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Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 13

Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 14
Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 15
Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 16
Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 17

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Transcript

Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (v) π‘₯=cos⁑𝑑, 𝑦=sin⁑𝑑 π‘Žπ‘‘ 𝑑= πœ‹/4At 𝒕= 𝝅/πŸ’ x = cos πœ‹/4 = 1/√2 y = sin πœ‹/4 = 1/√2 ∴ At 𝑑=πœ‹/4 , the point is (1/√2 " ," 1/√2) Finding 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=(𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Now, Hence, 𝑑𝑦/𝑑π‘₯=(𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 𝒙=𝒄𝒐𝒔⁑𝒕 Differentiating w.r.t.𝑑 𝑑π‘₯/𝑑𝑑=𝑑(cos⁑𝑑 )/𝑑𝑑 𝑑π‘₯/𝑑𝑑=βˆ’sin⁑𝑑 π’š=π’”π’Šπ’β‘π’• Differentiating w.r.t. 𝑑 𝑑𝑦/𝑑𝑑=𝑑(sin⁑𝑑 )/𝑑𝑑 𝑑𝑦/𝑑𝑑=cos⁑𝑑 𝑑𝑦/𝑑π‘₯=cos⁑𝑑/(βˆ’sin⁑𝑑 ) 𝑑𝑦/𝑑π‘₯=βˆ’cot⁑𝑑 At 𝒕=𝝅/πŸ’ 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_(𝑑=πœ‹/4)=βˆ’π‘π‘œπ‘‘(πœ‹/4) =βˆ’1 ∴ Slope of tangent at (1/√2 " ," 1/√2) is – 1 We know that Slope of tangent Γ— Slope of Normal =βˆ’1 βˆ’1 Γ— Slope of Normal =βˆ’1 Slope of Normal =(βˆ’1)/(βˆ’1) Slope of Normal =1 Hence Slope of tangent is – 1 & Slope of Normal is 1 Finding equation of tangent & normal Now Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (1/√2 " ," 1/√2) & having Slope βˆ’1 is 𝑦 βˆ’1/√2 =βˆ’π‘₯+ 1/√2 π‘₯+𝑦 =1/√2+ 1/√2 π‘₯+𝑦 =2/√2 π‘₯+𝑦 =√2 𝒙+π’šβˆ’βˆšπŸ=𝟎 Equation of Normal at (1/√2 " ," 1/√2) & having Slope 1 is (π‘¦βˆ’1/√2)=1(π‘₯βˆ’1/√2) 𝑦 βˆ’1/√2 =π‘₯βˆ’ 1/√2 𝑦 =π‘₯βˆ’1/√2+ 1/√2 𝑦 =π‘₯ 𝒙 =π’š

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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.