Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Slide1.JPG

Slide2.JPG
Slide3.JPG Slide4.JPG Slide5.JPG Slide6.JPG Slide7.JPG

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

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€—^2 If π‘₯=π‘Ž (cos⁑𝑑 + 𝑑 sin⁑𝑑)\ & 𝑦=π‘Ž (sin⁑𝑑 – 𝑑 cos⁑𝑑) We need to find (𝑑^2 𝑦)/〖𝑑π‘₯γ€—^2 First we find 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/𝑑π‘₯ . 𝑑𝑑/𝑑𝑑 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/𝑑𝑑 . 𝑑𝑑/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑦=π‘Ž (sin⁑𝑑– 𝑑 cos⁑𝑑 ) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 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) Using product rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 where u = t & v = cos t 𝑑𝑦/𝑑𝑑 = π‘Ž (cos⁑𝑑 βˆ’(𝑑𝑑/𝑑𝑑 . cos⁑𝑑+ (𝑑 cos⁑𝑑)/𝑑𝑑 . 𝑑 )) 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 ) 𝑑𝑦/𝑑𝑑 = π‘Ž .𝑑.sin⁑𝑑 Hence 𝑑𝑦/𝑑𝑑 = π‘Ž .𝑑.sin⁑𝑑 Calculating 𝒅𝒙/𝒅𝒕 π‘₯=π‘Ž (cos⁑𝑑+ 𝑑 sin⁑𝑑 ) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 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‘‘ ) 𝑑π‘₯/𝑑𝑑 = π‘Ž .𝑑.cos⁑𝑑 Now , 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 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/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = 𝑑(tan⁑𝑑)/𝑑π‘₯ (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" 𝑑π‘₯/𝑑𝑑 " = " 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 𝑦)/(𝑑π‘₯^2 ) = (sec^3⁑𝑑 )/(π‘Ž" " . 𝑑) Hence (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = (〖𝒔𝒆𝒄〗^πŸ‘β‘π’• )/(𝒂" " . 𝒕)

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.