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Example 34 (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 34 (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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.