Check sibling questions

Misc 17 - If x = a (cos t + t sin t), y = a (sin t - t cos t)

Misc  17 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  17 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Misc  17 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Misc  17 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Misc  17 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Misc  17 - Chapter 5 Class 12 Continuity and Differentiability - Part 7

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Misc 17 If π‘₯=π‘Ž (cos⁑𝑑 + 𝑑 sin⁑𝑑) and y=π‘Ž (sin⁑𝑑 – 𝑑 cos⁑𝑑), Find (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^We need to find (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 First we find π’…π’š/𝒅𝒙 Here, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑦=π‘Ž (sin⁑𝑑– 𝑑 cos⁑𝑑 ) Differentiating 𝑀.π‘Ÿ.𝑑. t 𝑑𝑦/𝑑𝑑 = 𝑑(π‘Ž (sin⁑𝑑– 𝑑 cos⁑𝑑 ))/𝑑𝑑 𝑑𝑦/𝑑𝑑 = π‘Ž 𝑑(sin⁑𝑑– 𝑑 cos⁑𝑑 )/𝑑𝑑 𝑑𝑦/𝑑𝑑 = π‘Ž (𝑑(sin⁑𝑑 )/𝑑𝑑 βˆ’ 𝑑(𝑑 cos⁑𝑑 )/𝑑𝑑) 𝑑𝑦/𝑑𝑑 = π‘Ž (cosβ‘π‘‘βˆ’ 𝑑(𝑑 cos⁑𝑑 )/𝑑𝑑) 𝑑𝑦/𝑑𝑑 = π‘Ž (cos⁑𝑑 βˆ’(𝑑𝑑/𝑑𝑑 . cos⁑𝑑+ (𝑑 cos⁑𝑑)/𝑑𝑑 . 𝑑 )) Using Product rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 𝑑𝑦/𝑑𝑑 = π‘Ž (cos⁑𝑑 βˆ’(cos⁑𝑑+(γ€–βˆ’sin〗⁑𝑑 ) . 𝑑)) 𝑑𝑦/𝑑𝑑 = π‘Ž (cos⁑𝑑 βˆ’(cosβ‘π‘‘βˆ’(sin⁑𝑑 ) . 𝑑)) 𝑑𝑦/𝑑𝑑 = π‘Ž (cos⁑𝑑 βˆ’cos⁑𝑑+𝑑 .sin⁑𝑑 ) 𝑑𝑦/𝑑𝑑 = π‘Ž (0+𝑑 sin⁑𝑑 ) π’…π’š/𝒅𝒕 = 𝒂 .𝒕.π’”π’Šπ’β‘π’• Calculating 𝒅𝒙/𝒅𝒕 π‘₯=π‘Ž (cos⁑𝑑+ 𝑑 sin⁑𝑑 ) Differentiating 𝑀.π‘Ÿ.𝑑. t 𝑑π‘₯/𝑑𝑑 = 𝑑(π‘Ž (cos⁑𝑑 + 𝑑 sin⁑𝑑)" " )/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = π‘Ž (𝑑(cos⁑𝑑 + 𝑑 sin⁑𝑑)/𝑑𝑑) 𝑑π‘₯/𝑑𝑑 = π‘Ž (𝑑(cos⁑𝑑)/𝑑𝑑 + 𝑑(𝑑 sin⁑𝑑)/𝑑𝑑) 𝑑π‘₯/𝑑𝑑 = π‘Ž (γ€–βˆ’sin〗⁑𝑑 + 𝑑(𝑑 sin⁑𝑑 )/𝑑𝑑) Using product rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 𝑑π‘₯/𝑑𝑑 = π‘Ž (γ€–βˆ’sin〗⁑𝑑+(𝑑𝑑/𝑑𝑑 . sin⁑𝑑+ 𝑑(sin⁑𝑑 )/𝑑𝑑 . 𝑑 )) 𝑑π‘₯/𝑑𝑑 = π‘Ž (γ€–βˆ’sin〗⁑𝑑+(sin⁑𝑑+cos⁑𝑑 . 𝑑)) 𝑑π‘₯/𝑑𝑑= π‘Ž (βˆ’sin⁑𝑑+sin⁑𝑑+𝑑 .cπ‘œπ‘ β‘π‘‘ ) 𝒅𝒙/𝒅𝒕 = 𝒂 .𝒕.𝒄𝒐𝒔⁑𝒕 Finding π’…π’š/𝒅𝒙 π’…π’š/𝒅𝒙 = (π’…π’š/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) 𝑑𝑦/𝑑π‘₯ = (π‘Ž" " .𝑑.sin⁑𝑑)/(π‘Ž" " .𝑑.cos⁑𝑑 ) π’…π’š/𝒅𝒙 = 𝒕𝒂𝒏⁑𝒕 Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝒅/𝒅𝒙 (π’…π’š/𝒅𝒙) = 𝒅(𝒕𝒂𝒏⁑𝒕)/𝒅𝒙 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑑(tan⁑𝑑)/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑑(tan⁑𝑑)/𝑑π‘₯ . 𝑑𝑑/𝑑𝑑 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) =sec^2⁑𝑑 . 𝑑𝑑/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) =sec^2⁑𝑑 Γ· 𝒅𝒙/𝒅𝒕 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = sec^2⁑𝑑 Γ· 𝒂.𝒕.𝒄𝒐𝒔𝒕 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (sec^2⁑𝑑 )/(π‘Ž" " . 𝑑.cos⁑𝑑 ) "We have calculated" 𝒅𝒙/𝒅𝒕 " = " π‘Ž" ".𝑑.π‘π‘œπ‘ β‘π‘‘ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (sec^2⁑𝑑 )/(π‘Ž" " . 𝑑 Γ— 1/sec⁑𝑑 ) (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = (〖𝒔𝒆𝒄〗^πŸ‘β‘π’• )/(𝒂" " . 𝒕) 2

Davneet Singh's photo - Co-founder, Teachoo

Made by

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.