Misc 17 - Chapter 5 Class 12 Continuity and Differentiability

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Misc 17 If 𝑥=𝑎 (cos⁡𝑡 + 𝑡 sin⁡𝑡) and y=𝑎 (sin⁡𝑡 – 𝑡 cos⁡𝑡), Find (𝑑^2 𝑦)/〖𝑑𝑥〗^We need to find (𝑑^2 𝑦)/〖𝑑𝑥〗^2 First we find 𝒅𝒚/𝒅𝒙 Here, 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) Calculating 𝒅𝒚/𝒅𝒕 𝑦=𝑎 (sin⁡𝑡– 𝑡 cos⁡𝑡 ) Differentiating 𝑤.𝑟.𝑡. t 𝑑𝑦/𝑑𝑡 = 𝑑(𝑎 (sin⁡𝑡– 𝑡 cos⁡𝑡 ))/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝑎 𝑑(sin⁡𝑡– 𝑡 cos⁡𝑡 )/𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝑎 (𝑑(sin⁡𝑡 )/𝑑𝑡 − 𝑑(𝑡 cos⁡𝑡 )/𝑑𝑡) 𝑑𝑦/𝑑𝑡 = 𝑎 (cos⁡𝑡− 𝑑(𝑡 cos⁡𝑡 )/𝑑𝑡) 𝑑𝑦/𝑑𝑡 = 𝑎 (cos⁡𝑡 −(𝑑𝑡/𝑑𝑡 . cos⁡𝑡+ (𝑑 cos⁡𝑡)/𝑑𝑡 . 𝑡 )) Using Product rule As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 𝑑𝑦/𝑑𝑡 = 𝑎 (cos⁡𝑡 −(cos⁡𝑡+(〖−sin〗⁡𝑡 ) . 𝑡)) 𝑑𝑦/𝑑𝑡 = 𝑎 (cos⁡𝑡 −(cos⁡𝑡−(sin⁡𝑡 ) . 𝑡)) 𝑑𝑦/𝑑𝑡 = 𝑎 (cos⁡𝑡 −cos⁡𝑡+𝑡 .sin⁡𝑡 ) 𝑑𝑦/𝑑𝑡 = 𝑎 (0+𝑡 sin⁡𝑡 ) 𝒅𝒚/𝒅𝒕 = 𝒂 .𝒕.𝒔𝒊𝒏⁡𝒕 Calculating 𝒅𝒙/𝒅𝒕 𝑥=𝑎 (cos⁡𝑡+ 𝑡 sin⁡𝑡 ) Differentiating 𝑤.𝑟.𝑡. t 𝑑𝑥/𝑑𝑡 = 𝑑(𝑎 (cos⁡𝑡 + 𝑡 sin⁡𝑡)" " )/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 𝑎 (𝑑(cos⁡𝑡 + 𝑡 sin⁡𝑡)/𝑑𝑡) 𝑑𝑥/𝑑𝑡 = 𝑎 (𝑑(cos⁡𝑡)/𝑑𝑡 + 𝑑(𝑡 sin⁡𝑡)/𝑑𝑡) 𝑑𝑥/𝑑𝑡 = 𝑎 (〖−sin〗⁡𝑡 + 𝑑(𝑡 sin⁡𝑡 )/𝑑𝑡) Using product rule As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 𝑑𝑥/𝑑𝑡 = 𝑎 (〖−sin〗⁡𝑡+(𝑑𝑡/𝑑𝑡 . sin⁡𝑡+ 𝑑(sin⁡𝑡 )/𝑑𝑡 . 𝑡 )) 𝑑𝑥/𝑑𝑡 = 𝑎 (〖−sin〗⁡𝑡+(sin⁡𝑡+cos⁡𝑡 . 𝑡)) 𝑑𝑥/𝑑𝑡= 𝑎 (−sin⁡𝑡+sin⁡𝑡+𝑡 .c𝑜𝑠⁡𝑡 ) 𝒅𝒙/𝒅𝒕 = 𝒂 .𝒕.𝒄𝒐𝒔⁡𝒕 Finding 𝒅𝒚/𝒅𝒙 𝒅𝒚/𝒅𝒙 = (𝒅𝒚/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) 𝑑𝑦/𝑑𝑥 = (𝑎" " .𝑡.sin⁡𝑡)/(𝑎" " .𝑡.cos⁡𝑡 ) 𝒅𝒚/𝒅𝒙 = 𝒕𝒂𝒏⁡𝒕 Again Differentiating 𝑤.𝑟.𝑡.𝑥. 𝒅/𝒅𝒙 (𝒅𝒚/𝒅𝒙) = 𝒅(𝒕𝒂𝒏⁡𝒕)/𝒅𝒙 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 𝑑(tan⁡𝑡)/𝑑𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 𝑑(tan⁡𝑡)/𝑑𝑥 . 𝑑𝑡/𝑑𝑡 (𝑑^2 𝑦)/(𝑑𝑥^2 ) =sec^2⁡𝑡 . 𝑑𝑡/𝑑𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 ) =sec^2⁡𝑡 ÷ 𝒅𝒙/𝒅𝒕 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = sec^2⁡𝑡 ÷ 𝒂.𝒕.𝒄𝒐𝒔𝒕 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = (sec^2⁡𝑡 )/(𝑎" " . 𝑡.cos⁡𝑡 ) "We have calculated" 𝒅𝒙/𝒅𝒕 " = " 𝑎" ".𝑡.𝑐𝑜𝑠⁡𝑡 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = (sec^2⁡𝑡 )/(𝑎" " . 𝑡 × 1/sec⁡𝑡 ) (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 ) = (〖𝒔𝒆𝒄〗^𝟑⁡𝒕 )/(𝒂" " . 𝒕) 2