Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

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

Transcript

Ex 5.6, 8 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯ = π‘Ž(cos⁑〖𝑑+log⁑〖 tan〗⁑〖𝑑/2γ€— γ€— ) , 𝑦=π‘Ž sin⁑𝑑Here 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑦 = π‘Ž sin⁑𝑑 𝑑𝑦/𝑑𝑑 " " = 𝑑(π‘Ž sin⁑𝑑)/𝑑𝑑 𝑑𝑦/𝑑𝑑 " " = π‘Ž 𝑑( sin⁑𝑑)/𝑑𝑑 𝑑𝑦/𝑑𝑑 " " = π‘Ž cos⁑𝑑 Calculating 𝒅𝒙/𝒅𝒕 π‘₯ = π‘Ž(cos⁑〖𝑑+log⁑〖 tan〗⁑〖𝑑/2γ€— γ€— ) 𝑑π‘₯/𝑑𝑑 = 𝑑(π‘Ž(cos⁑〖𝑑+log⁑〖 tan 〗⁑〖𝑑/2γ€— γ€— ))/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = a ((𝑑(cos⁑𝑑 ) )/𝑑𝑑 +𝑑(log⁑(γ€–tan 〗⁑〖𝑑/2γ€— ) )/𝑑𝑑) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +𝑑(log⁑(tan⁑〖 𝑑/2γ€— ) )/𝑑𝑑) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +1/tan⁑〖 𝑑/2γ€— . 𝑑(γ€–tan 〗⁑〖𝑑/2γ€— )/𝑑𝑑) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +1/tan⁑〖 𝑑/2γ€— . (sec^2⁑〖𝑑/2γ€— ).𝑑(𝑑/2)/𝑑𝑑 ) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +1/tan⁑〖 𝑑/2γ€— . sec^2 (𝑑/2). 1/2 ) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +cot⁑〖𝑑/2γ€— . 1/(cos^2 𝑑/2) . 1/2 ) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +γ€–cos 〗⁑〖𝑑/2γ€—/sin⁑〖 𝑑/2γ€— . 1/(cos^2 𝑑/2) . 1/2 ) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +1/(sin⁑〖 𝑑/2γ€— cos⁑〖 𝑑/2γ€— ) . 1/2 ) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +1/(2 sin⁑〖𝑑/2γ€— cos⁑〖𝑑/2γ€— )) 𝑑π‘₯/𝑑𝑑 = a (βˆ’sin⁑𝑑 +1/sin⁑𝑑 ) 𝑑π‘₯/𝑑𝑑 = a ((βˆ’sin^2⁑𝑑 + 1)/sin⁑𝑑 ) 𝑑π‘₯/𝑑𝑑 = a ((1 βˆ’ sin^2⁑𝑑)/sin⁑𝑑 ) 𝒅𝒙/𝒅𝒕 = a (〖𝒄𝒐𝒔〗^πŸβ‘π’•/π’”π’Šπ’β‘π’• ) We know that sin 2ΞΈ =2sin ΞΈ cos ΞΈ Replacing ΞΈ by 𝑑/2 sin t = 2 sin 𝑑/2 . cos 𝑑/2 Therefore, π’…π’š/𝒅𝒙 " =" (π’…π’š/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) 𝑑𝑦/𝑑π‘₯ = (π‘Ž cos⁑𝑑)/(γ€–π‘Ž cos^2〗⁑𝑑/sin⁑𝑑 ) 𝑑𝑦/𝑑π‘₯ = π‘Ž cos⁑𝑑 Γ— sin⁑𝑑/(π‘Ž cos^2 𝑑) 𝑑𝑦/𝑑π‘₯ = (π‘Ž cos⁑𝑑 . sin⁑𝑑)/(π‘Ž cos⁑𝑑 .cos⁑𝑑 ) 𝑑𝑦/𝑑π‘₯ = sin⁑𝑑/cos⁑𝑑 π’…π’š/𝒅𝒙 = π­πšπ§β‘π’•

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.