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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 34 (Method 1) Find 𝑑𝑦/𝑑𝑥 , if 𝑥^(2/3) + 𝑦^(2/3) = 𝑎^(2/3) . 𝑥^(2/3) + 𝑦^(2/3) = 𝑎^(2/3) Differentiating w.r.t. x (𝑑(𝑥^(2/3)))/𝑑𝑥 + (𝑑(𝑦^(2/3)))/𝑑𝑥 = (𝑑(𝑎^(2/3)))/𝑑𝑥 2/3 𝑥^(2/3 − 1) + (𝑑(𝑦^(2/3)))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 = 0 2/3 𝑥^((−1)/3) + (𝑑(𝑦^(2/3)))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 = 0 2/3 𝑥^((−1)/3) + 2/3 𝒚^(𝟐/𝟑 − 𝟏) × 𝑑𝑦/𝑑𝑥 = 0 2/3 1/𝑥^(1/3) + 2/3 𝒚^((−𝟏)/𝟑) × 𝑑𝑦/𝑑𝑥 = 0 2/3 1/𝑥^(1/3) + 2/3 𝟏/𝒚^(𝟏/𝟑) × 𝑑𝑦/𝑑𝑥 = 0 2/3 1/𝑦^(1/3) × 𝑑𝑦/𝑑𝑥 = −𝟐/𝟑 1/𝑥^(1/3) 1/𝑦^(1/3) × 𝑑𝑦/𝑑𝑥 = (−1)/𝑥^(1/3) 𝑑𝑦/𝑑𝑥 = − 𝑦^(1/3)/𝑥^(1/3) 𝒅𝒚/𝒅𝒙 = − (𝒚/𝒙)^(𝟏/𝟑) Example 34 (Method 2) Find 𝑑𝑦/𝑑𝑥 , if 𝑥^(2/3) + 𝑦^(2/3) = 𝑎^(2/3) . Let 𝒙 = 𝐚 〖𝒄𝒐𝒔〗^𝟑⁡𝜽 & 𝒚 = 𝐚 〖𝐬𝐢𝐧〗^𝟑⁡𝜽 Putting values in equation 𝑥^(2/3) + 𝑦^(2/3) = 𝑎^(2/3) 〖(𝒂 〖𝐜𝐨𝐬〗^𝟑⁡𝜽)〗^(𝟐/𝟑) + 〖(𝒂 〖𝐬𝐢𝒏〗^𝟑⁡𝜽)〗^(𝟐/𝟑) = 𝒂^(𝟐/𝟑) 𝑎^(2/3) 〖(cos^3⁡𝜃)〗^(2/3) + 𝑎^(2/3) 〖(〖si𝑛〗^3⁡𝜃)〗^(2/3) = 𝑎^(2/3) 𝑎^(2/3) . (sin^2⁡𝜃+cos^2⁡𝜃 )= 𝑎^(2/3) 𝑎^(2/3) = 𝑎^(2/3) So, our assumed values of x and y is correct Now, 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝜃)/(𝑑𝑥/𝑑𝜃) Calculating 𝒅𝒚/𝒅𝜽 𝑦 = a sin^3⁡𝜃 𝑑𝑦/𝑑𝜃 = 𝑑(a sin^3⁡𝜃 )/𝑑𝜃 𝑑𝑦/𝑑𝜃 = a (3 sin^2⁡𝜃) 𝒅(𝒔𝒊𝒏⁡𝜽 )/𝒅𝜽 𝑑𝑦/𝑑𝜃 = 3𝑎 sin2 𝜽 cos 𝜽 Calculating 𝒅𝒙/𝒅𝜽 𝑥 = a cos^3⁡𝜃 𝑑𝑥/𝑑𝜃 = (𝑑 (〖a cos〗^3⁡〖𝜃)〗)/𝑑𝜃 𝑑𝑥/𝑑𝜃 = a(3 cos^2⁡𝜃) (𝒅(𝒄𝒐𝒔 𝜽) )/𝒅𝜽 𝒅𝒙/𝒅𝜽 = 𝟑𝐚 〖𝒄𝒐𝒔〗^𝟐⁡𝜽 . (−𝒔𝒊𝒏⁡𝜽 ) Therefore 𝒅𝒚/𝒅𝒙 = 𝑑𝑦/𝑑𝜃 ÷ 𝑑𝑥/𝑑𝜃 = (3a sin^2⁡𝜃 cos⁡𝜃)/(−3a cos^2⁡𝜃 sin⁡𝜃 ) = (−sin⁡𝜃)/cos⁡𝜃 = −tan 𝜽 Finding value of tan 𝜽 𝑦/𝑥 = (𝑎 sin^3⁡𝜃)/(𝑎 cos^3⁡𝜃 ) sin^3⁡𝜃/cos^3⁡𝜃 = 𝑦/𝑥 (sin⁡𝜃/cos⁡𝜃 )^3 = 𝑦/𝑥 sin⁡𝜃/cos⁡𝜃 = (𝑦/𝑥)^(1/3) 𝒕𝒂𝒏⁡𝜽 = √(𝟑&𝒚/𝒙) Thus, 𝑑𝑦/𝑑𝑥 = −tan 𝜃 𝒅𝒚/𝒅𝒙 = −√(𝟑&𝒚/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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.