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

 

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.