Example 37 - Find dy/dx, if x2/3 + y2/3 = a2/3 - NCERT - Examples

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Example 37 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Example 37 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Example 37 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Example 37 - Chapter 5 Class 12 Continuity and Differentiability - Part 5 Example 37 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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

Transcript

Example 37 (Method 1) Find 𝑑𝑦/𝑑π‘₯ , if π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) . π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) Differentiating w.r.t. x (𝑑(π‘₯^(2/3)))/𝑑π‘₯ + (𝑑(𝑦^(2/3)))/𝑑π‘₯ = (𝑑(π‘Ž^(2/3)))/𝑑π‘₯ 2/3 π‘₯^(2/3 βˆ’ 1) + (𝑑(𝑦^(2/3)))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 = 0 2/3 π‘₯^((βˆ’1)/3) + (𝑑(𝑦^(2/3)))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ = 0 2/3 π‘₯^((βˆ’1)/3) + 2/3 π’š^(𝟐/πŸ‘ βˆ’ 𝟏) Γ— 𝑑𝑦/𝑑π‘₯ = 0 2/3 1/π‘₯^(1/3) + 2/3 π’š^((βˆ’πŸ)/πŸ‘) Γ— 𝑑𝑦/𝑑π‘₯ = 0 2/3 1/π‘₯^(1/3) + 2/3 𝟏/π’š^(𝟏/πŸ‘) Γ— 𝑑𝑦/𝑑π‘₯ = 0 2/3 1/𝑦^(1/3) Γ— 𝑑𝑦/𝑑π‘₯ = βˆ’πŸ/πŸ‘ 1/π‘₯^(1/3) 1/𝑦^(1/3) Γ— 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/π‘₯^(1/3) 𝑑𝑦/𝑑π‘₯ = βˆ’ 𝑦^(1/3)/π‘₯^(1/3) π’…π’š/𝒅𝒙 = βˆ’ (π’š/𝒙)^(𝟏/πŸ‘) Example 37 (Method 2) Find 𝑑𝑦/𝑑π‘₯ , if π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) . Let 𝒙 = 𝐚 〖𝒄𝒐𝒔〗^πŸ‘β‘πœ½ & π’š = 𝐚 〖𝐬𝐒𝐧〗^πŸ‘β‘πœ½ Putting values in equation π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) γ€–(𝒂 γ€–πœπ¨π¬γ€—^πŸ‘β‘πœ½)γ€—^(𝟐/πŸ‘) + γ€–(𝒂 〖𝐬𝐒𝒏〗^πŸ‘β‘πœ½)γ€—^(𝟐/πŸ‘) = 𝒂^(𝟐/πŸ‘) π‘Ž^(2/3) γ€–(cos^3β‘πœƒ)γ€—^(2/3) + π‘Ž^(2/3) γ€–(γ€–si𝑛〗^3β‘πœƒ)γ€—^(2/3) = π‘Ž^(2/3) π‘Ž^(2/3) . (sin^2β‘πœƒ+cos^2β‘πœƒ )= π‘Ž^(2/3) π‘Ž^(2/3) = π‘Ž^(2/3) So, our assumed values of x and y is correct Now, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Calculating π’…π’š/π’…πœ½ 𝑦 = a sin^3β‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑑(a sin^3β‘πœƒ )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = a (3 sin^2β‘πœƒ) 𝒅(π’”π’Šπ’β‘πœ½ )/π’…πœ½ 𝑑𝑦/π‘‘πœƒ = 3π‘Ž sin2 𝜽 cos 𝜽 Calculating 𝒅𝒙/π’…πœ½ π‘₯ = a cos^3β‘πœƒ 𝑑π‘₯/π‘‘πœƒ = (𝑑 (γ€–a cosγ€—^3β‘γ€–πœƒ)γ€—)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = a(3 cos^2β‘πœƒ) (𝒅(𝒄𝒐𝒔 𝜽) )/π’…πœ½ 𝒅𝒙/π’…πœ½ = πŸ‘πš 〖𝒄𝒐𝒔〗^𝟐⁑𝜽 . (βˆ’π’”π’Šπ’β‘πœ½ ) Therefore π’…π’š/𝒅𝒙 = 𝑑𝑦/π‘‘πœƒ Γ· 𝑑π‘₯/π‘‘πœƒ = (3a sin^2β‘πœƒ cosβ‘πœƒ)/(βˆ’3a cos^2β‘πœƒ sinβ‘πœƒ ) = (βˆ’sinβ‘πœƒ)/cosβ‘πœƒ = βˆ’tan 𝜽 Finding value of tan 𝜽 𝑦/π‘₯ = (π‘Ž sin^3β‘πœƒ)/(π‘Ž cos^3β‘πœƒ ) sin^3β‘πœƒ/cos^3β‘πœƒ = 𝑦/π‘₯ (sinβ‘πœƒ/cosβ‘πœƒ )^3 = 𝑦/π‘₯ sinβ‘πœƒ/cosβ‘πœƒ = (𝑦/π‘₯)^(1/3) π’•π’‚π’β‘πœ½ = √(πŸ‘&π’š/𝒙) Thus, 𝑑𝑦/𝑑π‘₯ = βˆ’tan πœƒ π’…π’š/𝒅𝒙 = βˆ’βˆš(πŸ‘&π’š/𝒙)

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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.