Misc 12 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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Misc 12 Find 𝑑𝑦/𝑑𝑥, if 𝑦=12 (1 –cos⁡𝑡 ), 𝑥=10 (𝑡 –sin⁡𝑡 ),−𝜋/2 " "<𝑥< 𝜋/2 Here, 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) Calculating 𝒅𝒚/𝒅𝒕 𝑦=12 (1 –cos⁡𝑡 ) 𝑦=12 −12 cos⁡𝑡 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑡 = 𝑑(12 − 12 cos⁡𝑡 )/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝑑(12)/𝑑𝑡 − 12 𝑑(cos⁡𝑡 )/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 0 − 12 (−sin⁡𝑡 ) 𝒅𝒚/𝒅𝒕 = 𝟏𝟐 𝐬𝐢𝐧⁡𝒕 Calculating 𝒅𝒙/𝒅𝒕 𝑥=10 (𝑡 –sin⁡𝑡 ) 𝑥=10𝑡 −10 sin⁡𝑡 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑥/𝑑𝑡 = 𝑑(10 − 10 sin⁡𝑡 )/𝑑𝑡 𝑑/𝑑𝑡 (10t − 10 sint) 𝑑𝑥/𝑑𝑡 = 𝑑(10 𝑡)/𝑑𝑡 − 𝑑(10 sin⁡𝑡 )/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 10−10 cos⁡𝑡 𝒅𝒙/𝒅𝒕 = 𝟏𝟎(𝟏−𝒄𝒐𝒔⁡𝒕 ) Therefore 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) 𝑑𝑦/𝑑𝑥 = (12 sin⁡𝑡)/10(1 −〖 cos〗⁡𝑡 ) 𝑑𝑦/𝑑𝑥 = (6 𝒔𝒊𝒏⁡𝒕)/(5 (𝟏 −〖 𝒄𝒐𝒔〗⁡𝒕 ) ) 𝑑𝑦/𝑑𝑥 = (6 . 𝟐 〖𝐬𝐢𝐧 〗⁡〖𝒕/𝟐〗 𝒄𝒐𝒔⁡〖 𝒕/𝟐〗)/(5 (𝟐 〖𝒔𝒊𝒏〗^𝟐⁡〖𝒕/𝟐〗 ) ) 𝑑𝑦/𝑑𝑥 = (6 cos⁡〖 𝑡/2〗)/(5 〖sin 〗⁡〖𝑡/2〗 ) 𝒅𝒚/𝒅𝒙 = 𝟔/𝟓 𝒄𝒐𝒕 𝒕/𝟐 𝑑𝑦/𝑑𝑥 = 6(2 sin⁡〖𝑡/2〗 − cos⁡〖 𝑡/2〗 )/(5 (1 −(1 − 2 sin^2⁡〖𝑡/2〗 )) ) We know that sin 2θ = 2 sin θ cos θ Replacing θ by 𝜃/2 sin θ = 2 𝒔𝒊𝒏⁡〖𝜽/𝟐〗 𝒄𝒐𝒔⁡〖𝜽/𝟐〗 and cos 2θ = 1 – 2sin2 θ Replacing θ by 𝜃/2 cos θ = 1 – 2sin2 𝜃/2 1 – cos θ = 2sin2 𝜽/𝟐