If x = a sec θ, y = b tan θ, find d^2y/dx^2 at x = π/6

Question 31 (Choice 2) If 𝑥 = 𝑎 𝑠𝑒𝑐 𝜃 , 𝑦 = 𝑏 𝑡𝑎𝑛 𝜃 𝑓𝑖𝑛𝑑 (𝑑^2 𝑦)/(𝑑𝑥^2 ) 𝑎𝑡 𝑥 = 𝜋/6 Given 𝑥=𝑎 sec ⁡𝜃, 𝑦=𝑏 tan⁡𝜃 𝑑𝑦/𝑑𝑥 = 𝑑𝑦/𝑑𝑥 × 𝑑𝜃/𝑑𝜃 𝑑𝑦/𝑑𝑥 = 𝑑𝑦/𝑑𝜃 × 𝑑𝜃/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝜃)/(𝑑𝑥/𝑑𝜃) Calculating 𝒅𝒚/𝒅𝜽 𝑦 = 𝑏 tan⁡𝜃 𝑑𝑦/𝑑𝜃 = 𝑑(𝑏 tan⁡𝜃 )/𝑑𝜃 𝑑𝑦/𝑑𝜃 = 𝑏 𝑑(tan⁡𝜃 )/𝑑𝜃 𝑑𝑦/𝑑𝜃 = 𝑏 .sec^2⁡𝜃 Calculating 𝒅𝒙/𝒅𝜽 𝑥=𝑎 sec ⁡𝜃 𝑑𝑥/𝑑𝜃 = 𝑑(𝑎 sec ⁡𝜃)/𝑑𝜃 𝑑𝑥/𝑑𝜃 = 𝑎 𝑑(sec ⁡𝜃)/𝑑𝜃 𝑑𝑥/𝑑𝜃 = 𝑎 (sec⁡𝜃.tan⁡𝜃 ) Therefore 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝜃)/(𝑑𝑥/𝑑𝜃) 𝑑𝑦/𝑑𝑥 = (𝑏 .〖 sec〗^2⁡𝜃)/(𝑎 (sec⁡𝜃.tan⁡𝜃 ) ) 𝑑𝑦/𝑑𝑥 = (𝑏 sec⁡𝜃)/(𝑎 tan⁡𝜃 ) 𝑑𝑦/𝑑𝑥 = (𝑏 . 1/cos⁡𝜃 )/(𝑎 (sin⁡𝜃/cos⁡𝜃 ) ) 𝑑𝑦/𝑑𝑥 = 𝑏 × 1/cos⁡𝜃 × cos⁡𝜃/〖a sin〗⁡𝜃 𝑑𝑦/𝑑𝑥 = 𝑏/𝑎 (1/sin⁡𝜃 ) 𝒅𝒚/𝒅𝒙 = 𝒃/𝒂 𝒄𝒐𝒔𝒆𝒄 𝜽 Now, (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 ) = 𝒃/𝒂 (𝐝(𝒄𝒐𝒔𝒆𝒄 𝜽))/𝒅𝒙 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 𝑏/𝑎 (d(𝑐𝑜𝑠𝑒𝑐 𝜃))/𝑑θ ×𝑑θ/𝑑𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 𝑏/𝑎 (d(𝑐𝑜𝑠𝑒𝑐 𝜃))/𝑑θ ×𝟏/(𝒅𝒙/𝒅𝜽) (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 𝑏/𝑎 (−𝑐𝑜𝑠𝑒𝑐 θ cot⁡θ) ×𝟏/(𝒂 𝒔𝒆𝒄 θ 𝒕𝒂𝒏 θ) (𝑑^2 𝑦)/(𝑑𝑥^2 ) = (−𝑏)/𝑎^2 1/sin⁡〖θ 〗 cot⁡θ × 𝒄𝒐𝒔 θ 𝒄𝒐𝒕 θ (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 ) = (−𝒃)/𝒂^𝟐 〖𝒄𝒐𝒕〗^𝟑⁡𝜽 We need to find (𝑑^2 𝑦)/(𝑑𝑥^2 ) 𝑎𝑡 𝑥 = 𝜋/6 Putting 𝑥 = 𝝅/𝟔 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = (−𝑏)/𝑎^2 〖𝑐𝑜𝑡〗^3⁡〖𝜋/6〗 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = (−𝑏)/𝑎^2 (√3)^3 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = (−𝑏)/𝑎^2 3√3 (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 ) = (−𝟑√𝟑 𝒃)/𝒂^𝟐

Question 31 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

If 𝑥 = 𝑎 𝑠𝑒𝑐 𝜃 , 𝑦 = 𝑏 𝑡𝑎𝑛𝜃 find d ² y / dx ² 𝑎𝑡 𝑥 = π/6

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Question 31 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

If 𝑥 = 𝑎 𝑠𝑒𝑐 𝜃 , 𝑦 = 𝑏 𝑡𝑎𝑛𝜃 find d 2 y / dx 2 𝑎𝑡 𝑥 = π/6

Transcript

If 𝑥 = 𝑎 𝑠𝑒𝑐 𝜃 , 𝑦 = 𝑏 𝑡𝑎𝑛𝜃 find d ² y / dx ² 𝑎𝑡 𝑥 = π/6