Slide5.JPG

Slide6.JPG

  Slide7.JPG

Slide8.JPG
Slide9.JPG Slide10.JPG

Go Ad-free

Transcript

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)

Davneet Singh's photo - Co-founder, Teachoo

Made by

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 and Computer Science at Teachoo