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Misc 12 - Find dy/dx, if y =12 (1 - cos t), x = 10 (t-sin t)

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


Transcript

Misc 12 Find 𝑑𝑦/𝑑π‘₯, if 𝑦=12 (1 –cos⁑𝑑 ), π‘₯=10 (𝑑 –sin⁑𝑑 ),βˆ’πœ‹/2 " "<π‘₯< πœ‹/2 Here, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑦=12 (1 –cos⁑𝑑 ) 𝑦=12 βˆ’12 cos⁑𝑑 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑𝑑 = 𝑑(12 βˆ’ 12 cos⁑𝑑 )/𝑑𝑑 𝑑𝑦/𝑑𝑑 = 𝑑(12)/𝑑𝑑 βˆ’ 12 𝑑(cos⁑𝑑 )/𝑑𝑑 𝑑𝑦/𝑑𝑑 = 0 βˆ’ 12 (βˆ’sin⁑𝑑 ) π’…π’š/𝒅𝒕 = 𝟏𝟐 𝐬𝐒𝐧⁑𝒕 Calculating 𝒅𝒙/𝒅𝒕 π‘₯=10 (𝑑 –sin⁑𝑑 ) π‘₯=10𝑑 βˆ’10 sin⁑𝑑 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑π‘₯/𝑑𝑑 = 𝑑(10 βˆ’ 10 sin⁑𝑑 )/𝑑𝑑 𝑑/𝑑𝑑 (10t βˆ’ 10 sint) 𝑑π‘₯/𝑑𝑑 = 𝑑(10 𝑑)/𝑑𝑑 βˆ’ 𝑑(10 sin⁑𝑑 )/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = 10βˆ’10 cos⁑𝑑 𝒅𝒙/𝒅𝒕 = 𝟏𝟎(πŸβˆ’π’„π’π’”β‘π’• ) Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 𝑑𝑦/𝑑π‘₯ = (12 sin⁑𝑑)/10(1 βˆ’γ€– cos〗⁑𝑑 ) 𝑑𝑦/𝑑π‘₯ = (6 π’”π’Šπ’β‘π’•)/(5 (𝟏 βˆ’γ€– 𝒄𝒐𝒔〗⁑𝒕 ) ) 𝑑𝑦/𝑑π‘₯ = (6 . 𝟐 〖𝐬𝐒𝐧 〗⁑〖𝒕/πŸγ€— 𝒄𝒐𝒔⁑〖 𝒕/πŸγ€—)/(5 (𝟐 γ€–π’”π’Šπ’γ€—^πŸβ‘γ€–π’•/πŸγ€— ) ) 𝑑𝑦/𝑑π‘₯ = (6 cos⁑〖 𝑑/2γ€—)/(5 γ€–sin 〗⁑〖𝑑/2γ€— ) π’…π’š/𝒅𝒙 = πŸ”/πŸ“ 𝒄𝒐𝒕 𝒕/𝟐 𝑑𝑦/𝑑π‘₯ = 6(2 sin⁑〖𝑑/2γ€— βˆ’ cos⁑〖 𝑑/2γ€— )/(5 (1 βˆ’(1 βˆ’ 2 sin^2⁑〖𝑑/2γ€— )) ) We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by πœƒ/2 sin ΞΈ = 2 π’”π’Šπ’β‘γ€–πœ½/πŸγ€— π’„π’π’”β‘γ€–πœ½/πŸγ€— and cos 2ΞΈ = 1 – 2sin2 ΞΈ Replacing ΞΈ by πœƒ/2 cos ΞΈ = 1 – 2sin2 πœƒ/2 1 – cos ΞΈ = 2sin2 𝜽/𝟐

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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.