Ex 5.6, 8 - Find dy/dx, x = a (cos t + log tan t/2) - Ex 5.6

Ex 5.6, 8 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.6, 8 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.6, 8 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.6, 8 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo