Last updated at April 19, 2021 by Teachoo

Transcript

Example 19 Find the equations of the tangent and normal to the curve ๐ฅ^(2/3) + ๐ฆ^(2/3) = 2 at (1, 1).Given curve ๐ฅ^(2/3) + ๐ฆ^(2/3) = 2 Differentiating both sides w.r.t x 2/3 ๐ฅ^(1 โ 2/3)+2/3 ๐ฆ^(1 โ 2/3) ๐๐ฆ/๐๐ฅ = 0 2/3 ๐ฅ^((โ1)/3)+2/3 ๐ฆ^((โ1)/3) ๐๐ฆ/๐๐ฅ = 0 2/3 ๐ฆ^((โ1)/3) ๐๐ฆ/๐๐ฅ = (โ2)/3 ๐ฅ^((โ1)/3) 1/๐ฆ^(1/3) ๐๐ฆ/๐๐ฅ = (โ1)/๐ฅ^(1/3) ๐ ๐/๐ ๐ = โ (๐/๐)^(๐/๐) Thus, Slope of tangent to the curve = โ (๐ฆ/๐ฅ)^(1/3) At point (1, 1) Slope = โ (๐/๐)^(๐/๐) = โ1 Hence, Equation of tangent at point (1, 1) and with slope โ1 is ๐ฆโ1=โ1 (๐ฅโ1) ๐ฆโ1=โ๐ฅ+1 ๐+๐โ๐ = ๐ Also, Slope of Normal = (โ1)/(๐๐๐๐๐ ๐๐ ๐ก๐๐๐๐๐๐ก) = (โ1)/(โ1) = 1 Thus, Equation of normal at point (1, 1) and with slope 1 is ๐ฆ โ 1 = 1 (๐ฅ โ 1) ๐ฆ โ 1 = ๐ฅ โ 1 ๐ฆ =๐ฅ ๐ โ๐=๐

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.