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Ex 5.6, 1 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, x = 2γ€–π‘Žπ‘‘γ€—^2, y = γ€–π‘Žπ‘‘γ€—^4Here 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑦 = γ€–π‘Žπ‘‘γ€—^4 𝑑𝑦/𝑑𝑑 " =" γ€–4π‘Žπ‘‘γ€—^(4 βˆ’1) "=" γ€–4π‘Žπ‘‘γ€—^3 Calculating 𝒅𝒙/𝒅𝒕 π‘₯ = 2γ€–π‘Žπ‘‘γ€—^2 𝑑π‘₯/𝑑𝑑 " =" 2 Γ— 2π‘Žπ‘‘ "=" 4π‘Žπ‘‘ Now, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Putting values 𝑑𝑦/𝑑π‘₯ = (4π‘Žπ‘‘^3)/4π‘Žπ‘‘ π’…π’š/𝒅𝒙 = t2

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.