If π‘₯ = π‘Ž 𝑠𝑒𝑐 πœƒ , 𝑦 = 𝑏  π‘‘π‘Žπ‘›πœƒ  find  d 2 y / dx 2    π‘Žπ‘‘ π‘₯ = π/6

 

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

Question 31 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 31 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 3 Question 31 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 4 Question 31 (Choice 2) - CBSE Class 12 Sample Paper for 2021 Boards - Part 5

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Transcript

Question 31 (Choice 2) If π‘₯ = π‘Ž 𝑠𝑒𝑐 πœƒ , 𝑦 = 𝑏 π‘‘π‘Žπ‘› πœƒ 𝑓𝑖𝑛𝑑 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘Žπ‘‘ π‘₯ = πœ‹/6 Given π‘₯=π‘Ž sec β‘πœƒ, 𝑦=𝑏 tanβ‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/𝑑π‘₯ Γ— π‘‘πœƒ/π‘‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/π‘‘πœƒ Γ— π‘‘πœƒ/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Calculating π’…π’š/π’…πœ½ 𝑦 = 𝑏 tanβ‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑑(𝑏 tanβ‘πœƒ )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑏 𝑑(tanβ‘πœƒ )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑏 .sec^2β‘πœƒ Calculating 𝒅𝒙/π’…πœ½ π‘₯=π‘Ž sec β‘πœƒ 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž sec β‘πœƒ)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Ž 𝑑(sec β‘πœƒ)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Ž (secβ‘πœƒ.tanβ‘πœƒ ) Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯ = (𝑏 .γ€– secγ€—^2β‘πœƒ)/(π‘Ž (secβ‘πœƒ.tanβ‘πœƒ ) ) 𝑑𝑦/𝑑π‘₯ = (𝑏 secβ‘πœƒ)/(π‘Ž tanβ‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = (𝑏 . 1/cosβ‘πœƒ )/(π‘Ž (sinβ‘πœƒ/cosβ‘πœƒ ) ) 𝑑𝑦/𝑑π‘₯ = 𝑏 Γ— 1/cosβ‘πœƒ Γ— cosβ‘πœƒ/γ€–a sinγ€—β‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑏/π‘Ž (1/sinβ‘πœƒ ) π’…π’š/𝒅𝒙 = 𝒃/𝒂 𝒄𝒐𝒔𝒆𝒄 𝜽 Now, (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = 𝒃/𝒂 (𝐝(𝒄𝒐𝒔𝒆𝒄 𝜽))/𝒅𝒙 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑏/π‘Ž (d(π‘π‘œπ‘ π‘’π‘ πœƒ))/𝑑θ ×𝑑θ/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑏/π‘Ž (d(π‘π‘œπ‘ π‘’π‘ πœƒ))/𝑑θ Γ—πŸ/(𝒅𝒙/π’…πœ½) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑏/π‘Ž (βˆ’π‘π‘œπ‘ π‘’π‘ ΞΈ cot⁑θ) Γ—πŸ/(𝒂 𝒔𝒆𝒄 ΞΈ 𝒕𝒂𝒏 ΞΈ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’π‘)/π‘Ž^2 1/sin⁑〖θ γ€— cot⁑θ Γ— 𝒄𝒐𝒔 ΞΈ 𝒄𝒐𝒕 ΞΈ (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = (βˆ’π’ƒ)/𝒂^𝟐 〖𝒄𝒐𝒕〗^πŸ‘β‘πœ½ We need to find (𝑑^2 𝑦)/(𝑑π‘₯^2 ) π‘Žπ‘‘ π‘₯ = πœ‹/6 Putting π‘₯ = 𝝅/πŸ” (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’π‘)/π‘Ž^2 γ€–π‘π‘œπ‘‘γ€—^3β‘γ€–πœ‹/6γ€— (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’π‘)/π‘Ž^2 (√3)^3 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’π‘)/π‘Ž^2 3√3 (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = (βˆ’πŸ‘βˆšπŸ‘ 𝒃)/𝒂^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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.