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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Example 19 Find the equations of the tangent and normal to the curve ๐‘ฅ^(2/3) + ๐‘ฆ^(2/3) = 2 at (1, 1). The curve is given as ๐‘ฅ^(2/3) + ๐‘ฆ^(2/3) = 2 2/3 ๐‘ฅ^((โˆ’1)/3)+2/3 ๐‘ฆ^((โˆ’1)/3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 2/3 ๐‘ฆ^((โˆ’1)/3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’2)/3 ๐‘ฅ^((โˆ’1)/3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’2)/3 ๐‘ฅ^((โˆ’1)/3) \ ร—๐Ÿ‘/๐Ÿ ๐’š^(๐Ÿ/๐Ÿ‘) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’ (๐‘ฆ/๐‘ฅ)^(1/3) Thus, Slope of tangent to the curve = โˆ’ (๐‘ฆ/๐‘ฅ)^(1/3) At point (1, 1) โ”œ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ](1, 1) = โˆ’1 Hence, equation of the tangent at point (1, 1) and with slope โˆ’1 is ๐‘ฆโˆ’1=โˆ’1 (๐‘ฅโˆ’1) ๐‘ฆโˆ’1=โˆ’๐‘ฅ+1 ๐‘ฆ+๐‘ฅโˆ’2 = 0 Also, Slope of the normal = (โˆ’1)/(๐‘ ๐‘™๐‘œ๐‘๐‘’ ๐‘œ๐‘“ ๐‘ก๐‘Ž๐‘›๐‘”๐‘’๐‘›๐‘ก) = (โˆ’1)/(โˆ’1) = 1 Hence, equation of the normal at point (1, 1) and with slope 1 is ๐‘ฆ โˆ’ 1 = 1 (๐‘ฅ โˆ’ 1) ๐‘ฆ โˆ’ 1 = ๐‘ฅ โˆ’ 1 ๐‘ฆ =๐‘ฅ ๐’š โˆ’๐’™=๐ŸŽ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.