘₯ 𝑑((π‘₯ – π‘Ž)^2 )/𝑑π‘₯ +" " 𝑑((𝑦 – 𝑏)^2 )/𝑑π‘₯ = 0 2(π‘₯ – π‘Ž). 𝑑(π‘₯ βˆ’ π‘Ž)/𝑑π‘₯ + 2 (𝑦 – 𝑏). 𝑑(𝑦 βˆ’ 𝑏)/𝑑π‘₯ = 0 2 (π‘₯ – π‘Ž) (1 βˆ’0) + 2(𝑦 – 𝑏) . (𝑑𝑦/𝑑π‘₯ βˆ’0) = 0 2 (π‘₯ – π‘Ž) + 2(𝑦 – 𝑏) . (𝑑𝑦/𝑑π‘₯) = 0 2(𝑦 – 𝑏) . 𝑑𝑦/𝑑π‘₯ = βˆ’2 (π‘₯ – π‘Ž) 𝑑𝑦/𝑑π‘₯ = (βˆ’2 (π‘₯ – π‘Ž))/2(𝑦 – 𝑏) π’…π’š/𝒅𝒙 = (βˆ’(𝒙 βˆ’ 𝒂))/(π’š βˆ’ 𝒃) Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = 𝑑/𝑑π‘₯ ((βˆ’(π‘₯ βˆ’ π‘Ž))/(𝑦 βˆ’ 𝑏)) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ 𝑑/𝑑π‘₯ ((π‘₯ βˆ’ π‘Ž)/(𝑦 βˆ’ 𝑏)) (𝑑^2 𝑦)/(𝑑π‘₯^2 )= βˆ’ ((𝑑(π‘₯ – π‘Ž)/𝑑π‘₯ (𝑦 – 𝑏) βˆ’ 𝑑(𝑦 – 𝑏)/𝑑π‘₯ . (π‘₯ – π‘Ž))/(𝑦 βˆ’ 𝑏)^2 ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ (((1 βˆ’ 0) (𝑦 – 𝑏) βˆ’ (𝑑𝑦/𝑑π‘₯ βˆ’ 0)(π‘₯ – π‘Ž))/(𝑦 βˆ’ 𝑏)^2 ) Using Quotient rule As (𝑒/𝑣)β€² = (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 where u = x βˆ’ π‘Ž & v = y βˆ’ b (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ (((𝑦 – 𝑏) βˆ’ (𝑑𝑦/𝑑π‘₯)(π‘₯ – π‘Ž))/(𝑦 βˆ’ 𝑏)^2 ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ (((𝑦 – 𝑏) βˆ’ (βˆ’ (π‘₯ – π‘Ž))/((𝑦 – 𝑏) ) (π‘₯ – π‘Ž))/(𝑦 βˆ’ 𝑏)^2 ) (𝑑^2 𝑦)/(𝑑π‘₯^2 )= βˆ’ (((𝑦 – 𝑏)^2 + (π‘₯ – π‘Ž)^2)/((𝑦 βˆ’ 𝑏)^2 (𝑦 βˆ’ 𝑏) )) (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 )= (βˆ’π’„^𝟐)/(π’š βˆ’ 𝒃)^πŸ‘ Now, finding value of γ€–[𝟏+ (π’…π’š/𝒅𝒙)^𝟐 ]/((𝒅^𝟐 π’š)/〖𝒅𝒙〗^𝟐 )γ€—^(πŸ‘/𝟐) (Given (π‘₯ – π‘Ž)^2+ (𝑦 – 𝑏)^2= 𝑐2) γ€–[𝟏+ (π’…π’š/𝒅𝒙)^𝟐 ]/((𝒅^𝟐 π’š)/〖𝒅𝒙〗^𝟐 )γ€—^(πŸ‘/𝟐) Putting values = γ€–[1+ ((βˆ’(π‘₯ – π‘Ž))/(𝑦 – 𝑏))^2 ]/((βˆ’π‘^2)/(𝑦 βˆ’ 𝑏)^3 )γ€—^(3/2) = βˆ’ γ€–[((𝑦 βˆ’ 𝑏)^2 + (π‘₯ – π‘Ž)^2)/(𝑦 – 𝑏)^2 ]/(𝑐^2/(𝑦 βˆ’ 𝑏)^3 )γ€—^(3/2) = βˆ’ γ€–[𝑐^2/(𝑦 – 𝑏)^2 ]/(𝑐^2/(𝑦 βˆ’ 𝑏)^3 )γ€—^(3/2) = βˆ’ [𝑐^2/(𝑦 – 𝑏)^2 ]^(3/2) Γ— (𝑦 βˆ’ 𝑏)^3/𝑐^2 = βˆ’ (𝑐/(𝑦 – 𝑏))^(2 Γ— 3/2) Γ— (𝑦 βˆ’ 𝑏)^3/𝑐^2 "= βˆ’" (𝑐/(𝑦 – 𝑏))^3 " Γ— " (𝑦 βˆ’ 𝑏)^3/𝑐^2 "= βˆ’" 𝑐^3/𝑐^2 Γ— (𝑦 βˆ’ 𝑏)^3/(𝑦 βˆ’ 𝑏)^3 = βˆ’π’„ = Constant Which is constant independent of a & b Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.